How exactly can an increase temperature be used to break the chemical bonds of a gas [SO2, for example]? So I understand the ideal gas law at a basic level, and that a rise in temperature can break bonds, however, I am unclear of how this works in a practical sense.

Say I have a container full of $\ce{SO2(g)}$ and I would like to break $\ce{SO2}$ into Sulfur ($\ce{S}$) and molecular oxygen ($\ce{O2}$). Would simply increasing the heat of the container to $n~\pu{K}$ be sufficient to break the bonds to get $\ce{S}$ and $\ce{O2}$ by themselves?

As we provide heat to any molecule, the energy of the molecule increases. Specifically, the energy of atoms increases. Using the fact that stability is inversely proportional to energy, we can conclude that as energy increases, the bonds between the atoms become more unstable and eventually break and constituent atoms disassociate.

This heat provided is known as bond disassociation enthalpy, or bond disassociation energy [BDE].


Why does the gas from dry ice flow down? I am reading Dan Broughton's answer in replying to the question "why does the gas from dry ice flow down" here.

According to the answer, the fog is water's color and water is condensed and suspended in the CO2.
Do you agree with this answer? If so then should water have color? I see some documents saying that it is nearly colorless.

  Why does the gas from dry ice flow down?

  Dry ice is simply the common term used for solid carbon dioxide (CO2),
  so the gas coming off of the dry ice is CO2 gas. CO2 gas has a density
  of 1.96 kg/m^3 and air (78% N, 21% O2) is 1.29 kg/m^3. Since the CO2
  is more dense, it sinks in air.

  But CO2 is colorless, so what makes it appear as a white mist? As the
  gas sublimes (change directly from solid to gas), it comes off of the
  dry ice at -78.5 C (-109.2 F).  Being well below the freezing point of
  water, any moisture in the air surrounding the dry ice condenses and
  is suspended in the CO2, creating the cloud/fog effect that we can
  see.

  These two phenomena combined are what create the fog that flows along
  the ground.

The gas flows downwards because it is more dense. It is more dense both because $\ce{CO2}$ is heavier than $\ce{N2}$ and $\ce{O2}$ and because the gas is colder than the ambient air.

The mist is water in the surrounding air condensing into microscopic droplets due to the lowered temperature.  The amount of mist depends on the humidity of the air; in perfectly dry air there will be no mist.

All of the gases involved are colourless. The mist appears white because it is in small particles and so it scatters any light falling on it.  It is the same colour as "natural" mist and for the same reason.


Why is it that every snowflake is unique? It's pretty hard to believe in around 4.5 billion years of this planet's existence, no snowflake pattern was repeated, and that's not including potential earth-like planets.

Is there a proof to the structure or is the probability even more statistically insignificant than I thought, or what?

It turns out that 2 identical snowflakes have been observed, but...

Two Identical Snowflakes

Although when we think "snowflake" we usually picture an object with 6-fold radial symmetry, snowflakes actually come in many different shapes (reference 1).  Many sites report that in 1988 an NCAR researcher found 2 identical snowflakes of the hollow columnar variety (reference 2 - see "Uniqueness" section, reference 3).  This pattern is much simpler than the 6-fold radial design, so the odds of finding 2 identical flakes are vastly improved.

But

If you wanted to confine the question to snowflakes of the 6-fold radial variety, then an argument can be made that 2 identical snowflakes will not be found.

Due to the vapor pressure difference between ice and water, vapor is pushed from water towards the ice (hence, the ice grows) in an ice-water droplet falling through the atmosphere.  As it falls through different temperature zones, the rate of ice crystal growth will change leading to different dendritic patterns along the radial arms.  Further, due to perturbations caused by temperature and wind gradients across a single crystal, each radial arm will have a slightly different pattern.

The mathematical argument against identical (6-fold radial) snowflakes is, as Jon Custer suggested in his comment, based on factorial probabilities.  The argument goes something like this.  If you have 10 different pictures to put in an album you could arrange them in 10! (10 factorial) or 3,628,800 different ways.  If you look at a snowflake under magnification, there are easily over a hundred different features that are identifiable.  In this case there are over 100! or greater than $10^{158}$ possible arrangements of these identifiable features.  Estimates are that there are around $10^{80}$ atoms in the universe.  Each visible feature in a snowflake contains a large number of water molecules and an even larger number of atoms.  Hence, the number of possible arrangements is far, far greater than the number of available atoms, making it extremely improbable that a duplicate of a complex snowflake would be created.

If you're still not convinced, we can add another level of probabilistic protection.  Remember that there are isotopes of hydrogen and oxygen.  The probability of finding two 6-fold radial snowflakes with the same dendritic pattern and the same isotopic substitution pattern is even closer to zero.


Why is bond-breaking endothermic? When electrons gain energy they become more excited and move to a higher energy level, which increases the tendency of an atom to form a bond with another atom. Thus, surely bond making requires the absorbance of energy?

Equally when electrons lose energy they drop to a lower energy level, which would break a bond. Therefore, why isn't bond breaking exothermic?

Could someone point out the flaw in my reasoning, because I don't get why bond-breaking is possibly endothermic?

When electrons gain energy they become more excited and move to a higher energy level, which increases the tendency of an atom to form a bond with another atom.

Let me assure you that this statement is incorrect. I’m not fully sure where you extracted it from, but I assume it stems from the way many schools teach hybridisation at the beginning of organic chemistry classes; requiring an $\ce{s\bond{->}p}$ excitation in carbon from $\mathrm{[He]\ 2s^2\,2p^2}$ to $\mathrm{[He]\ 2s^1\,2p^3}$, after which the s- and p-orbitals can form $\mathrm{sp^3}$ hybrid orbitals. This idea is nothing more than a school level simplification used to get around teaching of more complex molecular orbital theory and symmetry.

Nothing prevents you from constructing e.g. a methane molecule without initial hybridisation, i.e. starting from an unhybridised carbon atom and four hydrogen atoms in a tetrahedric arrangement. I refer you to the following scheme posted in a different question and originally taken from Professor Klüfers’ internet scriptum for basic and inorganic chemistry at the university of Munich:

As you can see on the right, carbon enters this scheme in the unhybridised ground state. There is no need to invoke a previous hybridisation before mixing orbitals; rather, it is necessary to determine the symmetry of orbitals and thereafter combine symmetry-equivalent orbitals in a bonding-antibonding fashion. Finally, fill in electrons from bottom to top.

This method will always result in stabilisation of bonding orbitals; the trade-off always being the destabilisation of antibonding orbitals in such a way that the (actual) energy gained is lower than the (virtual) energy lost.

Therefore, assuming a positive bond order, forming a bond typically liberates energy while breaking one will typically require energy. I am not aware of any counterexamples, but the sentence is phrased so that it remains true when the obligatory counterexample is posted as a comment.


How to calculate the rotational constants for different states from a stick spectrum? Given a stick spectrum for an unknown diatomic molecule, how do you determine the rotational constants $B_\nu$ for both the excited and ground states given that

$$E_{j}=B_{\nu}j(j+1)$$

\begin{array}{cc}
\text{Line position}\, \mathrm{(cm^{-1})} & \text{relative intensity} \\
\hline
27128.20 & 1.47 \\
27171.74 & 2.35 \\
27211.40 & 3.13 \\
27247.18 & 3.34 \\
27279.08 & 2.62 \\
27331.24 & 1.00 \\
27351.50 & 2.62 \\
27367.88 & 3.34 \\
27380.38 & 3.13 \\
27389.00 & 2.35 \\
\hline
\end{array}

In your question, it's not clear whether the quantum number assignments for the transitions are known in advance, so for generality I'll assume that you don't know what they are, and help you find them.

Combination-differences is the most common paper-based method to determine line assignments (for more detailed analysis, a computer-aided fitting program such as PGopher is best). From the line assignments, it's easy to find the $B$-values.

Rotational energy levels are given by
$$F_\nu(J)\approx B_\nu J(J+1)+D_\nu [J(J+1)]^2$$
where $B_\nu$ and $D_\nu$ are the rotational constant and centrifugal distortion constant associated with the manifold $\nu$.

Denoting the initial and final states by $1$ and $2$ respectively, the $P,R$ transition energies are given by
$$P(J)=F_2(J-1)-F_1(J)
\\R(J)=F_2(J+1)-F_1(J).$$
Combination-differences is can be done in two ways:

If your rovibronic transition originates from an initial manifold with well-documented spectral characteristics (including $B_1$), note that
$$R(J-1)-P(J+1)=2 (2 J+1) \left(B_1-2 D_1 \left(J^2+J+1\right)\right)\approx2B_1(2J+1)$$
when $D_1$ is small, and so then you can just try to find pairs of $P,R$ peaks with spacing $2B_1(2J+1)$ for various small values of $J$. This allows you to number the peaks as to their identity.
If instead the upper state has well-known $B_2$, then note that
$$R(J)-P(J)=2 (2 J+1) \left(B_2-2 D_2 \left(J^2+J+1\right)\right)\approx2 B_2 (2 J+1)$$
and so then you can just try to find pairs of $P,R$ peaks with spacing $2B_1(2J+1)$ for various small values of $J$. Again, this allows you to number the peaks as to their identity.

Once the quantum numbers are assigned, finding both $B_1$ and $B_2$ is easy. For example, $R(J)-R(J-1)=2 \left(B_2 (J+1)-B_1 J\right)$, so if you measure it for two values of $J$, then you have two unknowns and two datapoints, so just solve for $B_1$ and $B_2$.

Fun Pictures

Just to give a simple idea of how spectra are simulated, the following Mathematica code does a basic rotational spectrum assuming Boltzmann statistics at a reasonable temperature (hit Cell>Convert To>StandardForm to get it to render the subscripts correctly):

Subscript[F, k_][J_] :=
  Subscript[B, k] J (J + 1) - Subscript[D, k] J^2 (J + 1)^2;
R[J_] := Subscript[F, 2][J + 1] - Subscript[F, 1][J];
P[J_] := Subscript[F, 2][J - 1] - Subscript[F, 1][J];
ListPlot[Transpose[
  Table[{{R[J], (2 J + 1) Exp[-(1/30) Subscript[F, 1][J]]}, {P[
       J], (2 J + 1) Exp[-(1/30) Subscript[F, 1][J]]}} /. {Subscript[
      D, k_] :> 0, Subscript[B, 1] -> 1, Subscript[B, 2] -> 0.9}, {J,
    0, 15}], {2, 1, 3}], Filling -> Axis]


Why is trichloromethane more acidic than trifluoromethane? In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 78 under the topic "Back Bonding" it is given:

  $\ce{CHCl3}$ is more acidic than $\ce{CHF3}$ which is explained as follows:

  $$
\begin{align}
\ce{CHCl3 &<=> H+ + ^-CCl3}\\
\ce{CHF3 &<=> H+ + ^-CF3}
\end{align}
$$

  The lone pair on the $\ce{C}$ atom gets delocalised through 2pπ–3dπ bonding in $\ce{^-CCl3}$ which is not possible in the case of $\ce{^-CF3}.$

My doubt is, fluorine being the most electronegative element exerts a greater negative inductive effect (compared to that of chlorine) which stabilises the conjugate base formed. So I feel trifluoromethane must be more acidic than trichloromethane. But the reverse order is observed.

Please explain why is this so. Should we want to prefer the conclusion from back bonding over inductive effect?

We know that fluorine ($\ce{F}$) is more electronegative than chlorine ($\ce{Cl}$). Therefore, -I effect (Inductive Effect) of $\ce{F}$ would be more and trifluoromethane ($\ce{CHF3}$) would be more acidic.

But this is not the case. As -M (Mesomeric Effect) effect also plays an important role in determining the acidity. We also know that -M is considered over -I (Inductive Effect).

Now, If we compare the conjugate base of both the given molecules, $\ce{Cl3C-}$ and $\ce{F3C-}$, Chlorine has vacant $\mathrm{3d}$ orbitals whereas $\ce{F}$ does not have any vacant $\mathrm{d}$ orbitals (in fact, it does not even have $\mathrm{d}$-orbitals). Therefore, $\ce{Cl}$ can exert -M effect due to availability of vacant $\mathrm{d}$ orbital but $\ce{F}$ cannot exert the -M effect. Generally, -M effect stabilises the negative charge and increases acidic strength. Therefore, trichloromethane ($\ce{CHCl3}$) is more acidic than trifluoromethane ($\ce{CHF3}$).


Lemon juice, the ultimate anti-foaming agent? As an amateur soda maker, I have noticed an odd phenomenon: lemon juice sodas do not foam. Virtually every juice I have tried foams to some degree when mixed with carbonated water. Lemon juice does not. I have read "folk" explanations for this as being due to the "acidity" of the lemon juice, but this is not correct because other acidic juices like lime and orange juice foam. Why is it that lemon juice has an anti-foaming action?

Further information: it should be noted that "pasteurized" (ie, boiled) lemon juice does foam. Only fresh-squozen lemon juice is anti-foaming. Therefore, what is doing the anti-foaming may likely be some volatile compound.

My understanding of anti-foaming is that it is caused primarily by localized depression of surface tension, and that liquids having heterogeneous species promote anti-foaming because of the surface tension differential between those species at the surface. Therefore, one possible theory is that the lemon juice has some volatile oil which populates the surface of the water only in spots and is causing the surface tension breakage. Even if this theory is correct, it remains to determine what this mysterious volatile oil is that is the cause.

A lemon has an oil that is present mostly in the skin. For instance if you squeeze lemon into hot tea in a styrofoam cup and let the tea sit, then you'll see a surface roughness at the liquid level. The oil actually dissolves the styrofoam. The oil acts a defoamer. So with a styrofoam cup you're suppose to use "commercial" lemon juice which has no oil (lest the cup dissolve and spill the liquid).


Brillouin's Theorem: How do other determinants mix with the ground state? I am a bit confused about Brillouin's Theorem. I get that a singly excited determinant can not mix with the ground state wavefunction because that translates to an off-diagonal matrix element ($\langle \chi_a \chi_b|\mathcal H |\chi_r \chi_b \rangle$) and the Hartree-Fock eigenvalue problem requires off-diagonal elements to equal 0, but how about when a doubly excited determinant mixes with the ground state ($\langle \chi_a \chi_b|\mathcal H |\chi_r \chi_s \rangle$)? Both of them do not have $i=j$ thus both should be off-diagonal entries, yet a doubly excited determinant can mix with the ground state so what am I missing?

Brillouin's theorem is the direct consequence of the so-called Slater–Condon rules which for the case of the HF reference determinant $\Phi_{0}$ and a singly excited determinant $\Phi_{a}^{r}$ show that the Hamiltonian matrix element between these determinants reduces to the Fock matrix element between the corresponding orbitals
$$
  \newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
  \newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}
  \newcommand{\op}[1]{\hat{#1}}
  \newcommand{\el}{_{\mathrm{e}}}
  \newcommand{\core}{^{\mathrm{core}}}
  \newcommand{\oneeintphys}[3]{\bracket{#1}{#2}{#3}}
  \newcommand{\twoeintphysrantisymm}[2]{\langle{#1}\,||\,{#2}\rangle}
  \bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }
  =
  \oneeintphys{a}{\op{H}\core}{r} + \sum\limits_{j=1}^{n} \twoeintphysrantisymm{aj}{rj}
  =
  \bracket{ \psi_{a} }{ \op{F} }{ \psi_{r} } \, .
$$
Now taking the HF equations and orthogonality of spin orbitals into account we indeed get zero for $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }$ since
$$
  \bracket{ \psi_{a} }{ \op{F} }{ \psi_{r} }
  =
  \varepsilon_{r} \braket{ \psi_{a} }{ \psi_{r} }
  =
  0 \, .
$$
Now try to use the Slater–Condon rules for the case of the HF reference determinant $\Phi_{0}$ and a double excited determinant $\Phi_{ab}^{rs}$ to convince yourself that $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{ab}^{rs} }$ unlike $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }$ won't vanish. This is a very well-known result which manifests itself in many different ways. For instance, this is the reason why in order to improve the HF energies in configuration interaction scheme we include doubly excited determinants first (CID) rather than singly excited once (CIS): the later won't have any effect on electronic energy.


How to make a sample in ppm? (not a specific compound, only the sample) Is soaking 1000 mg tea powder in 0.1 L water for about 12 - 24 hours, can be considered as 10000 ppm tea sample?
Could anyone give me reference from a journal or something similar, about my question topic?
Really appreciate any help you can provide.

In layman's terms, ppm is similar to percentage, i.e., how many parts of X are there in a million. By that standard, 1000 mg tea / 0.1 L is indeed 10000 ppm tea "powder" in water. We are assuming that everything dissolves completely.
Scientifically, or in analytical chemistry, you question is meaningless. Tea powder is mixture so one cannot talk about the ppm of tea powder. This is why you will not find this in journals or books. Concentration is specified for a single component.
Tea powder is a very complex mixture. If we have a mixture of say K2SO4 and NaCl, it is not appropriate to mention the concentration of "salts" in water. It will be an ambiguous statement. A better way would be expressing the concentration of K2SO4 as x% and that of NaCl as y %. It is okay in analytical chemistry to mention "total carbonates" or "total acidity"; here the species being analytes are closely related.


Byproducts of the neutralisation of hydrochloric acid with sodium hydroxide Given a 1 M concentration of $\ce{HCl}$ and $\ce{NaOH}$ here are the pHs:

pH of $\ce{HCl}$: 1

pH of $\ce{NaOH}$: 13–14

and of course water has a pH of 7.

Here is a reaction that is most likely to occur:

$\ce{HCl + NaOH + H2O <=> H2O + NaCl(aq)}$

However I think that other reactions are theoretically possible.

Like for example theoretically sodium oxides could form from the $\ce{OH-}$ taking a proton from another hydroxide anion thus giving you $\ce{O^2-}$ which could react with the sodium to give you $\ce{Na2O}$ which is a sodium oxide.

I also think that theoretically $\ce{Cl-}$ could steal electrons from the $\ce{OH-}$ since $\ce{Cl}$ can exceed an octet of electrons. This would give you neutral $\ce{OH}$ which could then react with another chloride anion to give you $\ce{OHCl-}$(not sure if formula is written right or if it would be negatively charged) or in other words chlorine hydroxide. The proton could then be taken by another hydroxide anion giving you $\ce{OCl}$ also known as chlorine monoxide which could react with another chlorine to give you $\ce{OCl2}$ which is dichlorine monoxide.

Do these theoretical reactions that lead to oxyanions($\ce{O^2-}$ to be specific), chlorine oxides, chlorine hydroxides, and sodium oxides actually happen to some extent when you have $\ce{HCl}$ and $\ce{NaOH}$ in aqueous solution

Given a 1M concentration of $\ce{HCL}$ and $\ce{NaOH}$ here are the pHs:

  pH of $\ce{HCL}$: 1

  pH of $\ce{NaOH}$: 13-14

First, a 1 molar concentration of hydrochloric acid solution, $\ce{HCl}$, has a pH of 0.0 if we take ${-log[H_3O^+]=pH}$ to be true. (And it is mostly true for dilute solutions).

  and of course water has a pH of 7.

Pure water at 25 degrees Celsius has a pH of 7.0. Boil the water and the water will no longer have a pH of 7.0. Cool the water and pH will similarly change. The only condition for neutrality is that the concentration of hydronium ion equals the concentration of hydroxide ion. It just so happens that water at normal lab conditions has a hydronium ion and hydroxide ion concentration of ${1.0*10^{-7}}$ M, which corresponds to a pH of 7.0. Your statement only holds at a specific temperature. At other temperatures, this reaction - the auto-ionization of water - which controls the pH of pure water - will go to varying extents:

$\ce{2H_2O <=> H_3O^+ + HO^-}$

  Here is a reaction that is most likely to occur:

  $\ce{HCl + NaOH + H2O <=> H2O + NaCl(aq)}$

Since we are talking about solutions of HCl and NaOH, your equation isn't reflective of what's actually going in on the system. HCl and NaOH both dissolve in solution (unless the solution is saturated with either). So you don't have HCl or NaOH in solution, but rather hydronium ion and chloride anions in the case of HCl and sodium ions and hydroxide ions in the case of NaOH solution. A more accurate equation would be:

$\ce{H_3O^+ + HO^- ->2H_2O}$

Regarding:

  $\ce{NaCl(aq)}$

Sodium chloride has a very high solubility in water and dissociates unless the solution is saturated. So it would be more accurate to write simply sodium ions and chloride ions, instead of "aqueous" sodium chloride (unless you are referring to the ion-pair effect). Ion pairs often occur in solution; the name describes exactly what is implied - two oppositely charged ions are attracted to each other. For sodium chloride, ion pairing is likely minimal, but in other cases, such as between a carbocation and a halide anion, the ion pair effect has noticeable effects on chemical reactions (i.e. partial racemization rather than full racemization).

  Like for example theoretically sodium oxides could form from the
  $\ce{OH-}$ taking a proton from another hydroxide anion thus giving
  you $\ce{O^2-}$ which could react with the sodium to give you
  $\ce{Na2O}$ which is a sodium oxide.

So hydroxide ion reacts as an acid and abstracts a proton from another hydroxide ion.

This is at least theoretically possible, but thermodynamically unfavorable. The oxide anion is very unstable in water solution to the point of being able to react with the solvent (water) itself to a large extent.


Reversible elementary reaction I have a question regarding a reversible elementary reaction:

If we have an equilibrium where $$\ce{A<=>[k_1][k_{-1}]B}$$ , then $$-d[A]/dt= k_1[A]-k_{-1}[B]$$.

If we take $$[B]=[B]_0 + ([A]_0 - [A])$$, the equation above becomes $$d[A]/dt= -k_1[A]+k_{-1}([B]_0+([A]_0 -[A])$$

So far, I've taken the integrating factor to be $$u=\mathrm{e}^{(k_1+k_{-1})t}$$

and emerged with the following equation:

$$[A]=\frac{k_{-1}}{k_1+k_{-1}}([B]_0+[A]_0)(1-\mathrm{e}^{-t(k_1+k_{-1})}) + C \mathrm{e}^{-t(k_1+k_{-1})}$$

I have tried to put $$[A](t=0) = [A]_0$$ and the solution appears to be $$C=[A]_0$$

Are there further approximations required?

My lecturer has produced the final equation as

$$[A]=\frac{k_{-1}+k_1\mathrm{e}^{-t(k_1+k_{-1})}}{k_1+k_{-1}}([B]_0+[A]_0)$$

When we take $$t->\infty$$ our equations match, however.

I went through the derivation myself and I can reproduce the equation you got. But after using $C = [\ce{A}]_0$ and some rearrangements I get the following formula

\begin{equation}
\ce{[A]} = \frac{k_{-1}}{k_{1} + k_{-1}} \left( \ce{[A]_{0}} + \ce{[B]_{0}} \right) + \frac{\mathrm{e}^{-(k_{1} + k_{-1})t} }{k_{1} + k_{-1}} \left( \ce{[A]_{0}} k_{1} - \ce{[B]_{0}} k_{-1} \right)
\end{equation}

and not the one your professor got. But your professor's formula might be incorrect because for $t=0$ his formula yields the incorrect result $\ce{[A]}_{t = 0} = \ce{[A]_{0}} + \ce{[B]_{0}}$ instead of the correct result $\ce{[A]}_{t =  0} = \ce{[A]_{0}}$, so it doesn't satisfy the given boundary conditions.


Why are there two structures of L-Glucose in the Fischer projection? We are told that L-Glucose is an enantiomer of D-Glucose, which would mean that they are mirror images of each other. But then we're also told that the L and D forms is determined by the spacial relationship to glyceraldehyde.

Which of the structure is correct? Are they the same?

It looks to me like the left one is just wrong.
It looks like who made the image took D-glucose and just flipped the 5-OH.

The result? The molecule at left is not glucose, but L-Idose!


Calculating time taken for a set percentage of a reaction to occur based on half-life The half-life of a first order homogenous gaseous reaction: $\ce{SO2Cl2 -> SO2 + Cl2}$ is 8.0 minutes. How long will it take for the concentration of $\ce{SO2Cl2}$ to be reduced to 1% of the initial value?

I solved it this way:

Let initial mass be 100, so that the final mass is 1.
$$100 \cdot (\frac{1}{2})^n=1$$
I solve this for n, and multiply it by 8.0 to get the time taken.

Answer is spot on, and the method is identical to the one used in half life questions in Physics. Yet my Prof insists that I solve for the rate constant first, and then find half-life from that, and that my answer is wrong because n, the value for number of half-life periods, cannot be a decimal value.

The questions I found online all explicitly ask for the rate constant, which the question I got doesn't ask for.

Is my method right/wrong? Can you guys provide proof?

Ryan showed that the rate constant is given by $k=\frac{ln(2)}{t_{1/2}}$.  So the concentration at time t is given by
$$C=C_0\exp(-kt)=C_0e^{-ln(2)\frac{t}{t_{1/2}}}$$But,
$e^{-ln(2)}=\frac{1}{2}$.  Therefore, $$C=C_0\left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}$$Substituting $n=\frac{t}{t_{1/2}}$ gives
$$C=C_0\left(\frac{1}{2}\right)^n$$


Calculating moles of Hydrogen gas in multiple gas container If there was a 500 mL container filled with oxygen and hydrogen so it burns completely (2H:1O), how many moles of hydrogen gas would there be if there if pressure is 101.3 kPa and the temperature is 293 K?

Using pv = nrt$$1atm\times(0.5L) = (total moles)(.08206\frac{atm\cdot L}{mol\cdot k})(293K) \\ \text{total moles} = .02079mol $$ There are 3 total moles of gas with 2 of those being hydrogen and 1 being oxygen. This means 66% of the total moles is hydrogen and 33% is oxygen. Soooooo 66% times .02079 is the amount of hydrogen moles and 33% times .02079 is the amount of oxygen. TADA!


Why is the equilibrium constant unaffected by a change of the initial concentration? In my class, it was taught that the equilibrium constant does not vary with
the intial concentration of the reactants? Why is it so?

Clearly the math tells you that the equilibrium constant is independent of the initial compositions:

For $a\mathrm{A}+b\mathrm{B}\rightarrow c\mathrm{C}+d\mathrm{D}$,
$$K_c=\frac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$$
where $[\mathrm{C}]$, $[\mathrm{D}]$, $[\mathrm{A}]$, and $[\mathrm{B}]$ are concentrations at equilibrium.

This makes some sense intuitively: the rate of the reaction is dependent only on the present concentrations of compounds (and some other stuff like pressure and temperature), not previous concentrations. And equilibrium is defined as a position where the rate of forward and backward reactions are the same.


Naming phosphonium cations in ionic liquids In Rennie et al., J. Phys. Chem. Lett., 2013, the following cation and anion structures are given in Figure 1.  Figure 1(f) is captioned as butyltrimethylphosphonium, which seems to match the given structure.  However, in the caption, the abbreviation $\text{P}_{2225}$ is given, which seems strange:

Rennie et al. repeat this in their Supporting Information:

But, other authors, such as Tsuchida et al., J. Mol. Liq., 2018, use $\text{P}_{2225}$ as an abbreviation for triethyl-n-pentylphosphonium, which makes sense if the subscript numbers represent the number of carbon atoms in the substituent alkyl groups:

Does it seem to you that there is a typo in the Rennie et al. paper? Or, what are the standard abbreviations for phosphonium cations?

Skepticism is an integral part of a good scientist. Remember the motto "Paper never refused ink" when reading papers and its Supporting Information. You will find incorrect references (very frequent), occasionally wrong calculations, wrong formulae, or wrong names as you read more and more papers just because scientists, like normal human beings, make mistakes.
The strategy in such cases is not to stop at two papers, because the odds are 50:50, but find at least 6-7 papers to confirm the truth.
Your notation is correct for $\ce{P2225}$.

Reference: Another paper from RSC Two phosphonium ionic liquids with high Li+ transport number, Phys. Chem. Chem. Phys., 2015,17, 23041-23051
The odds are in your favor now.


Why does pasta become sticky after being cooked? This question was posed by my biochemistry professor at the end of a class and then never answered.

What I've gathered is that:

It has something to do with the denaturing of a protein, since we were on the subject of denaturing that lecture.
It most likely involves gluten, because that's the main protein in wheat. And, from personal experience of making glue, gluten is very sticky.
The answer might involve hydrophilic/hydrophobic interactions, because those interactions are disturbed during the denaturing process.

Any ideas?

After cooking, pasta has a lot of free surface starch ready to bond with things. This is why we rinse well after cooking to remove extra starch and cool down the pasta to prevent it from producing more surface starch.


What is the difference between "atomic hydrogen" and "nascent hydrogen"? My book (Comprehensive Chemistry by Dr. N . K. Verma, S. K. Khanna, Dr. B. Kapila) mentions two forms of hydrogen — "atomic" & "nascent".

It says that these two forms of hydrogen are more or less the same except that the former is produced by passing $\ce{H2}$ gas through an electric arc struck between two tungsten filaments and the latter is formed during chemical reactions in aqueous solutions. For example, $$\ce{Zn + H2SO4 -> ZnSO4 + 2[H]}$$

The book also said that the reactivity order of these hydrogens is

$$\ce{H2} \lt \ce{[H]} \lt \rm atomic~hydrogen$$

But I think  that the reactivity of $\ce{[H]}$  must be same as atomic hydrogen. So please explain the difference between them and also their reactivity order.

Is your book by chance very old? From the Wikipedia entry for "nascent hydrogen":

  Nascent hydrogen is purported to consist of a chemically reactive form
  of hydrogen that is freshly generated, hence nascent. Molecular
  hydrogen ($\ce{H2}$), which is the normal form of this element, is unreactive
  toward organic compounds, so a special state of hydrogen was once
  invoked to explain certain kinds of hydrogenations. Mechanistic
  understanding of such reactions is now available, and the concept of
  nascent hydrogen is discounted, even ridiculed.

Then, by example of how this concept came to be:

  Reductions of esters to give alcohols using a mixture of sodium and
  alcohols is called the Bouveault–Blanc reduction. It is an old
  reaction that has largely been superseded by alternative methods. At
  the time of popularity, the process caused much puzzlement because
  esters are unreactive toward hydrogen. It was also known that sodium
  reacts with alcohols to release $\ce{H2}$. it was concluded that some freshly
  generated ("nascent") hydrogen was responsible for this remarkable
  reaction. Subsequent studies have shown that this reaction proceeds
  via electron-transfer from metallic sodium to the ester substrate
  followed by protonation of the reduced intermediate. The evolution
  of hydrogen by the reaction of sodium and alcohol is purely a
  competitive reaction, the sole benefit being that in the presence of
  sufficient alkoxide, the sodium/alcohol reaction slows.

So in general, at least toward organic compounds, your relative reactivities of hydrogen species are correct if you take out "nascent" hydrogen so that:

$$\ce{H2} \lt \rm atomic~hydrogen$$


Finding OH- concentration given the H+ concentration - diverse methods There was this exercise which asked to find the OH- concentration given the H+ concentration, i firstly calculated the pH as -log[H+], did 14-pH to find the pOH and then did 10^(-pOH) to find the concentration of OH-.

My textbook used the fact that Kw (water equilibrium constant) is essentially
Kw = [OH-] * [H+] so it did    [OH-] = Kw/[H+]

The fact is...isn't Kw temperature dependent? My textbook says that it is

25 celsius --> Kw --> 1*10^-14

whereas

60 celsius --> 9.61*10^-14

So my question is...why is my procedure temperature independent while the book one temperature dependent? My procedure would always give the same concentration of OH whereas changing the Kw the textbook would also get other ones.

Thank you.

It is because you assumed the number 14 as a constant while it is actually $-\log(K_\mathrm w)$.
It is $-\log(1 \times 10^{-14}) = 14$ at 25 degrees Celsius but changed to $-\log(9.61 \times10^{-14}) = 13.02$ at 60 degrees Celsius.


at constant volume why the change of enthalpy doesn't equal the change of heat At constant pressure $p=\text{const.}$, change of internal energy equals heat subtract work(PV). So, $q=\Delta E - pV$. So is the the change of enthalpy. So we say at constant pressure, the change of enthalpy equals the change of heat. But when the volume is constant $V = \text{const.}$, no work is done. $\Delta E= q$, $\Delta H = \Delta q$. So why not in this situation too?

At constant volume, the pressure is not necessarily constant. The pressure will usually increase when the system is heated. This makes the $PV$ term not constant. For this reason, you get:

$$\Delta H = \Delta (E + PV) $$
$$\Delta H = \Delta E + \Delta (PV) $$
$$\Delta H = \Delta E + V\Delta P $$

Now heat is just $\Delta E$, because there is no work. So,

$$\Delta H = q+ V \Delta P $$


The respective speed of five molecules are 2, 1.5, 1.6, 1.6 and 1.2 km/s. The most probable speed in km/s will be? The respective speed of five molecules are 2, 1.5, 1.6, 1.6 and 1.2 km/s. The most probable speed in km/s will be?
  1. 2
  2. 1.58
  3. 1.6
  4. 1.31

I think the answer should be 1.6 km/h but the answer given in book is 1.31 km/h.
Most probable speed should be the mode of the given speeds, which is equal to 1.6 km/h. Am I correct?  If I am wrong, then what's the definition of most probable speed and how do I solve this problem?

Following up on user andselisk's answer
...

The problem is weird for several different reasons.

First the solution depends on assuming that the velocity of the gas molecules follow the Maxwell-Boltzmann distribution which is weird for only 5 molecules. The problem also doesn't state that the 5 molecules are a sample from a much larger number of molecules.
Second the answer depends on how you calculate to get a result.

For the Maxwell-Boltzmann distribution there are relationships between the most probable speed $v_p$, the average speed $\langle v \rangle$, and the root-mean-square speed $v_{rms}$. Following the Wikipedia article:

\begin{align}
\langle v \rangle &= \frac{2}{\sqrt{\pi}}v_p\\
v_\mathrm{rms} &= \sqrt{\frac{3}{2}}v_p
\end{align}

Rearranging these two equations we get two different ways to calculate the most probable speed.

\begin{align}
v_p &= \frac{\sqrt{\pi}}{2}\langle v \rangle\\
v_p &= \sqrt{\frac{2}{3}}v_\mathrm{rms}
\end{align}

Thus using the mean (average) speed we get:

\begin{align}
\langle v \rangle &= \frac{2 + 1.5 + 1.6 + 1.6 + 1.2}{5} = 1.58\\
v_p &= \frac{\sqrt{\pi}}{2}\langle v \rangle = 1.40
\end{align}

However using the rms speed we get:

\begin{align}
v_\mathrm{rms} &= \sqrt{\frac{2^2 + 1.5^2 + 2\cdot 1.6^2 + 1.2^2}{5}} = 1.60\\
v_p &= \sqrt{\frac{2}{3}}v_\mathrm{rms} = 1.31
\end{align}

The difference is due to small sample statistics and not really a statistically significant result. However the difference is absolute proof that the sample doesn't follow the Maxwell-Boltzmann distribution.


Where does the molar mass difference between N₂ and CO come from? The molar masses of $\ce{N2}$ is 28.0134 g/mol, for $\ce{CO}$ it's 28.0101 g/mol (WolframAlpha computation).

Of course, they do have the same mass in atomic units (28). My question is, where does this small discrepancy come from? Different binding energies in the respective nuclei? I know that oxygen is a "double-magic" nucleus from an atomic physics POV, that is, it's binding energy per nucleon is higher than what would be expected. But since normally, you equate mass with energy, shouldn't $\ce{CO}$ be heavier then?

The differences in molecular mass stem from two sources:

  Nuclear binding energy

The definition of the atomic mass unit and Avogadro's number (and thus the g/mol), is the mass of one atom of the carbon-12 isotope $\ce{^{12}_6 C}$. 1 amu = $\frac{1}{12}$ the mass of one atom of $\ce{^{12}_6 C}$ and the g/mol unit is based on the definition of Avogadro's Number, which is the number of $\ce{^{12}_6 C}$ atoms in 12 grams of $\ce{^{12}_6 C}$.

Because of the differences in binding energies in the various nuclei, the monoisotopic mass of individual isotopes are not integers. For example, the mass of $\ce{^{16}_8 O}$ is 15.99491463 amu (and not 16). The monoisotopic masses of the most common isotopes of oxygen, nitrogen, and carbon is shown below:

           mass number   % absundance    isotope mass
Carbon     12            98.930          12 (defined)
Nitrogen   14            99.632          14.003074
Oxygen     16            99.757          15.99491463

  Isotopes

Generally, molecular masses (unless otherwise specified) are average molecular masses composed of average molecular masses of the elements based on their isotopes. For example, even though the most common isotope of carbon has a mass defined as 12 amu, there is another naturally occurring isotope $\ce{^{13}_6 C}$ with 1.07% natural abundance, leading to carbon having a non-integer average atomic mass of 12.01 (to 4 significant figures). There are two isotopes of nitrogen, and three isotopes of oxygen.

            Average atomic mass
Carbon      12.011
Nitrogen    14.007
Oxygen      15.999

Thus, $\ce{CO}$ has a molecular mass of $12.011 + 15.999 = 28.010$ and $\ce{N2}$ has a molecular mass of $14.007+14.007=28.014$. Even their monoisotopic masses are different:

                    average molecular mass       monoisotopic mass
dinitrogen          28.014                       28.006148
carbon monoxide     28.010                       27.99491463


Can CH3COOH be written as C2H4O2? I was reading a chapter on acids and bases .It said something about acetic acid.And then I thought whether I could write CH3COOH as C2H4O2? please explain your answer.

Thanks!!

$\ce{C2H4O2}$ is the molecular formula of acetic acid. As such, it's a correct method to write the molecule. However, you should note that, as Ivan Neretin mentioned, molecular formulae are not necessarily unique for a given molecule.


Can benzene ring be the part of a acyl group Can benzene ring attached to c=o i.e carbonyl group be considered as a acyl group?

Yes, acyl groups can be derived from all kinds of oxoacids.

According to the Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995), acyl groups are defined as follows:

  Groups formed by removing one or more hydroxy groups from oxoacids that have the general structure $\ce{R_$k$E(=O)_l(OH)_m}\ (l\neq0)$, and replacement analogues of such acyl groups. (…) In organic chemistry an unspecified acyl group is commonly a carboxylic acyl group.

Especially concerning acyl groups derived from carboxylic acids in organic chemistry, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) defines:

  Carboacyl groups are $\ce{R-CO-{}}$, $\ce{-OC-R-CO-{}}$, or $\ce{-OC-R-[R'-CO-]_$x$-R''-CO-{}}$ groups and their functional replacement analogues, where $\ce{R}$, $\ce{R'}$, and $\ce{R''}$ are chains, rings, or ring systems, derived from carboxylic acids by the removal of the hydroxy group from each carboxylic acid group that is expressed by the suffix, and $x = 1 ,2,3$, etc.
  (…)

In particular, benzoyl is explicitly mentioned as an example for an acyl group.


PCC with substituted pyridines I was wondering if substituted pyridines can work the way pyridine does in order to oxidize alcohols to aldehydes.
Of course, I'm referring to pyridines without any functional group which can undergo oxidation reactions (eg: I was thinking methylpyridine, or trimethylpyridine, with the nitrogen untouched.)

I guess that the reactivity is the same, but is there any reference to the use of this kind of reagent?

Collidinium chlorochromate and gamma-picolinium chlorochromate are known oxidising agents. e.g. Synth Comm. 31 2001 1253

I see no reason why other methyl pyridines should not also work.


Stability of meta ethyl and meta methyl benzene carbocations respectively It is known that Methyl has a greater +I effect than Ethyl, but here, in this case, the condition is satisfied in the para benzene carbocations whereas, in the meta forms, it is the opposite. Shouldn't meta methyl benzene carbocation be more stable than meta ethyl benzene carbocation? Please, Can anyone explain?

We know that effect of neither hyperconjugation nor resonance is observed if the groups are attached in meta position. Hence, stability of structures I and IV will be decided on the basis of inductive effect only. Alkyl groups exert +I effect, and ethyl group exerts greater +I than methyl, hence stabilizing the electron deficient benzylic carbocation more. So, IV>I.
Now note that both of these structures must be less stable than II and III, because these are stabilized by both hyperconjugation and inductive effect, and hyperconjugation is a more stabilizing phenomenon than inductive effect. Since II has more $\alpha$-H than III(3 and 2, respectively), there would be more hyperconjugative structures for II, hence more stability. Hence, finally, the order is I<IV<III<II. This is the first option.


Thermodynamics and equilibrium constant I just wanted to make sure how we can know whether it is $K_c$ (equilibrium constant of concentrations) or $K_p$ (equilibrium constant of pressures) which "comes out" of the equation, $\Delta G^o= -RT \ln(K)$.

Until now, I have always taken $K_c$ when the reaction was in a solution or $K_p$ if it was a gas. But as there is a difference between their values, I'd like to know exactly which one to take.

What you are doing is right. The value to use depends on the definition of the standard state. The values of $K_c$ and $K_p$ are indeed different, but what you are not accounting for is the fact that $\Delta G^o$ also changes with the equilibrium constant used.

The naught over $G$, representing "standard state" establishes some rules. For homogeneous gaseous reaction, the standard $\Delta G$ corresponds to the standard state of all gases, which is defined in terms of pressure $P=\pu{1atm}$, and hence $K_p$ needs to be used with the quoted value of $\Delta G^o$.
For reactions involving solutes and pure liquids, the standard states are defined differently, in terms of their concentrations, and hence $K_c$ is to be used. Since the definition of standard state affects the value of $\Delta G$, we must choose that form of the equilibrium constant which uses the activity of the reacting species in terms of the quantity used to define the standard state of that species. This maintains uniformity.


Is Geologic Mineral Trapping of Chlorine Plausible? One strategy for sequestration of $\ce{CO2}$ is mineral trapping arising from geologic sequestration.  Another strategy is mineral trapping in concrete via the Mineralization via Aqueous Precipitation or MAP process.  In particular, Alkalinity Based on Low Energy or ABLE process offers an energy efficient MAP process.  However the ABLE process produces $\ce{Cl2}$ (or $\ce{HCl}$), which, at scale, produces its own environmental impact.

One way of dealing with $\ce{Cl2}$ (or $\ce{HCl}$) might be mineral trapping via geologic sequestration.

Is this plausible?

One of the problems of geological sequestration of chlorine is that most chloride minerals are very soluble.

The number of non-soluble (or very little) minerals is limited, but there are several options. The obvious ones are sodalite ($\ce{Na8(Al6Si6O24)Cl2}$) and chlorapatite ($\ce{Ca5(PO4)3Cl}$). There are other options though.

Webmineral lists a lot of chlorine-bearing minerals, and most of those are soluble. However, there could be several minerals with high percentages of chlorine that are insoluble enough under the conditions of your reservoir. But in general, your best bet are anhydrous silicates and phosphates, because those are the most likely to be stable over a long time.

That said, this stuff could be dumped in the ocean. The $\ce{HCl}$ could be neutralised with something like feldspars, turning it into clay and a solution of harmless sodium chloride.

Alternatively, you can oxidise the HCl and make chlorine gas, a useful product.


How to determine the potential of a cadmium zinc cell under standard conditions? I write the half-equations:
\begin{align}
\ce{Zn &<=> Zn^2+ + 2e^-}& E^\circ &= +0.763~\pu{V}\\
\ce{Cd^2+  + 2e- &<=> Cd}& E^\circ &= +0.403~\pu{V}
\end{align}

In order to find the $E_\mathrm{cell}$ is it right to say:
$$E_\mathrm{cell} = 0.763~\pu{V} + 0.403~\pu{V} = 1.166~\pu{V}?$$

Or should I say $0.403~\pu{V} - 0.763~\pu{V}$?

I will like to correct you that there is one problem with your question. The correct standard reduction potential is:
\begin{align}
\ce{Cd^2+  + 2e- &<=> Cd}& E^\circ &= -0.403~\pu{V}.
\end{align}
 And another thing is, you have used symbol $E^\circ$ for oxidation potential of $\ce{Zn}$ which should not be done. You can use there just $E = 0.7628~\pu{V}$ or in a better way, you can just write it with it's reduction potential value as:
\begin{align}
\ce{Zn^2+ + 2e- &<=> Zn}& E^\circ &= -0.7628~\pu{V}.
\end{align}
Now coming to your question, for these two reactions,
\begin{align}
\ce{Zn^2+ + 2e- &<=> Zn}& E^\circ &= -0.7628~\pu{V}\tag1\label{Zn}\\
\ce{Cd^2+ + 2e- &<=> Cd}& E^\circ &= -0.403~\pu{V}\tag2\label{Cd}
\end{align}
$E_\mathrm{cell}$ of any cell reaction is given as
$$E^\circ(\text{cell}) = E^\circ(\text{reduced species}) - E^\circ(\text{oxidized species})$$
or
$$E^\circ(\text{cell}) = E^\circ(\text{cathode}) - E^\circ(\text{annode}).$$

Here as $E^\circ$ value of reaction $\eqref{Cd}$ is greater (i.e. towards more positive). Hence $\ce{Cd}$  has greater tendency to accept electrons and get reduced. So the $\ce{Zn}$ is going to give electrons and gets oxidized. Now the half cell reaction will be,
\begin{align}
\ce{Zn &<=> Zn^2+ + 2e^-}\tag{at anode}\label{anode}\\
\ce{Cd^2+ + 2e- &<=> Cd}.\tag{at cathode}\label{cathode}
\end{align}

So according to formula
\begin{align}
E^\circ(\text{cell}) &= E^\circ(\text{cathode}) - E^\circ(\text{annode})\\
E^\circ(\text{cell}) &= -0.403~\pu{V} - (-0.7628~\pu{V}) = 0.3598~\pu{V}\\
\end{align}

The correct answer is $0.3598~\pu{V}$.


How to recover C18 column after prolonged exposure to phosphate buffer? A colleague of mine who occasionally does HPLC in a biochemistry lab discovered that a C18 column was not properly flushed after the previous experiment.
It appears that the column was stored for several weeks capped with aqueous phosphate buffer inside.

The main concern is partially crystallized phosphate as well as bacterial growth as there is no clear evidence whether antibiotics were added to the buffer or not.

Is there a way to recover the column, and if so, what procedure is typically recommended (solvents, temperature, flow rate etc., preferably with a reference for the lab report)?

The bad news is that HPLC columns often die because this ill-treatment. We have had the same issues and eventually several expensive columns was discarded due to the negligence of a student. Acetonitrile, when used above 90% with a phosphate buffer, leads to precipitation inside the column. I have great deal of experience with column manufacturing technology. You never add antibiotics to mobile phases. So don't worry about that. The solvents themselves, like ACN, and MeOH ensure that bacteria don't survive inside the column. This is the typical storage solvent after washing the column and ensuring all the buffer is out.

The salvaging procedure can be the following:

Run the column at really low flow rates say, 0.1 mL/min with a high aqueous content (90/10 H2O/ACN). If the pressure is extensively high change the flow direction. Columns have a flow direction marked on them. Try the direction which shows a lower pressure.
Record the pressure, if the pressure slowly goes down, this is a very good news.
If the step 2 is going well, wash overnight at 0.1 mL/min. If all went well, pressure is back to normal in the morning, you are good to go.
Never sonicate the column. It will damage the bed permanently.

P.S. Phase collapse has nothing to do with problem of precipitation inside the column. This is a completely different story. Please don't use that suggestion from LC-GC magazine (link in the comment above). Ron Majors is very experienced but here he talks about another issue.

Phase collapse is not a well understood phenomenon, a better word is de-wetting. Since the C18 phase is very hydrophobic, with highly aqueous mobile phases, the pores (where all the retention occurs) exclude water. Chromatographers, especially those who write in commercial magazines for public readership, tend to do a lot of imaginations.

If the column pressure becomes normal, repeat and old separation. If the retention time is similar, no need to worry. If it is reduced, then try the LCGC suggestion. Pressure is the main test to confirm if all ppt is gone or not.


A good book for learning paper and thin-layer chromatography What book would you recommend for learning paper and thin-layer chromatography? I don't have access to fancy equipment (just to these two basic techniques that are relatively cheap), so it would be nice if the book focused on them in addition to being easy to understand etc. It does not matter if advanced techniques are discussed in a chapter or two but if the book is solely about them, it would not offer me much.

I think an entire book on TLC/paper chromatography would be quite overkill and your best bet is on a more general practical organic chemistry book. I can recommend The Organic Chem Lab Survival Manual: A Student's Guide to Techniques by Zubrick if you are only interested in technique or need a introductory text. Else Vogel is of course the book for practical organic chemistry. TLC is covered in detail.

I'm not so sure about paper chromatography, I don't believe it is often used in the lab (at least I have never worked with it). TLC is much more common.


Determination of the iodine value of a fatty oil: Does the concentration of Na2SO3 effect the iodine value? I want to determine the iodine value of a fatty oil. For this I dissolve $\ce{1.0 g}$ of the oil in chloroform. Iodine bromide is then added and placed in a dark room for at least $\pu{30 min}$. After this, $\ce{KI}$ is added. Then, I am going to titrate with $\pu{0.1 M}$ $\ce{Na2SO3}$. Based on the added volume, I can calculate the iodine value of the unknown fatty oil.

Now I was wondering what the effect should be using $\pu{0.02 M}$ $\ce{Na2SO3}$ instead of $\pu{0.1 M}$? And is there a reason why the solution has to stay $\pu{30 min}$ in a dark room? I think there's no difference in outcome when it will be in a dark room, for example, say $\pu{3 h}$.

The reaction of $\ce{I2}$ or $\ce{IBr}$ with double bonds is slow. You have to wait half an hour for the reaction to be finished. But of course, you may wait a longer time, maybe $\pu{3 h}$ as you suggest. Or one night. Usually we are in a hurry, and we are looking for ways of avoiding to spend too much time for a given operation. That is why the procedure recommends half an hour.

And of course you may carry out the titration with a solution 5 times more diluted than recommended. The only drawback is that you may have to refill many times your burette before going to the end of the titration. Suppose $\pu{42 mL}$ of the $\pu{0.1 M}$ solution is required for performing the titration. This could be done with a usual $50$-$\pu{mL}$ burette. With a $\pu{0.02 M}$ solution, you will use $5 \times \pu{42 mL} = \pu{210 mL}$. In this case you will have to refill four times your burette. Four times! Bad luck!


Can I claim that a hot dip Aluminum coating in steel provides cathodic protection to a substrate since Al can act as sacrificial anode? I have a school experiment wherein I will put aluminum coating to a steel substrate, wherein the Al will act as a corrosion protector for the steel. We know that aluminum is also classified as a sacrificial anode. Can the aluminum coating on the steel provides a cathodic protection?

In practice, no -- at least, not at ambient temperature.  Aluminum passivates and this passive film breaks the potential electrochemical circuit.  The passivated aluminum can't react to release electrons to the steel.  For cathodic protection you need something that will corrode instead of the steel, meaning it can't passivate.

Usually, cathodic protection is offered by zinc which has less driving force, but it does not passivate and thus the electrochemical circuit remains intact.

Addendum:

Aluminizing does offer cathodic protection at high temperature where the passivation is less effective.


Does formal charge affect bond polarity? Bond polarity, as far as I understand, is a measure of the degree  to which shared electron density is distorted, and thus solely depends on the electronegativity difference.
Up until now, I had learnt that the dipole moment (charge on each atom x separation) is used as a measure of the polarity. However, for molecules with a formal charge like CO, even though the shared electron density is distorted towards the more electronegative atom, the dipole moment points in the opposite direction. For other molecules like Ozone, even though there is approximately no distortion of bonding electron density, the formal charges alone result in a dipole moment.
For a general molecule, am I supposed to take into account the formal charges for bond and thus molecule polarity, and it it even a measure of degree of distortion/ionic character at that point?

When you compute formal charges you split bonding electrons evenly between bonded atoms. This does not account for differences in electronegativity. It is similar to the difference between formal charge and oxidation state, as well explained in the Wikipedia.
A better picture (closer to the real electronic distribution) is, as Pauling would probably suggest, intermediate to the two, assuming a hybrid between possible Lewis structures (covalent and ionic).
In any case, when considering bond polarity, electronegativity overrules formal charge.    On the other hand inductive effects should not be dismissed.
The OP presented an excellent counterexample: CO. Here the carbon has a very slight negative charge relative to oxygen, as predicted based on the computation of formal charges.
I am not an expert on the development and application of electronegativity scales, but suffice it to say that these scales are (1) empirical and (2) provide useful guidelines. The original Pauling formulation of electronegativity cannot account for all possible bonding schemes. Case in point is the modification of the electronegativity scale to account for orbital hybridization, the following data via the Wikipedia:

Hybridization
χ (Pauling)1

C(sp3)
2.3

C(sp2)
2.6

C(sp)
3.1

'generic' C
2.5

Based on this modified scale the electronegativity of oxygen, 3.44, is no longer much larger than that of the carbon atom in carbon monoxide.
References

Data found in Wikipedia, original reference: Fleming, Ian (2009). Molecular orbitals and organic chemical reactions (Student ed.). Chichester, West Sussex, U.K.: Wiley.


Why do we assume that it is 1 mol? A question in my book was as following:

  Benzine and chloroform are considered very toxic organic solvents that dissolve in each other and in a solution of benzine and chloroform there is 0.45 mol of benzine. Find the  the percent of mass( percent concentration solutions) in this solution.

In the answer, It assumed that the number of moles in the solution is 1 mole then subtracted 0.45 to get the number of moles of chloroform.

Why did he do that? Why specifically 1? It will obviously change if I assume any other number. (Unless I multiply the number of moles of benzine by that number)
Or should we just say that for every a certain amount of moles of that solution has a percentage and every other percentage is right if it follows the same instructions?

Sorry, but the accepted answer is wrong. The problem is indeterminate. Is there 1 liter of solution, or 5 liters, or 10 liters, or some other amount?

If the problem was to be solved using mole fraction, then the problem should have been written that way. e.g.:

  Benzine and chloroform are considered very toxic organic solvents that dissolve in each other and in a solution of benzine and chloroform the mole fraction of benzine is 0.45. Find the the percent of mass( percent concentration solutions) in this solution.

There is no point in posing such sloppy problem statements.


What hinders SN2 more: geminal alkyl group or alkyl group on the aromatic ring? Which among these two (1-chloro-1-phenyl-ethane and 2-(chloromethyl)-1,3-dimethylbenzene) is more reactive towards SN2?

I strongly believe that it should be Y because of lesser hindrance on the electrophilic carbon.
However, the answer key of my exercise book states that X > Y. What's the reason behind that?

The correct answer is X because the 2 methyl groups on the Y are bulky enough to cause steric hindrance. Surprisingly, from what I have seen, if you only had one methyl group on Y, the SN2 reaction would proceed.


Is there any way to carry out voltammetry without a potentiostat? I'm trying to do a voltammetry experiment, but my high school doesn't want to have to buy a potentiostat (understandably, I know they are very expensive). Is there any way to carry out a voltammetry experiment, either linear sweep or cyclic, without a potentiostat? Or am I doomed to have to go back to the drawing board...

I should mention that the physics lab has some equipment like variable power sources, but not much--my school kind of under-appreciates science--and I was planning on creating a cruddy makeshift three-electrode cell.

This is a valid question. This project is entirely possible, but I would not advise it for a high school research project based on my experience building these. Much better projects are available.

A nonreactive electrode, for example gold or platinum, is required for good results.

Knowledge of op-amp design, or some other feedback control system, which is taught in advanced electronics courses is needed.  The control system fixes the potential between the the reference and the working electrode by drawing current from the third electrode.  The current drawn is measured to give some information about the system.

The Wikipedia article shows the circuit need, https://en.wikipedia.org/wiki/Potentiostat.

The open source cheapstat is available: http://web.chem.ucsb.edu/~kwp/cheapstat/

Integrated versions of the circuit are available. http://www.ti.com/product/LMP91000.  Evaluation kit ~$100 http://www.ti.com/tool/LMP91000EVM.

Information on making electrodes: https://publiclab.org/notes/JSummers/01-09-2014/potentiostat-notes-5-how-to-make-low-cost-electrodes

It is possible to build a potentiostat without advanced knowledge of electronics; however, this runs into costs. Getting experiments done with the equipment, resources, and funding available is something every scientists faces.


Energy of hydrogen bonds vs. kinetic energy of water molecules I watched an extremely engaging video "Why does ice float in water?" by George Zaidan and Charles Morton, https://www.youtube.com/watch?v=UukRgqzk-KE. In particular, video says "Below 4oC kinetic energy of water molecules falls below the energy of hydrogen bonds. So hydrogen bonds form much more frequently then they break". I decided to verify this statement, and on my way to do so I found energy of hydrogen bonds: "6-30 kJ/mol" according to this source http://book.bionumbers.org/what-is-the-energy-of-a-hydrogen-bond/. However I was unable to find a way to calculate kinetic energy of water molecules. Your help on this matter would be highly appreciated.

It looks like they calculated the average thermal energy of water molecules, which by classical equipartition theorem is $3RT/2$ kJ mol$^{-1}$. This works out at approx 4 kJ mol$^{-1}$ at 40 C. This amount of energy includes translational, vibrational and rotational energy. The vibrational quanta are large in water and almost all molecules will have only zero point vibrational energy. Rotational quanta are smaller and many will be excited at these temperatures. The smallest rotational quantum value for water is approx 0.2 kJ mol$^{-1}$ the next 6 times larger and the next 12 times (increasing as $J(J+1)$ for quantum number J) so a few rotational levels will be populated. Energy will flow between translation and rotations as random collisions occur between molecules of water. A full calculation involves using the partition functions. Calculating these and their contribution to heat capacity is treated in most undergraduate phys. chem. textbooks.


How to measure and transfer sticky solids What is the best way to measure and transfer a sticky compound in the lab, in particular for air/moisture sensitive reactions? I have a really sticky solid Wittig salt which I'm going to use in a Wittig reaction.

Ideally, you want your sticky solid to already be in a Schlenk flask or similar. If it is unfortunately still in a normal bottle, take the weight of a Schlenk flask, fill in enough of the sticky solid and re-weigh. Note the difference, this is what you have in your flask.

Assuming that the sticky solid has a low enough vapour pressure (spoilers: practically all solids do), evacuate the Schlenk flask under high vacuum and backfill with argon (or nitrogen, if that is your inert gas of choice. Add a known quantity of your (dry, stored over molecular sieve and under inert atmosphere below a septum) solvent. Ideally, you either want an amount of solvent that gives you an easy-to-handle concentration (e.g. $10~\mathrm{M}$) or an amount so that what you need is in an even volume of liquid (e.g. $1~\mathrm{ml}$).

Finally, transfer the solution into your reaction flask.

If you need the sticky solid in different solvents and it is air-stable, you can evaporate the dry solvent at the rotavap. If not, keep it in solution under argon (or nitrogen). If the latter is the case, do your colleagues a favour and choose a standard concentration, not a good filling amount for exactly your reaction.


IUPAC Nomenclature for aldehyde I was trying to name the following compound and I thought that if both the aldehyde group carbons should be included in the parent chain or I should go with the longest chain . I saw the solution for this and got two answers as I) 3-(Formylmethyl)hexanal and 2) 3-Propylpentanedial.
Can someone please help me in naming it correctly??

IUPAC (2013) rule

  P-44.1.1 The senior parent structure has the maximum number
  of substituents corresponding to the principal characteristic
  group (suffix) or senior parent hydride …

(more e.g. here) is applied in the phase of selecting "parent structure", before than the "longest chain rule" (P-44.3.2, e.g. here) in the phase of selecting the "principal chain".

Therfore the latter name, 3-propylpentanedial, is correct.


Why are 'table salt and sugar are both terrible candidates for recrystallization'? I received a comment on one of my other chemistry questions stating that

  table salt and sugar are both terrible candidates for recrystallization, albeit for different reasons.

Why are salt and sugar terrible candidates for recrystallization?

The solubility of $\ce{NaCl}$ has very poor temperature dependency. If you dissolve it at $100^\circ\rm C$ and precipitate at $20^\circ\rm C$, you'll essentially lose about 35 g to purify 3 g. Even using a fridge, you'll still lose more than half of the product, which is quite a bad business. Industrial process relies on evaporation, I think.

Sugar is another story. It does have a nice temperature dependence, but it dissolves too well, even when cold. Just imagine dealing with that awfully viscous syrup, like molasses. It would take forever to filter, and quite a while to crystallize. Industry can handle that, but for home chemistry I'd suggest something more pleasant.


Nucleophilic Substitution with ethyl bromide and NaOH Given is following nucleophilic reaction with product A. Draw the mechanism of the reaction and give the structure formula of A.

Here is my solution, I am sure it is not right. Could you help me, step by step to understand this reaction.

With the help from Waylander (s.o.) we get the following mechanism.


Can Substances on Titan be Reacted at a Thermodynamic Gain? Titan, a moon of Saturn, has readily available water ice ($\ce{H2O}$) and methane ($\ce{CH4}$) on its surface, at a frigid $\pu{93.7 K}$.

It would be significant if these (or possibly, other available hydrocarbons) could be reacted in a self-sustaining way, releasing usable energy.  I presume the best way of doing this would be to react by the simple following two step process:

Split water into hydrogen and oxygen:
$$\ce{2 H2O → 4 H2 + O2}$$
Burn oxygen and methane:
$$\ce{CH4 + 2 O2 → CO2 + 2 H2O}$$

Keeping in mind the starting temperature, does this sequence of reactions operate at a net energy release (that is, is it theoretically possible to use this reaction to power itself, with energy leftover)?  If not, could other hydrocarbons* found on Titan fit the bill?

*Methane, ethane, propane, and benzene exist. Others may as well.

The thermodynamic problem is not so much with the hydrocarbons reacting with O2 as it is with forming O2 from water.  That reaction is energetically uphill.  A big hill.  In fact water is a waste product of your second reaction.  The reason the second reaction is energetically favorable is due to the high stability of CO2 and water relative to the reactants, methane and oxygen.

Would it make sense that you should be able to then convert the water from your second reaction into O2 and H2 in a way that would consume less energy than it took to make the water in the first place?  That is almost equivalent to what you are proposing by starting out with the water already present on Titan.

The prohibitive energetics of your reactions apply on Earth as well as Titan.  You need a process to split the water to burn with hydrocarbons that will consume less energy than the energy you will get back from burning the hydrocarbons.  You could conceive of using solar power as the energy source for converting water into O2 and H2.  Or other energy sources.  But that wasn't your question, and you definitely won't get the energy back that you would require via your second reaction.  And good luck with my solar energy idea on Titan ;)


Are Pulay forces expensive to compute? In a thesis I am reading, it is said that one of the reasons for using plane-wave basis sets for first-principles molecular dynamics (aka ab initio MD) is that the Pulay forces[1,2] that arise from an MD using atomic basis sets are computationally expensive to calculate.

While I understand that having additional terms means more code to write, i.e. they make writing the software harder to write, is it true that they are CPU-intensive to compute? The criterion I would use to quantify this subjective statement is:

  Given that you have already calculated the energy and forces at that point, you have already computed a large number of  integrals required for this task and involving basis functions: overlap integrals, terms of the form $\left\langle\phi_\alpha\left|\hat{A}\right|\phi_\beta\right\rangle$ where operator $\hat{A}$ is either the hamltonian, its gradient, or any other operator necessary for the calculation of energy or forces. Would the calculation of Pulay forces require any more integrals to be evaluated, or can it be straightforwardly computed from those previously-calculated integrals alone?

[1] P. Pulay, Molec. Phys. 19, 197 (1969)
[2] See also slide 6 of this

Yes, you need additional quantities beyond the minimum necessary for calculating energies and the Hellmann-Feynman piece of the force when the wavefunction is not variational.

Here is a very rough sketch of why:

Puláy forces arise from applying the chain rule for calculating forces and was first discussed in the context of applying the Hellman-Feynman theorem.

Recall that the force is the change of energy with changes in (nuclear) coordinate and the energy is the expectation value of the (electronic) Hamiltonian, $E = \left<\psi\left|H\right|\psi\right>$. Applying the chain rule to this:

$$-F = \nabla E = \left<\psi \left\vert \nabla H \right\vert \psi\right> +  2\left< \nabla\psi \left\vert H \right\vert \psi\right>$$

The first term is what you get from Hellman-Feynman and is the expectation of a one-electron operator of the first you have listed. The second term goes away in Hellmann-Feynman only because it assumes the wavefunction is variational.

If you expand out the orbitals in the Slater determinant in some AO basis $\chi$:

$$ \psi(r_1, r_2, ..., r_N) = \phi(r_1) \wedge \phi(r_2) \wedge \dots \wedge \phi(r_N) $$
$$ \phi(r_1) = \sum_i c_i \chi_i(r_i) $$

then it is clear that $\nabla\psi$ generates terms which are MO coefficient derivatives (usually denoted $c_i^x$) and also AO derivatives ($\chi_i^x$). If the wavefunction were variational, then by definition all these derivatives are zero. If you work through the algebra, you will find that there are some new quantities that need to be calculated in order to work out the non-Hellmann-Feynman theorem, most notably the one-sided AO overlap derivative matrices $\left<\chi_i^x\vert\chi_j\right>$ which do not normally show up otherwise. These terms turn out to be critical for getting correct ab initio molecular dynamics since the wavefunction is rarely (if ever) variational in time.

There are tricks one can do to reduce the cost of calculating MO coefficient derivatives , but they cannot be obtained for free in general. And there's really no way around calculating the overlap derivative terms.


What gives milk its characteristic white color? The other day, my kid brother asked me this:

  What pigment gives milk a white color?

Now, I found that particular question quite interesting since, as a student of Biology, I already had a rough idea as to what constitutes (cow's) milk: Water, sugars, fats, and a whole bunch of vitamins and minerals. Yet as far as I know, none of them play a "pigment"-like role anywhere else in nature... which has left me deeply perplexed.

So what pigment is responsible for milk's characteristic white color?

There is no "pigment" that makes milk white.

According to Wikipedia:

  A pigment is a material that changes the color of reflected or transmitted light as the result of wavelength-selective absorption.

Going by that definition of a "pigment", then there is no (white) pigment in milk. In fact (as @Ivan points out in the comments) there is no such thing as a "white pigment", because that would require the "pigment" to absorb no visible radiation (if it absorbed all visible radiation, it'd appear black)

That's because the real reason for the white color of milk is a physical phenomenon called Scattering. (Source: Optics, 4th edition, E. Hecht)

(Diagram's from Wikipedia)

Milk is an emulsion of fat dispersed as these really tiny droplets in water. These fat globules (and some milk proteins) scatter light of all wavelengths in the visible spectrum. This is why milk appears white

This is the same reason (albeit different cause) as to why clouds often appear white; except here, the scattering is due to suspended water droplets and dust particles.


Carbon dioxide sequestration in Limestone The chemical reaction where dissolved carbon dioxide forms carbonic acid and then reacts with various calcium containing minerals to form calcium carbonate and bicarbonate is rather slow. I know that several years ago there was research into possible ways to speed up the reaction, possibly as a candidate for a truly long term way to scrub $\mathrm{CO}_2$ from the atmosphere. The inspiration for this idea is to speed up the part of the geologic carbon cycle that regulates atmospheric $\mathrm{CO}_2$ concentration.

Does anyone know how the research into catalysts or processes to speed this up are going?

This process does not rely on catalysts, but it addresses the essence of the question. The process described is to inject liquefied carbon dioxide into basalt formations, even if the basalt is very old. This makes the it plausible that we could scrub the excess $\mathrm{CO}_2$ from the atmosphere by:

growing plants,
burning those plants, capturing the CO2, and
burying that carbon using, partly, the energy from the burned plants.


Why does the anode solution contain Sn2+ in a Sn-Cu voltaic cell? Suppose you have the following voltaic cell: $\ce{Sn_{(s)}|Sn^{2+}_{(aq, 1.0 M)}||Cu^{2+}_{(aq, 1.0 M)}|Cu_{(s)}}$ and the salt bridge is $\ce{KNO_3}$. What I don't understand is why you need to have $\ce{Sn^{2+}}$ ions initially in the $\ce{Sn}$ half cell. As $\ce{Sn}$ is oxidized, electrons are supposed to travel through the wire connecting the two half cells and reduce the $\ce{Cu^{2+}}$ ions in solution, but if we have $\ce{Sn^{2+}}$ ions initially in the $\ce{Sn}$ half cell, the electrons might just as well reduce the $\ce{Sn^{2+}}$ ions. Why would the electrons travel all the way through the wire when there is $\ce{Sn^{2+}}$ ions right next to them that could be reduced?

I also don't understand why the $\ce{Cu}$ cathode has to exist. All we need is $\ce{Cu^{2+}}$ ions in solution that will be reduced, so why can't we just dip the wire directly into $\ce{Cu^{2+}}$ solutions. As the reaction proceeds, a $\ce{Cu}$ cathode will be created anyways as $\ce{Cu}$ metal precipitates on the wire.

I am obviously misunderstanding something, because according to my understanding the following voltaic cell should work: In one half cell, we have a $\ce{Sn}$ anode in water connected to a wire. The wire is submersed in the other half cell which is simply $\ce{Cu^{2+}}$ solution. The $\ce{Sn}$ gets oxidized and produces $\ce{Sn^{2+}}$ ions in the water. The electrons travel through the wire and reduce the $\ce{Cu^{2+}}$ ions in the cathode solution. The anode solution will become more positive and the cathode solution will become more negative, so we still need a salt bridge.

A voltaic circuit needs to be a complete circuit first an foremost. So, to answer your 2 main concerns, think of a voltaic cell as a electric circuit:

  In one half cell, we have a $\ce{Sn}$ anode in water connected to a wire.

Water is not a good conductor of electricity, dissolved ions are needed, such as the $\ce{Sn^{2+}}$ ions are needed to facilitate the movement of the electric current in the first place. Also, the presence of the ions allows the overall redox reaction be spontaneous to provide the electric current (ChemWiki).

  The wire is submersed in the other half cell which is simply $\ce{Cu^{2+}}$ solution

According to the ChemWiki Voltaic Cells, the metals plates provide a surface for the reactions to occur on, a wire just submersed into the solution provides a very small surface area for the reaction to occur.

A diagram from the UC Davis ChemWiki page Electrochemical Cells helps explain this (the example uses zinc instead of tin, but the principle is the same):


Can I safely use a separatory funnel to extract the precipitate from a solution? Can a separatory funnel be used to collect a precipitate, or would that risk clogging the stopcock on it?

I'd like to recoup the precipitate for filtering without passing the entire solution through a filter (to avoid contaminating it, and also for convenience / saving time).

Good thought, but Waylander is correct.

However, you might consider centrifugation to consolidate the ppt, then pour off the supernatant. If you don't have a centrifuge, and the quantity of liquid is not too large (100 mL or less), you could make your own, since it seems that your ppt settles fairly fast.

Put the solution in a tough plastic bottle, cap it, and get a pretty strong string or light rope, tie the bottle SECURELY and twirl it over your head. Maybe outdoors is better for this, unless you are very careful. A few gentle swings (~20) should be enough, if it is going to work at all. I have had some success with this method a few times.

Secure tying is key, since the extra Gs will put a strain on your knots. I did something like (vertical swing, about 4 foot circle) this inside a room with 12 foot ceilings, and the plastic bottle hit the ceiling and splashed (I did not have the cap on). Fortunately, the stain was not very noticeable, but I did lose some of the material.


What is the carbon content, by weight, of vegetable oil? I have only a high school understanding of chemistry, so please bear with me and my ham-fisted question.  I’m trying to understand the carbon content in vegetable oil.  I get that various oils have differing densities, so the measure needs to be “by weight”.  Molecular weight doesn’t help, I need the measure in useful measures like “grams per liter” (preferred).
(Also of note, I’m not certain I got the right tags)
UPDATE: Based on feedback, I looked up oil production per hectare to determine an oil type.  It appears that the highest production per hectare (which feeds into the analysis I'm working on) for grain-based oils is rapeseed.  For tree-based fruits, it's oil palm, and then there's algae oil.
I'm trying to establish how much carbon is embodied in the oil production of one acre of crop (a bit of information that probably would have been helpful to say first).

Vegetable oils are mainly made of glycerol esters of oleic acid, linoleic acid and palmitic acid. For example, the fatty acids extracted from olive oil are a mixture of $74$% oleic acid, $11$% palmitic acid, and $10$% linoleum acid, plus 5% other acids. These proportions are average values that may change from one sample to the next one.
The mass proportion of carbon atom is $77.2$ % in glycerol trioleate, $78.3$ % in glycerol trilinoleate, and $75.8$ % in glycerol tripalmitate


Cannizarro reaction of chloral When chloral undergoes reaction with a concentrated alkali, will it undergo Cannizarro reaction to form the corresponding alcohol and acidic salt, or should it form chloroform instead by departure of trichloro carbanion instead of the hydride ion?

I understand that the trichlorocarbanion is stabilized by the $-I$ effect of chlorine, so why not form that instead?

In the Cannizarro reaction an aldehyde with no alpha-hydrogens is treated with $\ce{OH^-}$.  Through a series of equilibria the aldehyde goes on to disproportionate into the corresponding carboxylate and alcohol as shown in the following mechanism.

Again, the only requirement for the Cannizzaro reaction to occur is that the aldehyde not have any alpha-hydrogens.  Formaldehyde and benzaldehyde are two very common aldehydes that undergo the reaction when treated with $\ce{OH^-}$.  Chloral ($\ce{CCl3CHO}$), having no alpha-hydrogens also undergoes the Cannizzaro reaction (reference)

Edit: Response to OP's comment

A key part of the Cannizzaro mechanism is shown in the second line of the above mechanism.  Here a hydrogen is transferred in a concerted step to the second molecule of aldehyde.  Free $\ce{H^-}$ is not generated in the process, rather it is a smooth, concerted transfer of a hydrogen atom.

[Note: This has been proven experimentally, when the reaction is run in $\ce{D2O}$, no deuterium winds up attached to the alcohol carbon.  This shows that free $\ce{H^-}$ was never involved, but rather a concerted transfer of the aldeyhdic hydrogen takes place].

Thanks to Mithoron for noting that, with chloral, the haloform reaction can apparently compete with the Cannizarro reaction.  Presumably, the pathway followed depends upon the reaction conditions.


When does nucleophilic attack and proton transfer happen? from the reaction, the 3rd molecule (carboxylic acid) formed undergo neutralization with OH- (proton transfer to a base).

However, is it possible for the 3rd molecule to behave as the 1st molecule (ester) and undergo nucleophilic attack by the OH- as they both have an electrophilic centre (carbon)?

If yes, will this reaction form a cycle (never forms carboxylic salt). If no, may I ask why? Is it related to stability or other effects? Thank you very much

Because you are a beginner to organic chemistry, I'll tell you this: Both esterification (e.g., Fischer esterification) and de-esterification reactions are reversible. The question you had at the end is why the last step of base catalyzed de-esterification reaction is only forward reaction. As you have learned in general chemistry class, acid-base reaction is one of the fastest reactions in chemistry (not counting nuclear reactions). Thus, the last step is governed by preference for acid base reactions:

$$\ce{R-COOH + OH- -> R-CO2- + H2O}$$

The $\mathrm{p}K_\mathrm{a}$ of $\ce{R-COOH}$ is about $4.7$ ($\mathrm{p}K_\mathrm{a}$ of $\ce{CH3-COOH}$ is $4.75$) while $\mathrm{p}K_\mathrm{a}$ of $\ce{H2O}$ is $15.7$. Therefore, forward reaction is about $10^{10}$ times faster than backward reaction.

Also remember, this reaction is in aqueous medium, therefore there are much more $\ce{H2O}$ molecule available than $\ce{R-OH}$ molecules. Thus, going back to ester reaction is much slower than becoming acid (governed by Le Chatelier principle).


How can normal potentials be explained? For these two metals, the standard reduction potentials are:
$$
\ce{2e-} + \ce{Cd^{2+}} \rightarrow \ce{Cd}:\mbox{ } E^0 = -0.403 V\\
\ce{2e-} + \ce{Ni^{2+}} \rightarrow \ce{Ni}:\mbox{ } E^0 = -0.25 V
$$

But the simple ionization energies are:

$$ \ce{Ni} \rightarrow  \ce{Ni^{2+}} \mbox{   } \Delta H=\Delta H_{i_{1}} + \Delta H_{i_{2}} = 2490 \ce{kJ. mol^{-1}}$$ $$ \ce{Cd} \rightarrow  \ce{Cd^{2+}} \mbox{   } \Delta H=\Delta H_{i_{1}} + \Delta H_{i_{2}} = 2500\ce{kJ. mol^{-1}}$$
The difference in $E^\theta$ is greater than the difference in ionization energy of both elements.
How can this be explained?

You are acting under two false assumptions

Electrode potentials are measured relative to real zero. They are not, there is no practical procedure to measure absolute electron potential. All standard potentials are measured relative to standard hydrogen electrode (well, actually some more pragmatic standard electrode, but recalculated to be against standard hydrogen electrode)
That values got for gas has anything to do with values for condensed matter. Hydration energy can and does shift electrode potentials greatly, as well as energy of crystal lattice.


What is the known cheapest chemical solution to extract CO2 from air on industrial scale? As far as I know, the cheapest non-chemical solution is: grow grass, and burn it. The resulting $\ce{CO2}$ is from the air. But I am interested on chemical industrial solutions now.

As far as I know, the main difficulty of the process is that the ratio of the $\ce{CO2}$ in the air is very low. Thus, extracting it means a large entropy difference.

My layman's intuition is that maybe some substance exists - probably an organic compound - which binds especially strongly to $\ce{CO2}$ and it is easily regenerated. One would pump air through it, and then regenerate it to get the $\ce{CO2}$ back. If such a compound exists, what is it?

The process should be powered by electrical energy only. Thus, solutions such as "buy gas, burn it, and you get much cheaper $\ce{CO2}$" are not acceptable.

On "cheapest" I understand, what is cheapest in industrial size, not in a laboratory? (for example, in a large chemical factory)

Your intuition is correct, and such compounds are plenty. The cheapest are probably $\ce{NaOH}$, $\ce{Ca(OH)_2}$ and the like, but these are not easily regenerable. Then there are organic amines, like monoethanolamine, which can do the same in a reversible manner. Indeed, they are used in industry.


Why doesn't fluorine harm the ozone layer the way chlorine and bromine do? I don't understand the chemistry here: fluorine is a halogen too.
I read that fluorine in the atmosphere readily forms HF, which is (somewhat) stable and doesn't catalytically break down ozone.
But, shouldn't loose chlorine in the atmosphere form HCl similarly?

TL;DR: The mechanism is different for fluorine and chlorine atoms. Fluorine atoms react and forms stable HF molecules while chlorine atoms turns into a radical by the action of UV which helps in the destruction of ozone.
Long answer:
Ozone depleting substance(ODS) are gases that take part in ozone depletion process. Most of the ODS are primarily chlorofluorocarbons(CFC) and halons that contain chlorine and bromine atoms respectively which reach the stratosphere and reacts with ozone and lead to ozone depletion. In this context, the chemical component responsible are chlorine radical (Cl·) and bromine radical (Br·). The ODS travel to the stratosphere without being destroyed in the troposphere due to their low reactivity and once in the stratosphere, the Cl and Br atoms are released from the parent compounds by the action of ultraviolet light, e.g.
$$\ce{CFCl3 ->[UV] Cl· + ·CFCl2}$$
Ozone is a highly reactive molecule that easily reduces to the more stable oxygen form by the catalytic action from the halogen radicals. Cl and Br atoms helps in destruction of ozone molecules through a variety of complex cycles. A simple cycle is as follows:
$$\ce{Cl· + O3 → ClO + O2}$$
$$\ce{ClO + O3 → Cl· + 2 O2}$$
Chlorine radical is released and the process is continued and the overall effect result in ozone depletion. Halon gases contains bromine and the process is similar to that of what CFCs does but the action is more potent and is more efficient in this ozone depletion process.

What about fluorine and iodine?

CFCs also contains fluorine atoms which gets released along with chlorine and bromine on photodissociation in the stratosphere. But this fluorine by itself does not contribute to ozone depletion. Fluorine atoms released are quickly sequestered into carbonyl compounds and subsequently into hydrogen fluoride by reacting with water and methane molecules which is very stable in the stratosphere. As such, CFCs are now being replaced with Hydrofluorocarbons (HFCs) according to Montreal protocol (but they do contribute to global warming).
Iodine containing organic molecules react so rapidly in the lower atmosphere that they do not reach the stratosphere in significant quantities and does not play role in ozone depletion.
References

https://en.wikipedia.org/wiki/Ozone_depletion
https://csl.noaa.gov/assessments/ozone/2010/twentyquestions/Q7.pdf
https://hal.archives-ouvertes.fr/hal-00256296/document


Is the current situation possible for recreaction in a lab? I want to convert cellulose to another form (crystalline => amorphous). Cellulose requires a temperature of 320 °C and pressure of 25 MPa to become amorphous in water.(Is the pressure achievable in the lab?)

Is this doable in the lab and how should I begin on reproducing that situation (which tools would I need) ?

I'm guessing that a bunsen burner would produce high enough temperatures (but I'm not quite sure about the pressure)

EDIT to my question, following up the advice of bobthechemist...

Crystalline substance to amorphous - Other ways?
You can convert crystalline cellulose to the amorphous, more reactive form by heating it to a temperature of around 320°C under a pressure of 25MPa.

But are there other, less extreme ways, to achieve that amorphous form of the compound?
So because I can't achieve these high temperatures, how can I achieve it in another way?

25MPa is just north of 3600psi for those of us that haven't converted to SI units; 320*C ~= 608*F. Your average lecture bottle is rated to about 3000psi, while compressed gas cylinders can go as high as 6000psi. Trouble is, most of them don't stand up well to being heated by an open flame. Temperature is another beast; your Bunsen burner can easily achieve a flame temperature well in excess of 600*F (and a simple variation, the Meker burner, can hit more than twice that), but your Pyrex glassware is only rated to 500*F, and that for short periods. Bunsen burners are often used for glassworking on the lab bench, creating custom tubing rigs by softening straight tubing and then bending or stretching to create bends and bottlenecks. If the glass can be molded over a Bunsen burner, it's not going to be able to sustain the temperatures you're talking about.

The short of it is that both the temperature and pressure are possible, but neither are achievable with a benchtop apparatus in a fume hood; you'll need an HTHP reactor vessel rated for your experimental conditions. They exist, but whether you can get access to one is another question only your superior can answer. If you were planning on visually studying the change in the cellulose, you'll need a sapphire viewport, and these get very expensive very quickly.

Once you have the right tool for the job, it becomes a rather academic task of calculating how much volume of reactants are required to achieve the desired pressure in the vessel when heated to temperature, then filling the vessel and evenly heating it. Only two of the vessels in the linked column hold any appreciable amount of material at the proper temp and pressure, and one of them will just barely meet your needs; if you unevenly heat any of these vessels, by the time the far end reaches the proper temperature the spot being heated will have failed.


Enthalpy change vs. change in internal energy in a system I've been trying to get my head around the differences between enthalpy change and internal energy change for a system. Here's my understanding of it so far. I'd appreciate any feedback on my understanding so far on the concepts - especially if there are misconceptions (and there probably are)

Internal energy is the 'intrinsic' energy of some system, can be measured with reference to something within the system, e.g. chemical bonds, vibrational energy states, etc.
Enthalpy change is the heat (i.e. energy transferred due to temperature difference) change at constant pressure due to some chemical process.

$\Delta H = \Delta U + P \Delta V$

$\Delta H$ is characteristic of a given chemical process - i.e. this value is the same regardless of what conditions it is run under (e.g. varying pressures, volumes, starting temperatures). As long as it's the same reaction, $\Delta H$ will be the same. It's the overall energy input required to make a process work, and is split between influencing $U$ and performing $PV$ work.

$\Delta U$ is a function of the specific conditions (i.e. volume, pressure) in which a specific instance of a chemical process is run.

So it's theoretically possible to set up a chemical process (say vaporisation) to run in such a way so that $\Delta U$ is zero, and all the enthalpy change is accounted for in the $P\Delta V$ term?

And what about cases where P is not constant? What happens then? Can we calculate the value of $\Delta H$? And does $\Delta H$ even have a meaning in this case?

Both $U$ and $H$ are not directly related to any particular process.  They are physical properties of the material being processed, and are unique functions of the thermodynamic state of the material, and not the process.  They are directly proportional to the amount of material, and its thermodynamic state is characterized by its temperature and pressure (and composition for a multicomponent system).  The internal energy is a fundamental property, but the enthalpy is a defined property given by $H = U + PV$, and is just a convenient parameter to work with is solving many kinds of thermodynamics problems.

For a process carried out at constant pressure in a closed system, the amount of heat $Q$ added to the system turns out to be equal to the change in enthalpy between the initial state of the material and the final state of the material.  For a chemical reaction carried out at constant pressure, the amount of heat $Q$ that must be added to the system in order for the final temperature of the products to equal the initial temperature of the reactants turns out to be equal to the enthalpy of the pure products minus the enthalpy of the pure reactants at that temperature.  Both these results follow from the first law of thermdynamics.

For a chemical reaction, generally both the internal energy and the enthalpy change between the initial and final equilibrium states.

  And what about cases where $P$ is not constant? What happens then? Can we calculate the value of $ΔH$? And does $ΔH$ even have a meaning in this case?

Since $H$ is just defined as $U + PV$, we can always calculate the change in enthalpy between two thermodynamic equilibrium states of a system.  $\Delta H$ certainly has a value in this case, even if it is not equal to the amount of heat added for a process between the two states.

The critical thing to remember is that $U$ and $H$ are physical properties of the materials comprising a system, and are independent of any specific process.


Retrosynthesis of a cyclopropane type molecule I'm trying to work out a retrosynthesis of the following molecule:

The issue I'm having is regarding the building up of the carbon skeleton. I know I can build the cyclopropane group using a carbene (diiodomethane, zinc) on the alkene. However, the alkene is proving difficult to build, I can't see why an obvious way around building this (Witting/from the equivalent alkyne) wouldn't work here. Is there something I am missing here like a rearrangement?

How about this? Cyclopropane from carbene + maleic anhydride. Add 2 eq methyl Grignard and trap with methyl iodide.

Possibly, you could do the exhaustive addition of Grignard first, then cyclopropanate and methylate in one shot with diazomethane. Thanks to Beerhunter for this suggestion in comments.


Reaction between Copper nitrate and ascorbic acid I am confused on the products of reaction between copper nitrate and ascorbic acid. When I searched this online, one of the reactions I got was: $$\ce{2Cu(NO3)2 + C6H8O6 <=> 2CuO + 4H2O + CO2}$$

I am really confused as to how they determined that to be the product. My thinking was that the ascorbic acid breaks into $$\ce{HC6H7O6 <=> H+ + C6H6O6-}$$ and so the hydrogen ion, whose activity is higher than copper, replaces the copper to form nitric acid. I was thinking that this could be a double displacement reaction and so the copper ascorbate would become a precipitate. What am I thinking wrong here?

It is always good to cite the link when you mention that you saw a redox reaction online. It is wrong anyway.

Aqueous Cu(II) is not an strong enough oxidizing agent that it will convert organic matter into carbon dioxide. In other words, it is a weak oxidizing agent. Ascorbic acid is oxidized to dehydroascorbic acid and Cu(II) should be reduced to Cu(I). If you heat the solution strongly, Cu(I) may convert into Cu$_2$O.

Ascorbic acid is indeed an acid, but there is nothing which prevents its conjugate base to behave like a reducing agent. Many organic acids can behave like that.

  I am really confused as to how they determined that to be the product.

By conducting the experiment. There is no other simple way to predict what will happen in a complex redox reaction.


The structure of sodium cuprous thiosulphate While studying about copper sulphate, I came across a reaction involving sodium thiosulphate (Hypo) and copper sulphate.

Although I don't know the utility of this reaction, I felt that it was important to understand the structure of $\ce{Na4[Cu6(S2O3)5]}$, which is formed as a the major product of the overall reaction.

The full set of reactions is given as follows:

$$
\begin{align}
\ce{CuSO4 + Na2S2O3 &-> CuS2O3 + Na2SO4}\\
\ce{2CuS2O3 + Na2S2O3 &-> Cu2S2O3 + Na2S4O6}\\
\ce{3CuS2O3 + 2Na2S2O3 &-> Na4[Cu6(S2O3)5]}
\end{align}
$$

  What is the structure (3D, not Lewis structure) of $\ce{Na4[Cu6(S2O3)5]}$ (or the ion $\ce{[Cu6(S2O3)5]^{-4}}$)?

Reference: Allen JEE-Mains+Advanced Chemistry, Unit 10, Chapter: Transition Elements:

For knowing these three dimensional complex structures, which are hard to predict theoretically, we have to use X-Ray diffraction or spectroscopic techniques (IR, NMR, or, mass). I don't have any such facilities to verify the complex structure, but I may logically predict a  logical (possibly incorrect) structure.

In the complex ion $\ce{[Cu_6(S_2O_3)_5]^4-}$, all the $\ce{Cu}$  is in +1 oxidation state, and thus can form a tetrahedral geometry around it with the help of other $\ce{Cu^+}$ ions and $\ce{S_2O_3^2-}$ ions. On the other hand,  thiosulfate can act as both a bidentate and a monodentate ligand.

So, in the predicted structure, all the $\ce{Cu^+}$ are in tetrahedral geometry. Three $\ce{S_2O_3^2-}$ act as bidentate ligands, whereas the other two $\ce{S_2O_3^2-}$ ions act as monodentate ligands.

Note that the donor atom of the thiosulfate anion can be either all $\ce{S}$, all $\ce{O}$, or a combination of $\ce{S}$ and $\ce{O}$. It is indeed hard to predict (because in some complexes, $\ce{Cu}$ binds with $\ce{O}$ regardless of being soft acid to preferably bind with $\ce{S}$ - as in $\ce{CuCl_2.4dmso}$) and thus is best to determine through IR spectroscopy by looking at $\ce{S-S}$ stretching frequency. Thus, the donor atoms may be different from what is shown in the possible structure.

The possible structure is:


Why is the volume in this compress gas cylinder constant throughout? I am stuck on this gas question and its really stressing me out.

  A compressed cylinder of gas contains $2.25 \cdot 10^3~\mathrm{g}$
  of $\ce{N2}$ gas at a pressure of $4.25 \cdot 10^7~\mathrm{Pa}$ and a temperature of
  $19.4~^\circ\mathrm{C}$. What volume of gas has been released into the atmosphere
  if the final pressure in the cylinder is $1.80 \cdot 10^5~\mathrm{Pa}$?
  Assume ideal behaviour and that the gas temperature remains unchanged.

I know I have to use the ideal gas law for this.
I've seen the solution for this problem, but something still doesn't make sense to me.
In the solution they used $$\frac{nRT}{p} = \frac{n_\mathrm{f}RT}{p_\mathrm{f}},$$ to find the final amount of nitrogen gas in moles, $n_\mathrm{f}$, remaining in the gas cylinder. But why is the initial volume and final volume the same? Since gas is being released into the atmosphere, shouldn't the final volume change?

It is about the volume of the cylinder, which of course does not change. You do not also consider the volume of the escaped nitrogen gas.