Cases when Raoult's law is unsolvable Setup

Consider a closed binary mixture of known total molar composition $n_i$, held at volume $V$ and temperature $T$.  The equilibrium phase composition is determined by $\mu_i^{vap} = \mu_i^{liq}$. Assume the vapor is an ideal gas mixture, the liquid is an ideal mixture, Poynting's correction is negligible, and $V_{liq} \ll V$. Then Raoult's law holds:

$$y_i P = x_i P_i^{sat}(T)\quad \tag{1}\label{1}$$

Simplify

For convenience, denote vapor quantity = $v_i$, liquid quantity = $l_i$, total quantity = $n_i=v_i+l_i$, and "saturation quantity" = $s_i=\frac{P_i^{sat}(T)\,V}{R T}$.  A closed system means $n_i$ are constant, and $s_i$ are constant at constant $T$.  For simplicity, choose units such that $R T = V = \sum_i n_i=1$, so that Raoult's law becomes

$$ n_i - l_i = s_i \frac {l_i}{\sum_i l_i} \tag{2}\label{2}$$

where $l_i$ are the only independent variables.  (The solution/result is independent of these choices. It holds for any Raoult's law system at constant $N, V, T$. It generalizes to infinite systems with fixed $\sum n_i/V$.)

Question

Eq. ($\ref{2}$) can be algebraically solved.  But it has no solution when $s_1>1$ and $s_1 - n_1 ≥ n_2 \frac{s_1}{s_2}$.  For example, if $s_1=2$ and $s_2=1/2$, then no solution exists when $n_2≤1/3$ ($x_2=y_1$ and $y_2=x_1$ at $\lim n_2\to \frac{1}{3}^+$).

This seems to predict that any generic species can form an uncondensable mixture if one is uncondensable (as a pure component), and there are few enough of the condensable species.  Is this correct?

As you have not provided a source for your definition of $s_i$, I shall redefine it in terms of Raoult's Law and the Ideal Gas Law. To do this, we shall first take Raoult's Law and solve for $P$.

$$\begin{align}y_iP&=x_iP_i^{sat}(T)\\P&=\frac{x_iP_i^{sat}(T)}{y_i}\end{align}$$

we shall then take this form, and plug it into the ideal gas law, considering only the gas phase of the equilibrium system. Note that because the system is at equilibrium, $P_{bulk}=P_{liq}=P_{vap}$ and $T_{bulk}=T_{liq}=T_{vap}$. If the volume of the liquid phase is assumed to be significantly lower than that of the vapor, then $V_{bluk}=V_{vap}$. After plugging in this, we will then work towards the form of $s_i$ as shown.

$$\begin{align}P_{vap}V_{vap}&=n_{vap}RT_{vap}\\PV&=n_{vap}RT\\n_{vap}&=\frac{PV}{RT}\\n_{vap}&=\frac{x_iP_i^{sat}(T)}{y_i}\frac{V}{RT}\\\frac{n_{vap}y_i}{x_i}&=\frac{P_i^{sat}(T)V}{RT}\end{align}$$

From this, and the definition of $s_i$ as you stated, we can show that

$$s_i=\frac{P_i^{sat}(T)V}{RT}=\frac{n_{vap}y_i}{x_i}$$

which is subject only to the reasonable assumptions of ideality as stated above, and the bounds per the definition of each term. Namely, $\left\{P_i^{sat}(T),V,T\right\}$ must be positive, real numbers and that $\left\{x_i,y_i\right\}$ must be positive, real numbers with bounds of $\left\lbrack0,1\right\rbrack$.

No further inequalities are required to describe the system.

To solve such a system, you would then set up a system of equations and numerically backsolve for an answer, in terms $n_{vap},x_{i},y_{i}$. For example you might feed a set of equations the seed values of $n_{vap}=1$ and $x_i=.5=y_i$ such as seen below, then optimize your values to minimize the error between the calculated and target values.

The above seed values could then optimize to a particular solution that looks like. Note that this does not represent the whole of the solution space.

If you wish to further control where your particular solution ends up, you can restrict the solution space by hold one of the values to be a known constant. Below is one such example as described per your comments.

In this case, it is not possible to find a reasonable solution, as $n_{vap}=0.25~\text{mol}$ lies outside the solution space for this system of equations. Without going too heavily into the derivation, it can generally be stated that this value is a bit too low to garuntee that you will be able to find a solution. This more says that if you had this system, it would likely evolve off more of its liquid into the gas phase at in order to reach equilibrium than it says the system could not exist, though yours is not an unreasonable reading of the result.

A good rule of thumb for this would be that you can expect to find a reasonable solution for $\left\{x_i,y_i\right\}$ for any system in which $s_1 \leq n_{vap} \leq s_2$ holds, however this is by far the most complicated way to solve out this system of equations. A simpler approach to this would be to hold either $x_i$ or $y_i$ constant to restrict the solution space, as any constant value on the bounds of $\left\lbrack0,1\right\rbrack$, as already assumed will garuntee a unique solution to the system of equations.


What are we seeing in this AC salt-water copper-wire experiment? There's a highly watched video on YouTube by "ElectroBoom" on the effects of passing AC (not DC) current through salt water. The water gets cloudy and at the end of the video we can see one of the copper wires has been eaten away by the process.

Is there really "a reaction of the copper with the salt in water" as the author of the video claims? Or is it a pure physical process where the copper "boils off" the wire?

Is there really "a reaction of the copper with the salt in water" as the author of the video claims? Or is it a pure physical process where the copper "boils off" the wire?

Yes it reacts, no it doesn't boil. Essentially the wire corrodes at an accelerated rate where during the part of the cycle where the wire is the positive terminal the copper reacts as follows:

$$\ce{Cu^0 + 2OH- -> Cu^2+ + 2e- + 2OH- -> Cu(OH)2 + 2e- -> CuO + 2e- + H2O }$$
(expanded for clarity, not a reaction mechanism)

There is obviously oxygen generated too but I am only focusing on the copper part of the reaction. During the part of the cycle where the wire is the positive terminal the reverse of this reaction does not occur on the other half of the cycle and over time more and more copper will be consumed eventually consuming the entire wire.


How to choose the best method and basis set for a calculation in computational chemistry? In Quantum chemical calculations, it is not always the case that the more complex (correlated) the method or the basis set the more accurate the results.

For example, it's sometimes found that MP2 results better fit the experimental results than that of CI calculation.

In other instances diffuse functions negatively impacts the accuracy of the calculation.

Are there any general rules for the choice of a method and the basis sets?

What is the degree of accuracy of a calculation with no experimental reference ?

The experimental equivalent of your question would be: what kind and level of impurity am I going to accept in my experiment?

In terms of basis sets: in principle you keep increasing the complexity of your basis set until you reach convergence in terms of electronic energy. What you consider to be converged depends of course on your preferences.

With regard to the choice of method. Again, this depends on what you consider to be accurate enough. In heterogeneous catalysis, typically density functional theory calculations are performed using a PBE functional. This is accurate to about 5-15 kJ/mol which is fine in this field. This type of accuracy would be completely unacceptable in other fields. If high accuracy is required, people typically use post-Hartree-Fock methods such as MP2 or higher or CI.

My recommendation in picking a method and a basis set is the following:

Decide what you consider accurate enough.
Look into the literature if there are calculations being performed on the same or equivalent type of system and analyse their results.
Experiment using different methods on a model system. Look how the energy converges with respect to basis set and methodology. Pick the method that you think is best, both in terms of computational speed and accuracy. (it is OK to pick a couple of methods)
Perform the bulk of your calculations using this method.
If you are going to report your calculation. Clearly explain the reasoning why you picked a certain method. The potential readers of your report really appreciate to hear your motivation.


Entropy change and heat transfer in non-ideal heat engines Please help me, a mathematician, to make sense of entropy. I know it's a topic with about a million questions already, so I understand if this gets ignored. Yet for the life of me, I cannot make sense of the previous answers.
So apparently the definition of entropy change in a system is $\Delta S = \int_\mathrm{rev} \frac{\mathrm d Q}{T}$ and this is a state variable. I also know that for irreversible paths, you have to find a reversible one that links the same thermodynamic states to calculate the entropy. Yet in my professor's notes it says
$$ \mathrm dS = \oint \frac{\mathrm dQ}{T} < 0 \text{ for irreversible processes }$$
(is this even still an entropy difference? The notation $dS$ instead of $\Delta S$ confuses me) and elsewhere, talking about the Sterling process, it also says
$$ \oint_{irrev} \frac{\mathrm dQ}{T} < 0 \text{ , so the entropy of the machine (not the total entropy) decreases } $$
so it really must be talking about the entropy change in the system (the machine) and not about the total entropy change of system + surroundings. But how can this be when for a system in a circular process the difference in any state variable is 0?

So apparently the definition of entropy change in a system is $\Delta S = \int_\mathrm{rev} \frac{\mathrm d Q}{T}$ and this is a state variable.

This definition is for the surrounding if it is boring, e.g. a temperature bath. The $_\mathrm{rev}$ subscript is just for the temperature bath, i.e. there could be irreversible processes elsewhere (e.g. in the system).

Yet in my professor's notes it says $$\mathrm dS = \oint \frac{\mathrm dQ}{T} < 0 \text{ for irreversible processes }$$

The inequality is correct. Equating it to $\mathrm dS$ is not. The entropy change for a cyclic process is zero because entropy is a state function.

[OP...] elsewhere, talking about the Sterling process, it also says
$$ \oint_\mathrm{irrev} \frac{\mathrm dQ}{T} < 0 \text{ , so the entropy of the machine (not the total entropy) decreases } $$

This is the Clausius inequality. If a heat engine is not ideal, it will not extract the maximal energy possible from the heat transfer from hot to cold reservoir. The claim about the entropy of the machine is incorrect. Indeed, the entropy of the machine stays the same (use your argument of taking a reversible path from one cycle to the next, i.e. do nothing). Overall, the entropy increases because the entropy loss in the hot reservoir is no longer matched by the entropy gain of the cold reservoir. Thus, we are not extracting the maximal work in the process.
I found this explanation helpful. They used $\delta Q$ instead of $\mathrm d Q$ to remind us that there is a path-dependency.


Number of possible microscopic states and their probabilities I'm having trouble with the following problem:

Boltzmann defined entropy as a measure of the number of possible microscopic states of a system, $W$,
as:
${S = k_BlnW}$
Here, $k_B$ is the Boltzmann constant. Hereafter, it is assumed that neither energy nor particles are
exchanged between the system of interest and the surrounding environment, and that the system is in
equilibrium at a constant temperature. In addition, the gas atoms and molecules (henceforth referred
to as ‘molecules’) are assumed to behave as an ideal gas.

Figure 1. Example of a microscopic state (4,6) in which ten molecules are distributed in two chambers separated by a wall.
Removing the boundary wall that separates the two chambers in Figure 1 allows the molecules to move
throughout the entire container. Assume that it is possible to measure the arrangement of the molecules
instantly after the removal of the wall. If this measurement is performed a sufficiently high number of
times, the probability that a state with an arrangement (n,m) will be found can be calculated.
1) Calculate the arrangement ($n^*$, $N$ − $n^*$) with the highest probability of occurrence for $N = 10$ and $N = 100$, as well as the corresponding probability of occurrence for each state.
2) Calculate the probability of observing a state in which $n^*$ is in the range $n^* – 0.02 N ≤ n^* ≤ n^* + 0.02 N$ for
$N = 10$ and $N = 100$. Here both $N$ and $n^*$ are integers.

I was able to calculate the first part and got the following results using a spreadsheet.
For $N = 10$:

$W(5,5) = 252$ and $P(5,5) = 0.226$
For $N = 100$:

$\ce{W(50,50) = 1.0*10^{29}}$ and $P(50,50) = 0.0788$
On the other hand, I don't understand what the mentioned range in the second part of the problem means and I'm therefore unable to solve it:

$n^*$ is in the range $n^* – 0.02 N ≤ n^* ≤ n^* + 0.02 N$

Some notes I had easily at hand may help. You have already calculated the probability of obtaining the chance with a number $k$ of type of ball (or molecule) out of a total on $n$ is
$$\displaystyle p=\frac{n!}{k!(n-k)!}\frac{1}{2^n} \tag{25c}$$
This distribution is a maximum when $k=n/2$. This can be seen with a straightforward argument. The factorial terms are symmetric, $k!(n-k)!=(n-k)!k!$ and always positive.  When $k=0$ or $k=n$ the probability is very small tending to zero; i.e $1/2^n$ is small so there must be a maximum somewhere in the range $0\to n$. The symmetric nature ensures that this will be at $k = n/2$. This can also be determined by differentiation. It is not possible to differentiate a factorial, as $n$ is discrete, but replacing $n!$ with the Sterling approximation $(n!=n^ne^n)$ and setting the derivative in $n$ to zero allows the maximum to be found.
The second feature of this probability distribution is that it becomes extremely narrow as $n$ increases. This is shown in the figure which shows the normalised probability $p/p_{max}$ vs $k/n$ which is a fraction between $0$ to $1$. At large $n$ the distribution becomes so narrow, or peaked at $n/2$, that almost all the probability is described by that at $n/2$, i.e. this value is so great compared to others that $p_{max}$ is in effect greater than all others combined.

The normalised probability function  $\displaystyle p=\frac{n!}{k!(n-k)!}\frac{1}{2^n}$ vs $k/n$ for different $n$. The the data only exists for integer values as shown as circles for $n=20$, at larger $n$ a line is drawn as the points become too congested. The curve for $n=10000$ is in red. At large $n$ the distribution becomes very narrow and the maximum value at $n/2$ is greater than all others combined and the distribution can be approximated by its maximum value.
You next need to choose $k/n$ corresponding to 2% either side of $k/n=0.5$ and calculate $p/p_{max}$.


Identifying spontaneous redox reactions Which one of the processes is NOT spontaneous?

  \begin{align}
\ce{Ni + Sn^2+ &-> Ni^2+ + Sn} \label{1} \tag{1} \\
\ce{Cu + Fe^2+ &-> Cu^2+ + Fe} \label{2} \tag{2} \\
\ce{Ni + 2Ag+ &-> Ni^2+ + 2Ag} \label{3} \tag{3}
\end{align}

I failed this class last semester so I'm redoing the tests to see if I can get this over with, however I'm confused with this question. To find out if a process is spontaneous I need to identify the oxidizing and reduction agents.

\eqref{1} $\ce{Ni}$ is the reduction agent and $\ce{Sn^2+}$ is the oxidizing agent;
 \eqref{2} $\ce{Cu}$ is the reduction agent and $\ce{Fe^2+}$ is the oxidizing agent;
 \eqref{3} $\ce{Ni}$ is the reduction agent and $\ce{Ag+}$ is the oxidizing agent.

For a redox reaction to be spontaneous the reduction potential of the oxidizing agent needs to be higher than the reduction potentials of the reduction agent, therefore \eqref{1} and \eqref{3} are not spontaneous, but only one of the reactions can be not spontaneous. Where did I go wrong?

For starting purposes, you should write the half cells, either with their oxidation/reduction potentials.
Now if you want to compare the oxidizing species then you have to consider the oxidizing potentials of all the species and the one with highest value will be the spontaneous half cell and the reverse of the other one will occur.
Remember you should consider only one type of potential at a time(i.e. either oxidation or reduction) for comparison.


How to calculate the entropy change of benzene when it changes from vapour into liquid? Given information:

  The boiling point of benzene at atmospheric pressure is $353~\mathrm{K}$
  the enthalpy of vaporization of benzene is $30.8~\mathrm{kJ~mol^{−1}}$ at this temperature.
  The molar heat capacities of the liquid and vapour are $136.1~\mathrm{J~K^{−1}~mol^{-1}}$ and $81.7~\mathrm{J~K^{−1}~mol^{-1}}$,
  respectively, and may be assumed temperature independent.

  Calculate the entropy change of the system, the surroundings and hence the universe when $1~\mathrm{mol}$ of benzene vapour at $343~\mathrm{K}$ and atmospheric pressure becomes liquid benzene at $343~\mathrm{K}$. Also, will this process occur spontaneously?

I know that $\mathrm{d}S = \frac{\mathrm{d}H}{T}$ therefore,
$\displaystyle\mathrm{d}S = \frac{-30.8 \times 10^3~
\mathrm{J}}{343~\mathrm{K}}=-89.8~\mathrm{J~K^{-1}}$ which is the entropy of the system.

How do I continue from there, utilising the molar heat capacities given?

You just used $\mathrm{d}S=\mathrm{d}H/T$. Also you didn’t told whether the heat capacities are $C_{\mathrm{m},p}$ or $C_{\mathrm{m},V}$. Anyways I found that you are using $C_{\mathrm{m},p}$ from the benzene data page on Wikipedia. You need to do something like this:
$$\Delta S=\left(1\;\mathrm{mol}\cdot81.7\;\mathrm{\frac{J}{mol\cdot K}}\cdot\ln\frac{373\;\mathrm{K}}{343\;\mathrm{K}}\right)+\left(\frac{30.8\;\mathrm{\frac{kJ}{mol}}}{373\;\mathrm{K}}\right)+\left(1\;\mathrm{mol}\cdot136.1{\;\mathrm{\frac{J}{mol\cdot K}}}\cdot\ln\frac{343\;\mathrm{K}}{373\;\mathrm{K}}\right)\\\approx78.012\;\mathrm{J/K}$$


Chemical tests given by glucose (biomolecule) We know that glucose gives Tollen's test because it's a reducing sugar, and having a hemiacetal group, on hydrolysis can be converted into an aldehyde group, giving the positive test.

But, when we come to tests like the Schiff's test or with sodium bisulphite, it does not show any reaction which are shown by aldehydes. Why? Is the answer that hemiacetal group cannot be converted into aldehyde or anything else? Probably there shouldn't be any water used in these reactions, so no hydrolysis, no conversion from hemiacetal to aldehyde. Is there any water present in Tollen's reagent?

This are two different types of reactions.

Tollens is a redox reaction in which silver is reduced and glucose is oxidized.
This reaction depends on the redox potential of glucose only.

Schiff and bisulphite tests are addition reactions. The reagents condense with the aldehyde group.
This occurs also in the case of glucose but it is not sufficient to give a positive result.
In the case of bisulphite, the adduct must precipitate, but it is too soluble.
In the case of the Schiff test, the adduct must dehydrate, but it forms the glucosamine derivative instead.


Differentiation of H = U + pV I was asked to use direct differentiation of "H = U + pV" to find a relationship between:

$$ \left(\frac{\partial H}{\partial U}\right)_{\!P}$$   and   $$\left(\frac{\partial U}{\partial V}\right)_{\!P} $$

So my first step was to take a differential of the equation twice, once in terms of U and once in terms of V:

$$ \left(\frac{\partial}{\partial U}H\right)_{\!P}=\left(\frac{\partial}{\partial U}(U +PV)\right)_{\!P}\rightarrow\left(\frac{\partial H}{\partial U}\right)_{\!P}= 1 + P\left(\frac{\partial V}{\partial U}\right)_{\!P}$$
and

$$ \left(\frac{\partial}{\partial V}H\right)_{\!P}=\left(\frac{\partial}{\partial V}(U +PV)\right)_{\!P}\rightarrow\left(\frac{\partial H}{\partial V}\right)_{\!P}= \left(\frac{\partial U}{\partial V}\right)_{\!P} + P$$

When I tried to substitute one equation into the other, they would simply end up all canceling out. Is there another way?

A way to do it would be:

\begin{align}
\left( \frac{\partial H}{\partial U} \right)_P \left( \frac{\partial U}{\partial V} \right)_P &= \left( \frac{\partial H}{\partial V} \right)_P = \left( \frac{\partial U}{\partial V} \right)_P + P\\
\left( \frac{\partial H}{\partial U} \right)_P &= \frac{\left( \frac{\partial U}{\partial V} \right)_P + P}{\left(\frac{\partial U}{\partial V} \right)_P}\\
\left(\frac{\partial U}{\partial V} \right)_P &= \frac{P}{\left( \frac{\partial H}{\partial U} \right)_P - 1}\\
\end{align}


Total number of stereoisomers of cyclobutane-1,2,3,4-tetracarboxylic acid Question

Find all stereoisomers of:

My attempt
I got four isomers, and none of them seem to be optically active as they either have a plane of symmetry or a point of symmetry:

However, my sir keeps insisting there are five isomers. Could anybody help me identify the fifth one?

You can view 1,2,3,4-tetrasomethingcyclobutane as a square with black and/or white corners (black representing e.g. up, then white representing down configuration of the substituent). Two squares represent the same object when after applying some of the following operations on the first you get the second one:

rotation in the paper plane (of course)
mirroring AND color inversion

By drawing all 24 = 16 possible squares and using these two rules, you quickly find that they represent only 4 different objects, as seen on the following sketch.

Note that the second "controversial" operation mirroring AND color inversion is in our scenario, because of D2 symmetry of all squares, equivalent with more proper, "mirror-free" operation - rotation around some in-plane axis, considering that obverse of the square has inverse colors. (Because of the symmetry, the operation can even be reduced to color inversion only.)


Does rubber use one or two sulfur atoms to crosslink? Does rubber use one or two sulfur atoms to crosslink?

According to the Polymer Database,
the sulfur bridges formed in the vulcanization process of natural latex are usually comprised of 4-6 sulfur atoms, although this polyssulfide chain can be shorter if the temperature is increased.

  http://polymerdatabase.com/polymer%20chemistry/Vulcanization.html


Why do certain signals in proton NMR experience extensive splitting despite only having 1 neighbour? I came across the compound 3-methylpentan-2-ol. Looking at its proton NMR spectra, it seems there is a lot of splitting for the signal at 4 ppm.

I'm assuming that the 4 ppm represents the proton in the hydroxyl group (since it clearly is much more deshielded than the other protons), but this would mean it only has one neighbouring hydrogen and hence should exist as a doublet.

I just wanted to clarify, is there any reason why the 4 ppm peak has so much splitting despite only having one neighbour? Or otherwise, I have also considered, could I be incorrect in assuming the 4 ppm peak represents the hydroxyl group, and if so why?

This question has a few different parts to it.

Part 1 - what's that peak at 3.6ppm?
Well, it's not the -OH peak.  The peak at 3.6 is from the CH(OH).

Part 2 - where is the -OH peak if that isn't it at 3.6ppm?
Peaks from labile protons are frequently not observed, and have variable chemical shifts, unless great care is taken in sample preparation. This sample was run in CDCl3, and the -OH peak exchanges with H2O, and comes at ~1.6ppm.

Part 3 - can we account for the splitting of the peak at 3.6ppm now we know what it is?

The CH(OH) proton should couple to proton on the adjacent carbon (split to a doublet) and also the protons of the methyl group (split to a quartet). It won't couple to the -OH for the same reason we don't readily observe the -OH; it is rapidly exchanging with water. So, we expect that the peak at 3.6ppm should be a doublet of quartets. Looks more complicated than that to me.

Part 4 - It looks  more complicated that a doublet of quartets to me. What's happening there?
Of course, 3-methylpentan-2-ol has 2 stereocentres, and so we have here a mixture of diastereomers in solution in the sample. Both diastereotopic CH(OH) protons have very similar chemical shifts, and so there are in fact 2 sets of doublets or quartets. It's also why the rest of the spectrum looks such a mess.


How do I dissolve sodium oleate in distilled water I wanted to know how to dissolve sodium oleate powder in distilled water?  Does it have to be at a certain temp?  I need a 2% solution

A product information by Sigma-Aldrich, a chemical supplier, cites Merck Index regarding the solubility of sodium oleate.  And states "It is
also soluble in water (approximately 100 mg/ml)", in other words, 10 mass%.  Since this value is larger than the 2 mass% requested by you, and provided sodium oleate is a surfactant anyway, just go ahead.  A gentle warming, like 30 to 40 Celsius, may accelerate the process, though.


Le Chatelier's principle with pure solids and liquids In the following reaction:

$$\ce{A(s) <=> B(g) + C(g)}$$

since adding a pure solid is not supposed to shift the equilibrium in any way, does that mean that I can essentially remove all of the solid except for 1 molecule, and the equilibrium will not shift at all since the concentration of the solid stays the same? Similarly, does this mean that I can essentially fill up the container with the solid, and no extra B and C will be produced?

Your first statement is right.
You just need a certain amount of solid or pure liquid for the reaction to occur, just like how putting in more solid in a saturated solution doesn't increase dissolution, while you need solid in there to call it an equilibrium between aqueous and solid phases.
However, the amount of solid or liquid does change the kinetics of the reaction: the rate at which it proceeds to equilibrium.

But if you put in some solid A, the Q is 0, so some B and C will form. It's the same reason why putting in a sugar cube in water increases aqueous sugar concentration: there is an equilibrium between solid and aqueous components, it's just that the constant means the maximum amount of aqueous concentration.


Why does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site? I am currently studying the textbook Mass Spectrometry, third edition, by Jürgen H. Gross. Chapter 2.4.3 Bond Dissociation Energies and Heats of Formation says the following:

  The heat of formation of organic radicals and positive ions decreases with their size and even more importantly with their degree of branching at the radical or ionic site. A lower heat of formation is equivalent to a higher thermodynamic stability of the respective ion or radical. The corresponding trends are clearly expressed by the values given in Tables 2.2 and 2.3, This causes the fragmentation pathways of molecular ions proceeding by formation of secondary or tertiary radicals and/or ions to become dominant over those leading to smaller and/or primary radical and ionic fragments, respectively (Sect. 6.2).

Why does the heat of formation of organic radicals and positive ions decrease with their size and degree of branching at the radical or ionic site?

I would greatly appreciate it if people would please take the time to explain this.

Here is a rather simple structure explanation presented on this site. To quote, where 'pieces' refer to an organic molecular with a non-symmetric branching structure:

You’ll notice something – the simpler the pieces are, the easier they are to stack together, which provides a tighter fit with fewer spaces. Here, by putting a kink in the block, we make them harder to stack.
What’s this got to do with chemistry?
When compounds freeze, the process is a lot like stacking bricks. The more symmetrical the molecules are, the easier it will be  and the fewer spaces there will be between the molecules. Fewer spaces = better stacking. Hence, when you compare hexane to its structural isomer, 2-methylpentane, hexane has a much higher melting point due to the regular arrangement of its structure.
Better stacking, higher melting point. Case closed. Right? Not quite.
It’s also about surface area [...]
It looks like as we increase branching, we’re increasing melting point and decreasing boiling point. What’s going on? [...]
Going from “branched” to “highly branched”  makes a molecule more compact and sphere-like. As the surface area of the molecule decreases (remember that spheres have the lowest surface area/volume ratio of any shape) they will become more compact and thus easier to pack. This explains the melting point phenomenon [...]
Boiling point is related to the forces between molecules, which in the case of hydrocarbons is Van Der Waals interactions. If you’ve ever seen microscope images of a gecko’s feet – which allow it to climb walls – you’ll see that there is no adhesive but the pads contain a tremendous amount of surface area. It’s all about the Van der Waals interactions.
As we decrease surface area, we are going to decrease intermolecular Van Der Waals interaction and therefore decrease boiling point.

The summary relationships are presented as follows:

linear versus branched —> higher melting/boiling points due to better stacking and surface area contact.
highly branched vs. branched —> more sphere-like –> better stacking –> higher melting point
highly branched vs. branched  —>more sphere-like – -> lower surface area —> lower boiling point.

The above connects boiling point to branching structure, and this paper "Hypothetical Thermodynamic Properties: The Boiling and Critical Temperatures of Polyethylene and Polytetrafluoroethylene" found here links BP to thermodynamics. Per the opening reference statement: "A lower heat of formation is equivalent to a higher thermodynamic stability".
This completes the argument as to why the heat of formation decreases for organic radicals with decreased size and degrees of branching.


Is it possible to write a redox reaction with two oxidations and one reduction together? Let's take the Redox reaction between ${MnO_4}^-, {C_2O_4}^{2-}$ and  ${SO_3}^{2-} $

Is it possible to write the complete reaction as below or do I need to consider the two reactions separately?

$\ce{4MnO4- + 5C2O4^{2-} + 5SO3^{2-} -> 4Mn^2+ + 10CO2 + 5SO4^{2-}}$

I wouldn't try to combine these reactions. You would want to consider both reactions and maybe look up the standard reduction potentials of all reactants and products involved to give yourself an idea of what is oxidizing what and what is reducing what. In general, you should let the SRP's be a guide for trying to figure out what actually happens in these reactions.

The reaction you wrote doesn't really describe anything chemically. We have to make sure not only the math works out, but also the chemistry. Also, your charge balance is wrong, so even if the chemistry did work out, the reaction would be impossible. You seem to have 24 (-) on the reactants side and net 2 (+) on the products side.

Remember to always check BOTH your mass (MB) and charge balance (CB) work out when you're doing REDOX problems.

The following two reactions are the acidic balanced REDOX reactions for the the species you have written. Of course, these may not actually be the real reactions that would occur. For that, you would have to look things up in tables in your book.

$\ce{6 H+ + 2MnO4- + 5SO3^2- -> 5SO4^2- + 2Mn^2+ + 3H2O} \quad\text{(CB: -6 : -6)}$

$\ce{16 H+ + 2 MnO4- + 5C2O4^2- -> 10 CO2 + 2Mn^2+ + 8 H2O} \quad\text{(CB +4 : +4)}$


Why is the common magnesium ion Mg(II) and not Mg(I) when the second ionization energy is higher than the first ionization energy? The first ionization energy for magnesium is given as $\pu{737.7 kJ/mol}$, and the second ionization energy is $\pu{1450.7 kJ/mol}$.

Given this information, doesn't it take $\pu{737.7 kJ/mol}$ to form $\ce{Mg^1+}$, while $\pu{737.7kJ/mol} + \pu{1450.7kJ/mol} = \pu{2188.4 kJ/mol}$ is required to form $\ce{Mg^2+}$?

Why is it then, that the $\ce{Mg^2+}$ ion is more common than the $\ce{Mg^1+}$ ion, while more energy is required to form the $\ce{Mg^2+}$ ion?

As mentioned by Zhe, we have to look at the entire process by which an ionic compound is formed, not just the energy for a single part.
For example, consider the formation reaction of $\ce{MgO}$
$$\ce{Mg(s) +1/2O2(g)->MgO}$$ To find the enthalpy of this process, we could follow a Born-Haber cycle:
This demonstrates that while the ionization of $\ce{Mg}$ to $\ce{Mg^{2+}}$ might be unfavorable, the overall process to form the compound is not. So the ionization of $\ce{Mg}$ is driven by the favorability of the overall process.2
It is also worth noting why we do not instead form $\ce{Mg2O}$. If we were to construct a Born-Haber cycle for this compound, it would take less energy to form $\ce{2Mg^{+}(g) +O^{2-}(g)}$ than it would to form $\ce{Mg^{2+} (g) + O^{2-}(g)}$. However, the lattice energy of $\ce{Mg2O}$ would be sufficiently smaller than the lattice energy of $\ce{MgO}$ that $\ce{MgO}$ is still more favorable to form. Lattice energies are heavily dependent on the charge of the ions involved and having a $+2$ ion will lead to twice as large a lattice energy as an ion with a $+1$ charge all else being equal.
Even in this simple case, all else might not be equal. To make a more direct comparison between we can utilize the Born-Lande or Born-Mayer to obtain a calculated value of the lattice energy.


Washing stainless steel separator from organic residue In my lab we have a stainless steel separator for liquid products of one of our hi-pressure catalytic testing assembly. It is a steel cylinder with two short pipes and one another short pipe with a valve. The problem is, the separator cannot be disassembled and washed. It accumulated organic residues for unknown time, presumably years (it happened before it moved into our responsibility). Some residues were washed with acetone and other solvents, but we want it much cleaner.

Personally, I would love to threat it repeatedly with a strong oxidizer, but I'm afraid that stainless steel might disagree. Also, heating above 200-300 Celsius unfortunately is out of question.

The residue is from Fischer-Tropsch synthesis, so it probably consists of heavy hydrocarbons and possybly some oxygenates and products of their reactions.

Any suggestions?

First try commercial cleaning solutions such as Alconox or Pierce RBS-35, with ultrasonic agitation (if you can sneak a transducer into the reactor).

Second, an oxidizer such as Godax Labs' Nochromix might work, but test it on an unimportant part of the reactor to see if it safe and effective. Read and understand its safety data sheet before use.

A last choice is chromic acid, $\ce{H2CrO4}$, which in some concentrations does not damage some types of stainless steel.  N.B. Chromic acid is toxic, carcinogenic and harmful to the environment. With chlorides, it can produce toxic and volatile chromyl chloride. In high concentration it can cause ignition of materials. Check a material safety data sheet before trying it! Also prepare to dispose of this as hazardous waste.
Even if the reference states it is safe for the particular steel in use, try it on an unimportant area... perhaps with some of the organic "gunk", to see if the gunk is removed and the reactor is unharmed.


Why is there only a slight change in pH when soda loses its fizz? I did an experiment for my chemistry class and did a control with sparkling water and the pH increased by a significant amount, but when done with sodas like cream soda, coke, or 7 Up, the pH change is so minimal that I can't say that there is a significant change in the pH level. Why do the pH level of these sodas to stay the same when they lose their fizz?

In addition to fizz (carbonic acid), Coca-Cola contains phosphoric acid (for an additional tangy taste). According to this leaflet put out by Coca-Cola, a 250 mL can of Coca-Cola contains 43 mg of phosphorus. With some quick calculation, you can find that this translates to $\ce{5.5538 \times 10^{-3} M}$ $\ce{H3PO4}$, which would have a pH of approximately 2.44. Pretty acidic! The other sodas are sure to contain other acids as well.


How to find the oxidation numbers of individual ions of a transition metal with an average fractional oxidation state In the classical example compound of $\ce{Fe3O4}$, we know that $\frac 23$ of Fe ions display +3 oxidation state and $\frac 13$ of Fe ions display +2 oxidation state fulfilling charge neutrality condition. However, this is the only solution considering the most common oxidation states of Fe are +2 and +3.

Now, consider the compound $\ce{La(Ni_{0.1}Mn_{0.9})O3} \rightarrow \ce{La10NiMn9O30}$,

Assume, the oxidation numbers of La, Ni and O are +3, +2 and -2. Therefore, under charge neutrality condition, the charges of 9 Mn ions should add up to 28. Knowing that the most common oxidation states of Mn are +2, +4, +7, we will have two solutions where charges add up to 28.

4 Mn ions showing oxidation number of +2 and 5 Mn ions showing oxidation number of +4
7 Mn ions showing oxidation number of +2 and 2 Mn ions showing oxidation number of +7

How could I decide what is actually happening?

If it helps, I need to know this to calculate the average ionic radius of a B-site cation of a perovskite structure.

EDIT

Edited the typo by changing ox. state of La to +3.

Furthermore, I am looking for an answer that makes sense theoretically (real world formation may be different). It would be great if somebody could guide me for a relevant publication, if there is any.

To accurately determine oxidation states in compounds, you need to have a good idea of what the actual underlying bonding model is like. Only in rare cases you can resort to shortcuts.

One such shortcut which is often very helpful for binary compounds is to take one element’s ‘obvious’ oxidation state and use maths to derive the other one. That, for example, works well with ethane (hydrogen must have $\mathrm{+1}$, so carbon has $\mathrm{-III}$), iron(III) chloride ($\ce{Fe^{+III}/Cl^{-I}}$) or manganese dioxide ($\ce{Me^{+IV}/O^{-II}}$).

In some cases, such a shortcut will lead to a fractional oxidation state such as in $\ce{Fe3O4}\ (\ce{Fe^{+\frac83}/O^{-II})},$ $\ce{HN3}\ (\ce{H^{+I}/N^{-\frac13})},$ or $\ce{S4O6^2-}\ (\ce{S^{+\frac52}/O^{-II})}.$ One might be tempted to take the closest common or known oxidation states of the element in question and use maths to derive a potential distribution. That works by accident in $\ce{Fe3O4}$ ($\frac83$ is $2\frac23$ which suggests $\ce{Fe1^{+II}Fe2^{+III}}$) and $\ce{HN3}$ ($-\frac13$ suggests $\ce{N2^{\pm0}N1^{-I}}$) but it already fails with $\ce{S4O6}$ because although $\ce{S3^{+II}S1^{+IV}}$ gives an overall oxidation state of $+\frac52$ so does the correct solution of $\ce{S2^{\pm0}S2^{+V}}$.

Since the simple model already broke down for binary compounds, there is little hope it will be of much use in ternary or even quarternary compounds—especially if there are less leads to go on. So while it might be possible to use the simple shortcut to derive the oxidation states in aluminium potassium sulphate or potassium alum ($\ce{KAl(SO4)2}$) and by extension that of chromium in chromium potassium sulphate or chrom alum ($\ce{KCr(SO4)2}$), it is a lot harder if you have two redox-promiscuous elements (nickel and manganese) to consider.

In short, there is no simple way to derive the oxidation states just by looking at the formula and I myself would have to research what publications say and how they arrived at this conclusion.


why doesn't alcohol react with sodium hydroxide but react with potassium hydroxide? I have read that ROH+ NaOH doesn't form alkoxide ions because water is a stronger acid than all alcohols except methanol. However while reading xanthate test I found that alcohol reacts with KOH to form alkoxide ions. Why is it so?

The reaction of alcohol + NaOH producing $\ce{C2H5ONa}$ + water is an equilibrium which is usually strongly pushed to the left hand side. But in the presence of $\ce{CS2}$, the alkoxide $\ce{C2H5ONa}$  is transformed into xanthate. So the equilibrium is driven to the right hand side.


When there is more than one amino group in a molecule, which group is more likely to react when forming an imine? Why does the third group react in this example and not the first?

Imine formation essentially involves an nucleophilic amine and an electrophilic carbonyl group. To decide which is the nucleophilic $\ce {N}$ atom which performs the nucleophilic attack on the electron-deficient $\ce {C}$, we would have to first consider each $\ce {N}$ atoms nucleophilicity, which we can interpret here to be the ability of the $\ce {N}$ atom to donate its lone pair of electrons.

For 1 and 2, the lone pair is delocalised into the carbonyl group and they can be seen as behaving as amides. This delocalisation significantly reduces their nucleophilicity. As discussed by Ron here, the delocalisation is so strong that restricted rotation about the $\ce {C-N}$ bond is observed. For 3, there is no such delocalisation present, thus the lone pair of the $\ce {N}$ is readily donated to electrophiles, making it the likely nucleophilic atom for the imine formation.

Another possible effect that raises the nucleophilicity of that $\ce {N}$ atom is the alpha effect, also discussed here. The following diagram (p. 513) from Clayden, Warren, & Greeves (2012) illustrates the molecular orbital theory basis of the effect. Although it is used to illustrate the effect for peroxide, the same concept can be extended to our case of nitrogen atoms. The adjacent $\ce {N}$ atom with the lone pair can increase the nucleophilicity of the $\ce {N}$ atom. However, the alpha effect may be less prominent for our case since the lone pair of the adjacent $\ce {N}$ is also being strongly delocalised into the carbonyl group. Nonetheless, it is an interesting consideration, further justifying why 3 should be most nucleophilic among the three.

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.


Treating a sample with water, how to deduce present compounds given precipitate and solution parameters? A sample consists of one or more of the following salts: $\ce{Cu(CH3CO2)2}$ (acetate), $\ce{BaCl2}$, $\ce{(NH4)2SO4}$, $\ce{SrCO3}$, $\ce{KNO3}$. The sample, when treated with water, gives a white precipitate that is filtered off and a blue solution. The precipitate is a white solid that is completely insoluble, both in water and in $\pu{6M HCl}$ solution.

Which compounds MUST be present? Explain why.

But $\ce{Cl}$ is soluble with all the given metals! Another question gave the parameter that the precipitate is insoluble in water and $\ce{HI}$, but that was easy because there was a salt with Hg given, so it was obvious that $\ce{HgI}$ must be the precipitate. Please help. Thank you very much.

You correctly point out that Cl "is soluble with all the given metals". This statement is a little unclear, though, which might be why you're confused. What this means, exactly, is that no precipitate that incorporates Cl can form. For example, the solid $\ce{KCl}$ cannot precipitate, because the combination of $\ce{K}$ and $\ce{Cl}$ is soluble.

This fact alone doesn't forbid precipitates of other species. For example, the solid $\ce{BaSO4}$ is insoluble in water. We know that $\ce{BaCl2}$ alone would be soluble in water, but the barium atoms in $\ce{BaSO4}$ don't care that there are $\ce{Cl-}$ ions in the solution; all that matters is that the combination of $\ce{Ba}$ and $\ce{SO4}$ prevents the two from separating. As a consequence, when $\ce{Ba^2+}$ and $\ce{SO4^2-}$ meet in solution, they merrily form the insoluble product $\ce{BaSO4}$, regardless of the other ions present in the solution.


How to calculate melting/boiling points at different pressures I know water boils at below room temperature in a vacuum, and want to know if there's an equation to calculate the melting or boiling points of elements or compounds at a given pressure.

How would I go at attempting this?

The main reason I ask is because I'd like to know if Gallium, a metal which melts at $29.77^\circ C$ at atmospheric pressure would be in a liquid state at room temperature if stored under a vacuum.

Thanks!

I don't believe there is an equation that you can use for melting points of a general substance as a function of pressure (since the melting phase transition has a lot to do with the geometry of the molecule and the structure of the solid), but there is one for the boiling point of any pure substance when you are not near the critical point.

The liquid-vapor transition follows a very specific curve on the pressure-temperature plane of a PVT diagram. This is given by the Clausius-Clapeyron relation, which at temperatures and pressures that are not close to the critical point, can be approximated as:

$$
\ln\frac{P_2}{P_1} = \frac{-\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)
$$

This is the two-point form, which lets you predict the pressure of the vapor phase at a given temperature if you know the temperature and pressure at another point, and the enthalpy of vaporization.

You can use this to predict boiling points as a function of pressure by recognizing that:

  The boiling point is the temperature at which the vapor pressure at equilibrium exceeds atmospheric pressure

This means that if you know the boiling point of a substance at 1 atm (the normal boiling point), and you know $\Delta H_{vap}$ for that substance, you can predict the boiling point at another pressure by plugging in values for $P_1$ and $T_1$ (the normal boiling point), $P_2$ (the pressure you are interested in), and $\Delta H_{vap}$, the enthalpy of vaporization (also sometimes called the latent heat), then solving for $T_2$ (the new boiling point.)

Like I said, this only works when you aren't near the critical point, but for most applications that's not a concern.

For your specific case, especially since you are probably more interested in the liquid-solid transition at this temperature, you are better off looking at a phase diagram.

Here's one for gallium (source) (original source):

The solid-liquid line has a negative slope at low pressure (as ron pointed out) which means the melting temperature increases as pressure decreases. Depending on how high the vacuum is, and on what temperature "room temperature" is, you might be in the solid phase - but the slope is steep, so even at 0 atm the melting point is still right around 300 K.


How can we measure the atomic radius of an atom? The definition of atomic radius is half the average internuclear distance of an atom.

But many times I've been told that it is not possible to handle a single atom. Thus how can we measure the internuclear distance of an atom?

As you told it is not possible to measure the inter nuclear distance of single atom. But here internuclear distance does not mean diameter of a single atom but it means distance between nucleus of two atom of same element.

This internuclear distance can be determined by two methods:

X-rays method
Spectroscopy method

Note: Atomic radius is not a set value for a given atom. Eg. $\ce{Fe^{+2}}$ ion have different atomic radius than  $\ce{Fe+}$. Moreover,  $\ce{Fe^{+2}}$<  $\ce{Fe+}$


Find oxidation number of a metal in a compound Calculate the oxidation number of $\ce{Fe}$ in $\ce{Fe2O3}$.

Alright. So we have a metal and an non metal. This means that iron will lose electrons and give them to oxygen. So it will be a positive oxidation number for iron I guess.

The question is, how much? A wild guess is $2+$ because, well, each iron would is bound to two oxygens so I imagine each oxygen will take one electron each?

Oxidation number of Fe is +3. This is because oxygen has a charge of -2. There are 3 oxygen atoms. So net charge becomes -6. Thus to balance it, each of the 2 Fe atoms must have a charge of +3 (-6 + 6 = 0).


Why is electronegativity increasing with oxidation state increase? I may be a rudimentary fact or something really "not-that-hard-to-figure-out", but

  In inorganic chemistry it is common to consider a single value of the electronegativity to be valid for most "normal" situations. While this approach has the advantage of simplicity, it is clear that the electronegativity of an element is not an invariable atomic property and, in particular, increases with the oxidation state of the element. Wikipedia

Really? That means for example $\ce{C}$ in $\ce{CO3^{2-}}$ is more electronegative than a "bare" C atom? Why is that so?

Interestingly, I'm not able to find relevant info in the Net (and the wikipedia citation that should have hopefully got me results is behind a big bad paywall), so am leaving "reference-request" for the case of additional study.

Yes absolutely, electronegativities are hardly static values when you start combining elements and forming ions or molecules.

Electronegativity is simply a measure of an ability of an atom to attract electrons in a covalent framework.

So keeping in mind that like charges repel (electrons repel electrons) ... it follows that elements in a high oxidation state are more electronegative - i.e. more able to attract and stabilize electrons. High oxidation state elements have essentially been "robbed" of their electrons and likely bear some degree of partial positive charge ... this is "good" for attracting and stabilizing nearby electrons.

Conversely consider a lone pair - which is defined to have an electronegativity of 0. Lone pairs cannot stabilize other electrons and don't attract other electrons to themselves ... the electron-electron repulsion would be too great.


Why is the yield of substitution reactions decreased at higher temperatures? This is according to my text :

  Increasing temperature favours elimination over substitution. This is because energy of activation for elimination is higher than that of substitution since higher degree o charges in bonding is involved in elimination reaction. [sic!]

Well I understand why more elimination occurs at high temperatures (I guess this is due to $\Delta G$ being more negative due to multiple product formation as well), but why do substitution reactions have their yields decreased as temperature goes up?

This is according to the master of organic chemistry; (Even though it says that there is a relative increase, the graph shows a decrease in yield of the substituted products):

Aren't they supposed to have their output increased as well, as the activation energy barrier has already passed and rate increases with increase in temperature, or am I missing something fundamental here?

In short, this is because both elimination and substitution occur parallely and are competing; amount of elimination product + substitution product = constant. If the elimination rate increases more than the increase in substitution rate, then the ratio of elimination product to substitution product will increase. This is what happens on heating, though the rate of both elimination and substitution increase with temperature.

Broadly, there are two types of reactions: those that are kinetically controlled, and those that are thermodynamically controlled. Most elimination-substitution reactions are exothermic and are kinetically controlled. That means that the proportion of products (elimination or substitution) depends only on the rate of elimination and substitution, and not on the thermodynamic favour for the reaction.

Given that the reaction proceeds to completion, the ratio of elimination product to substitution product is equal to the ratio of elimination rate to substitution rate. Suppose elimination and substitution rates are equal, then the products form in a 50:50 proportion.

The ratio of elimination rate to substitution rate is

$$
\frac{e^{\frac{-\Delta G_{e}^{*}}{RT}}}{e^{\frac{-\Delta G_{s}^{*}}{RT}}} = e^{\frac{\Delta G_{s}^{*}-\Delta G_{e}^{*}}{RT}}
$$

where $ \frac{-\Delta G_{e}^{*}}{RT}$  and  $\frac{-\Delta G_{s}^{*}}{RT}$ are the energy barriers for elimination and substitution respectively.

Since the barrier for elimination is more than the barrier for substitution, the above ratio increases with increase in temperature, and hence more elimination product is formed at higher temperatures.


How to estimate the equilibrium constant of the reaction between NaF and HCl Since $\ce{HCl}$ is more acidic than $\ce{HF}$, the left hand side of a reaction has a stronger acid than the right hand side.

Since $\ce{NaF}$ is more basic than $\ce{NaCl}$, the left hand side has a stronger base relative to the right hand side.

$$\begin{align}\text{Stronger Acid/Base} & \ce{->} \text{Weaker Acid/Base}\\ \text{(Less Stable)} & \ce{->} \text{(More Stable)}\end{align}$$

So shouldn't the equilibrium constant be larger than $1$?

My textbook says the equilibrium constant is less than $1$.

We can entirely ignore the sodium cation; it is only a spectator ion. This means, the reaction we are observing is:

$$\ce{HCl + F- <=> Cl- + HF}\tag{1}$$

The equilibrium constant for this reaction is:

$$K = \frac{[\ce{Cl-}][\ce{HF}]}{[\ce{HCl}][\ce{F-}]}\tag{2}$$

We can expand this equation:

$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]}\tag{2'}$$

And then we realise that that is nothing else than:

$$K = \frac{[\ce{Cl-}][\ce{H+}][\ce{HF}]}{[\ce{HCl}][\ce{H+}][\ce{F-}]} = \frac{[\ce{Cl-}][\ce{H+}]}{[\ce{HCl}]} \cdot \frac{[\ce{HF}]}{[\ce{H+}][\ce{F-}]} = \frac{K_\mathrm{a}(\ce{HCl})}{K_\mathrm{a}(\ce{HF})} \tag{2''}$$

So the equilibrium constant is the fraction of the acidity constants of $\ce{HCl}$ and $\ce{HF}$. We know that $\ce{HCl}$ is a much stronger acid than $\ce{HF}$. This is reflected by the inequation:

$$K_\mathrm{a}(\ce{HCl}) > K_\mathrm{a}(\ce{HF}) \tag{3}$$

Since the numerator is larger than the denominator, the value of the fraction must be larger than $1$. Therefore, the product side is preferred.

If your book arrives at any other conclusion, it is disregarding experimental results and should be replaced.


MO-Scheme of SCN- and its bonding properties when used as a ligand I tried to figure out the MO-scheme of the tetragonal-bipyramidal complex trans-$\ce{[Co(en)2(NCS)2]SCN}$ in which the isothiocyanate ligands are bound to the $\ce{Co^3+}$-Ion in $\eta^{1}$-mode (en = ethylenediamine). While undertaking this task I noticed that I don't know the MO-scheme of the isothiocyanate ion, which I need to determine whether it acts as a $\sigma$ and/or $\pi$ acceptor or donor.

I tried to construct the MO via a symmetry-based treatment but ran into the problem that the isothiocyanate ion - belonging to thet $C_{\infty v}$ point group - lacks symmetry elements perpendicular to its principal axis. Thus, I wasn't able to construct the bonding and antibonding sigma-orbitals with this method.

I could construct those $\sigma$-interactions by hand or use $\ce{CO2}$ as a model but I don't know whether this will give me the right results. I haven't found the MO-scheme in the internet so I would be happy if someone could advise me on how to construct a qualitative MO-scheme of $\ce{SCN-}$ or simply show it to me. I would also be very grateful for getting some information on how exactly the isothiocyanate ion acts as a ligand (which $\pi$-interactions and how strong are they, is there also $\sigma$-donation as with carbonyl ligands, etc.).

Edit:
I forgot to mention explicitly that the $\ce{SCN-}$ ion is bound to the metal center via its nitrogen atom.

One issue with SCN-, is that it can bind either at the nitrogen or at the sulfur. The way you presented the formula, $\ce{[Co(en)2(NCS)2]SCN}$, suggested that the SCN- ligand is bound via the nitrogen. If it were bound by the sulfur, the formula might be written  $\ce{[Co(en)2(SCN)2]SCN}$.

SCN- should have a similar MO diagram to OCN-, which is isoelectronic with CO2, so your idea to compare to CO2 is not  a bad one. However, since SCN- has a different symmetry group, you cannot trust that comparison too much. Your comparison to CO is also a good one, since SCN- and CO have the same point group.

However, what you really need to do is determine the symmetry group for $\ce{[Co(en)2(NCS)2]SCN}$. Then, assign symmetry labels to the SCN- ligands. The MO that is most important for SCN- is, qualitatively, the "lone pair" on the atom pointing at the metal center, which can form a "$\sigma$-bond" with cobalt. I would only consider using those $\pi$ orbitals on SCN- if the complex needs more electrons to get to 18. When you assign symmetry labels to the SCN- ligands, consider them as a pair: i.e. assign a symmetry label to the in-phase pair and then assign a symmetry label to the out-of-phase pair. The in-phase pair should probably have the same symmetry label as the cobalt $d_{z^2}$ orbital.


What is the best way to make a substantial amount of soot in a high school laboratory? I need to make about 50 grams of soot (carbon black particles) in a high school laboratory. What would be the best way to do this?

The purpose of needing 50 grams of soot is because I would like to carry out an experiment that investigates how the percentage soot in oil (e.g. diesel) affects the viscosity of the mixture. To measure the viscosity, I would have to put 500 ml of oil in a measuring cylinder and do the ball-bearing method. The highest percentage I would like to test is 10% soot and that calls for around 50 g of it.

There is an extensive literature on producing carbon black see for instance Donnet, Jean-Baptiste, ed. Carbon black: science and technology. CRC Press, 1993.

Instead of adapting the Roman recipe for producing lamp black explained on page 19, I would use olive oil rather than resin or turpentine (that is quite a nasty substance). For doing it on a small scale maximizing the yield maybe I would try (under the fume hood)  to heat the oil in a flask attached to a water-cooled Liebing condenser after reaching the smoke point the soot should start to accumulate on the walls of the condenser. It should be quite easy to recollect the carbon black from the wall of the condenser with nylon bristle and maybe from a little funnel with a slight offset fixed on the exit of the condenser. All this should be carried out using heat-resisting glassware in a lab under the supervision of an adult.

P.S.

  The highest percentage I would like to test is 10% soot and that calls
  for around 50 g of it.

This means that you will need more than 50 g for the whole experiment so if you test 10 different proportion you will need at least 80 g performing the experiment only once. It could require a lot of oil to produce 80 grams.


How do I chill water without a refrigerator? How can we chill water in a plastic bottle without a freezer or refrigerator?

Scenario:

I'm in the office and I have two plastic bottles of water (brought from home). I'm in the office for 3 to 4 hours and this means the bottles I brought from home will get hot (and won't be drinkable in a hot season). Now I want to make it cool so that I can drink chilled water, without access to ice with salt or a refrigerator.

Is it possible?

I'm really not sure this is a chemistry question.

Having said that, evaporative cooling will do the job for you. Here on the Indian sub-continent a clay/earthenware water container works wonders using the technique of evaporative cooling.

In the absence of earthenware, wrap two pieces of thick cloth of the appropriate size around your containers. Keep both cloth wet/moist, and let nature do the rest.

You may also consider taking a bit of pot luck


Why is a reaction more favorable if one of the products is not soluble/sparingly soluble? I have heard people go so far as to say that the "driving force" of a reaction is the formation of the precipitate. Why is this so? Whilst the process is almost certainly exothermic (I think - please clarify) the entropic effects are surely unfavorable.

If the reaction product is removed from the phase (liquid solution for example) in which the equilibrium reaction is occurring, significant amount of product can be obtained even if the equilibrium constant for the forward reaction is relatively small ($K<1$).

Aside from precipitation, product could also be removed by vaporization, with subsequent condensation (distillation) to collect the product.  Anything that reduces the product concentration in the phase where the equillibrium occurs favors the forward reaction.


Number of optically active compounds among the products of ozonolysis The number of optically active compound(s) obtained upon complete ozonolysis of
  the following optically active compound is

In the question, I broke all the double bonds and added a $\ce{-CHO}$ group according to the question (ozonolysis).
Checking for the optically active compounds I found two of them:

But the solution has given only one compound! Can you please help me with this?

Ozonolysis of the given compound gives $\ce{I}$ , $\ce{II}$ , $\ce{III}$ and $\ce{IV}$.

$\ce{I}$ and $\ce{II}$ are Homomers and have same configuration "S" and therefore identical (as shown in the figure).
$\ce{III}$ and $\ce{IV}$ are achiral . Therefore , the total number of optically
active compounds formed after ozonolysis is $\ce{= 1}$.


Do only outer electron shells take part in forming chemical bonds? Do only outer electron shells take part in forming chemical bonds? Or could an inner shell create a bond under some conditions?

Great Question.
From this article, it seems that extreme conditions of pressure can lead to new type of molecules,bondings and how the electrons participate in the bonding.

  Under very high pressures, it appears, electrons in the atom's inner shells can also take part in chemical bonds.

  “It breaks our doctrine that the inner-shell electrons never react, never enter the chemistry domain,” says Mao-sheng Miao, a chemist at the University of California, Santa Barbara, and the Beijing Computational Science Research Center in China. Miao's calculations show that under extreme pressures cesium and fluorine atoms can form exotic molecules with inner-shell bonds.

  Miao identified two molecules that, at high pressure, would involve cesium's inner electrons as well. To form cesium trifluoride ($\ce{CsF3}$), a cesium atom would share its single valence electron and two inner-shell electrons with three fluorine atoms. Four inner electrons would go into making cesium pentafluoride ($\ce{CsF5}$).

  “That forms a very beautiful molecule, like a starfish,” Miao says. He reported his findings in Nature Chemistry. Both the shape of the resulting molecules and the possibility of their formation are “very surprising,” says Nobel Prize–winning chemist Ronald Hoffmann, a professor emeritus at Cornell University.

Found this in the comments about the stability of the exotic molecules at normal pressures

  "Since the compound is thermodynamically unstable it would decompose to the more stable form at lower pressures. Since this means that they are going to higher states of enthalpy I would expect that this decomposition would be a endothermic reaction which means that they would get cold during the process."


Explanation of the strength of phosphorus–oxygen bond When discussing the Wittig reaction, Clayden's Organic Chemistry (2nd ed.) cites the strength of the P=O bond formed in triphenylphosphine oxide as a driver of the reaction through enthalpy:

  The P=O bond, with its bond energy of 575 kJ mol−1, is one of the strongest double bonds in chemistry, and the Wittig reaction is irreversible and is driven forward by the formation of this P=O bond. No need here for the careful control of an equilibrium necessary when making acetals or imines.

According to this table of bond dissociation energies, the value for the P=O bond is similar to the bond energy values for bonds such as the Be–F bond (577 kJ mol−1) and the H–F bond (568.6 kJ mol−1).

However, taking into account the large differences in the p-orbital sizes of phosphorus and oxygen atoms, and the sizes of the phosphorus and oxygen atoms, I fail to understand the similarity in the strength of the P=O and other bonds listed above.

The oxygen atom is larger than the hydrogen, fluorine, while the phosphorus atom is larger than hydrogen, fluorine, and beryllium. Thus, I would imagine that the sigma orbital overlap would be weaker due to a larger distance of interaction in P=O than in the H–F and Be–F bonds.
Due to the differences in the sizes of the p-orbitals of oxygen and
phosphorus, I would naturally infer that the p-orbital interactions and, thus, the pi bond in P=O would be weak.

Are there other factors that I am failing to consider that strengthen the phosphorus–oxygen double bond?

References

Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.

The phosphorus oxygen double bond is a relatively short bond. The oxygen p-orbital lone pairs are donated to the phosphorus antibonding orbitals further stabilizing and strengthening the P-O bond. This idea is supported by molecular orbital theory and ab inito calculations. Consequently, this suggests there is no d-orbital involvement in the phosphorus(V) atom.


Raoult's Law and Mole Fraction Raoult's Law states that the partial vapor pressure of each component of an ideal mixture of liquids is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

As we know that the liquid and vapour in a container are at dynamic equilibrium,

So the mole fraction which is mentioned in Raoult's Law is measured at equilibrium or initially?

Raoults Law says that, for an ideal liquid mixture, the partial pressure of each component in the gas phase is equal to the vapor pressure of the pure component (at the temperature of the system) times the  mole fraction of that component in the liquid phase.  If the system is at equilibrium, the mole fraction to use in the equation is the current mole fraction.


What makes Gold so different from Platinum? Looking at Wikipedia, Gold & Platinum are in adjacent seats.

    Platinum                                Gold
[Xe] 4f14 5d9 6s1                     [Xe] 4f14 5d10 6s1
Atomic number of 78                   Atomic number of 79
Atomic pass of 195.084u               Atomic mass of 196.96657 ± 0.000004 u

How does that single electron make such a huge difference?

That one electron is making sure that Gold has a complete d-orbital, and therefor is much more stable. Elements want to achieve noble gas configurations, and if they can't, then they want to have a half-filled orbitals. In the case of gold the d-orbital is filled completely and the s-orbital is half filled. Whereas Platinum doesn't have that complete d-orbital filled.

Chemical properties (such as reactivity) are influenced by electrons. The amount of neutrons will determine if it can be radioactive or not(,...) look at Tritium : that is a hydrogen atom with 2 neutrons in it and it is radioactive. Whilst protium (the most common isotope of Hydrogen with 1 proton and 0 neutrons) is not radioactive.


Orbital size of Carbon and Oxygen in Methanal I was given an introductory question into Carbonyl chemistry today and I wanted to ask a question about Methanal.

The following picture was shown below of Methanal:

and I noted that the orbital sizes differed for both Carbon and Oxygen in the top and bottom diagrams. I thought one of these diagrams must be wrong.

I then knew that as Oxygen has more protons than Carbon, then its atomic radius is smaller, meaning the orbital size must be smaller too, so I thought that the bottom diagram represented the size of the p orbitals better both for anti bonding and bonding.

Is the diagram for $\pi$ bonding drawn correctly in terms of orbital size?

The orbital size in the picture shows the coefficient in the linear combination of atomic orbitals, i.e the pi bonding MO has a larger contribution from the oxygen atomic orbital, and the pi anti-bonding MO has a larger contribution from the carbon atomic orbital.

Here is another picture for a different molecule where they use the same convention, but also show the coefficients explicitly (source):

  Is the diagram for π bonding drawn correctly in terms of orbital size?

An orbital is not a "thing" so it does not have a size. It is a 3-dimensional function. The picture just shows the symmetry of that function and the node (where it is zero). For a slightly more realistic depiction, you can check out molcalc.


Preparation of iodine solution How to prepare $\mathrm{0.1M}$ $\ce{I_2}$ solution.
I have found in books that it should be prepared in $\ce{KI}$ solution but none of them explains why? Please explain in detail not just the procedure.

Since $\ce{ I_2}$ is a non-polar covalent molecule, it does not ionize in water. It  cannot also be soluble in the polar water.

$\ce{KI}$, which is a polar, ionic compound, will ionize and dissolve in water. When KI dissolves in water, it ionizes to $\ce{K^+}$ and $\ce{I^-}$.

The $\ce{I^-}$ will react with $\ce{ I_2}$ to form the complex ion $\ce{I_3^-}$. $\ce{I_3^-}$ being negatively charged will dissolve in water.


Why are there 3N-6 degrees of freedom in molecules? Count the degrees of freedom using the minimum number of coordinates required to specify a position.

Let's say there are 3 atoms, 3 coordinates for center of mass, and 3 for an axis. But we need 6 coordinates to specify a line in a three dimensional space: $(x,y,z)=(x_0,y_0,z_0)+t(a,b,c)$. Which line do we choose? Even if we have a line, how do we use the remaining three coordinates to specify the atoms?

You have N atoms, each atom can be described by a set of cartesian coordinates (x,y,z). So for each atom we have 3 degrees of freedom which means we end up with 3N degrees of freedom. If we look at an isolated structure without any external field or anything like that then only the relative position of each atom to the others matters. So we can get rid of some degrees of freedom. Translating every atom in the same way into x, y or z direction won't change anything, so we can reduce the degrees of freedom by 3. Rotating it around x, y or z doesn't change anything either, so we are reducing by another 3, so we end up with 3N-6.


How to calculate the mass of hydrogen in a borane sample? In a $\pu{45.9 g}$ sample of borane, a compound containing only boron and hydrogen, that's $85.63\%$ boron, what is the mass of hydrogen in this sample?

I was just going to take $\pu{45.9 g}$ and multiply by $0.1437$ to get the mass of hydrogen, but I don't think it works that way because they don't equally contribute to the mass of the compound. Then I thought, maybe I should consider the molar masses of both atoms, hydrogen being $\pu{1.008 g mol-1}$ and boron being $\pu{10.811 g mol-1}$. I am stuck at this point. How to solve the question.

The problem does not need any difficult approach, not even relative atomic masses, if you trust the one who made this exercise. Since borane (whatever type it is)* consists (by exercise's hypothesis) only of hydrogen and boron and since boron is $85.63\%$ by mass, hydrogen's mass percentage would be the remaining $100\% - 85.63\% = 14.37\%$. Therefore, we have $\pu{6.60 g}$ of hydrogen in this sample.

The compound borane, $\ce{BH3}$, is about $78.1\%$ boron per mass. However, there is a class of compounds called boranes which also only consist of boron and hydrogen. From this class pentaborane(9), $\ce{B5H9}$, with about $85.7\%$  boron by mass would fit the exercise.


Buffer Solution with Strong Acid? I was thinking: Could we could make a strong acid-strong base buffer solution? Take, for instance, a mixture of 1 L $\ce{HCl}$ (0.1 M) and 1 L of aqueous salt, $\ce{Na_2SO_4}$ (also 0.1 M), note that $k_2=0.012$ for the polyprotic acid. How can the pH of the mixture be calculated? Additionally, would it be "stable" like a buffer?

Buffers can be made two ways.

Buffers

Weak Acid/Base + Salt of the Conjugate Base/Acid

Let’s take acetic acid ($\ce{HC2H3O2}$) and sodium acetate ($\ce{NaC2H3O2}$).

$$\ce{HC2H3O2 + H2O <=> H3O+ + C2H3O2-~~~~~~~~~~K_a=1.8 x 10^{-5}}$$

Acetic acid has a small $\ce{K_a}$ value, meaning that it does not dissociate much. Conversely, sodium acetate readily dissociates as sodium is an alkali metal. By having equimolar concentrations of both weak acid and conjugate base, the addition of an acid adds hydronium, which reacts with the acetate ion to shift the equilibrium to the left, and the addition of a base absorbs hydronium to shift the equilibrium to the right. Either way, the hydronium concentration, which determines the pH, remains close to the same.

Weak Acid/Base + Strong Base/Acid

This relies on the same principle as above but supplies the conjugate base in a different manner. Continuing with our acetic acid example, here, this would be added, alone. Then you would add half of the moles of acid in strong base, such as $\ce{NaOH}$. This would neutralize any available hydronium, shifting the equilbrium to the right until there was the same half-and-half of acid and conjugate base.

Your Solution

Strong Acid?

Unfortunately, strong acids do not make for good buffers as they fully dissociate -- the equilibrium is so far to the right that it is very unstable.

Salt of the Conjugate Base?

$\ce{Na2SO4}$ shares no common ion with $\ce{HCl}$, so it provides no conjugate base ($\ce{{SO4}^2-}$ is the conjugate base of the second dissociation of $\ce{H2SO4}$). Even if you used $\ce{NaCl}$, it still would not work because of the reason outlined above.

pH?

If you're still interested, the pH of the solution above is easily calculated. As $\ce{HCl}$ fully dissociates, its concentration (0.1 M, as you say above) is equal to the hydronium concentration. However, the addition of another liter of neutral solution halves the hydronium concentration to 0.05 M. Therefore,

$$\ce{-log(0.05)=pH=1.3}$$

Hope this helps!


How to turn acetaldehyde into acetic acid using one of the five chemicals on the list? From the Russian State Chemistry Test:

It's easy to see $X$ is $\ce{H2O}$: water turns acetylene into acetaldehyde.

But what can turn acetaldehyde into acetic acid?

I checked: $\ce{NaOH}$ would turn it into sodium acetate, $\ce{CuO}$ would not react (you'd need $\ce{Cu(OH)2}$), ethanol would turn it into either an acetal or hemiacetal.

Would acetaldehyde react with formic acid to produce acetic acid? I tried to google for it, and found no mention.

Would water in the presence of a heat source somehow react with acetaldehyde? I googled and discovered that it might form hydrates, not acetic acid.

So I'm at a loss. If you consider the question too easy, you could give just a hint.

Oxidation of aromatic aldehydes to acids by air oxidation catalysed by CuO is reported in Molecules 2008 13 948

None of the other reagents can oxidise aldehyde to acid.


Why does water split into ions? What is the property of a water molecule that allows it to split into $\ce{OH-}$ and $\ce{H+}$ ions?

And subsequently, why - when these ions are attracted towards each other in a neutralisation reaction, for example - can these ions then become a single covalently bonded molecule?

The splitting described doesn't actually happen in bulk water. There is just the autoionization reaction:

$$\ce{H2O + H2O <=> H3O+ + OH-}$$

The net transfer of a proton from one water to another creates two ions. This is somewhat disfavored. Hence, the autoionization constant is quite small:

$$\mathrm{K}_{w} = \ce{[H3O+][OH-]} = 10^{-14}$$

Neutralization is a somewhat related process, whereby one species is able to create an excess of hydronium, and another is able to create an excess of hydroxide. This creates an imbalance in the autoionization equilibrium, and it rapidly shifts left to consume excesses of both ions.


Vapour-Liquid Equilibrium I have been studying flash distillation, and I am reading the VLE tables my textbook has provided.
I am looking at these tables and I am not understanding the data, I think.

Why is the vapour mole percent ($y_{benzene}$) and liquid mole percent of benzene ($x_{benzene}$) zero at 110.62 $^0 C$? Does that mean benzene just disappear from the flash distiller at that temperature?

Absolutely not. It means that for the mixture to be boiling at 110°C, it must not contain any benzene.


Why does an acid salt possess acidity as well as basicity? For example, $\ce{BaHPO_{4}}$ has acidity of 2 and basicity 1.

Why does this so? Its basicity is 1 because it has 1 replaceable hydrogen ion.
But why does this have acidity?

This is because Barium Hydrogen Phosphatev ($\ce{BaHPO4}$) is an amphoteric species:

  In chemistry, an amphoteric species is a molecule or ion that can
  react as an acid as well as a base

(Wikipedia)

Whether it acts as a base or as an acid depends on the medium the salt is dissolved. In a pure water solution, it's an acid, since the following occurs:

$$\ce{BaHPO4 + H2O <=> Ba^2+ + PO4^3- + H3O+}$$

The phosphate ion also has its reactions with water forming other species:

$$\ce{PO4^3- + H3O+ <=> HPO4^2- + H2O}$$

$$\ce{HPO4^2- + H3O+ <=> H2PO4^- + H2O}$$

Since these are all weak bases, you won't "consume" all $\ce{H3O+}$ and the solution will be acid.

If, however, you add a strong acid, the scenario changes:

$$\ce{BaHPO4 + H3O+ <=> Ba^2+ + H2PO4^- + H2O}$$

You'll still have all phosphate species as above, but now Barium Hydrogen Phosphate has acted as a base, since it received some hydrogen atoms, which is a Brønsted-Lowry base definition.

Remember the same kind of behaviour happens with water, even if pure, because of self ionization:

$$\ce{2H2O <=> H3O+ + OH-}$$

Here, water acted as both the base and the acid, which is why it's an amphoteric species.


Are alkenes or alkynes reduced first with diimide? I know that $\ce{NH2NH2}/\ce{H2O2}$ is used as an reducing agent for selective reduction of alkenes and alkynes. $\ce{NH2NH2}$ reacts with $\ce{H2O2}$ to form diimide ($\ce{N2H2}$), which then undergoes a percicylic mechanism with the alkene/alkyne to give the reduced product.

My textbook had this question: what does the reaction of but-1-en-3-yne (vinylacetylene) with 1 equivalent of $\ce{N2H2}/\ce{H2O2}$ give?

Now my question is which reacts first? I know that in electrophillic reactions the alkene has greater reactivity than the alkyne but this is a pericyclic mechanism so i am not sure about it. The given answer is that alkene part reacts, not the alkyne.

Your given answer to the textbook question is seemingly correct. I said seemingly, because this reaction is not well understood to date, even though its first introduction was given at time as far as 1929 (Ref.1). The reaction was mostly hidden without getting attention in literature until to the middle of 20th century. This is mostly because the reactive species of the reaction, the unstable molecule, diimide ($\ce{HN=NH}$) was not well understood.

By 1960's, the unstable $\ce{HN=NH}$ was generated in a variety of ways such as oxydation of hydrazene (by $\ce{H2O2}$ or atmospheric $\ce{O2}$), or decarboxylation of azodecarboxylic acid, or thermal decomposition of p-toluenesulfonylhydrazene. It reacts with alkynes (e.g., diagram A below; Ref.2) and unhindered alkenes (e.g., diagram B below; Ref.3) to transfer two hydrogens in a cyclic process (diagram C below; Ref.4), resulting in a strictly cis reduction. This is the mildest way of performing such reductions. As shown in the diagram A below, one generally useful application of this reduction by diimide is it did not cleave the weak $\ce{C-I}$ bond during the reduction process. Most other reducing agents will cause competing or complete cleavage of the $\ce{C-I}$. The most important advantage in this case is in the conversion of iodoalkyne to vinyl iodide, it gives predominantly the cis-isomer. In addition, it is note worthy that this method of reduction with diimide does not lead to the cleavage of other sensitive functions such as $\ce{O-O}$ and $\ce{N-O}$ bonds.

To my understanding, the type of functions reduced by diimide are depend on the way it was generated. Besides, the rate determining step is always the rate of diimide generation. In general, trans-alkenes react more rapidly than cis-alkenes. During the mechanistic study of reaction, it was reported that diimide reduction of symmetrical multiple bonds proceeds readily. Those bonds included $\ce{C=C}$, $\ce{C#C}$,$\ce{N=N}$, and $\ce{O=O}$. On the other hand, more polar functions such as $\ce{C=O}$, $\ce{C#N}$,$\ce{C=N}$, and $\ce{O-N=O}$ are either very slowly reduced or not reduced at all (Ref.4).

The table 1 has given some uniques alkene substrates used in this process (Ref.5). In this research, 2-nitrobenzenesulfonylhydrazene has been used to generate diimide in situ.

  Abstract of Ref.5: A one-pot protocol for the formation of 2-nitrobenzenesulfonylhydrazide (NBSH) from commercial reagents and subsequent alkene reduction is presented. The transformation is operationally simple and generally efficient for effecting diimide alkene reductions. A range of 16 substrates have been reduced, highlighting the unique chemoselectivity of diimide as a reduction system.

A recent publication has shown that diimide can be generated by refluxing haydrazene in air (with solvent). The reduction of diimide generated by this method gives somewhat different results. For example, a norbornene derivative gives 100% conversion in Table 1. However, similar substrate used in Ref.6 has given only 42% yield (Entry 15; Ref.6). Further, a styrene derivative (cis-1-phenylpropene) gives 61% conversion in Table 1. Yet, styrene substrate used in Ref.6 (should be better since it is a terminal alkene) has given only 16% yield (Entry 17; Ref.6). Therefore, it is very convincing that the outcome is highly depend on the metheod of generating diimide,

Based on all, the result of the reaction in the question is uncertain.

References:

J. Hanus, J. Voříšek, "Action de L'hydrazine sur Quelques Acides Incomplets $\ce{C_nH_{2n-2}O_2, \ C_nH_{2n-4}O_2,\ C_nH_{2n-6}O_2}$," Collect. Czech. Chem. Commun. 1929, 1, 223-227 (https://doi.org/10.1135/cccc19290223).
Scott E. Denmark, Shyh-Ming Yang, “Total Synthesis of (+)-Brasilenyne. Application of an Intramolecular Silicon-Assisted Cross-Coupling Reaction,”  J. Am. Chem. Soc. 2004, 126(39), 12432-12440 (https://doi.org/10.1021/ja0466863).
James D. White, Rich G. Carter, Kurt F. Sundermann, Markus Wartmann, “Total Synthesis of Epothilone B, Epothilone D, and cis- and trans-9,10-Dehydroepothilone D,” : J. Am. Chem. Soc. 2001, 123(23), 5407-5413 (https://doi.org/10.1021/jo9710137).
E. E. van Tamelen, R. S. Dewey, M. F. Lease, W. H. Pirkle, “Selectivity and Mechanism of Diimide Reductions,” J. Am. Chem. Soc. 1961, 83(20), 4302-4302 (https://doi.org/10.1021/ja01481a057).
Barrie J. Marsh, David R. Carbery, “One-Pot o-Nitrobenzenesulfonylhydrazide (NBSH) Formation−Diimide Alkene Reduction Protocol,” J. Org. Chem. 2009, 74(8), 3186-3188 (https://doi.org/10.1021/jo900237y).
Hao Chen, Jianmin Wang, Xuechuan Hong, Hai-Bing Zhou, Chune Dong, "A simple and straightforward approach toward selective C=C bond reduction by hydrazine," Canadian Journal of Chemistry 2012, 90(9), 758-761 (https://doi.org/10.1139/v2012-057). Here is the available PDF of the article through Science Gate.


Why can acidification of solution lead to selective precipitation of ions such as Fe2+ and Zn2+? Given $K_\textrm{sp}(\ce{ZnS})= \pu{2.5E-22}$ and $K_\textrm{sp}(\ce{FeS})= \pu{3.7E-19}$, why would $\ce{Fe^{2+}}$ ions remain in solution but not $\ce{Zn^{2+}}$ with acidification of the solution?
I have little understanding of this, so I don't have a great attempt at the solution.

FeS has 1500 times the solubility of ZnS.  Given the slow addition of a small molar fraction of acid that solubilizes both Zn(II) and Fe(II), that fraction will retain at equilibrium.  It gets complicated if soluble polysufides exist. Iron gives pyrite and marcasite ($\ce{FeS2}$) via mackinawite ($\ce{FeS}$) and greigite ($\ce{Fe3S4}$), but they are insoluble.


Redox reactions: identifying what is oxidized vs reduced $$\ce{SO4^2- + H2O2 + 2H+ -> SO2 + O2 + 2H2O}$$

In the above chemical reaction, I'm having trouble identifying what is oxidized and what is reduced.

My guess is: sulfur has reduced but I can't work out what has oxidised. My guess is the middle oxygen as it has lost two hydrogen atoms to $\ce{2H2O}$.

Yes sulfur of $\ce{SO4^2-}$ has undergone reduction as the oxidation number reduces from +6 to +4 and Oxygen of $\ce{H2O2}$ has undergone oxidation as its oxidation number increases from -1 to 0.

$$\ce{\overset{\color{red}{+6}}{S}O4^{2-}\+H2\overset{\color{red}{-1}}{O2}\+2H+->\overset{\color{red}{+4}}{S}O2\+\overset{\color{red}{0}}{O2}\+2H2O}$$

NOTE: Oxidation sate of oxygen in peroxide is -1. This because in the neutral molecule hydrogen peroxide the oxidation state of $\ce{H}$ is +1 and as the sum of oxidation numbers of the two hydrogen atoms and the two oxygen atoms must be zero so each oxygen in $\ce{H2O2}$ has -1 oxidation state.


Determining the nature of common pH indicators (methyl orange) using Le Chatlier's principle I got some good information about this topic yesterday, and I understood what I was told, but I have come up with a specific way of phrasing the problem I'm having. It is slightly different than before, and I'm looking for someone to tell me if what I'm doing makes any sense or not, rather than just rote learning facts like 'methyl orange indicator is a weak acid/base'. I want to be able to use this process for any weakly acidic/basic indicator I come across.

(Note: Methyl orange is taken to be orange when it has a moderate pH (HIn, not very acidic or basic, and red between a pH of 4.4 and 3.1 - i.e. when it is In- or In+ depending on the scenario below)

Assume methyl orange, an indicator, is a weak acid. It dissociates as follows in aqueous solution;

We know that if an acid is added to this system at equilibrium, the concentration of hydrogen ions will increase. Using Le Chatlier's principle; to oppose this stress, the reaction shifts to the right, and the colour of the initial indicator (in this case methyl orange, so orange) will dominate and become more intense.

The solution will not go red, like it would if a base was added and proton concentration was decreased, as the equilibrium would shift in the other direction.

The problem is, that if you add an acid to this system in real life, a colour change occurs at a pH of 4.4, from orange to red. This means that methyl orange indicator cannot be an acid. Right?

If you instead assume that methyl orange is a weak base, it dissociates like so in aqueous solution;

So here, if you add an acid, the concentration of hydroxide ions will decrease (due to bonding between the hydrogen ions in the acid and these hydroxide ions to form water), so the equilibrium shifts to the right, to compensate.

This means that the red colour will dominate.

This result is in line with observation, as methyl indicator will change colour from orange to red at a pH of 3.1-4.4, which is acidic.

Does my reasoning therefore demonstrate that methyl indicator must be a weak base?

Methyl Orange is the common name for Sodium 4-[(4-dimethylamino)phenyldiazenyl]benzenesulfonate. So the compound is a salt or a conjugate base. It is the sodium salt for the indicator since that form is more soluble than the acid form.

The pKa is 3.47 in aqueous solution, so at pH values of 3.1 or lower the molecule is protonated.

At pH values greater than 4.4 all the molecules in solution are deprotonated as the sulfonate anion.

Linked two images from ChemWiki.

The confusion is caused by rather poor wording in a question.

I'd prefer the wording to be:

      ... In aqueous solution methyl orange indicator behaves like a weak acid. Letting HX represent the protonated form of the indicator, it dissociates as follows:

It is splitting hairs, but methyl orange is the name of the sodium salt, not the protonated form of the indicator.


Can citric acid create a polyester? Citric acid has both alcohol and carboxylic acid groups, so could it "esterify" with the other citric acid molecules in solution?

I notice that it is a tertiary alcohol. I don't suppose that would significantly hinder esterification, would it? Then again, it's not as if citric acid simply polymerizes with other citric acid  molecules in solution anyway.

The esterification of carboxylic acids with tertiary alcohols usually occurs by an SN1 mechanism wherein the oxygen of the alcohol is protonated and a tertiary carbocation is formed. It in turn reacts with the carboxylic acid to form the tertiary ester. The hydroxyl group in citric acid, while tertiary, is alpha to a carboxyl group. A cation alpha to a carboxyl group (-CO2H) is a destabilized entity. This mechanism would be precluded. Traditional Fischer esterification is likely sluggish owing to steric hinderance.


How to calculate the mass of water to add to copper(II) sulfate to form a 4% solution? How many grams of water are required to make a $4.00\%$ solution with $15.0~\mathrm{g}$ of copper(II) sulfate pentahydrate?

I calculated the percent composition of water in a molecule of $\ce{CuSO4.5H2O}$ and it came out to be $36\%$. From that I calculated that there was a total of $5.4~\mathrm{g}$ of water in the sample. I'm guessing that we have to calculate how much water is required in order to make it a $4.00\%$ solution, but I don't understand why it wants it in grams. When
$$\text{Percent by volume} = \frac{\text{grams of solute}}{\text{mL of water}}?$$

A $4~\%$ solution is one that contains $4~\mathrm{g}$ of the solute in $96~\mathrm{g}$ of solvent to give $100~\mathrm{g}$ of solution. Just to get that out of the way.
To calculate the mass of the solute ($m(\ce{CuSO4})$ — not to be confused with $m(\ce{CuSO4 . 5 H2O})$) we can use at least two ways which are mathematically equivalent:

Calculate the amount of $\ce{CuSO4 . 5 H2O}$ — this is equivalent to the amount of $\ce{CuSO4}$. Then calculate the latter’s mass.
$$M(\ce{CuSO4 . 5 H2O}) = 249.72~\mathrm{g\ mol^{-1}}\\[1.2em]
n(\ce{CuSO4 . 5 H2O}) = \frac{m}{M} = \frac{15.0~\mathrm{g}}{249.72~\mathrm{g\ mol^{-1}}} = 6.01 \times 10^{-2}~\mathrm{mol}\\[1.5em]
M(\ce{CuSO4}) = 159.62~\mathrm{g\ mol^{-1}}\\[1.2em]
m(\ce{CuSO4}) = n \times M = 6.01 \times 10^{-2}~\mathrm{mol} \times 159.62~\mathrm{g\ mol^{-1}} = 9.59~\mathrm{g}$$

Use the molar weights of copper(II) sulphate and water to determine the mass percentage of copper(II) sulphate in the salt.
$$\frac{M(\ce{CuSO4})}{M (\ce{CuSO4 . 5 H2O})} = \frac{159.62~\mathrm{g\ mol^{-1}}}{249.72~\mathrm{g\ mol^{-1}}} = 63.9~\%\\[1.5em]
\Longrightarrow m(\ce{CuSO4}) = 15.0~\mathrm{g} \times 63.9~\% = 9.59~\mathrm{g}$$

In any case, we will arrive at a mass of the solute of $9.59~\mathrm{g}$. This is equivalent to the $4~\%$ mentioned at the very top. Therefore:
$$m_\mathrm{tot} = \frac{m_\mathrm{solute}}{0.04} = 240~\mathrm{g}$$
$$m_\mathrm{solvent} = 0.96 \times 240~\mathrm{g} = 230~\mathrm{g}$$
You noted that you already have $5.4~\mathrm{g}$ of water in the copper(II) sulphate hydrate present. Thus, you need to add the difference:
$$m_\mathrm{water} = 230~\mathrm{g} - 5.4~\mathrm{g} = 225~\mathrm{g}$$


Bleach in drain cleaners I have a clogged pipe in my bathroom so I went to the store to get something to unclog it. Basically there are two types that you can find in my area:

Sodium hydroxide pellets
"Liquid plumber" mixtures. The components for those are mainly surfactants, sodium hydroxide and sodium hypochlorite.

As far as I recall from the water treatment courses I took in college, hypochlorite's oxidizing activity is heavily dependent on pH, as the active form is hypochlorous acid, not the hypochlorite ion. At high pH values the oxidizing activity becomes really low whereas low pH will make it stronger.

Does it make sense to buy a product that mixes a strong base, such as sodium hydroxide, with sodium hypochlorite? Wouldn't it be a waste of money?

Lye is very corrosive, just like acid, except opposite (hope that makes sense). Although lye is a very powerful chemical, and it is sufficient for many blockages, oxidizing agents (like bleach) may be helpful for breaking down paper/cellulose blockages.

Mixing bleach with acid will cause chlorine gas evolution... bad, bad, bad.

EDIT- The oxidative powers of hypochlorous acid and hypochlorite ion(s) compared:

Source: Descriptive Inorganic Chemistry, Third Edition (By Geoff Rayner-Canham, Tina Overton)

So in terms of power, yes hypochlorous acid is a stronger oxidizer, but it's not wildly different.

On a related note (see comments below) hypochlorous acid is a much better microbicide than the hypochlorite ion, because it penetrates cell walls more easily. However, regarding a drain cleaner, the oxidative difference is much less pronounced. And it's worth mentioning that at above pH 12, a cell will lyse (burst open) and become saponified, so that obviates whatever microbial effect hypochlorite or hypochlorous acid might have on bacteria... like being thrown into a poisonous wood chipper.


Is Cracking of Every HydroCarbon Family possible? (Alkane ,Alkenes,Arenes) I know Cracking is usually used to crack alkanes to get smaller hydrocarbons (as they are easier to ignite) and alkenes(as they have more industrial uses)...but can we crack arenes and alkenes as they are also included in hydrocarbons definition.
But I really doubt about cracking of benzene as lot of energy is required to destroy the delocalised ring. But given infinite energy that is surely possible.... but the product won't be stable... :/

Simple arenes (benzene, toluene, xylenes) are a typical product of a steamcracker. https://en.wikipedia.org/wiki/Cracking_(chemistry)#Thermal_methods

Catalytic hydrocracking otoh. (with added hydrogen at high pressure) leads to an increase in the aliphatic fraction and lowers the aromatic content. https://en.wikipedia.org/wiki/Cracking_(chemistry)#Hydrocracking

in itself, benzene is quite stable. It's a questions if it has sth. to react with.


A chemical compound with these conditions? I am searching for a chemical compound which meets these five conditions:

slightly soluble in water
layered strutcture
ionic conductor
thermochromism
lead must be a part of it.

I will satisfy 4 of the five conditions, but you will have to research the fifth. Lead(II) iodide has a layered structure, is slightly soluble in water, is ionic (but only weakly). I hope that this helps.


How to calculate the average quadratic velocity of carbon dioxide from given data? I am working on this question, and I am unsure of how to proceed. The question is:

  There is $31.7~\mathrm{g}$ of $\ce{CO2 (g)}$ in a container. The partial pressure of $\ce{CO2}$ is $2.62~\mathrm{atm}$ and the volume of the container is $26.8~\mathrm{L}$. What is the average quadratic velocity ($\mathrm{m/s}$) of the $\ce{CO2}$ in the container?

I think that the formula I should use is this one: $V = \sqrt{3RT/M}$, but I am unsure on how to find the temperature.

Use the Ideal Gas Law: $PV=nRT \to T = \frac{PV}{nR}$.  From the mass you can find the number of moles, $n$. $P$ and $V$ are given, and $R$ is the universal gas constant.

Even better you could rearrange the equation to $RT = \frac{PV}{n}$, substitue the value for $RT$ and do less arithmatic.


Reaction of solid metals with a solution of metal ions Explain the following experiments and their results. What is their chemical equations as well?

When immersing a Magnesium (Mg) chunk inside aqueous solution of Copper ions (Cu(II)), a red-brown sediment of Copper appears. Why?
When immersing a Silver (Ag) stripe inside aqueous solution of Calcium ions (Ca(II)), no chemical reaction takes place. Why?

You can explain these experiments with the Standard Electrode Potential. With these experimental values you know which element/compound releases more energy upon receiving a comparable amount of electrons. As a rule of thumb, alkalimetals and alkaline earth metals release less energy than, for example, elements from the halogen-group.

With this table at hand, you can predict that $\ce{Mg}$ will be oxidized and the $\ce{Cu^2+}$ reduced, because of the potential for the two following reactions:
$$\ce{Mg^2+ + 2e^- -> Mg(s)} \quad E^\circ=−2.372\,\text{V}$$
$$\ce{Cu^2+ + 2e^- -> Cu(s)} \quad E^\circ=0.337\,\text{V}$$

Reducing copper-ions will release more energy, so the following chemical reaction will apply:
$$\ce{Mg +Cu^2+-> Cu + Mg^2+}$$

In an analogous fashion, for the second question:
$$\ce{Ag^2+ + 2e^- -> Ag(s)} \quad E^\circ=+1.98\,\text{V}$$
$$\ce{Ca^2+ + 2e^- -> Ca(s)} \quad E^\circ=−2.868\,\text{V}$$

Reducing $\ce{Ag^2+}$ will release more energy - since silver is already in its solid state, the following reaction will not take place:

$$\ce{Ag +Ca^2+-> Ag^2+ + Ca}$$

Note: To see how those potentials or potential difference relate to energy see, for example, the Nernst Equation / Gibbs free energy