How explosive is benzene compared to TNT? Benzene is explosive within the explosive limits of minimum $\pu{1.2 \%}$ and maximum of $\pu{7.8 \%}$ in air. Source: Wikipedia More accurate, benzene itself is not explosive, but the fuel-air mixer or, fuel–air explosive (FAE), is. This fuel-air explosive is used, because of its power in Thermobaric weapons https://en.wikipedia.org/wiki/Thermobaric_weapon Benzene does burn and could be a fuel in a fuel-air explosive. So it would be, more specific, a benzene-air explosion.
I would like to know what is the explosive power of benzene measured in TNT-equivalent. Benzene can explode like in China in 2005 which was 'only' 100 ton benzene.
The reason why I want to know is that there was a huge explosion in Beirut on August 4th 2020, with the equivalent of 1.1 kiloton TNT. In my country, the Netherlands, benzene is transported in huge ships of 2000 ton. This could explode, spreading the benzene gas over the area, which is heaver than air. Benzene is highly carcinogen. I would like to know how a possible explosion could be compare with the explosion in Beirut. Benzene burns very easy, so may be only a fraction of benzene would explode, the rest would burn. I just want to know what we, as citizens, might expect when things goes seriously wrong.
An approach could be a list with the relative effectiveness of benzene compared to TNT. All kind of benzene derivatives are mentioned in that list but not benzene itself. I could not find  a list in which benzene itself is mentioned.
So my (much to simple) calculation is:

1 ton TNT: $\pu{4184 \times 10^{12} J}$ see: TNT equivalent
Energy [(Heat of combustion) / mol]: $\pu{3267.6 kJ/mol}$ see: Benzene
Energy / kg:   $\pu{41831 \times 10^{6}J/kg}$ ($\pu{\frac{3267.6 ~ kJ/mol}{78114 ~ g/mol} \times 1000 g/kg}$)
Mass: $\pu{2000 kg}$
Total energy: $\pu{83662 \times 10^{12} J}$ ($\pu{41831 \times 10^6 J/kg \times 2000 kg}$)
TNT-equivalent: $\pu{~ 20 kiloton}$ TNT ($\pu{\frac{83662 \times 10^{12}~  J}{4182 \times 10^{12}~  J/ton}}$ TNT)

My questions are:

Could all benzene explode at once? Because benzene might be mixid in
air within the explosive limits range, but this will be only for a
limited ammount of benzene. But because the explosion the rest of the
benzene will be blown into the air and possible then be again be
within the explosive limites ranges. So causing a chain-reaction. The
explosion in Jilin in China took 1 hour so that seems to be the case.
If not is it possible to calculate what is the maximum of benzene that
could explode?

Note: I have very limited chemical knowledge. I can passive read chemical formulae and make very simple calculations like above.

All flammable organic liquids could, in principle, create large fuel-air explosions but the conditions are very hard to achieve accidentally
The first important point to note is that benzene should not be a particular worry. There are plenty of other, widely used, chemicals and mixtures that could, in principle, cause large explosions. Standard automotive petrol (gasoline) for example is present in most vehicles (non-diesel ones anyway) and is transported in 40 tonne tankers on most roads. Fires and explosions happen, but rarely, and they are never known to be of the thermobaric type you are worried about. Most motor vehicles have 40kg or so of fuel that is just as flammable as benzene but after accidents they rarely catch fire or explode and never do so with the power of a fuel air thermobaric explosion (the hollywood trope that all crashed cars explode is pure nonsense).
But big fuel air explosions are known to be possible and have been developed into some very effective military weapons. These are known as thermobaric bombs and the largest are just about the biggest non-nuclear explosions possible in existing military arsenals. But a quick read of how the weapons work (see this wikipedia entry) suggests that they require some fairly notable technology to get right. Largely this is because the fuel has to be dispersed at just the right concentration in the air to get a detonation or deflagration rather than a fiery but unimpressive "phut". It isn't easy to achieve this by accident with a large volume of fuel.
Bad maintenance and safety procedures in chemical and oil plants can lead to accidental fuel-air explosions. The Buncefield explosion in a UK oil storage depot was caused by one. It was a huge explosion probably involving about 300 tons of fuel (but that is only about 0.1% of the fuel stored in the depot a lot of which caught fire but didn't explode). Another famous UK example is the Flixborough disaster caused by a fuel-air explosion caused by a leaky pipe in a cyclohexane plant. In both incidents poor management and engineering contrived to make a dangerous event possible and standards were rewritten to prevent future mistakes. That such event are rare should be of significant comfort if you are worried about the dangers of transporting tankers full of benzene. It is worth noting that none of these (very bad0 accidents came close to involving the majority of the flammable compounds stored nearby. It is hard to get the right fuel-air mixture with very large volumes of fuel, thank goodness.
A more common form of fuel air explosion might also put the rare chemical plant incidents in perspective. This is explosions caused in flour mills and grain elevators. a mixture of finely divided flour or grain dust is explosive and plants that handle flour and grain have to be specially designed to avoid dangerous dust explosions. These are far more common that explosions in chemical plants. While this may not be much comfort, it does suggest that worrying about benzene is not that big a real world risk.


Defining the Turbomole named BP86 functional in Gaussian According to Turbomole, the BP86 density functional has the following definition:

$$
\mathrm{Exchange = LDA + B88\\
Correlation = LDA(VWN(V)) + P86}
$$

What is the equivalent definition in Gaussian using IOp keywords?

From the Gaussian 09 manual:

  User-Defined Models. Gaussian 09 can use any model of the general form:
      P2EXHF + P1(P4EXSlater + P3ΔExnon-local) + P6EClocal + P5ΔECnon-local

  The only available local exchange method is Slater (S), which should be used when only local exchange is desired. Any combinable non-local exchange functional and combinable correlation functional may be used (as listed previously).

  The values of the six parameters are specified with various non-standard options to the program:

  IOp(3/76=mmmmmnnnnn) sets P1 to mmmmm/10000 and P2 to nnnnn/10000. P1 is usually set to either 1.0 or 0.0, depending on whether an exchange functional is desired or not, and any scaling is accomplished using P3 and P4.
  IOp(3/77=mmmmmnnnnn) sets P3 to mmmmm/10000 and P4 to nnnnn/10000.
  IOp(3/78=mmmmmnnnnn) sets P5 to mmmmm/10000 and P6 to nnnnn/10000.

  For example, IOp(3/76=1000005000) sets P1 to 1.0 and P2 to 0.5. Note that all values must be expressed using five digits, adding any necessary leading zeros.

  Here is a route section specifying the functional corresponding to the B3LYP keyword:

  #P BLYP IOp(3/76=1000002000) IOp(3/77=0720008000) IOp(3/78=0810010000)

  The output file displays the values that are in use:

IExCor=  402 DFT=T Ex=B+HF Corr=LYP ExCW=0 ScaHFX=  0.200000
ScaDFX=  0.800000  0.720000  1.000000  0.810000

  where the value of ScaHFX is P2, and the sequence of values given for ScaDFX are P4, P3, P6 and P5.

For BP86 specified normally, this looks like:

IExCor=  404 DFT=T Ex=B Corr=P86 ExCW=0 ScaHFX=  0.000000
ScaDFX=  1.000000  1.000000  1.000000  1.000000

Those 3 IOps don't completely define the functional, however, IOp(3/74) is also needed. See the IExCor field, which corresponds to Becke 1988 exchange (400) combined with Perdew 81 + Perdew 86 correlation (4).


Can aldehydes be reduced to alcohols efficiently using LiAlH4 (excluding cost effectiveness)? In my textbook (NCERT 12th class Chemistry, Part 2, India), $\ce{H2}$ with $\ce{Pd}$ catalyst is the only method mentioned.

Yes, $\ce{LiAlH4}$ can be used for the reduction of an aldehyde to an alcohol. Compared to $\ce{NaBH4}$ it is a stronger reducing agent due to the electronegativity differences (Pauling electronegativity of $\ce{Al}$ is 1.61 & $\ce{B}$ is 2.04) making the metal-H bond more polar in case of $\ce{LiAlH4}$. Therefore, it reacts very violently with water and other acidic group containing compounds, and careful addition of $\ce{LiAlH4}$ is required.

The mechanism is rather straightforward, with alkoxide formation and protonation steps to get the final product.

In the lithium aluminum hydride reduction, water is usually added in a second step, whereas in the sodium borohydride reduction the methanol solvent system achieves this protonation immediately.


How to calculate the enthalpy change from the reaction of silver nitrate and sodium iodide? When $\pu{20.00 mL}$ of $\pu{1.00 M}$ $\ce{AgNO3}$ solution is added to $\pu{20.00 mL}$ of $\pu{1.00 M}$ of $\ce{NaI}$ at $\pu{25 ^\circ C}$ in a calorimeter, a white precipitate of $\ce{AgI}$ is formed. The temperature of the aqueous mixture increases to $\pu{40 ^\circ C}$.

I am trying to calculate the $\Delta H$ for the reaction per mole of $\ce{AgI}$.
The specific heat of the aqueous mixture is $\pu{4.184 J//g K}$, the density of the mixture $\rho = \pu{1.00 g//ml}$. And I assume that the calorimeter absorbs a negligible amount of heat.

How am I suppose to go about finding the $\Delta H$ per mole of $\ce{AgI}$?
I don't really know where to start here as I am unclear about the whole problem here. In order to use the formula
$$Q = m C_p \Delta T$$
I need to find the mass. But which mass do I take? $\ce{AgI}$, or $\ce{NaNO3}$,  or is it the mass of the entire product of the reaction?

This is the balanced equation
$$\ce{AgNO3 + NaI -> AgI + NaNO3}$$

When the reaction takes place and releases its heat, everything in the calorimeter  heats up: the water, the $\ce{AgI}$ and the $\ce{NaNO3}$. But of these the water is by far the most massive and for that reason, as a reasonable approximation, we consider that mass that absorbs the heat to be the water, i.e. $40.00\ \mathrm{g}$.

Use this to calculate $Q$.

Then determine how many moles of $\ce{AgI}$  was formed, by stoichiometry. The quotient of these two numbers gives you the reaction enthalpy per mole of $\ce{AgI}$


Reaction mechanism of redox reaction So on the data, it said that we can reduce potassium permanganate by utilizing oxalic acid, thus producing carbon dioxide.
But can anyone propose a reaction mechanism of how the whole reaction happens, and how did oxalic acid break up into $\ce{CO2}$?
Thanks in advance!

I think it would go through a manganese-oxalate complex (oxalate is a good ligand for many metals), and that would fragment via a reaction such as:

The Mn(V) species would likely disproportionate to manganese dioxide and permanganate.


Electrolysis of Potassium Iodate/Hydrogen Peroxide in Briggs-Rauscher Oscillation I would like to investigate the effect of adding a voltage in the Briggs-Rauscher oscillation reaction. I will examine the oscillations visually.

The ingredients I will use include potassium iodate, sulfuric acid, acetone, manganese sulfate, starch indicator and hydrogen peroxide.

There are a number of complex reactions which have not to this day been completely analysed. Hence I will not attempt to explain all of them.

To the point. What electrodes should I use for electrolysing potassium iodate? Just like the saltwater circuit, would electrolysis will proceed as iodine carries charges towards the aluminium anode, and as potassium obtains them in the copper cathode? Could I electrolyse hydrogen peroxide alternatively?

I would like to examine reduction potentials, comparing them to what extent the electrolysis of potassium iodide prevents the behavior of potassium iodide in the other equilibria, as possibly seen visually.

Is it also possible for the manganese metal to conduct with copper just as potassium might?

The experiment has been attempted. Electrolysis through the use of copper electrodes in fact "kills" the reaction. This means that the whole system is immediately turned dark-blue, and eventually made more clearer, and eventually light blue indicating the presence of copper ions.

It is suspected that not only is the sulphate ion electrolysed, but perhaps the copper iodizes the potassium and other compounds.


Reaction mechanism of combustion of hydrogen After a discussion about the usefulness of hydrogen fuel for cars with a friend of mine, I wondered what the reaction mechanism for combustion of $\ce{H2}$ was. I only study (well almost) organic chemistry, so I have no idea if the mechanisms from organic chemistry can be applied at all to non-organic molecules.

So we have the following reaction:

$$2\ce{H2 + O2 -> 2 H2O}$$

In most textbooks this would be described as a redox-reaction with the two half reactions:

$$2\ce{H+ + 2 e- -> H2}$$
$$\ce{O2 + 4 H+ + 4 e- -> 2 H2O}$$

But I wonder what the real mechanism is. Based on my organic thinking I thought the following (it's probably completely wrong, but I think it is important to state because someone can correct me much better if I express my thinking than if he has no idea where the misconception started):

If the temperature is high enough a long pair of oxygen could break the $\ce{H-H}$ (the lone pair will be donated to the $\sigma^{*}$ bond) bond and form $\ce{2 H-}$ (each H gets one electron because of their equal electronegativity). $\ce{H-}$ is very unstable and one will immediately react with the empty oxygen orbital, the other will break the $\ce{O=O}$ $\pi$ bond. The electrons in the $\pi$ bond will go to the positively charged oxygen atom (the one that donated its electrons in the first step). Now one oxygen atom is neutral and the other has a positive charge (because the electrons of the $\pi$ bond went for the other oxygen), where the $\ce{H-}$ will attack forming hydrogen peroxide ($\ce{HO-OH}$). The previous process will now repeat, but instead, the $\sigma$ bond between the oxygen atoms will be broken, resulting in two water molecules, $\ce{2 H2O}$.

This post deals with the mechanism that is observed in the gas phase. It is of course not as simple as the equation might suggest and you did suspect that already.
$$\ce{2H2 + O2 -> 2H2O}$$

This will be divided into many different elementary sub reactions. Any mixture of oxygen and hydrogen is metastable (stable as long as you do not change the conditions). If you provide sufficient energy to overcome the activation barrier, the reaction will take its cause:
$$\ce{H2 <=>[\Delta T] 2 H.}$$

The resulting hydrogen atoms are very unstable and will react with anything in reach, but most importantly with $\ce{O2}$ (which is a triplet biradical), forming hydroxyl radicals. This reaction is endothermic (it requires energy).
$$\ce{H. + O2 -> HO. + O}$$

The resulting oxygen radicals can again react with $\ce{H2}$ to form more hydrogen radicals. This reaction is also endothermic.
$$\ce{O + H2 -> HO. + H.}$$

The hydroxyl radicals can also react with $\ce{H2}$ to form more hydrogen radicals, which is a slightly exothermic (releases energy) reaction.

The net result of those equations leads to a slightly exothermic sum, with high potential:
$$\ce{3H2 + O2 -> 2H2O + 2H.}$$

In principle, the hydrogen radicals react as a catalyst. However, this reaction is a highly branched chain reaction, including a lot of radical reactions. Due to this, more and more hydrogen radicals will be produced.

This scheme will eventually end when the concentrations of $\ce{O2,H2}$ will become lower forcing the excess radicals to react with each other.
$$\ce{HO. + H. -> H2O}$$

Another possibility is forming as a byproduct hydrogen peroxide, which is a very exothermic reaction:
$$\ce{H. + O2 -> HO2.}\\
\ce{2HO2. -> H2O2 +O2}$$

(Also happening but not as important: $\ce{HO. + HO. <=> H2O2}$)

As long as there are hydroxyl and hydrogen radicals present, the peroxide will take part in the chain reaction. However, this reaction also provides the necessary energy to cleave more $\ce{H2}$. peroxide is also easily cleaved again, or reacts with each other:
$$\ce{2H2O2 -> 2H2O + O2}$$

This all results in the main product water.

Please keep in mind, that there are many factors, that influence these reactions. It is strongly dependent on pressure, temperature, and surroundings. Surfaces and/ or catalysts involved in this reaction may change it completely.

Further reading:

How explosive is hydrogen gas?
Eugene P. Dougherty1 and Herschel Rabitz1, J. Chem. Phys. 72, 6571 (1980) most comprehensive (theoretical) study on the $\ce{H2-O2}$ system.
Wiberg, Nils: Lehrbuch der Anorganischen Chemie Kap. 3.2 (german) The mechanism is in part taken from this source.
http://arrow.utias.utoronto.ca/~ogulder/ClassNotes3.pdf Thanks to Silvio Levy


Which is more stable Ethyl carbocation or Neopentyl carbocation? As per my knowledge, hyperconjugation is more dominating over inductive effect, so due to three alpha hydrogen ethyl carbocation should be more stable. In some books, while solving questions I observed that answers can match if I take neopentyl carbocation as more stable.

so which one of them is comparatively more stable?

In that case ethyl carbocation has two alpha hydrogen atoms as CH2 is bonded to C+.And in case of neopentynyl carbocation CH3 is attached to C+ .more alpha hydrogen atoms ,more hyperconjugative structures,and more stability.and there are three donating groups in neopentynyl carbocation.these may be the reasons


Comparing acid strength of oxoacids derived from oxides Acidic strength has been compared before here and here, but the first question is about organic chemistry (and I am talking about inorganic chemistry), and the second does not discuss molecules such as $\ce{Cl2O7}$, $\ce{P4O10}$, $\ce{CO2}$, $\ce{N2O5}$, $\ce{Al2O3}$ etc. These molecules do not contain hydrogen. So what are the factors which affect their acid or base strength?
The following question brings me this doubt:

Which of the following is the correct order of acidic strength?
(A) $\ce{Cl2O7}$ > $\ce{SO2}$ > $\ce{P4O10}$
(B) $\ce{CO2}$ > $\ce{N2O5}$ > $\ce{SO3}$
(C) $\ce{Na2O}$ > $\ce{MgO}$ > $\ce{Al2O3}$
(D) $\ce{K2O}$ > $\ce{CaO}$ > $\ce{MgO}$

The question quoted in the OP is designed to test a periodic table trend concept.

Particularly, within a given row of the periodic table, excluding noble gases, the element oxides form increasingly acidic aqueous solutions from left to right across the row.

See the lecture by Dr. Pinkington The Acid-Base Character of Oxides and
Hydroxides in Aqueous Solution for additional information.

Only answer (A) is increasingly acidic left to right across the periodic table.


Derivation of van 't Hoff equation for temperature dependence of equilibrium constant While I was reading about the usefulness of the quantity $\Delta H$, I found that it can be used to calculate the how the equilibrium constant varies with temperature. How can this be done?

Does it agree with the predictions of Le Chatelier's principle (that for an exothermic reaction, increasing the temperature disfavours product formation, and vice versa)?

The equation linking $\Delta H^\circ$ and $K$ is called the van 't Hoff equation. Since Philipp's comment on your question already links to a very thorough discussion of where the equation $\Delta G^\circ = -RT\ln{K}$ comes from, I won't repeat it.

The definition of the Gibbs free energy, $G$, is $G = H - TS$. Using $\mathrm dG = V\,\mathrm dp - S\,\mathrm dT$ we obtain the Maxwell relation

$$\left(\frac{\partial G}{\partial T}\right) = -S$$

and hence the Gibbs–Helmholtz equation (derivation here)

$$\left(\frac{\partial (G/T)}{\partial T}\right) = -\frac{H}{T^2} \quad \Leftrightarrow \quad \left(\frac{\partial (\Delta G^\circ/T)}{\partial T}\right) = -\frac{\Delta H^\circ}{T^2}$$

Since $\ln K = -\Delta G^\circ/RT$, we have

$$\frac{\mathrm d(\ln{K})}{\mathrm dT} = -\frac{1}{R}\frac{\mathrm d}{\mathrm dT}\left(\frac{\Delta G^\circ}{T}\right) = \frac{\Delta H^\circ}{RT^2}$$

This is the differential form of the van 't Hoff equation; it's not the most useful thing to us though because it only tells you the slope of a plot of $\ln{K}$ against $T$ at one given point. We usually separate the variables and integrate with respect to both sides:

$$\int_{\ln{K_1}}^{\ln{K_2}}\!\mathrm d(\ln{K}) = \int_{T_1}^{T_2}\!\frac{\Delta H^\circ}{RT^2}\,\mathrm dT$$

$$\ln{K_2} - \ln{K_1} = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2} \right) $$

So, if you know the equilibrium constant $K_1$ at a certain temperature $T_1$ and you want to find the equilibrium constant $K_2$ at a different temperature $T_2$, you can just plug in your values into the equation and solve for $K_2$.

Note that this equation supports what you know of Le Chatelier's principle; if the reaction is exothermic, $\Delta H^\circ < 0$, and if you increase the temperature from $T_1$ to $T_2 > T_1$ then $(1/T_1 - 1/T_2) > 0$. The RHS of the equation is therefore negative, and that means that $\ln{K_2} < \ln{K_1} \Rightarrow K_2 < K_1$ which implies that the equilibrium position has shifted to the left.

Note that the last step (the integration) makes the assumption that $\Delta H^\circ$ is a constant over the temperature range $T_1$ to $T_2$. Note that this is, in general, not true but if the temperature range isn't too huge you will get pretty accurate results from the use of this equation.


Determine reaction rate constant from the reaction time? I was doing some research and I came across an experiment similar to mine (potassium iodate with sodium bisulfite) from the University of Massachusetts (http://www.chem.umass.edu/genchem/chem122/122_Experiment_4.htm), but with the reaction of potassium permanganate with oxalic acid.

The rate equation is given as:
Rate = k[KMnO4][H2C2O4].

It was stated that the value of k could be determined by dividing the concentration of the potassium permanganate [KMnO4] by the reaction time to find the rate and then dividing the rate by [KMnO4] multiplied by the concentration of the oxalic acid [H2C2O4] to find the reaction rate constant.

They provide the following equations:

Rate = [KMnO4]/t

k = Rate/([KMnO4][H2C2O4])

Would this be applicable to my reaction (potassium iodate with sodium bisulfite), because this would allow me to find the activation energy without performing trials to find k experimentally?

Assuming the reaction is a one step reaction, and does not have any elementary reactions, this could potentially work. However, if there are elementary reactions, this would change the rate law, and it would not be Rate = k[KIO3][NaHSO3].
You would also need to know the reaction order for both the potassium iodate and the sodium bisulfate, as well as the coefficients for both in the original reactions, yes you would be able to do this.
Knowing the coefficients, you could figure out the rate using rate -dA/adT. You could plug this rate into the rate law, Rate = k[KIO3]^m[NaHSO3]^n. However, you would need to know the reaction orders for KIO3 and NaHSO3, to plug in for m and n, and these can only be determined experimentally. If you know these, however, this equation would work for you.


If x and y are 2 extensive properties of a system then is x/y always intensive? I have a case to contradict my own question but in few books, it is given that x/y is always intensive. So
I got confused. My case is as follows :
Let a wire be our system whose resistance is R,
area of cross section is A,
length is L
and constant of resistance be ρ.
V,R are extensive
V(volume) = AL
R  = ρ(L/A)
So R/V = [ρ(L/A)]/AL = ρ/A^2
ρ is a constant but, A depends on the shape and size of system. So ρ/A^2 depends on shape and size of system. So it is an extensive property.
Therefore x/y can be extensive as well.
But in the below picture which is of G.R.B Op Tandon physical chemistry, it is given x/y is always intensive. Where did I make a mistake ?
Please note that I did not understand this is related more to physics or chemistry, so I asked it here as
chemistry is where I have first come across this problem. If I should ask this in the physics website, please tell me and I will do so.

Extensive properties are those that are directly proportional to the mass of the system, under the restriction that all intensive properties are held constant.
The restriction that "all intensive properties are held constant" means we are not changing the inherent nature of the system; rather, we are only changing its size.
For instance, if we keep all the intensive properties constant, and double the mass of the system, all the extensive properties will double.
Thus extensive properties all have the form $X=nZ$, where n is the number of particles in the system, and Z is some intensive property.  [We can use n in place of mass because the restriction that all intensive properties are held constant means that the composition doesn't change.]
Consequently, if we have two extensive properties, X1 and X2, their ratio is:
$$\frac{X1}{X2}=\frac{nZ1}{nZ2}=\frac{Z1}{Z2}$$
Since the n's cancel out, the ratio of two extensive properties becomes the ratio of two intensive properties.  And since intensive properties are independent of n, their ratio is also independent of n.
Based on the above definition of extensivity, volume is only extensive in 3-dimensional systems, because it's only in 3-D systems that $V\propto n$.  Area is only extensive in 2-D systems, because it's only in 2-D systems that $A\propto n$.
I have never before considered resistance in the context of extensivity, but I believe the reason you are running into trouble is that resistance depends both on the size of the system and its geometry.  Given that, I would conclude that resistance is neither intensive nor extensive.
You could make resistance extensive by constraining the geometry, specifically by constraining A to be constant, in which case resistance would be a 1-D extensive property that depends upon L: $R \propto L \propto n$.
However, I don't believe you could turn resistance into an extensive property by constraining L to be constant, since then you would have: $R \propto A^{-1} \propto n^{-2}$.


Why does equilibrium favour weak acid or weak base? For example, in the reaction $\ce{NaOH +HCl->NaCl +H_2O}$, the products are favoured by a factor of approximately $10^{24}$.
Is this a general rule? Is it because the products are more stable than the reactants?

Here is a simple intuitive explanation. In your above example, there is actually 2 reactions occurring, the forward and the reverse reaction. Let us consider the forward reaction. Here, both $\ce{NaOH}$ and $\ce{HCl}$ are strong bases and acids. What this means is that HCl has a strong ability to donate a proton and NaOH has a strong ability to accept protons. This means that when NaOH and HCl react together, the reaction will basically go to completion. So you will end up with up with very little NaOH  and HCl and the equilibrium will favour the products.

Now lets look at the reverse reaction which is: $$\ce{Na+ + Cl- + H2O -> NaOH + HCl}$$ Here $\ce{Cl-}$ is the conjugate base of HCl which is  a strong acid. This means that $\ce{Cl-}$ is a really weak base, meaning that it is pretty bad at pulling protons from molecules. Also water is the conjugate acid of NaOH, which is a strong base. This means that water is a really weak acid, meaning that it is also pretty bad at kicking out one of its protons. This means that in this reaction, barely any of the products will react and you will mainly end up with NaCl and water.

As you can see in both reaction the equilibrium will favour the weak acid and base as they aren't able to react that well together. This effect can also be explained by using equilibrium constants but I won't go into that.


Does the triplet sigma state of a diatomic molecule experience spin-orbit coupling? I know that states with spin S=0 in a diatomic molecule have no spin orbit coupling, independent on the value of the projection of the total electronic angular momentum.

I expect the same is true if the absolute value of $\Lambda$ is equal to zero independent on the spin of the diatomic molecule.

Is it correct that a diatomic molecule with a $^3\Sigma$ state has no spin orbit coupling?

Yes I would think so because the projection of orbital angular momentum is zero in forming $\Omega=\Lambda + \Sigma$. This would seem to correspond to Hund's case (b) where $\Lambda =0$ but $S \ne 0$, so no spin orbit coupling (unlike Hund's case (a)(c)(d)). The spin angular momentum remains fixed in space and the molecule rotates under it. If $R$ is the whole body (molecular) angular momentum this couples with the spin so that each rotational level splits into $2S+1$ components and the total angular momentum $J$ takes values from $R+S$ to $R-S$. The splitting is scaled with a spin rotation constant rather than spin-orbit constant but in $^3\Sigma$ there is also a $s_1\cdot s_2$ spin-spin coupling term.


Stability of complexes based on number of rings $\ce{[Co(edta)]^-}$ has 5 rings in total and in $\ce{[Co(en)_3]^{3+}}$, there are 3 rings. The stability of the former is more than that of the latter. The reason I was told was that a complex with more number of rings is more stable.
Why is this so?

When polydentate ligands binds to a metal in a ring form, it is called chelate ring and the corresponding complex is called chelate complex.
Experimentally, it is observed that metal complexes of polydentate ligands with more rings are significantly more stable than the corresponding complexes of similar ligands but with less rings. Because, the larger the number of ring closures to a metal atom, the more stable the compound. This phenomenon is called the chelate effect.
But, do note that ring number is not the only factor to determine the stability of a complex. There are several other factors like nature of central metal ion or the ligands, ring size, ionic size. For this question, let's consider ring size.
The more the ring size, the less stable is the complex. The entropy of complex is changed if the size of chelating ring is increased or decreased. Four membered ring compounds are unstable, whereas five-membered are more stable because of negligible ring strain. Likewise, ligands with a flexible organic backbone like ethylenediamine are significantly more stable than complexes with six-membered chelate rings, which are in turn much more stable than complexes with four- or seven-membered rings.
Refer to the following articles for more details:

https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Stability_of_Metal_Complexes_and_Chelation
https://old.amu.ac.in/emp/studym/99996027.pdf


Materials with solid volume greater than liquid volume Water has a solid volume greater than its liquid volume.

This question is in two parts.

1) Are there other materials that have solid volumes greater than liquid volumes? 2) What about the structure of a molecule makes its solid volume greater than its liquid volume?

There is a bunch of such materials; among the elemental compounds, they include silicon, gallium, germanium, and bismuth. As for the properties... well, it just so happens that their crystal structure contains a little too much of empty space. Crystal packing is a tricky thing, it can't be predicted with certainty even with the most sophisticated methods.


Solvent effects on acidity of ethanoic acid Is ethanoic acid most acidic in DMSO, water or methanol?

I suspect that the conjugate base would be most stabilized by a polar protic solvent (meaning it's weakest in DMSO), however, water and methanol are both protic and polar. Which causes ethanoic acid to be most acidic lowest $\mathrm{p}K_\mathrm{a}$? This was asked in an exam question where the following $\mathrm{p}K_\mathrm{a}$ values were meant to be assigned to the acid in each of the solvents: $9.5$, $12.6$, $4.8$.

Ethanoic acid has $\mathrm{pK_a}$ values of 4.76, 9.63 and 12.3 in water, methanol and DMSO respectively (source here and here). Comparing this with the $\mathrm{pK_a}$ values given in the question we can see that ethanoic acid is most acidic in water and least acidic in DMSO. $$\ce{pK_{a}(H2O) < pK_{a}(CH3OH) < pK_{a}(DMSO)}$$

In general the acidity of a species in different solvents is affected by three factors:

The dielectric constant of the solvent. Solvents with larger dielectric constants favour charge separation because the force pulling the charges together as they are separated is lower. This is evidenced by the Born equation which estimates the Gibbs free energy of solvation of gaseous ions: $$\Delta G = -\frac{N_Az^2e^2}{8\pi\epsilon_0r_0}\Bigg(1-\frac{1}{\epsilon_r}\Bigg)$$
The strength of solvation of the conjugate base of the acid.
The strength of solvation of the conjugate acid of the solvent.

Dielectric constant: The dielectric constants of water, methanol and DMSO are 80.1, 32,7 and 46.7 respectively (source). This gives some support to the greater acidity in water but contradicts the observed acidity in methanol and DMSO. This suggests that dielectric constant may not be the major factor that determines the relatives acidities.

Anion solvation: As you suggested in the question, water and methanol are protic solvents but DMSO is aprotic and so the ethanoate ion will be more strongly solvated in water and methanol, shifting the acid-dissociation equilibrium in favour of dissociation. This is considered to be the main factor which explains the greater acidity of ethanoic acid in protic solvents as opposed to aprotic solvents. When comparing water and methanol the strength of anion solvation will be broadly similar because both form $\ce{O-H\bond{~}Anion}$ hydrogen bonds.

Cation solvation: The difference in the stability of the conjugate acid of the solvent may be the factor that causes the difference between water and methanol. The $\mathrm{pK_a}$ of $\ce{CH3OH2+}$ is approximately -2.5 (several sources, none giving exact answers) and the $\mathrm{pK_a}$ of $\ce{H3O+}$ is -1.7. Therefore protonated methanol is less stable than protonated water and so this disfavours dissociation of the acid in methanol. This page claims that this can be attributed to oxygen $p$-orbital donation into $\ce{C-H}~\sigma^*$ orbitals but I don't have enough knowledge of MO theory to verify this (maybe someone else can).

Overall the greater acidity of ethanoic acid in water than methanol seems to be down to the greater dielectric constant of water and the greater stability of the solvent cation formed.

This page says some good things about the subject (some of my ideas were developed from reading this).

Additional reading: This paper looks to have some detail on the anion solvation aspect but I can't access the full text so I'll leave it as a link.


Alkene production by dehydrohalogenation What is the difference in the reaction mechanism when alkyl halides are reacted with aq.koh and alcoholic koh along with products

The fundamental reason lies in the difference between the action of a base and a nucleophile.

There are two types of mechanism through which substitution proceeds:-

SN1 Reaction - Here, the substrate undergoes hetrolytic cleavage to produce a carbocation and an anion. The thing which solvents do is to stabilise the newly formed ions. A Polar solvent(One case-Water) stabilises the intermediate stage more than the reactant stage because the reactant state forms weak dipole-dipole bonds with the solvent where as the intermediate stage has full fledged proper charges on then to interact with the solvent and forms strong ionic bonds.
SN2 Reaction - Same thing but because of the changes in mechanism type of solvent changes but the essence is same.

This tells us that the solvent is crucial to generate a strong electric field which can help the reaction proceed further by ionising the substrate whereas a base is not so capable of handling this type of work

In elimination reaction, the solvent generates a strong electric field around the substrate which in turn helps in the cleavage of C-H bond so that it can stabilise the carbocation(in E1 mechanism) or help the transition state to get rid of the leaving group(In E2 mechanism).

Basic action is a matter of equilibrium which shows how readily it can accept protons and go forward whereas nucleophilicity is all about the rate how fast it can go which in turn shoes us how eager it is to form bond or get rid of its negative charge.


Why do strong polyprotic acids not dissociate completely, whereas strong bases with multiple (OH)- groups do? $\ce{H2SO4}$ is a strong acid, which follows the reaction
$$\ce{H2SO4 -> HSO4- + H+}$$
Not
$$\ce{H2SO4 -> SO4^2- + 2 H+}$$

However,

$$\ce{Ba(OH)2 -> Ba^2+ + 2OH-}$$

Is there a reason why these "polyprotic analogs" in bases completely dissociate but polyprotic acids do not?

Note: This is in context to titrations, where $\pu{0.10M}$ of $.\ce{Ba(OH)2}$ is $\pu{0.20M}$ of $\ce{OH-}$, and leads to different values of titrant needed.

In simplistic terms, Arrhenius acids are molecules while Arrhenius bases are ionic compounds. Much like $\ce{BaCl2}$ – another ionic compound – it is a good first rule of thumb to assume it completely dissociates into its respective ions: one $\ce{Ba^2+}$ and two single-charge anions. You could say that barium were never really connected to either hydroxide; ionic compounds are often shown as collections of balls simply in a certain arrangement (crystal structure) but easily separable.

Arrhenius acids, which are all molecules, are different. There is no inherent separation point; dissiociation in water actually requires the transfer of a proton, i.e. the breaking of an $\ce{X-H}$ covalent bond. Two such bonds don’t get broken simultaneously, instead in a first reaction one proton is removed – but then you end up with an anion. Removing another $\ce{H+}$ has to go against electrostatic attraction and that is a harder job to do so it happens less and the $\mathrm pK_\mathrm a$ values differ.

In actual fact, the picture I have drawn for Arrhenius bases is simplistic. Ionic bonds also involve orbital interactions; no bond is purely ionic or purely covalent. Cations and anions obviously also electrostatically attract each other. As barium is twice positively charged, it does tend to bind the hydroxide fragments rather tightly. Overall, upon dissolution it is much more likely to end up with what effectively is a coordination compound of the $\ce{[Ba(OH)(H2O)_n]+ + OH-}$ type.


Good book for foundational chemistry I need a good book that explains the basics of chemistry in a lucid and importantly, systematic manner. Most books on this subject are not comprehensive and assume that the reader is well acquainted with certain topics. Also, most chemistry books have references to later chapters. And later chapters, references to earlier ones. This makes it very confusing.

I would like a book that assumes (effectively) no prior knowledge and builds up the basics needed to understand chemistry (structural theory, acid-base theory, electrode potentials) to a level adequate to understand undergraduate level organic and inorganic chemistry.

The reason I need this book is because frequently, while studying inorganic and to a certain extent organic chemistry, I am unable to understand the rationale behind a certain reaction. And most textbooks do not take the effort of explaining it. So, chemistry, for me,  just reduces to memorizing a set of reactions. I think solid basics should remedy that.

For the concepts underlying General Chemistry find that Huheey's book gives very lucid explanations.  As far as Organic Chemistry goes, vol. A of Carey-Sundberg's opus might be what you are asking for.  What level of text are we looking at?


Why are the boiling points of aldehydes less than those of ketones? Both aldehydes and ketones have a carbonyl group, but since the carbonyl group of ketone is between alkyl groups, wouldn't it be more difficult for it to form hydrogen bonds than aldehydes? Why then, do aldehydes have a higher boiling points than ketones?

It isn't that good a generalization: always look at the data first.

Here is a table of most of the aldehydes and ketones with 6 or fewer carbons (the labels are used in the chart later):

Now plot this on a chart:

Branches is the number of branches in the carbon chain.

Note that while for 3 and 4 carbons the ketones do have higher boiling points, it is not clear this is true for 5 carbon compounds and certainly not true for 6-carbon ones.

So, I'd argue the pattern of boiling points is complicated and there is no simple pattern that needs to be explained.


Why is water taken in calculating K, equilibrium constant For the given question,

I was trying to solve it but I was not considering the concentration of water in the calculation (because water is a pure liquid and its concentration is supposed to be taken as 1), but then I needed the volume to be able to calculate the answer. I saw the solution,

Is it taken here because the reactants are organic and thus water would probably not be the solvent?

The question they gave you is somewhat flawed, in that there are four compounds and you don't know the range of their mutual solubility.

To really answer the question, you would need information such as from Solubility, liquid–liquid equilibrium and critical states for the quaternary system acetic acid–ethanol–ethyl acetate–water at 293.15 K Fluid Phase Equilibria vol. 313, pages 46–51.

The reactants, ethanol and acetic acid, are completely miscible; however, ethyl acetate only has limited solubility in water.

When you evaluate the equilibrium constant, all the reacting compounds, including water, which are in a single phase, must be included.


Why does burning wood produce methane and other hydrocarbon chains? This post kinda helps me understand and it sounds like it has to do with pyrolysis, but I'm having trouble putting the exact reactions together.

Under the proper conditions, the hydrocarbons present in wood are completely oxidized to $\ce{CO2}$ and water upon combustion.  In fact, this happens every day in analytical chemistry laboratories for example: a sample of wood is dried, ground and placed in a combustion chamber of oxygen with a combustion catalyst at ~$\mathrm{950^oC}$. The evolved $\ce{CO2}$ and $\ce{H2O}$ (along with other gases and ash) from the combustion can then be analyzed to quantitatively determine how much carbon and hydrogen were originally in the wood.

Now lets consider a campfire.  Do you dry and grind your firewood, mix it with a catalyst and apply a heat source of $\mathrm{950^oC}$ under an $\ce{O2}$ atmosphere? Of course not, so why do we go through all these rigors to get a quantitative burn?  So that it all gets burned, no partial combustion allowed. And partial combustion is the key to most of the unburned hydrocarbons that you refer to. Parts of the wood are barely hot enough to burn at all, but still can thermally break down the hydrocarbons into shorter, volatile organics which are are carried away by wind or convection from the fire before the complete combustion to $\ce{CO2}$ and water can be completed.

You also mentioned pyrolysis in your question, and the answer given in your link is certainly relevant here, and it represents one of the processes that I'm talking about. For pyrolysis, you need an oxygen-poor environment.  Going back to the campfire scenario, this largely describes the situation down in the smouldering embers on a calm night.  It's very hot, there is plenty fuel, but no process bringing in sufficient oxygen for combustion. Under these conditions, thermal decomposition rather than just partial oxidation plays a major role in producing unburned hydrocarbons.  Think about methane for example; it is certainly not partially oxidized,


How do the molecules know the future? How do they proceed to form "Most Stable" compound? I'm tutoring few students for Chemistry. During the course, I many times use the phrase "Out of these possibilities this one is stable. So it is formed". I get questions like, "How do the molecules know?"

Honestly I don't know. How do the molecules know the future? How do they proceed to form "Most Stable" compound?

Cross Post: https://physics.stackexchange.com/questions/51894/how-do-the-molecules-of-reacting-compounds-proceed-to-form-most-stable-molecul

Chemical reactions occur because the thermal energy in a substance makes molecules bump into each other, and each time they bump into each other there is some chance for a reaction to take place.  This is an oversimplified model, but it suffices to get some intuition for what's going on.

The molecules of course do not "know" that some configuration is more stable than others.  What happens is that the molecules react randomly, again and again; they take a sort of random walk through the available configurations.  Every now and then, a molecule will purely randomly find its way into the stable state.  Once it's there, it's "stuck" - it has a much smaller chance of getting out of the stable state.  That's more or less what it means for a state to be stable.

Over time, more and more molecules will find their way into the stable state and get stuck, so the result looks as if the molecules "knew" which state to go to and just went straight there.  But actually they only got there through the process of randomly bouncing around the configuration space, plus the stable state acting as a trap that molecules can enter but not (as easily) leave.


Chemistry of solids: allowed wavevectors I'm a complete beginner in the chemistry of solids, and many of the things I came across don't make sense. I have two subquestions, which I think it would be most logical to post as a single question.

The longest wavevector of a commensurate matter wave is $k_{max} = \pi / d$, where d – lattice-plane spacing, so that amplitude minima and maxima of the wave  coincide with adjacent lines / planes of atoms. I don't understand this explanation – why is, for example, $k = 2\pi / d$ inconsistent?
Bloch's equations dictate that the value of the electron wavevector isn't uniquely defined, and the vector is uncertain to modulo G, where G – a general Reciprocal Lattice (RL) vector. This means that wavevectors greater than $\pi / d$ are permitted and can be found in Brillouin's zones of order greater than 1 in the RL. This seems to be in contradiction to the point above – what am I missing?

In a crystal, electron waves are reflected off the periodic lattice and this (Bragg) reflection is the cause of the energy gaps also called a band structure. If there is a weak periodic potential in a 1D crystal then the energy gaps occur at wavevector $k=\pm \pi/d$ where d is the lattice constant.

If $\bar G$ represents the reciprocal lattice vector then the general Bragg condition for diffraction of a wave with wavevector $\vec K$ can be written as
$(\vec k + \vec G)^2=k^2 $

In one dimension this condition becomes

$$k=\pm \frac{1}{2}G = \pm n\pi/d$$

and $G=2\pi n/d$ is a reciprocal lattice vector and $n=1,2,3..$  an integer. This is where your factor of 2 appears.

The reflection at $k=\pm \pi/d$ occurs because one wave is reflected of an atom and interferes constructively with phase difference $2\pi$ with one reflected from the nearest neighbour atom. The Brillouin zone is the $\bar k$ space between $-\pi/d$ and $+\pi/d$.

(The reciprocal lattice vectors have the form $\vec G =h\vec A+k\vec B + l\vec C$ for integers h, k, l. The vectors are defined as

$$\vec A= 2\pi \frac{b~ \times~ c}{a\cdot b~ \times~ c}~~
\vec B= 2\pi \frac{c~ \times~ a}{a\cdot b~ \times~ c}~~
\vec C= 2\pi \frac{a~ \times~ b}{a\cdot b~ \times~ c}$$
for primitive lattice vectors a, b, c)


Fundamentals of mathematics and physics for physical chemistry Before I start studying chemistry, I want to gain some knowledge about the fundamentals of mathematics and physics that are important to better understand the principles of physical chemistry, especially since I haven't taken any maths or physics courses yet. Which fields of maths and physics should I take a look at? Suggestions for literature are welcome.

There are three levels of scholarship. An analogy is that someone has looked at the ocean from far away and he feels that he has seen it all (standard undergrad mindset). Next stage, is when you walk on the beach and your feet become wet and dirty with sand. You have a feeling how ocean water and the beach feels like (fresh MS/PhD). The higher level of scholarship is when you are deep diving into the ocean and realize there is too much to learn and the knowledge you have is quite limited (a mature scientist who has spent his life in research).

Since you are saying that you have not taken mathematics and physics courses, it may not be the right time to begin a proper physical chemistry course.

Couple of nice books:

Mathematics in Chemistry An Introduction to Modern Methods by Hecht

Mathematics for Physical Chemistry by Robert G. Mortimer

Get hold of them from a library and get a feel for it (the first level of scholarship). You don't need to master all to open an undergraduate physical chemistry book. You rather learn it along the way. See if you like the subject in general.


Bond lengths in hydrocarbons It seems like the first molecule has a bond order of $1$, the second and third have a bond order of $7/6$ and the fourth molecule has a bond order of $3/2$. Why is the answer not $4 < 2 = 3 < 1$?

Well, 2 cannot be equal to three to start off, because a double bond it way shorter than a single bond.

The Answer to this is D, 2 has the shortest bond because it is a double bond, next 4 is a benzene ring, which is a resonance structure, which means that all the carbons technically have partial bonds between each other, or you can imagine the double bonds to be continuously rotating between the different carbons. Lastly 3 and 1 both have normal single bonds, so would have an equal bond length.

Therefore, since the double bond is shortest, the resonance between a single and double is a bit less short, but not as long as a single bond, and a single bond is the longest, the answer it 2(double bond)<4(between double and single bond)<3&1(which are both single bonds)


Does borax do anything more for "boosting" detergent than adding active oxygen would? Borax, $\ce{Na2B4O7}$, is often marketed as a "laundry booster" under the brand "20 Mule Team Borax".  The unit crystal of borax can be seen below.

Other laundry products in the past have added sodium perborate (which can be produced from borax, hydrogen peroxide, and sodium hydroxide) as a bleaching additive.  The "peroxide" portion can readily be seen below.

Through what mechanism does the plain borax "boost" the detergent?  Would it be more effective to throw some weak peroxide and a small amount of base into the laundry to form the perborate instead of using plain borax?

Borax was used as a "builder" in washing formulations long before perborate was used as a bleach. "Builder" is the terminology for additives which reduce the influence of water hardness. Other examples of builders include trisodium phosphate, soda, polyphosphates (like Graham's salt) or today most often aluminosilicate ion exchangers (Phosphates are more or less forbidden nowadays because of eutrophication of rivers and seas).


How to draw the structure and find the oxidation number of Phosphorous in H2P2O7 Find the Oxidation number of Phosphorus in $\ce{H2P2O7^2-}$.
So I thought of drawing the lewis structure to find the oxidation numbers. Because I wasn't sure if Oxygen was bonded to a Hydrogen or not. Then I realised that the two Hydrogens could be ionised, so I'd be left with $\ce{P2O7^4-}$.
But then when I tried drawing this Lewis structure I got really confused, As I tried to draw it with the two Phosphorus as central atoms and I couldn't make it. (I know I can just use general rules to find oxidation numbers but I feel like lewis structures often give more accurate answers). So could you please tell me how to draw the structure and more importantly guide me through the process. Any help is appreciated, thanks!

If you didn't understand what  Poutnik has sain in his comment, this is how it should be done:
You have realized your parent compound is $\ce{H4P2O7}$ (adding two $\ce{H+}$). Correct? If so, concentrate on following hypothetical reaction:
$$\ce{2H3PO4 -> H4P2O7 + H2O} \tag1$$
I believe you know what $\ce{H3PO4}$ is looks like. Now, draw two $\ce{H3PO4}$ side by side. Choose two closest $\ce{OH}$ groups (that would be $\ce{-P-\color{red}{OH} \space \space \color{tequish}{HO}-P -}$). Then, remove one $\ce{\color{red}{-OH}}$ and the $\ce{\color{tequish}{-H}}$ from the second $\ce{\color{tequish}{-OH}}$ you chose to eliminate a $\ce{H2O}$ molecule. Finally combine two remaining molecules together such that final structure forms $\ce{-P\color{red}{-}\color{tequish}{O}-P -}$ bond. This is your $\ce{H4P2O7}$. Eliminate any two hydrogens to get $\ce{H2P2O7^2-}$ structure.
Note: Since above hypothetical equation $(1)$ does not require oxidation or reduction to form $\ce{H4P2O7}$, it is safe to assume that oxidation state of $\ce{P}$ in both $\ce{H3PO4}$ and $\ce{H4P2O7}$ is same.


Why are there different chemical structures for epinephrine (adrenaline)? The Wikipedia structure is:

Note the empty bond at far-right.

"Journals on the Web" has:

CH3 added at far-right.

And rxlist.com shows:

There's an extra H going into the page! Is one of these incorrect, or are they all consistent but w/ some implied bonds or something?

There are often shortcuts taken in drawing complicated molecules, particularly if there's not any interesting chemistry going on at those functional groups.

The $\ce{-CH3}$ is implied at the end of the "stick," just like how the carbon atoms aren't explicitly labeled in the benzene ring.

Hydrogens are rarely explicitly drawn for the same reasons, given that you can infer the hybridization by inspection.

See Skeletal formula for more details.  Briefly:

  Carbon atoms are usually depicted as line ends or vertices with the assumption that all carbons have a valence of 4 and carbon-hydrogen bonds, usually not shown explicitly, are assumed to complete each C valence. A skeletal formula shows the skeletal structure or skeleton of a molecule, which is composed of the skeletal atoms that make up the molecule.


Energy cycle centered around methane production / consumption Many car manufacturers have developed projects for hydrogen cars but I've always been wondering why they put so much effort into hydrogen, as it is hard to store, and fuel-cells use to restitute energy back are based on rare-earth-metal not suitable to produce many cars for everyone.

Nature mainly store energy with carbon, I'm not a chemist but is it that hard or not yet enough economically viable for not seing rising projects around simple hydrocarbure synthesis. Easy to store, easy to consume, but maybe hard to produce, right ?

Production : $\ce{CO_2 + 2H_2O}$ (+ Renewable Energy Source) $\ce{-> CH_4 + 2O_2}$

Consumption : $\ce{CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O}$ (+ Energy To Use)

Could even $\ce{CO_2}$ be simply captured from air with limewater in such cycle ?

Sorry if my question may seems too simplistic, but I'd really like to handle (in simple terms) what are the economic/technical constraints to elaborate such apparently simple round-trip.

Vehicles that run on natural gas already exist, and have done for many years - you can find CNG (compressed natural gas) cars and larger vehicles around the world.

But there's very little synthesising of methane because there are very few places with a sensible carbon price, so it's just cheaper to extract natural gas and burn it.

Research is ongoing on methane synthesis - this is usually done under the name "power to gas", which can refer to hydrogen or methane. Either way, it usually starts with the splitting of water to get hydrogen, and then is sometimes combined with methanation of the hydrogen. The latter is an exothermic reaction, so combining the two processes can give pretty good efficiencies (see e.g. this paper).

Hydrogen / methane for cars is looking like a lost battle: electric cars have almost certainly already won; it's just going to take a few years until they completely dominate. However, there is still the possibility that larger vehicles (heavy goods vehicles, boats, aircraft) might settle on hydrogen or renewable methane as their decarbonisation option.

By the way, just because a fuel-cell might use rare-earth metals, doesn't make it unsuitable for mass market. Most rare-earth metals aren't rare, despite the name. Fuel cells have lots of issues, and fuel-cell vehicles really haven't achieved much yet, but that's to do with lack of reliability and longevity, rather than scarcity of components.


Electrolysis of Water What is the mechanism for the electrolysis of water? I understand that hydrogen and oxygen gas are made, but how exactly does this happen when electrons are passed through water? A diagram would be very useful.

Would this have anything to do with electronegativities?

Nobody has given an answer yet, so I'll give it a crack.
In electrolysis, as you probably know, there are two electrodes – the cathode (negative) and the anode (positive). We can deal with what happens at each separately, then tie them together.
At the cathode:
$$\ce{2H2O + 2e- -> H2 + 2OH-}$$
You're interested in the mechanism of these reactions. We know that water exists in solution in an equilibrium:
$$\ce{H2O + H2O <=> H3O+ + OH-}$$
At the cathode, electrons are supplied to the $\ce{H3O+}$ ions, reducing them:
$$\ce{H3O+ + e- -> H2O + H* }$$
With many reactions happening at once, the $\ce{H*}$ radicals react together, forming $\ce{H2}$. By removing the $\ce{H3O+}$ ions, it shifts the equilibrium of water to the right, to oppose the change – removing $\ce{H3O+}$ ions and increasing the concentration of $\ce{H2O}$:
$$\ce{H2O + H2O <=>> H3O+ + OH-}$$
Because the concentration of $\ce{OH-}$ ions doesn't change, the local solution around the cathode becomes basic – this can be done in a school lab. Add indicator. Electrolyse. Observe the color change near the cathode.
At the anode:
$$\ce{2H2O -> O2 + 4H+ + 4e-}$$
Here I suspect the mechanism is a bit similar:
$$\ce{H2O + H2O <=> H3O+ + OH-}$$
We take the $\ce{OH-}$ and oxidise it:
$$\ce{OH- -> OH* + e-}$$
This results in a radical – the hydroyxl radical – being formed. The steps from here, to get to water being formed are probably quite complicated as a mechanism.


Confusion in solution for this Question regarding mole concept $\pu{48g}$ of graphite and $\pu{16g}$ of $\ce{O2}$ is given. The question is to count the no of molecules of $\ce{CO2}$ produced?
Now, the solution in book is this
No of $\ce{C}$ atoms = $\frac{\pu{48g}}{\pu{12g}}$ = $\pu{4mol}$. No of $\ce{O}$ atoms = $\frac{16}{16}$ =$\pu{1 mol}$ , and no of $\ce{O2}$ atoms = $\pu{0.5 mol}$.
The reaction given is
$\ce{C + 2O ->CO2}$.
$\ce{CO2}$ produced is $\pu{0.5 mol}$ here.
My question is why not to use the reaction $\ce{C + O2 ->CO2}$?
What would the difference be in the answer ?

16 g of dioxygen contains 16 g of oxygen atoms. To find out how much carbon dioxide is produced, you can either use a realistic description of the reaction, or you can go atom by atom. Either path gives the same solution, but the one chosen by the textbook is not modern chemistry, and makes some of us angry or sad.

My question is why not to use the reaction $\ce{C + O2 -> CO2}$?

You should because that chemical equation describes the reactants and products much better than what the textbook has. It does not describe the mechanism (i.e. the steps to arrive at that net reaction), though.

What would the difference be in the answer.

Same numerical answer (unless you get confused because the reaction does not match the actual reactants).


Titration of bases against HCl Suppose I take three bases in a solution, $\ce{NaOH}$, $\ce{Na2CO3}$ and $\ce{NaHCO3}$, and I titrate this solution with $\ce{HCl}$. Since the strengths of the bases are different, does it mean that $\ce{HCl}$ would first react with the strongest base completely and then start reacting with the next strongest base?

This is perhaps not the best example, since hydroxide ion will react with bicarbonate ion to form carbonate ion, and also as you titrate carbonate ion note that it will form bicarbonate again. But, generally, yes, the strongest base will be consumed first as you titrate. The titration curve will tell you the amounts of the bases present and the pKas of their conjugate acids.


Calculating the density matrix from a WFN file I want to calculate the density for the $\ce{HCN}$ molecule. This molecule has 14 electrons and thus 7 occupied molecular orbitals. I have a .wfn file which includes 8 parts, where each of them has 14 rows and 5 columns (due to 5 basis functions). The first part belongs to the basis functions (exponents) and the 7 other parts contain coefficients for each of seven MO's. How can I calculate the electron density using this file? Generally I have used the electron density formula based on the Roothaan expansion introduced in Szabo's textbook, namely:

$$
P_{\mu\nu} =2\sum_{a=1}^{N/2} C_{\mu a}C_{\nu a}
$$

where $a$ runs over the number of electrons, though I have seen in several codes that $a$ is considered as the number of occupied MOs. $\mu$ and $\nu$ stand for the number of basis functions (5 here).

Should I use all 7 parts of the coefficients to obtain the electron density of the $\ce{HCN}$ ground state, or is the first part which has the minimum energy sufficient?

where a runs over the number of electrons, though I have seen in several codes that a is considered as the number of occupied MOs.

If it is closed shell (all spatial orbitals are either doubly occupied or empty), then running up to $N/2$ is the same as running over all occupied orbitals. Note that your formula is for closed shell systems described by a single determinant.

There are two representations of the first order reduced density matrix, or simply density matrix: the discrete one and the continuous one.
What you have in the formula is the discrete one, represented in the space of primitive functions. The only reason why it is discrete is because the original Schrödinger equation has been projected on a finite dimensional space. You need to locate the Molecular Orbital Primitive Coefficients ($C_{\mu a}$) in the file. You could as well calculate the overlap matrix ($S$) of the primitives and then get the density matrix in the basis of molecular orbitals
$$ D =  C^t S P S C.$$
You will see that it is a diagonal matrix with 2's.

You could also represent in real space the general density matrix,
which is continuous and spans $\mathbb{R}^3$. You could discretize the space and compute the values there. Then, the general formula
$$ \rho(\pmb{r},\pmb{r}') = \sum_{a,b} D_{ab}\phi_a(\pmb r) \phi_b^*(\pmb r'),$$
where $\{\phi_a\}_{a=1}^M$ represent the spatial canonical molecular orbitals, and
becomes
$$ \rho(\pmb{r},\pmb{r}') = 2\sum_{a}^{N/2} \phi_a(\pmb r) \phi_a^*(\pmb r')$$
for a closed-shell single determinant wavefunction.
Setting $\pmb r' = \pmb r$ you get the density.

If you plan to do it you will have to deal with some nuances, like the normalization used by the program that generated the .wfn file.


Calculate mass concentration of Ca(OH)2 solution Can anyone help me calculate the mass concentration of $\ce{Ca(OH)2}$ solution with molar mass $\pu{74.1 g/mol}$ at the $\pu{pH of 12.8}$.

My work so far:

From the pH I get the molar concentration of $\pu{0.063 mol/L}$.   Assuming we have $\pu{1L}$ of substance we have $\ce{n(OH-) = 2 \times n(CaOH2) = 0.126 mol/L}$

I put the results in $m = M \times n$ to get 9.35 g which is incorrect. Where did I go wrong?

There is a general law about the solubility of salts containing doubly or triply charged ions, which is not often presented in textbooks. The law is the following : The measured solubility of such a salt or such a hydroxide is ALWAYS greater than the calculated values derived from solubility products, or from pH measurements. This is due to the fact that the solubility product and the pH data do not take into account the existence of intermediate ions.

In the case of $\ce{Ca(OH)2}$, there are of course ions like $\ce{Ca^2+}$ and $\ce{OH-}$ in its solution. But there are also ions like $\ce{Ca(OH)+}$, and non dissociated molecules $\ce{Ca(OH)2}$. Alas these ions and molecules are rarely mentioned in usual textbooks. But this explains why the solubility of $\ce{Ca(OH)2}$ is bigger than the values obtained from pH measurements. I know that the average reader is reluctant to believe me. I may be more persuasive if I use another example.

Let's speak about the solubility of $\ce{H3PO4}$. Any student in chemistry knows that the solubility is not related to the solubility product $\ce{[H+]^3[PO4^3-]}$, because of the existence of intermediated ions, like $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$. These ions are accepted in the large chemistry family. Everybody accepts that the solubility of $\ce{H3PO4}$ is much bigger that the value derived from the solubility products, or from pH measurements.

That is the same for $\ce{Ca(OH)2}$ and all substances containing a doubly charged ion like $\ce{CaSO4}$, etc. And here the activity is not sufficient to explain this discrepancy. I know that some chemists think that the activity may be introduced to explain this discrepancy. But in the case of $\ce{CaSO4}$ for example, the concentration is so low that the activity factor may be considered as equal to 1.

For $\ce{CaSO4}$, the effect is more pronounced, because of the presence of two doubly charged ions. The solubility calculated from the solubility product is 0.0051 M. The measured solubility, obtained from direct titration is 0.015 M. Three times more !


Why can't unimolecular reactions be 2nd order? Why can't unimolecular reactions be second order? For example, doubling the concentration of a single reactant would result in four times the original rate of reaction.

A first order reaction involves the behavior of individual particles, usually their decomposition into smaller, more reactive species.  The rate of such a reaction relies solely on the number of particles present, so if we double the number of particles then the rate doubles simply because there are twice as many particles available to decompose.

A second order reaction involves the interaction of two particles with one another. (Even if both particles are the same it is still a bimolecular reaction.)  If we double the concentration of either species we double the likelihood of a collision between those species.  If both species are the same (ex: $\ce{2A -> A2}$) then the result of doubling the concentration is a quadrupling of the rate because we simultaneously double the number of particles that can interact and the number of interactions that can occur.

An metaphorical example involving people:

Unimolecular:
A couple can spontaneously break up into two separate people because of its internal properties.  (Assume there are no intercouple issues involved here.)

$\ce{C -> M + W}$

If we double the number of couples we would expect breakups to occur twice as often because there are twice as many couples to be broken up.  However, since there are no intercouple issues in this example, adding additional couples to our consideration does not increase the likelihood of existing couples breaking up.

Bimolecular with Two Different Species:
A man and a woman can join together to make a couple if they collide in the right orientation with enough energy. :-)

$\ce{M + W -> C}$

If I increase the number of either men or women, it is more likely that I will form a couple simply because there are more interactions between men and women occurring.  Each species' increase is first order though, since they require a partner of the other gender to couple with.  i.e. Doubling the number of women only doubles the rate at which couples are formed because they still have to connect with a man, and the number of men has not changed.

Bimolecular with a Single Species:
Homosexual men can join together to form a couple.
$\ce{M + M -> C'}$

If I double the number of men, then each additional man can interact with every other man in the room, greatly increasing the likelihood of a couple forming.  This means that not only are there more men available to make couples (the first doubling of the rate), but that the existing men are also more likely to be involved in a connection that results in a new couple being formed (the second doubling of the rate.)


Why ammonium hydroxide precipitates metal salts? When I was going through the chapter on pnicotgens, I got to know that ammonia acts as a Lewis base and it dissociates in water giving out hydroxide ion and the further hydroxide of ammonia, which we refer to as ammonium hydroxide, precipitates metal salts in a form of metal hydroxides.

I want to know the reason why this thing happens. According to me, ammonium hydroxide should form a complex with the metal being a base after donating a pair of electrons. Correct me if I am wrong. A reaction is given below as an example of a reaction between ammonium hydroxide and zinc sulfate:

$$\ce{ZnSO4 + 2 (NH3 * H2O) → Zn(OH)2 + (NH4)2SO4}$$

It's simply because of the concentration. In aqueous solution, the formation of a precipitate is favored over the formation of a complex ion, and it can be shown using solubility equilibria and $K_\mathrm{f}$ (but it won't be pretty).
This is just like the QA test for copper(II). After a small amount of aqueous ammonia is added, $\ce{Cu(OH)2}$ is observed. Add more, and you get the deep blue $\ce{[Cu(NH3)4]^2+}$ complex.


Strength of intramolecular vs intermolecular hydrogen bonds Why are intramolecular hydrogen bonds weaker than intermolecular hydrogen bonds?

It is mainly due to kinetic motion in molecules

The kinetic energy of molecules follows the Boltzmann distribution. Therefore, at room temperature molecules are constantly moving.

Intermolecular interaction, such as intermolecular hydrogen bonding only take place for a short time when two molecules are in each other's vicinity. Shortly after, the molecules separate and the interaction does no longer exist.

Intramolecular interactions on the other hand are much stronger since the two atoms forming the hydrogen bond remain in each others proximity due to structure of the molecule. In other words, they exhibit strain.


Do the dimensions of the cathode and anode matter in electrochemistry? I'm curious whether the dimensions of the cathode and anode matter in an electrochemical process. Let's take copper electrorefining as an example. For instance, should the cathode and anode be of the same length and be in alignment with each other or does the process not care?

From my perspective, the cathode doesn't 'see' the anode, it only sees the $\text{Cu}^{2+}$ ions and therefore the anodes dimensions and placement relative to the cathode don't have an effect. Note, let's ignore the increase in resistance as the cathode and anode are placed further away and other similar effects. The essence of what I'm trying to understand is how their placement/alignment effects the occurrence of the reaction.

The dimensions of your electrodes should not effect the open circuit voltage, since this is determined solely by the identity and concentrations of the redox species.

Once the cell is running, the dimensions may affect how much the voltage drops off as you draw current, known as "overpotential". If your cell is limited by its ability to deliver ions to the surface of the electrode, or to transfer charge into the electrode, then the higher surface area of a large electrode is likely to result in a more efficient battery.


Acid Catalyzed Hydration of an Alkene Synthesis Problem Devise syntheses of the following three compounds, starting in each case with any alkene that contains four carbons or fewer. You do not have to write mechanisms, although at this point it may be very helpful for you to do so.

$\ce{(CH3)3COH}$

$\ce{(CH3)2CHOH}$

$\ce{CH3CH2CHOHCH3}$

This is the question I'm trying to tackle. I'm having a bit of trouble with the third part of it, which is the synthesis of butan-2-ol. The text solutions says that I should use but-2-ene, which I can understand but I don't quite understand why my answer is also not plausible. Can someone help me out with this?

You propose to use but-1-ene rather than but-2-ene as a starting material to synthesise butan-2-ol via addition of water to the double bond. There is no reason why it would not happen the way you drew.

However, using but-2-ene allows only one possible reaction pathway. With the relatively high symmetry of the molecule ($C_\mathrm{2h}$, assuming trans), attacking either side of the double bond will lead to an identical product: The molecule’s intrinsic $C_2$ axis will transform one into the other. Thus, it is only possible to generate butan-2-ol.

Starting from but-1-ene, you have two non-equivalent first steps, generating either the primary or the secondary carbocation. These obviously cannot be transformed into each other by symmetry (the alkene’s symmetry is now $C_\mathrm{s}$). Thus, you will get a side product: butan-1-ol. These products will be difficult to separate. If anything, careful distillation would do the job.

As RobChem already noted in the comments, (organic) chemists typically attempt to avoid side reactions and different products, hence why the book suggests using the more symmetric alkene.


Ionic factor of glucose? I am trying to find the $K_\mathrm{b}$ for 50 g of glucose dissolved in 1 kg of ethanol, given the change in boiling point is $\pu{2.2 °C}$. The molality is therefore 0.278. But what is the ionic factor $(i)$? I was told it is the number of ions, so I tried 24, but the answer did not match up with the accepted value.

A more appropriate name for the ionic factor is the van't Hoff factor of colligative properties. In the simplest terms, it is the number of particles created after being dissolved in the solution. Since glucose is not an ionic compound, it does not dissociate, therefore $i=1$ (meaning one glucose molecule stays one glucose molecule).


Can electrons be present outside orbitals. If yes how does this affect chemical reactions In my physical chemistry textbook it is written (Or typed) that orbitals are regions where probability of finding electron is high (90%-100%). But since orbitals are regions of probability is it possible for electrons to be present outside orbitals for just a small number of atoms in a reaction if so how does this affect reactions that take place with these small number of atoms

There are more levels of abstraction involved. At first, orbitals are simplification and they are best described as "one electron wavefunction", ie. area of space where one electron can reside (or two, if you think of spatial orbitals, as opposed to spin-orbitals).

Being expressed as a wavefunction, the orbital (and therefore electron residing in it) has nonzero probability in whole space (I mean, whole universe). But the probability decays very fast.

If you plot a probability density, see image below, you can clearly see the behaviour. If, on the other hand, want to plot the 3d surface, which are often shown in textbooks, you have to state some cutoff value. Often it is probability density of 0.9, ie. 90%. By this you obtain the typical pictures of orbitals.

I would deliberately leave your question without answer, as they stem from severe confusion caused by your textbook.


Synthesis of nitrocellulose I am trying to understand the synthesis of nitrocellulose.
A mixture of sulfuric acid and nitric acid is used to nitrate cellulose. The sulfuric acid acts as a catalyst, producing $\ce{NO2^{+}}$ ions from the nitric acid which bind at certain sites on the cellulose chain to nitrate it. So nitrate salts can be used instead of nitric acid because all you really need are the nitrate ions. The sulfuric acid acts by protonating nitric acid to yield nitronium and hydronium ions. If so, wouldn't any sufficiently strong acid, such as hydrochloric acid, do just as well.

As long as the Ka for that acid is higher than nitric acid's, it would work the same right?
Also, wouldn't very small amounts of nitronium ion be present without the sulfuric acid anyway?
Could one prepare nitrocellulose using only nitric acid if it were soaked for a much longer time?

The sulfuric acid acts by protonating nitric acid to yield nitronium and hydronium ions.

Electrophilic nitration reactions like the production of nitrocellulose require sulfuric acid because of the dehydrative properties of sulfuric acid. Sulfuric acid is a non-nucleophilic acid - all it can do is mediate proton transfers and dehydrate things (sulfuric acid has a large negative enthalpy change of mixing with water). Thus,$\ce{NO2+}$ is free to react with whatever you want it to.

$$1)\ \ \  \ce{H2SO4 + HNO3 <=> HSO4- + H2NO3+} $$
$$2)\ \ \ \ce{H2NO3+ <=> H2O + NO2+} $$
$$\text{overall)    }\ \ \ce{HNO3 + H2SO4 <=> H2O + NO2+ +HSO4-}$$

Hydrochloric acid and nitric acid produce aqua regia. Hydrochloric acid is a nucleophilic acid, and thus the decomposition products after proton transfer to nitric acid are different. The chloride ion is a good nucleophile, and reacts with the nitronium ion, eventually producing $\ce{NO2}(g)$ and $\ce{Cl2}(g)$. There is a big difference between $\ce{NO2+}$ and $\ce{NO2}$.

$$1) \ \ \ \ce{HCl + HNO3 <=> Cl- + H2NO3+}$$
$$2) \ \ \ \ce{H2NO3+ <=> H2O + NO2+} $$
$$3) \ \ \ \ce{2NO2+ + 2Cl- <=> 2NO2 + Cl2}$$
$$\text{overall)    }\ \ \ce{2HCl + 2HNO3 <=> 2H2O + 2NO2 + Cl2}$$

  So nitrate salts can be used instead of nitric acid because all you really need are the nitrate ions.

Yes. However, starting with nitric acid shifts the equilibrium towards nitronium a little.

  Also, wouldn't very small amounts of nitronium ion be present without the sulfuric acid anyway? Could one prepare nitrocellulose using only nitric acid if it were soaked for a much longer time?

Maybe, but not enough to be useful. Nitric acid is not available pure. It is only stable to a maximum concentration of 70-75% by weight (remainder water). The other reason that sulfuric acid must be used is that it is available "pure" (or at least 98% by weight) and it is so hydrophilic so as to sequester the water. Hydrochloric acid has a max concentration of 35-40% by weight remainder water. All the extra water gets in the way and shifts the equilibrium in the opposite direction (the reaction produces water).


Chemical reaction of radicals with different multiplicity I am studying the chemical reaction of radicals by computational chemistry software.
I just want to know what is the rules for multiplicity.
I have two radicals for which, the ground state of one is a doublet and the other is a triplet.
A chemical reaction between them is possible or not, if a chemical reaction is possible then in what multiplicity the reaction will proceed.
what multiplicity should I use in my input file?

With respect to the question "if a chemical reaction is possible or not", here is some background material on oxygen species.
In particular, Singlet oxygen apparently reacts with oxygen (also referred to as dioxygen) creating a very strong (but transient) oxidizing atomic oxygen.
Examples of possible reactions of interest, including atomic oxygen, can be found, per Equations (1) to (5), from a work discussing graphene patterning here: http://olab.physics.sjtu.edu.cn/papers/2017/29.Huan%20Yue_PCCP_2017.pdf , where apparently, O(3P) is created from severe collision quenching of O(1D) atom with air or oxygen and acts as the major oxidant in the work by Huan Yue and colleagues ‘Exploring the working mechanism of graphene patterning by magnetic-assisted UV ozonation’.
Note, O(3P) is also known as highly reactive ground-state 3P oxygen and a form of atomic oxygen.
Also be mindful, in the real world if any water vapor is present, Singlet oxygen reacts with H2O forming H2O2, where O(1D) has a half-life of around 45 minutes.


Trend in reducing property of dioxides I am not sure of the explanation for the following statement:

  The reducing property of dioxides of group 16 elements decreases from $\ce{SO2}$ to $\ce{TeO2}$.

Is the statement valid? If yes, how do I explain it theoretically?
If not, what would be the actual trend in that case?

The statement is actually valid.If you go from S to Te, the compounds of Se and Te are stable at lower oxidation states due to inert pair effect.
So, the tendency of these group 16 elements to be in higher oxidation states decreases down the group.That means, tendency of the elements getting oxidised decreases down the group and thus their reducing property(reducing others, getting oxidised itself) decreases frm S to Te.


Purpose of an Iodine Flask Is there any specific reason why Iodine Flasks are used instead of Erlenmeyer Flasks when doing Iodine Determination tests?

Iodine Flasks are quite expensive when compared to Erlenmeyer Flasks.

Does the funnel-shaped top form serve a specific function?

Iodine flasks differ from normal conical (Erlenmeyer) flasks in three main ways:

The ground glass joint.
The modified stopper.
The funnel shaped top.

Iodine itself, and iodine related molecules produced during the analysis are, to a certain extent, volatile (you can see this if you just put a bead of iodine in a sealed vial and watch the vial fill with vapour). The ground glass joint of the iodine flask prevents these vapours from escaping, improving the accuracy of your analysis.

As for the funnel, its simply there to help get your stuff into the flask in the first place, preventing loss through spillage and preventing the need for a separate glass funnel. Anything that ends up on the tapered neck is simply rinsed directly into the flask.


2-Methylpropan-2-ol and Iodoform test Can 2-methylpropan-2-ol be converted into methyl ketone the action of $\ce{I2}$ in the presence of alkali base and thus give iodoform test?

Answer given to this question is that it can't give iodoform test. Please could someone confirm this with an explaination.

No, 2-methylpropan-2-ol cannot give iodoform test, as it cannot be converted into a methyl ketone.

Generally, alpha hydroxy methyl type (this is not the actual nomenclature) groups can give iodoform on oxidation.


How to determine the proton concentration in the following solution in mol/L? A $\pu{0.8L}$ solution contains $\pu{0.60 g }\ce{ NaOH}$ in water. Then the above question follows.

It's been a long time that I made some chemistry exercises. Now I'm stuck with a couple of hesitations.

I know one needs to calculate the number of protons of the molecule sodium hydroxide. But do you need to take water into account or not? I'm thinking not but then the question asks for the solution. So I probably should do that which gets me to $\pu{29 g mol-1}$. Otherwise it is $\pu{20 g mol-1}$.

So that I would divide by $\pu{0.60g}$ which gives me for the first one $\pu{48 mol}$ and the second one $\pu{33 mol}$. Then I'd divide it by $\pu{8.0L}$.

I probably should have a larger answer than what I've done so it is not correct but I don't know exactly how I should work this out.

I'd appreciate a hint.

You need more than a hint. You need a hard shove...

The overall notion here is that water dissociates

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

and that the equilibrium is given by

$$K_w = 1.00\times10^{-14} = \ce{[H+][OH-]}$$

where $\ce{[H+]}$ is the molar concentration of $\ce{H+}$ ions and $\ce{[OH-]}$ is the molar concentration of $\ce{OH-}$ ions.

NOTE: The pH of a solution is defined as the negative log of the $\ce{H+}$ ion concentration, and pOH of a solution is defined as the negative log of the $\ce{OH-}$ ion concentration. So

$$14 = \text{pH + pOH}$$

Sodium hydroxide is a strong base, so the assumption is that it completely dissociates.

$$\ce{NaOH(solid) <=> Na+(aq) + OH-(aq)}$$

The molecular mass of sodium hydroxide is 40.00 grams/mole.

$$\ce{[OH-]} = \dfrac{0.60\text{ g}}{40.00\text{ g/mole}\times0.80\text{ liter}} = 0.01875\text{ mole/liter}$$

Rearranging the $K_w$ equation

$$\ce{[H+]} = \dfrac{K_w}{\ce{[OH-]}} = \dfrac{1.00\times10^{-14}}{1.875\times10^{-2}} = 5.3\times10^{-13}\text{ mole/liter}$$


Chemical structure of xenon heptafluoride anion When $\ce{XeF6}$ acts as a fluoride acceptor, $\ce{XeF7-}$ is formed. What is the hybridisation, geometry, and shape of $\ce{XeF7-}$?

About the geometry and shape:

Source: Housecroft, Sharpe, Inorganic Chemistry, Pearson Education Limited, 2005


Does Chemical Formula of Matter Influence on how would it be Charged? Firstly: Suppose we rub glass ($\ce{SiO2}$) against paper (which is made from cellulose $\ce{(C6H10O5)_{$n$}}$). We receive positively charged glass and negatively charged paper. Is that because of total ionization energy of molecule stronger than glass or crystal system too?

Secondly: When we rub atoms against other atoms, do electrons accelerate each other by the same electrostatic charge and change their orbitals to higher orbits due to synchrotron radiation they emit. Or is it because of the fact that protons are more close are ionizing electrons by electrostatic attraction? Am I correct?

Firstly: Suppose we rub glass ($\ce{SiO2}$) against paper (which is made from cellulose $\ce{(C6H10O5)_{$n$}}$). We receive positively charged glass and negatively charged paper. Is that because of total ionization energy of molecule stronger than glass or crystal system too?

Although atoms and molecules like to be uncharged there is a difference between the electron affinity and the ionization energy. Electron affinity is a measure of willing a substance is to accept an extra electron. Ionization energy is a measure of how willing a substance is to donate electrons.

When the paper is rubbed against the glass the surfaces of the two substances come into physical contact. So the electron affinity of the paper is greater than the ionization energy of the glass rod, so electrons are stripped from the glass and are transferred to the paper. Enough electrons are transferred to balance the affinity/ionization relationship. This is known as the Triboelectric effect.

To be totally honest the Triboelectric effect is not well understood. There are a lot of enigmatic interactions in how the surfaces contact, how many times the surfaces contact when rubbing, how specific contact points my heat up, or how some physical transfer may take place. Read the Wikipedia article for more details on these points.

    Secondly: When we rub atoms against other atoms, do electrons accelerate each other by the same electrostatic charge and change their orbitals to higher orbits due to synchrotron radiation they emit. Or is it because of the fact that protons are more close are ionizing electrons by electrostatic attraction? Am I correct?

There are a number of erroneous concepts in the setup/question. The first explains the phenomena.

Synchrotron radiation is electromagnetic radiation and has absolutely nothing to do with the electrostatic experiment that you are discussing.

The electrostatic experiment that you are discussing does not "accelerate electrons." So rubbing a glass rod with paper doesn't make a particle accelerator.

Neither the paper nor the glass has bare protons floating around.


Reaction between hydrogen peroxide and ozone What is the reaction mechanism of the reaction:

$$\ce{H2O2 + O3 -> H2O + 2O2}$$

I have tried searching this on Google but am not getting good answers.

This is basically an excerpt of the answer I have given to another quite similar question.

Ben is quite right, neither the reaction nor the mechanism is simple. It involves several different radical chains all resulting in the decomposition of ozone.[a]

It is most likely that the reaction will start with the formation of oxygen radicals in the gas phase ($\ce{O}$ atom transfer).$$\ce{O3 + O3 -> O3 + O2 + O}$$

In presence of Hydrogen donors like $\ce{H2O2}$, which "is expected to be an excellent $\ce{H}$ donor."[a] Those are forming the main chain carriers will be $\ce{HO, HO2}$.
 Finally stopping at some point by recombination of these species.
\begin{aligned}\ce{
O + H2O2 &-> OH + HO2\\
  HO + O3 &-> HO2 + O2\\
  HO2 + O3 &-> HO + 2O2\\
  2HO2 &->}\text{stable products}
\end{aligned}

In aqueous solution the decomposition is highly pH dependent, but may always work because of autoprotolysis. $$\ce{2H2O <=> H3+O + {}^{-}OH}$$

Especially important might also be protolysis of hydrogen peroxide. $$\ce{H2O2 + H2O <=> HO2- + H3+O}$$

Steahelin and Hoigné propose several hypotheses in which way ozone will be decomposed, but I will not go into absurd detail here.[b] The core assumptions involve $\ce{{}^{-}OH}$ as the initiator of a radical chain. Followed by radical (ionic) species like $\ce{.O2-, .O3-, HO2., .OH}$ resulting again in a chain reaction.
\begin{aligned}\ce{
  O3 + {}^{-}OH &-> .O2- + HO2.\\
  O3 + {}^{-}OH &-> HO2- + O2\\
  O3 + {}^{-}OH &-> .O3- + .OH\\
  HO2. &<=> H+ + .O2-\\
  O3 + HO2- &-> .OH + .O2- + O2\\
  O3 + .O2- &-> .O3- + O2\\
  .O3- + H2O &-> .OH + {}^{-}OH + O2
}\end{aligned}


Lewis Acid / Base reaction for SO2 and H2O I am having trouble with this question:

Identify the Lewis acid and Lewis base in each of the following
reactions:

$\ce{SO2(g) + H2O(l) = H2SO3(aq)}$

What I tried:

A Lewis acid is an electron pair acceptor.

A Lewis base is an electron pair donor.

The other problems of this type involved cations or anions, so I was able to identify the cation as the Lewis acid and the anion as the Lewis base.

However, this question does not involve ions.  So I tried to look at the lone pairs on the molecules.  Since they both have lone pairs, I figured since $\ce{SO2}$ has more lone pairs, it would act as the Lewis base.  Also in the final molecule, the hydrogen atoms would be bonded to two separate oxygen atoms, so I thought of them as migrating to the lone pairs on the $\ce{SO2}$ oxygen atoms.

However, the textbook says that $\ce{SO2}$ is the Lewis acid and $\ce{H2O}$ is the Lewis base.

Why would this be the case?

Molecules that contain polar multiple bonds can function as Lewis acid because the central atom is electron deficient with a vacant orbital that can accept an electron pair.

When $\ce{SO2}$ dissolves in water , it forms the weak diprotic acid $\ce{H2SO3}$ (sulfurous acid) from Lewis acid/Lewis base reaction:

$$\ce{O\bond{=}S\bond{=}O + H2O <=> H2SO3}$$

The $\ce{O}$ atom of $\ce{H2O}$ molecule donates a lone pair to the $\ce{S}$ of  $\ce{SO2}$,forming a new $\ce{S\bond{-}O}$ $\sigma$-bond and breaking $\ce{S\bond{=}O}$ $\pi$-bond.


Quantum mechanical model of atom While going through the quantum mechanical model of atom section in my chemistry textbook, the text states:

  The existence of quantized electronic energy levels is a direct result of the wave-like properties of electrons and {the energy levels} are allowed solutions of the Schrodinger wave equation.

What does 'direct result of wave-like properties of electrons' mean?

To choose a more basic example, let us consider the particle in a box with

\begin{equation}
V(x)={\begin{cases}0 & 0\lt x \lt L\\\infty & \text{otherwise}\end{cases}}
\end{equation}

To solve this we use the Schrödinger Equation

\begin{equation}
-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)=E\psi(x)
\end{equation}

subject to the boundary conditions
\begin{equation}
\psi(0)=\psi(L)=0.
\end{equation}

The origin of these boundary conditions is, that the wave function needs to be a) continuous and and b) zero outside of the potential well, since probability goes to zero for an infinite potential.

As an ansatz for the wave function we take the solution of the free particle, i.e. planes waves:
\begin{equation}
\psi(x) = \exp\left(-ikx\right)
\end{equation}
with
\begin{equation}
k^2=\frac{2mE}{\hbar^2}.
\end{equation}
What we now have is a periodic wave function, with period $L$ (or an integer multiple of that) due to the boundary condition, hence
\begin{equation}
kL = 2\pi n.
\end{equation}
At this point we have introduce the integer quantum number $n>0$. With this we can now write down wave function and the corresponding energy. Both depend on $n$ and are therefore quantized.

So the origin of the quantized nature here are the boundary conditions. The quantum number naturally show up in the solution of the Schrödinger Equation. Obviously this can happen for other system, such as the Hydrogen atom as well.

As a final side note: The Schrödinger equation does not always lead to quantized solutions, it depends on the system. There are for example continuous states when considering the dissociation of molecules.


Effective nuclear charge and electron repulsion To the best of my knowledge, the effective nuclear charge $Z_\mathrm{eff}$ is usually defined such that the potential energy of an electron in an atom, $\langle V \rangle$, can be expressed as

$$\langle V \rangle =
  -Z_\mathrm{eff}\langle\psi|\frac{1}{|r - r_a|}|\psi\rangle$$

The Hartree-Fock expression for $\langle V \rangle$, meanwhile, is

$$\langle V \rangle =
  - Z\langle\psi | \frac{1}{|r - r_a|} | \psi\rangle
  + \sum_i \langle\psi |\hat{J_i} |\psi\rangle
  - \sum_{i \in \sigma} \langle\psi | \hat{K}_i | \psi\rangle$$

Is there any relation between these expressions? It was taught in college Chemistry that $Z_\mathrm{eff}$ accounts for electron shielding, which is simply electron repulsion effects. Is the effective nuclear charge just a semiempirical value fitted to approximate the total potential energy of an electron in an orbital, or is there another meaning behind the number?

Based on what I know, Clementi et al’s values for $Z_\mathrm{eff}$ are not evaluated based on this principle. Could I inquire why this is not the case?

Is the effective nuclear charge just a semiempirical value fitted to approximate the total potential energy of an electron in an orbital, or is there another meaning behind the number?

No, that's pretty much it. The Wikipedia entry has a partial periodic table with some $Z_\mathrm{eff}$ values. If you go further down to larger atoms, it becomes obvious that $Z_\mathrm{eff}$ decreases quite a lot towards the outer shells, so you could argue that the concept has its merits in capturing a quite significant effect.


What is the chemical equation for the digestion/decomposition of cellulose? I'm trying to find the chemical equation/pathway from cellulose $\ce{(C6H10O5)_n}$ to methane $\ce{CH4}$. What are the other reactants or products in the reaction?

$$\ce{C6H10O5 + ? -> CH4 + ?}$$

Everything I've found talks about bacteria and enzymes (cellulase, mostly) in terms of the process of enteric fermentation, but I've yet to find a single reaction or series of reactions for this conversion. At least one product would have to have oxygen in it, but that really hasn't narrowed down my search.

I'm mainly interested in how methane is created from the diets of herbivores, so it's very possible that I've been researching the wrong reactants from the beginning.

Biologically, methane is produced by methanogens, which are Archeabacteria. They do not utilize carbohydrates such as cellulose or glucose directly. Instead, they typically consume acetate or a mixture of hydrogen gas and carbon dioxide. These stoichiometry of these two processes is

$\ce{4H2 + CO2 -> CH4 + 2H2O}$ or

$\ce{CH3C(O)OH -> CH4 + CO2}$.

To get overall conversion of cellulose to methane, the cellulose is first fermented by the enteric bacteria you read about, which can produce acetate, hydrogen and carbon dioxide by a variety of pathways. These products are then utilized by the methanogens. One example of an enteric fermentation would be

$\ce{C6H10O5 + 3H2O -> 2CH3C(O)OH + 4H2 + 2CO2}$.

Combining this with the methanogenesis reactions, we can write the net equation of

$\ce{C6H10O5 + H2O -> 3CH4 + 3CO2}$.


Electronic configuration of Ace-Ala-Nme Below is the structure of Ace-Ala-Nme. I have labelled just four of the carbon atoms (1-4 in blue). I need to identify (i) the hybridization state, and (ii) the number of electron withdrawing groups for each.

I believe the amine (-NH) groups are electron donating while the carbonyls (-CO) are electron withdrawing. I therefore suspect that all four labelled carbon atoms are sp3 with 0 withdrawing groups. But this is assuming that the donating/withdrawing tendency of the amine and carbonyls cancel each other out. Can anyone confirm this to be the case?

1: sp3 with 1 electron withdrawing group
2: sp3 with 2 electron withdrawing groups
3: sp3 with 0 electron withdrawing groups
4: sp3 with 1 electron withdrawing group

I would say your earlier analysis was correct.  If you look at the resonance structures that can be drawn for an amide linkage, you find there is considerable positive charge both on carbon and nitrogen.  This should serve to make them both electron withdrawing, relative to an attached carbon, in your structure.

EDIT: To clarify this answer with regards to some of the comments posted above, I'd like to add that typically when an amino nitrogen is situated on a double bond or aromatic ring it can donate electrons through resonance.  In fact, that is what it is doing in the resonance structure on the right - it is donating its lone pair of electrons via resonance.  However, the amino group is also electron withdrawing in an inductive sense relative to carbon.  Further the $\ce{-NH3^+}$ group can no longer interact via resonance and is simply a strong electron withdrawing group.  The resonance structure on the right is analogous to the $\ce{-NH3^+}$ situation and since it is a significant contributor to the overall description of the amide (cf. there is significant double bond character in the amide $\ce{C-N}$ bond) it acts inductively as a strong electron withdrawing group for the attached $\ce{sp^3}$ carbon.


What is the chemistry behind smell of human perspiration(sweat)? What is the chemistry behind  the bad smell of human perspiration? also what is its composition?

Why does sweat smell?!

Sweat released from sweat glands odourless. Bad smell of sweat is due to certain bacteria on your skin mix with sweat(That is why sweat smell bad after sometime) and also due to certain hormones released during puberty.

I have tried my best to keep this part chemistry point of view(Actually this belongs to biology).

  Chemical composition of sweat

The main reason people perspire is so evaporation of water can cool the body. Thus water is main component.

The chemical composition of perspiration varies between individuals and also depends on what they have been eating and drinking, why they are sweating, how long they have been perspiring and several other factors.

On average, the mineral composition is:

sodium (0.9 gram/liter)
potassium (0.2 g/l)
calcium (0.015 g/l)
magnesium (0.0013 g/l)

Trace metals that the body excretes in sweat include:

zinc (0.4 milligrams/liter)
copper (0.3–0.8 mg/l)
iron (1 mg/l)
chromium (0.1 mg/l)
nickel (0.05 mg/l)
lead (0.05 mg/l)

Source: ncbi and chemistry.about