text:
Enthalpy change for a van der waal gas

One mole of Van der Waal gas is compressed from $\pu {20 dm^3}$ to $\pu{10 dm^3}$ at $\pu{300 K}$. Given $a = \pu{3.60 dm^6 mol^-2}$ and $b = \pu{0.44 dm^3 mol^-1}$. Find $\Delta H$.

$\Delta H = -\mu C_p \Delta P + C_p \Delta T$
Since for VDW gas $ \displaystyle \mu C_p = {2a\over RT}- b$.
The process took place at $\pu {300 K}$ which means $\Delta T = 0$.
$\therefore  \Delta H = -\mu C_p \Delta P + C_p \Delta T = \left(-{2a \over RT} + b\right)\Delta P = \pu{0.4399 kJ/mol}$.
But the answer is $\pu {-30.5 J/mol}$
I think my mistake is taking $\Delta T = 0 $ but if that were the case what should I take as $\Delta T$, I don't even know the process.
Is the author's answer correct ?

@ChesterMiller I wonder what is wrong with what I am doing ? that will be more helpful.

You can start by telling us where you got our equation for $\Delta H$ from.

The units for a appear to be wrong for a vdw gas, should they not be $\pu{dm^6\cdot bar \cdot mol^{-2}}$ assuming molar volumes.?

@porphyrin Yes, I noticed that too.  Also, that value of a seems very low to me (if the correct units involve Pa).

@Chester Miller, yes, but if we assume numerical value is ok then 3.6 is approx a for $\ce{CO2}$ but then b is almost ten times too big for this gas. In fact it seems to be too large for most any gas in $\pu{dm^3 mol^{-1}}$ .

@ChesterMiller I used $dH = {dH\over dP}_T dP + {dH\over dT}_P dT$.

But you have to integrate, because there is a large volume change and pressure change.

If these constants are supposed to be for CO2, then the correct values are: $ a = 3.6\ bar\ dm^6\ mol^{-2}$ and $b=0.044\ dm^3\ mol^{-1}$

Using these values for the constants, I get a value of about -355 J/mol for $\Delta H$

@ChesterMiller Yes I doubt the credibility of the answer. I also tried your equation and got wrong answer.

and using the values in the question the enthalpy comes to $-301.3 \pu{J/mol}$  so it looks like a double typo in powers of $10$.

The equation for $\Delta H$ for a van der Waals gas should be:
$$\Delta H=C_p^{IG}\Delta T-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+b\left(\frac{RT_2}{V_2-b}-\frac{RT_1}{V_1-b}\right)$$where $C_p^{IG}$ is the heat capacity of the gas in the Ideal Gas (IG) region (low pressures and high specific volumes).  For constant temperature ($T_1=T_2=T$), this equation reduces to:$$\Delta H=-2a\left(\frac{1}{V_2}-\frac{1}{V_1}\right)+bRT\left(\frac{1}{V_2-b}-\frac{1}{V_1-b}\right)$$
Explanation:
In the link https://physics.stackexchange.com/questions/321968/internal-work-done-by-a-gas/322170?noredirect=1#comment725750_322170, I showed that, for a van der Waals gas, the internal energy U is given by:  $$U(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}$$
where $U(T_{ref},\infty)$ refers to the internal energy in the ideal gas reference state $T=T_{ref}$ and $V\rightarrow \infty$ (i.e., a reference state in which the gas exhibits ideal gas behavior).
The enthalpy H of the gas is equal to U + PV:
$$H(T,V)=\left[U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-\frac{a}{V}\right]+\left[-\frac{a}{V^2}+\frac{RT}{V-b}\right]V$$
Combining the terms on the right hand side gives:
$$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTV}{V-b}$$
But, $$\frac{V}{V-b}=\left[1+\frac{b}{V-b}\right]$$
So, substituting into the equation for H gives:
$$H(T,V)=U(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}+R(T-T_{ref})+RT_{ref}$$
But, $$U(T_{ref},\infty)+RT_{ref}=H(T_{ref},V)$$and$$\int_{T_{ref}}^{T}{C_v^{IG}(T')dT'}+R(T-T_{ref})=\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}$$where $C_p^{IG}(T)$ is the ideal gas heat capacity at constant pressure.  So, finally, we obtain:$$H(T,V)=H(T_{ref},\infty)+\int_{T_{ref}}^{T}{C_p^{IG}(T')dT'}-2\frac{a}{V}+\frac{RTb}{V-b}$$

How did you get this equation ?

@A---B See my explanation added to the Answer.

Thank you. It might take me some time to digest all this.

In addition to the method detailed by @Chester Miller, I offer an alternative derivation. It is often the case in thermodynamics that there are two or more ways of solving the same problem.
To calculate the change in enthalpy at constant temperature we can start with
$$dH=TdS+VdP$$
and then find the change with volume and integrate in V at constant T.
Thus
$$\left(\frac{\partial H}{\partial V}\right )_T=T\left(\frac{\partial S}{\partial V}\right)_T+ V\left(\frac{\partial P}{\partial V}\right)_T$$
using Maxwell’s equations
$$\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$
gives
$$\left(\frac{\partial H}{\partial V}\right )_T=T\left(\frac{\partial P}{\partial T}\right)_V+ V\left(\frac{\partial P}{\partial V}\right)_T$$
which nicely give  derivatives in P and V . Integrating means first working out the derivatives since
$$\int dH=\int \left[ T\left(\frac{\partial P}{\partial T}\right)_V+ V\left(\frac{\partial P}{\partial V}\right)_T\right ] dV $$
For the van der Waals gas
$$P=\frac{RT}{V-b}- \frac{a}{V^2}$$
the derivatives are
$$T\left(\frac{\partial P}{\partial T}\right)_V= \frac{RT}{V-b} $$
 and
$$V\left(\frac{\partial P}{\partial V}\right)_T = -\frac{VRT}{(V-b)^2}+\frac{2a}{V^3} $$
substituting and working out the indefinite integral gives
$$H=-\frac{2a}{V}+\frac{bRT}{V-b} + const$$
Now the constant is the standard state value, chosen, say, to be at infinite volume, but  we  do not need to know what the standard state is as we only need the difference in H.
Now take the difference to get the change in enthalpy
$$\Delta H = -2a \left (\frac{1}{V_2} -\frac{1}{V_1} \right) +RTb\left ( \frac{1}{V_2-b}  -\frac{1}{V_1-b}  \right) $$
( As an aside, as the gas expands the work done can also be calculated using $w=-\int pdV$. This is
$$w = -\int_{V_1}^{V_2} \left (\frac{RT}{V-b}-\frac{a}{V^2}\right )dv = -RT\ln \left(\frac{V_2-b}{V_1-b} \right )-a\left(\frac{1}{V_2}-\frac{1}{V_1} \right)$$
As the work done on an perfect gas is $w_{per}=-RT\ln(V_2/V_2)$ we can see that is the volume is large compared to b the work done by the van der Waals gas is less than that for  the perfect gas as a result of intermolecular interactions.)

Thanks for the answer. I prefer this derivation little more over that one.


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Find whether a gas cools or heats up on Joule-Thomson expansion

A gas that follows $P(V - nb)= nRT$ is subjected to Joule-Thomson expansion. Tell whether it cools or heats up.

$$\mu = {\partial T \over \partial P} = {\partial P(V - nb)/nR \over \partial P} = \frac1{nR}\left(V - nb + {\partial V \over \partial P}\right) = \frac1{nR}\left(V - nb - {nRT\over P^2}\right)$$
Now how do I determine whether $\mu >0 $ or $\mu < 0$ without knowing anything  about temperature or anything else ?

The equation you gave for $\mu$ is incorrect.  That partial derivative is supposed to be at constant enthalpy H.  Do you know the mathematical relationship between dH, dT, and dP?

Yes if you mean $dH = -\mu C_p dP + C_p dT$.

I set $dH = 0$ then I get $\mu dP = dT$.

Not that part.  The next part.  Do you really think that that expression you wrote  is equal to the partial derivative of T with respect to P at constant H?

There are some details in this answer http://chemistry.stackexchange.com/questions/71543/enthalpy-in-adiabatic-process/71567#71567 . You should find that the coefficient for your gas is $-B/C_p$.

The equation for dH is:  $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$

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text:
Why does crystalline solids have well defined 3d geometry
Crystalline solids have well defined geometry because the intermolecular forces are uniform throughout the solid unlike amorphous solid where it varies from place to place.
But still the question hold why does some solids have this property like why $\ce{NaCl}$ forms a crystalline solid but not glass or something else.
I thought it is because crystalline solids are ionic but diamond is an counterexample among many.
Question :
Why does some solids have crystalline structure while others don't ?

Keep two concepts separate: the solid crystalline state is not bound to occur on ionic compounds only.
In a simple view, crystalline describes a regular arrangement of matter, with predictable order both in short, as well in long range.  For the crystalline state, it is unimportant if the constituents of the crystal are atoms (like metal atoms), charged particles (like your example of NaCl), or neutral molecules (like benzoic acid) or even a mixture of neutral and charged molecules (many protein crystals contain water).
Some materials are easier packed in such a regular pattern, than others.  In the instance of metals, their (on first approximation) sphere like atomic geometry and similar radius eases such a settlement.  Inorganic salts tend to be a bit more complicated, as often smaller ions are localized in gaps built by larger ions (octahedral / tetrahedral gaps).
Organic molecules may equally build crystals, yet their shape and conformational flexibility around single bonds (for example in alkyl chains) and multiple intermolecular interactions may keep them over a long time in a glassy state (like supercooled benzoic acid, or some brands of honey).  More than for the aforementioned examples: formation of a nice crystal may be thermodynamically favourable, but it is not enough "just to cool down" below the melting temperature to reach this energetic minimum.  There must be seeds of crystallization, and typical conditions of their formation (temperature, concentration, etc.) are not necessarily the same where nicely shaped crystals growth further.
It is not surprising crystals attract attention, for example in material science and crystallography, and that there are journals with focus on the relationship between molecular properties, and macroscopic / crystallographic ones; like CrystEngComm or Crystal Growth & Design.

"some brands of honey", is honey a solid ?

@A---B While "honey" is a mixture of several compounds, it is save to assume it largely consists of sugars, with fructose and glucose as the most important contribution among them.  These, if pure, are solids.  The ease honey crystallizes out correlates with the ratio of fructose : glucose, because yes, if not solid, honey is a supercooled liquid (cf. https://en.wikipedia.org/wiki/Honey#Phase_transitions, for example).  It is one of the examples, outside the chem lab, that illustrate how thermodynamics is about energies, and kinetics about the rates reactions and phase transitions occur.

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Do catalysts increase the rate of exothermic reaction more in a reversible reaction?
I'm just learning chemistry for school and I've learnt that adding a catalyst reduces the activation energy for the forward and backward reaction by equal amounts, therefore the rate of forward and backward reaction increase by the same amounts so equilibrium stays the same but is reached more quickly.
I'm having some trouble understanding this. I know that the activation energy of the exothermic reaction is always smaller than that of the endothermic reaction. So thinking about the Maxwell Boltzmann distribution, a change in $E_a$ near to $0$ will have a greater effect on the rate of exothermic reaction than the endothermic reaction. Therefore, a catalyst should cause equilibrium to shift to the exothermic side?

Look at the form of the Arrhenius rate equation (this is fine since they mostly all have this form):
$$k = Ae^{-\frac{E_{a}}{RT}}$$
If you make the substitution, $E_{a}->E_{a}+\Delta E$, then you have:
$$k = Ae^{-\frac{E_{a} + \Delta E}{RT}} = Ae^{-\frac{E_{a}}{RT}}e^{-\frac{\Delta E}{RT}}$$
This means that you're just scaling the rate constant $k$ by another constant. This scaling applies to both the forward and reverse rate constants. If the concentrations were such that they were in equilibrium before, they're still in equilibrium now.
You'd have to prove this in general for complex reactions with complicated rate laws, but for simple reactions, this should suffice to help you see that the equilibrium is unaffected.

So the reaction rates don't increase by the same amounts, the increase by the same proportion?

Because rate dependence on energy is exponential, it is by proportion. In the end, the laws of chemistry work pretty well, so you can try to rationalize backwards from rate equivalence at equilibrium.

Oh, I see what i'm missing. There have to be 2 distributions, one for the reactants and one for the products. This will allow for both areas to become the same

wait but then it should favour the side of the reaction with the greater concentration of reactants?

Again, chemistry works, so no.

ok i guess i can take that. You see I have a very mathematical approach in scientific problems. I attempt to be very picky about every detail and different ways of coming to the conclusion. It helps so much in understanding physics but not so much in chemistry

Notice that what you're missing from the Maxwell-Boltzmann distribution is how this relates directly to the activation energy? Is it just a function of $v$? If so, how does decreasing the barrier translate to changing entire equation? You'd have to perform a total substitution. After that you need to integrate from some cutoff $v=v'$ (how does this depend on catalyst?) to $v=\infty$.


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Do catalysts shift equilibrium constant towards 1?
I want to be able to understand shifts in equilibrium from the maxwell boltzmann distribution. One thing I cannot get my head around is the effect of catalysts on the equilibrium position - supposedly it's none at all.
But consider these diagrams of the M.B distribution of both reactants and products of a reversible reaction.

The top diagram shows the distribution of the products of the reaction which equilibrium favours. It has a bigger concentration of particles. The line dividing the shaded region on these diagrams represents activation energy and the $dE$ represents the shift in activation energy due to the catalyst. We can see that the shaded regions, $A_1=A_2$ because forward and backward rates are the same at eqm. But clearly, the rate of the top reaction increases more as a greater fraction of particles have above activation energy. So adding a catalyst should shift equilibrium to the middle?
Basically, catalysts work better on reactants with higher concentration. So shouldn't the side with the lower concentration be favoured?

The energy difference between the reactants and products is unchanged by catalysis.

how is this related to the question?

re 'how is this related', everything! This is because the free energy is unchanged, since the reactants and products are the same, and as $\Delta G^0=-RT\ln(K_e)$  so $K_e$ is unchanged.

Oh ok I can understand that there's probably a link between free energy and changing equilibrium. I don't yet have much understanding of this area of statistical physics and chemistry (at all). Although my question was to specifically about approaching the problem using those graphs (albeit I am now aware that it's a terrible way to approach it in the first place).

It's not terrible, just a little bit flawed, and I wouldn't blame you for thinking that way. It is often taught that way in school, too, and my school was no exception: $A$ is the collision frequency and $\exp(-E_\mathrm{a}/kT)$ is the proportion of effective collisions, which can be represented by the M-B distribution. Qualitatively, the M-B diagram works, and personally I had always assumed that the maths would work out, until I tried it last night. So, good question, and it inspired (hopefully) a decent answer from me, although I can't say I'm anywhere near an expert on reaction dynamics.

Also, by the way, note also that the shaded area does not represent a concentration, in the conventional sense of the term (number of moles / volume). I get that perhaps you are referring to an "effective concentration" of particles with enough energy to react, though.

TL;DR Your Maxwell–Boltzmann diagram up there is not sufficient to describe the variation of rate with $E_\mathrm{a}$. Simply evaluating the  shaded area alone does not reproduce the exponential part of the rate constant correctly, and therefore the shaded area should not be taken as a quantitative measure of the rate (only a qualitative one).

There is a subtle issue with the way you've presented your drawing. However, we'll come to that slightly later. First, let's establish that the "proportion of molecules with sufficient energy to react" is given by
$$P(\varepsilon) = \exp \left(-\frac{\varepsilon}{kT}\right) \tag{1}$$
Therefore, for a reaction $\ce{X <=> Y}$ with uncatalysed forward activation energy $E_\mathrm{f}$ and uncatalysed backward activation energy $E_\mathrm{b}$, the rates are given by
$$k_\mathrm{f,uncat} = A_\mathrm{f} \exp \left(-\frac{E_\mathrm{f}}{kT}\right) \tag{2} $$
$$k_\mathrm{b,uncat} = A_\mathrm{b} \exp \left(-\frac{E_\mathrm{b}}{kT}\right) \tag{3} $$
The equilibrium constant of this reaction is given by
$$K_\mathrm{uncat} = \frac{k_\mathrm{f,uncat}}{k_\mathrm{b,uncat}} = \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \tag{4}$$
As you have noted, the change in activation energy due to the catalyst is the same. I would be a bit careful with using "$\mathrm{d}E$" as the notation for this, since $\mathrm{d}$ implies an infinitesimal change, and if the change is infinitesimal, your catalyst isn't much of a catalyst. So, I'm going to use $\Delta E$. We then have
$$k_\mathrm{f,cat} = A_\mathrm{f} \exp \left(-\frac{E_\mathrm{f} - \Delta E}{kT}\right) \tag{5} $$
$$k_\mathrm{b,cat} = A_\mathrm{b} \exp \left(-\frac{E_\mathrm{b} - \Delta E}{kT}\right) \tag{6} $$
and the new equilibrium constant is
$$\begin{align}
K_\mathrm{cat} = \frac{k_\mathrm{f,cat}}{k_\mathrm{b,cat}} &= \frac{A_\mathrm{f}\exp[-(E_\mathrm{f} - \Delta E)/kT]}{A_\mathrm{b}\exp[-(E_\mathrm{b} - \Delta E)/kT]}  \tag{7} \\[0.2cm]
&= \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \frac{\exp(\Delta E/kT)}{\exp(\Delta E/kT)} \tag{8} \\[0.2cm]
&= \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \tag{9}
\end{align}$$
Equations $(9)$ and $(4)$ are the same, so there is no change in the equilibrium constant.

The question then arises as to how eq. $(1)$ is obtained. The simplest way is to invoke a Boltzmann distribution, which almost by definition gives the desired form. However, since you have a Maxwell–Boltzmann curve, I guess I should talk about it a bit more. The fraction of molecules with energy $E_\mathrm{a}$ or greater is simply the shaded area under the curve, i.e. one can obtain it by integrating the curve over the desired range.
$$P(\varepsilon) = \int_{E_\mathrm{a}}^\infty f(\varepsilon)\,\mathrm{d}\varepsilon \tag{10}$$
where the Maxwell–Boltzmann distribution of energies is given by (see Wikipedia)
$$f(\varepsilon) = \frac{2}{\sqrt{\pi}}\left(\frac{1}{kT}\right)^{3/2} \sqrt{\varepsilon} \exp\left(-\frac{\varepsilon}{kT}\right) \tag{11}$$
At first glance, we would expect this to be directly proportional to the exponential part of the rate constant, i.e. $\exp(-E_\mathrm{a}/kT)$. Alas, it is not that simple. If you try to work out the integral
$$\int_{E_\mathrm{a}}^{\infty} \frac{2}{\sqrt{\pi}}\left(\frac{1}{kT}\right)^{3/2} \sqrt{\varepsilon} \exp\left(-\frac{\varepsilon}{kT}\right) \,\mathrm{d}\varepsilon \tag{12}$$
you don't get anything close to the form of $\exp(-E_\mathrm{a}/kT)$. Instead, you get some "error function" rubbish, and some nasty square roots and exponentials. (You can use WolframAlpha to verify this.)
Why is this so? Well, it turns out that there are other terms that also depend on $\varepsilon$ and therefore need to go inside that integral (they aren't constants and can't be taken out).
The simplest example is that faster molecules tend to collide more often, so even though the right-hand tail of the diagram seems to contribute very little to the "proportion of molecules with sufficient energy", it actually contributes more significantly to the overall rate because these molecules collide more often. In collision theory this is described using the "relative velocity" of the particles $v_\mathrm{rel}$.
There is also another complication, in that the Maxwell–Boltzmann distribution, the direction of the particles is not accounted for. (For more insight please refer to Levine Physical Chemistry 6th ed., p 467.) Therefore, there has to be yet another term that takes into account the direction of movement of the particles. The idea is that a head-on collision between two molecules is more likely to overcome the activation barrier than is a $90^\circ$ collision. The term that compensates for this is the "collision cross-section" $\sigma$.
If you go through the maths (and I don't really intend to type it out here, it's rather long, but I will give some references) then you will find that at the end you will recover the form $\exp(-\varepsilon/kT)$. Once you have arrived at this, it's very straightforward to see that the increases in rate of both the forward and backward reaction cancel each other out.
Now, as for the promised references, Pilling and Seakins's Reaction Kinetics pp 61-2 have a short outline of the proof. Atkins's Physical Chemistry 10th ed. has a slightly longer proof on pp 883-4.

In a catalysed reaction the activation barrier is not lowered but the catalyst causes the reaction to proceed by a different route which occurs more rapidly.
Thus there is a third party a 'chaperone' if you like that controls the way the two reactive species interact. This can be a surface, or a metal-organic or an enzyme, the details do not matter, but what matters is that the catalyst causes the reactants to undergo different reactions to that which would occur in homogeneous solution to produce the same product.
As the final outcome is the same so is the equilibrium. The kinetic approach is outlined by @orthocresol based on the Arrhenius equation. From a thermodynamic view, which is the most general, at equilibrium $\Delta G^{\mathrm {o}} = -RT\ln(K_e)$ and as $\Delta G^{\mathrm {o}}$ is unchanged in the reaction (the reactants and products are the same)  so is the equilibrium constant $K_e$.
Now to your approach using the Maxwell-Boltzmann (MB) distribution. It is not at all clear that Maxwell-Boltzmann distribution applies to a liquid. It is based on ideas around the perfect or ideal gas, and point like particles colliding but have no other interaction. In a liquid molecules do interact with one another since a liquid has a  meniscus and is distinct from its vapour.
However, there still has to be a distribution of speeds and hence energy. On collision the energy is converted into increasing vibrational/rotational energy in the colliding partners until by random chance enough energy is reached to cause reaction. As collisions can occur at at least $10^{12} \pu{s^{-1}}$ only very occasionally, perhaps only one in a million collisions, is enough energy present to cause reaction unless the activation energy is very low. In these cases diffusion limits the bimolecular reaction rate constant to $\lt 10^{11} \pu{dm^3 mol^{-1}s^{-1}}$ and often far less but depending on solvent viscosity. Thus in these cases it is the rate of approach that limits reaction not energy, so clearly MB energy distribution does not apply.
In solution it is normal to describe the rate constant of a (non-diffusion controlled)  reaction using transition state theory. This involves calculating the properties of the transitions state and reactants using statistical thermodynamics. The weakness of this approach is that the transition state properties have to be 'guestimated' in some way.
Appendix
[ Using the MB distribution an Arrhenius type expression can be obtained for a bimolecular gas phase reaction. The expression for the rate constant at energy E is
$$k(E) \approx \int_0^{\infty} Q(E)\sqrt{E} ~f(E)dE$$
where $f(E)$ is the MB distribution in terms of energy (by substituting $E=mv^2/2$), and $Q(E)$ is the reaction cross-section. Under conditions of thermal equilibrium the rate constant is the product of reaction cross section $Q(E)$, the magnitude of the relative velocity $\sqrt{E}$ and the thermal distribution of relative speed $f(E)$. The simplest form of $Q(E)$ is to suppose that it is a hard sphere of diameter d (like a billiard ball) in which case it has the form
$$Q(E)= \pi d^2(1-E_a/E)$$
when $E \ge E_a$ and zero otherwise. Performing the integration and converting to temperature produces and Arrhenious type equation
$$ k(T) \approx \exp(-E_a/k_BT)$$
where $k_B$ is the Boltzmann constant.]


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Experiment using hydroxides of alkaline earth metals
In three different test tubes we add a very small amount of:
1)In the first test tube, we add oxide of magnesium
2)In the second test tube, we add hydroxide of calcium
3)In the third test tube, we add hydroxide of barium
Then we add deionized water in the three test tubes. We shake them and after we filtrate them, we measure the pH of each filtrate.
I'm curious why we use hydroxides of calcium and barium but we use oxide of magnesium instead of hydroxide of magnesium?

I would attribute this two 2 reasons:
Reason 1: Insolubility of Magnesium hydroxide.
Magnesium hydroxide is almost insoluble in water and will not easily dissolve in water.
Reason 2: Safety is one of the reason of the choice of compounds.
The formation of hydroxides of alkaline earth metals are generally exothermic and this intensity increases down this group.

Reaction of magnesium oxide with water will produce the
corresponding hydroxide:
$$\ce{MgO + H2O → Mg(OH)2  (very slight reaction)}$$
(The reaction quickly stops because the magnesium hydroxide formed is almost insoluble in water and forms a barrier on the magnesium preventing further reaction).
The reaction of calcium oxide to produce its hydroxide is
exothermic:

$$\ce{CaO (s) + H2O (l) ⇌ Ca(OH)2 (aq) (ΔH_r = −63.7 kJ/mol of CaO)}$$

Barium oxide is  harmful to human skin and may cause irritation on exposure for prolonged periods or in large quantity.

Its reaction with water is also know to be violent.

So in my opinion it is much better to user hydroxides of Ba and Ca.

Probably a matter of availability/cost.  Magnesium oxide is easily obtained by calcining the hydroxide or carbonate and can be kept in its oxide state, whereas with calcium and heavier alkaline earth metals the anhydrous oxide is much harder to obtain; the hydroxides or carbonatesrequire higher temperatures to calcine in the first place.  Thus for the heavier alkaline-earth metals the hydroxide is more readily available instead of the pure oxide.
But do not fuss too much over this.  Both oxide and hydroxide should react the same way.  Magnesia should give a slightly alkaline solution, lime will give a more alkaline solution and the barium compound will act as a full-fledged strong base.

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Why does soda explode if artificial sweetener or other powders are added?
I opened a can of soda and spilled some generic artificial sweetener into the soda and it exploded. Why did this happen? I am not talking about mentos, I mean artificial sweeteners like Splenda.

Possible duplicate of How does the Mentos-Coke explosion work?

This answer is applicable to your question also.

I would suggest tbat the added fine powder provided a lot of nucleation sites for the dissolved CO2 in the soda very quickly hence the foaming.

I don't think a more general question should be considered dulicate.

Artificial sweetener powder causes nucleation of the carbonation in the soda/pop/soda-pop. The carbon dioxide concentration in the soda is higher than the equilibrium concentration for the partial pressure of $\ce{CO2}$ in the atmosphere thus over time the soda will off-gas $\ce{CO2}$ and if allowed to sit long enough will become "flat".
The off-gassing is spontaneous but not instantaneous, thus it is subject to kinetic rates. To form a bubble, enough $\ce{CO2}$ molecules need to coalesce at one time to form a large enough bubble to be stable and not redissolved in the soda (nucleating).
When a rough surface is introduced, the number of $\ce{CO2}$ molecules that must come together to form a stable bubble is far less that the pure drink, so the off-gassing occurs much faster causing the observed "explosion".
You can observe this when certain glasses (ussually beer glasses) etch a small ring in the glass, creating a rough surface and the formation of bubbles at the ring.  you can also observe this if you put a larger rough object in the soda. This could be a rock, a piece of wood (like a tooth pick), or even a Mentos.
To solve this you might be able to dissolve it in a small amount of water first. This may not completely solve issue though, as most sweeteners use silicon oxide or titanium oxide powder to prevent clumping, neither of which are soluble and will still nucleate bubbles.

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text:
PCC with substituted pyridines
I was wondering if substituted pyridines can work the way pyridine does in order to oxidize alcohols to aldehydes.
Of course, I'm referring to pyridines without any functional group which can undergo oxidation reactions (eg: I was thinking methylpyridine, or trimethylpyridine, with the nitrogen untouched.)
I guess that the reactivity is the same, but is there any reference to the use of this kind of reagent?

I think the main role of pyridine is to abstract the chloride ion from the intermediate formed leading to an elimination reaction. So yeah, the substituted ones should also work. However, its efficiency might be reduced as the +I effect of the substituents will decrease the net positive charge on nitrogen in pyridine.

Collidinium chlorochromate and gamma-picolinium chlorochromate are known oxidising agents. e.g. Synth Comm. 31 2001 1253
I see no reason why other methyl pyridines should not also work.

There may be steric issues with bulky primary alcohols.

Triggered by the two chemical names mentioned, I found  it interesting to browse before and after this page google books provides: https://books.google.de/books?id=z7KZFQvfM3gC&pg=PA12&lpg=PA12&dq=picolinium+chlorochromate&source=bl&ots=T8NSAgErlb&sig=zP3B5i1hWC75odrHJXx8_-AcC2o&hl=de&sa=X&ved=0ahUKEwjrks6htKfTAhVCCSwKHc8QBewQ6AEISzAF#v=onepage&q&f=false


text:
Why is the plateau on a heating curve in equilibrium?

From the interval of 100 to 140 minutes, why is the substance in equilibrium?
I believe it's because at this interval, freezing and melting begin to occur at the same rate and, when two opposing reactions occur at the same rate, equilibrium is present. However, since the potential energy is decreasing (since the substance is freezing), the equilibrium favors the exothermic reaction causing the substance to eventually become solid in the interval of 140 to 160 minutes.

Can you make it clear what the graph is ? I mean what is the substance you are measuring temperature of ?

@Raghav The substance was not given. Is that necessary? The graph is a heating curve.

You are right about the equilibrium existing because both freezing and melting occur at the same rate.
Imagine ice and water kept in a thermos flask under normal atmospheric pressure. You will observe that there is no change in the mass of both the phases.
But there is actually a lot of activity going on. Some molecules of ice enter the liquid phase and some molecules of water enter the ice phase. As both these processes occur at the same rate you don't see any change in the mass of both the phases.
In fact such equilibriums exist for all pairs of solids in their own liquids and gases above their own liquids at particular temperatures.
In your case I think there is a mechanism to extract heat from the system which shifts the equilibrium towards liquid changing to solid ie freezing.

The flat regions are where the solid is changing into liquid and liquid into gas at the higher temperature. These regions persist whenever there are still two phases in equilibrium. The heat necessary to cause the melting used to be called the Latent Heat of Fusion but nowadays is called the enthalpy of fusion (or melting) and similarly for vaporisation.
The solid is held together by a cohesive or lattice energy, generally called intermolecular energy (potential energy with attractive and repulsive parts) which has to be overcome for the solid to melt, and the heat absorbed by the solid is used to do this which means that the temperature does not rise until all the solid has melted. A liquid also has intermolecular forces but weaker than in the solid and thus has increased entropy. This increase in entropy has the effect of lowering the Gibbs free energy.
( The change of enthalpy on fusion can be calculated from the definition of enthalpy $H=U+PV$. Between the two phases, liquid to solid, the change is
$\Delta H = U_l-U_s +P(V_l-V_s)$
but as the change in volume between liquid and solid is small we can ignore the PV work term and so
$\Delta H = U_l-U_s $.
Using the heat capacity allow this to be evaluated since $U=U^{\mathrm{o}} +\int_0^T C_p(T)dT$
with the result  that
$ \Delta H = U_l^{\mathrm{o}}-U_s^{\mathrm{o}} +\int_0^T C_p(T)_l-C_p(T)_s dT$.
In vaporisation the PV term ignored in the solid-liquid change has to be added and becomes $PV_g$ for the gas and which for an ideal gas is RT  )


text:
How do stoichiometric coefficients effect the half-life of a reaction?
$$\ce{2A -> 3B + C + 2D}$$
This is a first order reaction. Rate constant $k$ for reaction is given to be (say) $\pu{1.386E-2 min^{-1}}$. From this we get the value of $t_{\frac{1}{2}}$ of reaction as $\frac{\ln(2)}{k} = \pu{50 min}$. Now this $t_{\frac{1}{2}}$ should mean half life time of reaction (and not for $\ce{A}$). However, in my textbook it is written that if concentration of $\ce{A}$ is $\pu{200 M}$ at $t=0$ then at $t = \pu{100 min}$ the conc. of $A$ would be $\pu{50 M}$ as $100$ minutes is two times of the half life (and thus the conc. of $A$ should reduce by $2^2$ times).
I think this is wrong as the half life of $A$ is half of the half-life of reaction (since rate of consumption of $A$ is double the rate of reaction as can be seen from the stoichiometric coefficients). According to me after $100$ minutes the concentration of $A$ should be $\pu{25 M}$. Is my logic correct? Or am I missing something?

see also the answer to a similar question https://chemistry.stackexchange.com/questions/74328/is-half-life-for-first-order-is-depend-on-the-stochiometric-coefficient-of-react/74338#74338

Consider what happens to your argument if I rewrite the equation in the following equivalent form:
$$\ce{A -> \frac{3}{2}B +\frac{1}{2}C +D}$$
That is, you write

... since rate of consumption of A is double the rate of reaction as can be seen from the stoichiometric coefficients

In contrast, the rate of the reaction, in this case, is defined to be the rate of decay of $\ce{A}$,
$$-\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = k[\ce{A}]$$
When you look at it like this, it is more clear that the stoichiometry should not have an effect on the decay process as you claimed.
Furthermore, if we integrate the equation above after separating by variables,
$$\int_{[\ce{A}_0]}^{[\ce{A_t}]}\frac{1}{[\ce{A}]}\mathrm{d}[\ce{A}] = \int_0^t-k\mathrm{d}t$$
we get,
$$\ln\left(\frac{[\ce{A_t}]}{[\ce{A}_0]}\right)=-kt.$$
If we try to find the time at which the original concentration has halved, we get the formula you have written above which is,
$$t_{\frac12}=\frac{\ln(2)}{k}.$$
The important point here, and this isn't true for other reaction-orders, is that the half-life is independent of initial concentration. This is not what one would intuitively expect, and indeed it's not true in general, but it is characteristic of all exponential-decay processes.
Indeed, this is a very deep result which doesn't tend to get emphasized enough in my opinion.
It is true that sometimes you can say that a reaction will be second order based on stoichiometric coefficients, but you are told this is first order, so that is probably the source of your confusion.
So, your textbook is right.

You cannot change the stoichiometric coefficient of the reaction. "but you are told this is first order"....order has nothing to do with stoichiometric coefficient.. The rate constant $k$ had been given for the first equation i.e. $2A \rightarrow 3B+C+2D$ and not for $A \rightarrow (3/2)B+(1/2)C+D$.

Stoichiometric coefficients only tell you the proportion of  reactant molecules to product molecules. For all chemical reactions, you can divide through by a constant and it changes absolutely nothing about the actual reaction taking place. If it happens that the reaction is two molecules of $A$ colliding, then it makes less sense physically to write it that way, but all rate equations, etc. derived from the reaction will be identical. In contrast, because this is a first order process, it makes more physical sense to write the equation down with a coefficient of $1$ in front of $A$.

The reason I say the whole thing about being told it is first order is because it is often tempting to guess that because there is a $2$ in front of the $A$, it is a bimolecular reaction, but it need not be one as we see here.

See the IUPAC definition of rate of reaction (https://goldbook.iupac.org/html/R/R05156.html). The rate of reaction (rate constant) changes if you change the stoichiometric coefficient i.e. divide by $2$. I don't see how dividing the equation by $2$ and use the same value of rate constant as before.

@blue, it may be useful to distinguish elementary steps from balanced reactions. Elementary steps are those that directly represent molecular interactions, and those for which coefficients cannot be so manipulated. Balanced reactions, however, simply provide a summary of the reaction, and make no attempt to specify the mechanism or interactions taking place.

@a-cyclohexane-molecule I get that point. But did you see the IUPAC definition of rate of reaction. If you divide the stoichiometric coefficients by $2$ then even the rate constant changes. You cannot use the same value of rate constant after dividing by 2.

@blue The division by two would have happened for all terms including the rate which doesn't change anything. The reason I rewrote the equation the way I did is because it is a decay process. If it were a second-order reaction then it would make sense to multiply by 2 from the way I wrote it. Your textbook shouldn't have written down the reaction the way that it did. It's not wrong. Just unclear and confusing.

@blue, I think I see your point. Your issue isn't with reaction order, but the rate constant definition. To summarize your argument: say our reaction is $\ce{A -> B}$ and $k$ = 10 s$^{-1}$, but if our reaction is $\ce{2A -> 2B}$, then $k$ would be 5 s$^{-1}$. The half-life of species A remains the same in either case, being $t_{1/2} = \ln 2/k$ in the former and $t_{1/2} = \ln 2/2k$ in the latter, but the reaction rate constant is different. Is that right? I think your argument is sound, but I'd like to see what others have to say.

@a-cyclohexane-molecule Exactly!!


text:
Why doesn't this β-hydroxycarboxylic acid undergo dehydration upon heating?

This is a β-hydroxyacid, but why doesn't it get dehydrated to give an α,β-unsaturated acid on heating?
The mechanism which I think should occur is:

What you've drawn isn't a dehydration, it's a retro-aldol reaction. You're losing formaldehyde instead of water. And your product isn't an alpha,beta-unsaturated acid.

@orthocresol I'm not sure of the mechanism that causes the conversion from beta hydroxy acid to alpha beta unsaturated acid. Could you give me a link? The mechanism in the image is just a guess...

Isn't the carbocation more likely to be captured by the carboxy group giving a 4-membered lactone?

@Waylander Which carbocation?

Depends on which O you think gets protonated first. If the carboxy group protonates the the alcohol OH can attack similar to acid-catalysed ester formation. The resulting B-lactone is a strained species,yet B-lactones exist.

@Waylander Why can't we just protonate the OH, and then decarboxylate leading to simultaneous loss of H2O, creating a new exocyclic double bond?

@SpaceSkiier  The decarboxylation will not proceed because there is no alpha proton to the carboxyl group

@Waylander To make things clearer, the final product I am talking about is methylenecyclohexane, which can be formed (at least theoretically) by simultaneous decarboxylation and dehydration of the last but one structure in Martin's answer. Kindly tell me why this cannot occur.

In order to obtain an α-β-unsaturated carbon acid from a β-hydroxy acid, you need a secondary or tertiary carbon in α position, because otherwise you would have to break a carbon-carbon bond, which requires a lot more energy then simply heating.
The mechanism is basically self catalysed, which is why this may happen easily with a little bit of heat. The acid may protonate intramolecular (or intermolecular) the hydroxy group. You have water as a nice leaving group, creating a carbocation, and abstraction of a proton in α-position leads to the olefin.
If you do not have that α-hydrogen, you are stuck with the carbocation. You might be able to obtain a lactone,like Waylander suggests, but I doubt this would easily happen with a little bit of heating, as spiro-compounds with four-membered rings tend to be quite unstable.

In your attempt you choose the less acidic hydrogen to transfer and then break a carbon-carbon bond resulting in a very unstable molecule which would rearrange to cyclohexanecarboxylic acid. I'd suspect you would need insane temperatures to overcome the activation barrier.

In the last but one structure you have drawn, we could've even removed CO2 and H2O in a single step, forming a new exocyclic double bond, leading to an alkene. Can you please explain why this would be incorrect?

@SpaceSkiier No I cannot. Sorry. You might want to [ask] a follow up question and someone else answers that.


text:
Reaction between bromobenzene and methyllithium to yeld toluene: possible?
I was solving some reactivity problems, and I was asked to prepare toluene from bromobenzene.
While the suggested solution of the problem involves the preparation of a Grignard reagent, followed by a reaction with methyl bromide, I was wondering:
Could the same compound be prepared with the straightforward reaction between bromobenzene and methyl lithium?
I tried to search some similar reactions, but I found nothing for both the possible ways.

Probably not, bromobenzene might not be electrophilic enough.

MeLi certainly does lithium-halogen exchange and in the presence of MeI takes bromobenzenes to toluenes   Synth. Comm. 2001 2323
If you're asking whether the MeBr produced by the exchange can act as the methylating agent? I have never seen this reported but maybe someone else can pull up a reference.


text:
Is DNA a base inside the cell?
DNA is deoxyribonucleic acid because it has phosphate groups in every one of it's nucleotides.
However, in aqueous solutions protons will ionize, leaving phosphates alone as negative groups, i.e., lewis bases.
In this case, which is seen in all of our cells, is DNA structure actually the conjugate base of the theoretically described crystal form of DNA?

Hello and welcome to Chemistry.SE. If you have any questions about the site, feel free to visit the [help]. Best of luck with your interesting question!

In nuclei DNA as coniugated base is bound to histons which work as coniugated acid.

Related: Why are lactate and lactic acid used synonymously in biochemistry?

In a solution of Hydrochloric acid, almost all of  $\ce{HCl}$ exists as $\ce{Cl-}$ which is a base. Would you call it Hydrochloric base instead?

By the way it's very related to pKa concept as I now understood it. DNA should be reduced to a bulk of phosphate groups for simplicity, of which the pKa value is ~3. Since we know that physiologic pH is ~7, ionized form of phosphate groups must have ten thousand times more concentration than the non-ionized form. That's how we know DNA is actually what makes the environment acidic. Otherwise I was thinking we might have very well defined DNA in it's ionized form, as DNB. But understanding why this would be wrong depends on understanding the pKa concept. So thanks @orthocresol.

The pH is 7, it's not really an acidic environment. In terms of charge it's simply balanced by a whole host of positively charged stuff, like histones (which Mithoron mentioned), but also cations like Mg2+, etc. The name DNA is, to some extent, a misnomer; DNA as it exists in the body is not an acid, since it exists mostly in the form of its conjugate base.

(As for whether it's different from the crystal structure, I am not sure whether Franklin crystallised DNA as the acid or as the salt, would have to look it up later. However you are certainly quite correct in saying that DNA exists as the conjugate base)

related https://chemistry.stackexchange.com/questions/24791/why-dna-is-negatively-charged-and-what-makes-it-so


text:
Strength of OH bond
This question might seem a little general, but how can we know the strength of the OH bond of third period element hydroxides? Does it increase across the period or does it decrease?But more importantly, on what should we rely in order to understand how the strength of the bond changes?


text:
How to estimate the enrichment of U-234 from the enrichment of U-235?
Fresh nuclear fuel made from naturally occurring uranium (i.e. neither MOX nor ERU fuel) is mainly characterised by the enrichment of $\ce{^235U}$. The mass and activity concentrations of $\ce{^235U}$ and $\ce{^238U}$ can be directly calculated from the given level of enrichment. If necessary, the mass and activity concentrations of relevant decay products can also be calculated easily.
However, the gas diffusion or gas centrifuge processes for $\ce{^235U}$ enrichment do not only lead to an increase in the $\ce{^235U}$ concentration in the fuel, but also to a more than proportional increase in the $\ce{^234U}$ concentration. This can be seen from the following table where the mass concentrations of uranium isotopes in natural uranium and typical fresh nuclear fuel (with $3.5\ \%\ \ce{^235U}$) are compared.
$$
\textbf{Mass concentration of uranium isotopes (in g/kg HM) in fresh nuclear fuel} \\
\begin{array}{lll}
\hline
 & \text{Natural U} & 3.5\ \%\ \ce{^235U} \\
\hline
\ce{^234U} & \hphantom{00}0.06 & \hphantom{00}0.4 \\
\ce{^235U} & \hphantom{00}7.2 & \hphantom{0}35.0 \\
\ce{^238U} & 992.8 & 965.0 \\
\hline
\end{array}$$
The corresponding activity concentrations are shown in the second table.
$$
\textbf{Activity concentration of uranium isotopes (in MBq/kg HM) in fresh nuclear fuel} \\
\begin{array}{lll}
\hline
 & \text{Natural U} & 3.5\ \%\ \ce{^235U} \\
\hline
\ce{^234U} & 13 & 93 \\
\ce{^235U} & \hphantom{0}0.58 & \hphantom{0}2.8 \\
\ce{^238U} & 13 & 12 \\
\hline
\end{array}$$
Is there a simple  and reliable calculation method or model that can be used to estimate the enrichment of $\ce{^234U}$ from a given enrichment level of $\ce{^235U}$?

It would depend on the enrichment method used, and how much somebody cared about the $^{234}$U.

Could a sequential off-line or continuous on-line monitoring by mass spectroscopy, discerning the isotopes (like $\ce{^{79}Br}$ / $\ce{^{81}Br}$  or $\ce{^{35}Cl}$ / $\ce{^{37}Cl}$ in small molecular organic chemistry) assist to establish a descriptive model about the enrichment of the uranium isotopes?


text:
Is cyanidin aromatic?
The structure of cyanidin is as follows:

There are eight double bonds in the entire compound, and hence $8 \times 2 = 16$ π-electrons, which does not satisfy the $4n + 2$ rule. However, the molecule is usually depicted as aromatic in literature. MarvinSketch (a software) also reports the entire molecule as aromatic.

Is this molecule aromatic? Why?
Are all anthocyaninidins aromatic?

I guess you have to count the electrons for the upper benzene ring and lower fused rings seperately. Then the lower one has 10 electrons (aromatic) and upper one has 6 (again aromatic)

Probably that's true. Upper ring is aromatic and lower ring is also aromatic if counted separately.

No doubt it is aromatic.
The $4n+2$ rule is intended for a single cycle of π-electrons, which may possibly include multiple rings but can't have any interior atoms or pendant π- bonds (which would not fit in a cycle). Here, as Kartik indicates in a comment, there are evidently two such cycles and you have to consider them separately. One cycle has six π-electrons and the other has ten. You have a molecule (or more accurately, a molecular ion) with two aromatic cycles.

Thanks for the reply. What do u mean by interior atoms?

Look at pyrene.  The presence of two conjugated atoms surrounded by the others means you can't fit them all into a single cycle.  Thus the $4n+2$ rule is not reliable.  And true to form it doesn't work.

Hückel's rules can only be applied to the compounds they were derived from: monocyclic, planar hydrocarbons. For most aromatic compounds they are a lot less reliable and only a very rough approximation. https://goldbook.iupac.org/html/H/H02867.html


text:
Does adding water to a reaction mixture shift equilibrium?
Just like decreasing volume in a gaseous reversible reaction shifts the position of equilibrium towards the side that produces more moles (increasing pressure), shouldn't adding water shift equilibrium to side that produces more moles in a liquid equilibrium mixture (increasing concentration)?
On research, I've seen an explanation that it does not because the concentration of both reactants and products decrease evenly, but I don't understand this. The side with more moles should have a greater decrease in concentration because the concentration is proportional to the number of moles. Also, by the same argument, decreasing volume in a gaseous reaction should not shift equilibrium.
My guess is that adding water does affect equilibrium for a reaction with different number of moles on each side. Because the equilibrium constant is affected. Is this right?

Apparently it is right (I asked at work, to be sure). If you consider a reversible reaction $\ce{A + B = C}$ as the effect of two opposite reactions with rates $\ce{r_f = k_f \cdot [A] \cdot [B]}$ and  $\ce{r_b = k_b \cdot [C]}$, equilibrium is reached when the forward and backward rates are the same: $\ce{r_f = r_b}$, which also gives you the definition of the equilibrium constant  $\ce{K=\frac{k_f}{ k_b}}$. If you suddenly double the volume, the two rates change, so all concentrations have to readapt, for the equilibrium to be re-established (as both rate constants stay the same).

The equilibrium constant is not affected. But the number of moles of products and educts may change to hold the equilibrium when the amount of solvent changes.

Yes, the equilibrium does indeed readjust after a change in concentration. This can be seen by careful examination of the formula of the equilibrium constant $K$. Consider an arbitrary reaction that produces two products from a single reactant and is in equilibrium:
$$\begin{align}\ce{A &<=> B + C}\tag{1}\\[0.5em]
K &= \frac{[\ce{B}][\ce{C}]}{[\ce{A}]}\tag{2}\end{align}$$
Let’s assume initial concentrations of $\pu{2M}$ for each species; that gives us $K = 2$ in dimensionless units (you can also use M as a unit here). Remember that $K$ is a constant so we can examine whether equilibrium has been reached by examining $K$.
Now let’s double the volume of solvent. The new concentration of each reactant is $\pu{1M}$. Therefore:
$$K' = \frac{[\ce{B}][\ce{C}]}{[\ce{A}]} = \frac{1\times 1}{1} = 1\tag{2'}$$
Since $1\ne2$, we are no longer at equilibrium. Thus the concentrations must readjust. How? Obviously, the value of the fraction is too low. To increase its value, we can increase the numerator or decrease the denominator — luckily for us, both of these mean that we must perform the reaction in a forward direction to get think right again. Since $[\ce{B}]=[\ce{C}]$, we can actually calculate the ratio of concentrations and from $2[\ce{A}]+[\ce{B}]+[\ce{C}]=\text{const.}=4$, we can calculate exact values.
$$\begin{align}&&K &= \frac{[\ce{B}]^2}{[\ce{A}]}\\[0.5em]
\Longrightarrow&&[\ce{A}] &= \frac{[\ce{B}]^2}2\\[1.3em]
&&2\times\frac{[\ce{B}]^2}{2} + [\ce{B}] + [\ce{B}]&=4\\
&&[\ce{B}]^2 + 2[\ce{B}]-4 &= 0\\
\Longrightarrow&&[\ce{B}]_{1/2} &= \frac{-2\pm\sqrt{2^2+4\times4}}{2}\\[0.5em]
&&[\ce{B}]_{1/2} &= -1\pm\sqrt{5}\\[0.5em]
&&[\ce{B}]_{1/2} &\approx 1.24\end{align}$$
Thus, the product side has increased.

A more real-world example is maybe the dimerisation of hydrogenchromate $\ce{HCrO4-}$ according to $(3)$:
$$\ce{2 HCrO4- <=> Cr2O7^2- + H2O}\tag{3}$$
This is also pH-dependent, since hydrogenchromate is a weak acid and can deprotonate to give chromate $\ce{CrO4^2-}$. However, it is fairly easy to try this experiment yourself: take an orange coloured dichromate solution and add water. A colour change to yellow should be observed indicating that the reaction has equilibrated to the reactants’ side.

So just to confirm, for the final example, this occurs because the reactant side produces a greater amount of moles, not because the amount of water has increased thus increasing the reaction rate? I take that it is the former since you're not actually increasing the concentration of water however I just wanted to make sure.

@bio The real underlying reason for the hydrogenchromate/dichromate example is a change in pH which corresponds to more or less water being present in the system. In fact, it is often written with $\ce{CrO4^2- + H+}$ instead of $\ce{HCrO4-}$.

Dilution of an aqueous ionic equilibrium favors the reaction that produces more ions. This changes the position of equilibrium. When in doubt, consider the expression of K and know that the equilibrium shifts to satisfy the constant K at a given temperature.


text:
How to name this compound
2 bromo 3 methyl-butane or 3 bromo 2 methyl-butane. In general, if we have a symmetrical situation I know the functional groups should be ordered alphabetically and I also know that the smallest possible number on the functional group should be used. But what if each f-group is equidistant from the ends of the chain. Which one gets the smallest number?

https://chemistry.stackexchange.com/questions/68624/alphabetization-rule-in-case-of-consecutively-placed-substituents

ok, does this take priority over alphabetical order then?

Sorry, I'm not really sure what you mean. Alphanumeric order dictates that the name is of the form x-bromo-y-methylbutane. That's a given, and you didn't seem to have an issue with that. Your question is whether (x,y) should be (2,3) or (3,2). As described in the linked question, rule P-45.2.3 would lead to (2,3) being the preferred locant set. Quoting myself: "The correct way to do it is [...] first, order the prefixes by alphabetical order, and then choose the locant set in order to minimise the locants."

Also related: Order of citation of alkyl and halide prefixes in IUPAC names


text:
Why would rinsing the flask with acid in this experiment affect its accuracy?
The following question is #21 from this test form (PDF link):

The apparatus in Figure 4 was set up to measure the time taken for $\pu{20.0 cm3}$ of sodium thiosulfate solution to react with $\pu{5.0 cm3}$ of hydrochloric acid in a $\pu{100 cm3}$ conical flask at $\pu{20 °C}$. The timer was started when the sodium thiosulfate solution was added to the acid in the flask. The timer was stopped when it was no longer possible to see the cross on the paper.

What is likely to decrease the accuracy of the experiment? [1 mark]
A) Rinsing the flask with acid before each new experiment.
B) Stirring the solution throughout each experiment.
C) Using the same piece of paper for each experiment.
D) Using different measuring cylinders to measure the volumes of acid and sodium thiosulfate

The answer given as correct is (A). Why is rinsing the conical flask with acid likely to decrease the accuracy of the experiment? I would imagine that stirring the solution, option (B), introduces error since the rate of stirring is difficult to control.

Probably because the use of hydrochloric acid (which is a reactant involved in this reaction!) could result in some residual aqueous HCl in the flask, which may add to the HCl added later.
Anything you use to clean the flask should be inert/not involved in that reaction. If you really want to use HCl, then you should later wash the flask with a lot of water, but that's not said in the answer.
Also note that the volumes involved here are not so large (only 5 ml of HCl are used), so a "small" residue might have a large effect on the result.
As for the stirring, the colour of the reaction is caused by the sulfur precipitate. But I think that, if you stir in each experiment, the times might probably decrease, but you would still observe a trend with temperatures.
That answer would be wrong if it involved stirring in some experiments and not in others.


text:
Clarification about the difference of internal energy formula
I am trying to understand the concept of internal energy and it's difference. My textbook said that the difference in internal energy of a system is
$$
\Delta U=Q+W
$$
where $Q$ is the heat absorbed or released by system and $W$ is the work done by system to the surrounding or by the surrounding to the system. After that, it mentioned that the signs of $W$ and $Q$ are determined like this:

The work done by surrounding to system: $(+)$. The work done by system to surrounding: $(-)$. Same on heat.

After that general definitions, it started to explain the internal energy differences in various systems like isochoric system, isobaric system. In the section of isobaric system it says that:

If a system were free that can expand against external pressure, when heat is given to that system, it will give some energy by doing some work. So the internal energy difference is:$$
\Delta U=Q_p-W
$$ where $Q_p$ is the heat given to a isobaric system.

It is completly OK till the new formula, but isn't the new formula bizarre? Because if we give heat to a system and it will do work. As a result the difference would be like:
$$
\Delta U_{example}=Q_{p_{example}}-(-W_{example})
$$
So, does the work done by system increase its internal energy?

Your book has driven you crazy with its stupid sign convention.  If the gas is expanding against constant pressure, then the equation is still $$\Delta U=Q+W$$ but, in this case, the work done on the system by the surroundings is negative (i.e., W is a negative number, because the system is doing work on the surroundings).  So not all the heat added to the system is being used to increase its internal energy.  The increase in internal energy is less than if the gas was not pushing back its surroundings.

I can't upvote it due to low reputation but your answer is explanatory.


text:
Why don't electrons simply stick to the nucleus on account of being *oppositely* charged, instead of moving around it?
Why don't electrons stick to the nucleus on account of being oppositely charged charged (the nucleus being positively charged)? But contrary to my intuition, electrons seem to follow a whole bunch of wacky paths (orbitals) around the nucleus (as demarcated by orbital wave-functions). Why is this?
Do note:
I'm not asking "What keeps the electron in orbit". All I want to know, is why electrons don't simply find themselves attached to the nucleus, and why do they even bother moving around the nucleus (in orbitals) at all? My textbooks, unfortunately, make no reference to this issue...they wholeheartedly adopt the "current" QM model of the (hydrogen) atom.

The cheeky answer to this is "because something keeps the electron in orbit around the nucleus."

https://en.wikipedia.org/wiki/Electron_degeneracy_pressure

In a way, electrons do stick to the nucleus to form what we call an atom. There are no paths around.

eh, orbitals, orbits, all that stuff is just a mathematical approach to press quantum mechanics in a canvas our primitive brains can understand

Richard Feynman alludes to this exact problem in the first volume of the Feynman Lectures on Physics. I'll present his argument here (based on the Uncertainty Principle), albeit, in my own words ;)

The German physicist [since he dealt with atomic chemistry/physics...by all means, go ahead and call him a "chemist" ;) ], Werner Karl Heisenberg came up with what we now know as "The Heisenberg Uncertainty Principle"... a revolutionary idea that points to an "inherent fuzziness", that exists within quantum systems (which is a fancy word that refers to any region of space that is sufficiently "small" enough to introduce wave-particle duality), and becomes apparent when we try to "measure" the various parameters that constitute them.
Basically, what (one version of) the Uncertainty Principle states is that:

The product of the uncertainty in measurement of velocity and the uncertainty in measurement of the position of a particle can never be less than a certain constant; i.e- ħ/2

Rephrasing this

If we were to know a particle's position to a very high precision (i.e- small uncertainty in measurement of position) then the corresponding uncertainty in the measurement of the particle's velocity (or momentum, if you know its mass) increases greatly. The same holds true vice versa.

In other words,

You cannot know both a particle's position and momentum in a quantum system to a great precision simultaneously.

Here's a good analogy for this (Part 3: Atomic Structure - Analogy with a ballon)
From this point on, I guess the question can best be dealt with by "proof by contradiction".

First part of the question: Why don't electrons stick to the nucleus?

We know for a fact, that Heisenberg's Uncertainty Principle holds true (yeah, I didn't prove it...but for now, just take my word for it).
So, let's assume that electrons do stick to the nuclei of their respective atoms. In that event, we would know the electrons' positions and momenta (with respect to the nucleus) with almost absolute certainty...something that Heisenberg's Uncertainty Principle does not allow.
Seeing that this situation is impossible, it is obvious that our little "assumption" must be wrong (Knowing the the Uncertainty principle is indeed, correct).
So yeah, electrons don't just go and simply "stick" to atoms.

Second part of the question: Why do electrons follow a whole bunch of wacky paths?

The Dane, Niels Bohr postulated that electrons move in certain, fixed orbits around a nucleus...a fairly convincing (but not very accurate) model that is still often brought up in high-school and under-grad physics and chemistry courses. The Bohr Model was developed before the Uncertainty Principle.
But electrons do not move in fixed orbits ...the Uncertainty Principle triumphs again!
So if electrons don't stick to the nucleus, and if they don't move in fixed orbits around the nucleus, then where exactly are they?
We can never exactly point out an electron's position in an atom. However, we can determine the probability of finding electrons in a given region of space around the nucleus. These regions of space with a very high electron probability density are called orbitals.

This answer has been greatly over-simplified. Thought it does deal with the gist of the topic, I'm personally looking forward to seeing more detailed answers myself.

I wouldn't say it's oversimplified, but incorrect. Uncertainty Principle has no importance here.

@Mithoron Eh? But I recall hearing Feynman using the same argument: http://www.feynmanlectures.caltech.edu/I_02.html (Read page 2-3, Quantum Mechanics, 2nd paragraph). I think I'll trust Feynman on this one. Thanks for the down-vote by the way ;)

Feynman can also be wrong...

Electron degeneracy pressure not Uncertainty Principle.

The electron degeneracy argument is more to do with pushing atoms together and then what happens to energy levels via the Pauli Principle than with an isolated atom and its electrons and nucleus. Feynman's argument seem sound to me.

@Mithoron 1) electron doesn't fall into the nucleus even in a hydrogen atom, where degeneracy pressure doesn't exist, 2) degeneracy pressure comes from the uncertainty principle, as the very link you give also briefly mentions

I agree that the uncertainty explanation is correct. This principle determines what happens because as the electron gets closer to the nucleus and the energy gets lower the electron is confined into a smaller space, i.e. is closer to the proton, this means that the kinetic energy increases and so does the uncertainty in the momentum. So the energy is a compromise between the kinetic and potential energy and this applies to all levels. The existence of atoms would then seem to be a consequence of the uncertainty principle.

The non-local nature of the electron is due in general to its small mass and QM, not to fermionic character.

The planetary view of a a small electron orbiting a nucleus does indeed not give a satisfactory explanation; Bohr simply postulated stable orbits in his original model.
In QM, you could argue that the bound electron is the wave function [1], and therefore much larger than the nucleus, despite being much lighter. The closest thing to sticking together in this view is for the nucleus to settle in the center of that electron wave function.
For a more deterministic picture you might adopt Mills' view [2] which suggests the bound electron is a spinning charge distribution (orbitsphere) which completely encompasses the nucleus — think something like a soap bubble around a grain of salt. Angular momentum is an inalienable property of this electron, with or without a nucleus at its core; i.e. it is not so much that the electron ‘bothers’ to move around the nucleus, but rather it spins around its own axis all the time. (The size of a free electron in this model is the De Broglie wavelength, making it indistinguishable from a point-charge at high energies.)

[1] See for example https://gravityandlevity.wordpress.com/2015/04/11/how-big-is-an-electron/ option #1
[2] Randell L. Mills, "Classical Quantum Mechanics" *(Physics Essays, Volume 16: Pages 433-498, 2003)
http://www.brettholverstott.com/annoucements/2017/8/5/summary-of-randell-millss-unified-theory

This is only a comment on the answer by Pallas, which I upvoted:

Fig. 3-4. The wave function and probability distribution as functions of r for the n = 1 level of the H atom. The functions and the radius r are in atomic units
https://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html
I visualize the electron as a ball of cotton candy (spun sugar), which is not a point, and which, as Pallas says, simply surrounds the nucleus. So the electron does, in fact, stick to the nucleus, as good as it can! An s-electron has a high probability of touching the nucleus. The only way to get closer would be to become a neutron - oops, but neutrons have a half-life of only about 15 minutes before decaying back to a proton and electron.
So the cotton ball with the nucleus in the middle is sticking, not travelling around. And maybe the uncertainty principle plays some part in why the electron isn't a point charge.
But I wonder why the electron is a trinity of charge: that is, why does it take the charge of 3 quarks to equal the charge of one electron? Surely, the lowest charge (a quark) must be the elementary unit. Mulliken would have agreed (I think).

To add to the plethora of recent answers: it's not that the electron cannot "crash" into the nucleus, it certainly can, the likelihood depending on the isotope in question. The process of fusion of proton and electron to form a neutron is called electron capture. None other than Luis Alvarez studied such processes early in his career. You might find people more knowledgeable on the subject over at physics SE.

A simplified explanation would be to think about a balance between minimising potential and kinetic energy. While an electron sticking to the nucleus would decrease potential energy due to electrostatic attraction, the spiked and extremely localised wave function would be high in kinetic energy (correlated with the curvature of the wavefunction).
I do understand that this is a simplified view but I believe it is a neat explanation.


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What will be the possible major product(s) on heating this primary alcohol i.e. 9-fluorenylmethanol with (conc) H2SO4?

When the above compound is heated with (conc.) $\ce{H2SO4}$ the $\ce{OH}$ will probably be protonated and it will leave, producing a carbocation. After that possibly there will be a hydride shift. Now, after this step what could be the possibilities. I was thinking of ring expansion (which could potentially form phenanthrene) but again I doubt that a negative charge on a $sp^2$ carbon of the benzene ring would attack the positive charge after C-C bond cleavage. Or another possibility is formation of a double bond to form a structure like:

Which is the correct (major) product formed?

Ring expansion is possible and we get  an aromatic compound I suppose which is quite stable....

@SurajS For ring expansion we need to form a negative charge on the benzene ring which has to potentially attack the formed carbocation due to leaving of the $\ce{OH2+}$ group. Is that feasible?

I think this process favors thermodynamics, Phenanthrene might be the major product...

Sorry I don't get why - ve charge on the benzene has to be formed... Isn't it th 3° carbon making way for the rearrangement so that 1° carbocation stabilizes to 2°?

Which  "$1°$ carbocation stabilizes to $2^o$" are you talking about? I can't understand. Can you draw it and show? @SurajS

@SurajS A negative charge is formed on the benzene ring when the C-C bond between $\beta$ carbon and the $\gamma$ carbon of the $\ce{OH}$ group cleaves. That negative charge can attack the $1^o$ $\alpha$ carbon so that it can expand.

What I feel is that forming phenanthrene would be possible because of its high stability due to resonance. Initially, $\ce{-OH2}$ leaves out forming a 1°carbocation. Now the $\ce{\beta C-\gamma C}$ cleaves so as to make way for the better 2°carbocation which is also thermodynamically favourable.

Related: https://www.researchgate.net/publication/231517152_Dehydration_of_9-Fluorenylcarbinol_a_New_Synthesis_of_Phenanthrene


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Relative reacting tendency of H2/Pd towards double bond and triple bond in straight carbon chain

If both double bond and triple bond is present in a straight carbon
  chain the which one (double bond or triple bond) will be reduced by
  $1$ equivalent of $\ce{H2(Pd)}$ ?

My guess is that the triple bond should react as it is more reactive but I'm not sure. Also does the answer depend on whether the double bond and triple bond are conjugated with each other or not?
Let us consider two examples (A) and (B):
(A)

(B)

Which bond will be reduced in case (A) and (B) ?

March Advanced Organic Chemistry 7th ed, p 1499 lists alkynes as being easier to reduce than alkenes (in general), although I guess it will probably depend on the exact substrate.

@orthocresol Thanks. Does it depend on whether the double bond and triple bond are conjugated or not? Any ideas?

Triple bond will be more reactive because it can be adsorbed easily on the surface of catalyst through its pi bonds. If double and triple bonds are in congugation then also triple bond is more reactive as in this situation product formed will be conjugated diene.


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Homemade copper etching solution
I'm trying to make an etching solution for copper (PCB) with easy to access materials. I'm using hydrochloric acid (9%), peroxide (common "oxygene" that contains many more ingredients), table vinegar, table sea salt, and bicarbonate soda. Now the recipe mentioned only hydrochloric acid and the peroxide, but it mentioned a hydrochloric acid with high concetrantion (31%). My solution with (the easily accessible) 9% hydrochloric acid didn't worked so I added the other elements and it worked fine. The problem is that I don't know what I'm doing, and what elements causes the reaction. I start with the hydrochloric acid, I put some peroxide inside. Alone it does nothing to the copper. I then add some vinegar, salt, and soda (I suppose soda helps to get oxygen inside the solution) and after a while the reaction starts and it works really quick. (the 5x5 cm copper piece gets dissolved as I like in around 5 minutes. (and whatever I have painted on it with my permanent marker remains).
So can someone explain me what reactions take place, and perhaps which is the crucial material and which I may omit?
Also when I work with that stuff my nose gets irritated, how harmful is this solution and how I may dispose it safely?

The addition of soda probably has a negative effect. I would try without it.


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Does temperature change pressure in isochoric system?
Consider the following equilibrium reaction:
$$\ce{2 NOCl (g) <=> 2 NO (g) + Cl2 (g)} \qquad  \Delta H > 0$$
My textbook has a statement that:

If you decrease the temperature when keeping volume constant, then reaction shifts to the reactants.

I know that since the enthalpy change $\Delta H$ is positive, we can think of the reaction as:
$$\ce{2 NOCl (g) + \text{heat} <=> 2 NO (g) + Cl2 (g)}$$
And according to Le Chatelier's principle we can say that if we decrease the temperature, then reaction favor the side have heat in it. This comes me so reasonable but there is one more thing we have to consider in my opinion: pressure.
While keeping volume constant, decreasing temperature will decrease also pressure. So reaction will favor the side that have more molecules in it. In this case, it is the opposite side of the previous one.
Is my reasoning true? If it is, how can I conclude the side determining job?

You need to look up the Van't Hoff Isochore; $d\ln(K_p)/dT = \Delta H^0/(RT^2)$. In an endothermic reaction ( $\Delta H^0 \gt 0$) , $K_p$ increases with increasing temperature, so more dissociation at higher temperatures and  less at lower temperatures as you initially state.

@porphyrin As far as i understand, even pressure decreases (means that $Q_p$ decreases), $K_p$ always decreases much more than $Q_p$, so, it ends up with always favoring reactants.(for $\Delta{H}>0$ of course.)

$K_p$ is constant with change of pressure  but does change with temperature.

The change in $K_p$ i mentiond is due to temperature change. So is it what is going on?

the answers here might help. https://chemistry.stackexchange.com/questions/73644/how-does-the-partial-pressure-of-gas-c-change-with-the-change-in-volume-of-the-c/73975#73975

Since it's given that ΔH > 0 (endothermic reaction) ,the reaction will proceed backward (towards the reactant). However as you know ,
Partial Pressure Of Gas= Total Pressure x Mole Fraction;
hence , even if the pressure is affected , the mole fraction increases/decreases in a way so that partial pressure remains constant.
If you calculate Kp ( Pressure Equilibrium constant) you will know that it depends on the partial pressure of the gases involved . Since partial pressure remains same no effect takes place. Hence , pressure  will have no effect on the equilibrium.
Also note that equilibrium constant is only affected by the change in temperature.
Hope this helps.
EDIT:-
In the light of the comments I would like to clarify ,
Partial pressure of a particular gas is the fraction of the gas multiplied by the total pressure which means it exerts a fraction of the total pressure. When I increase the total net pressure exerted by all gases , the fraction of moles of each gas decreases (according to the mathematical equation) thereby contributing lesser to the total pressure .
You can think of it as a ratio:-
If 1 mole X Gas exerts 10 bar pressure when total pressure is 20 bar , 0.5 moles of X gas will exert the same 10 bar pressure when total pressure is increased to 40 from 20 , thereby keeping the ratio same.

You don't even need to do all that ... Just looking at the fact that what is happening to the temperature/pressure/volume is enough to determine where the reaction will proceed. And as for calculation could you send me the Kp and moles intially or at equilibrium so that I can try as well?

@Abdullah I edited the answer


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In the mechanism of Cannizzaro reaction why does the hydride migrate rather than the benzene ring?

The hydride ion is not very stable. On the contrary a negative charge on a $\ce{sp^2}$ on the benzene ring would be much more stable. So why doesn't the benzene ring migrate instead of the hydride, as the benzene ring with a negative charge would be a better leaving group (as we know that less basic groups are better leaving groups)?
I've seen some examples where the benzene ring does migrate instead of the hydride for example when two nitro groups are attached at ortho positions. But I think the benzene migration should always happen instead of the hydride migration for reasons which I stated above.
P.S: Someone suggested me that the "real" mechanism is the transition mechanism. But even that does not explain why the transition state involving o-dinitro benzene would be more stable than the one involving $\ce{H}$ but the one involving just a benzene would be less stable than $\ce{H}$. How do we logically explain the discrepancy?

One possibility:  hydrogen has better overlap for forming the delocalized bond in the transition state.  You have to heavily withdraw electrons from the benzene ring (e.g. nitro groups) to make the hydrogen non-hydridic and force the benzene ring to migrate instead.

Actually I dont know why you think phenyl anion is very stable( not saying incorrect, just asking why?) . Its not resonance stabilized. If it had been Phenylacetaldehyde then maybe it would be stable anion but in that case the reactant may give Aldol condensation reaction. What am I missing?

@AlphaRomeo Because it has a $sp^2$ carbon which stabilizes the negative charge.

How? Pka of benzene is in 40s .. and I dont see much stabilization by inductive effect as resonance wont occur. Anyway, I will brush up my concepts :P

@AlphaRomeo Who's talking about inductive effect here? I'm talking about $sp^2$ hybridization.

AFAIK Inductive effect is also caused by $sp^2$ hybridisation.

@Mesentery Not in this case.

Has it got anything to do with the bond energies?

Why are we even talking about $\ce{H^-}$ and $\ce{Ph^-}$? Do they really form in solution? I don't think so.


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Can any other common reagent other than than H3PO4 cause this conversion?

Can any other common reagent other than than $\ce{H3PO4}$ cause this conversion? What about (conc) $\ce{H2SO4}$ or (conc) $\ce{HNO3}$ or (conc) $\ce{HCl}$?

This is an intramolecular Friedel-Crafts acylation.  Other sources call it Haworth reaction.
Concentrated phosphoric acid (in its viscous solution up to $\pu{85 m\%}$) is indicated as it is both strong (to generate the acylium cation), as well as removing the water generated in the course of the reaction (which could trap the acylium cation just generated).  Hydrochloric acid (up to $\pu{37 m\%}$), in comparison, is a strong Bronsted acid, too,  but is lesser hygroscopic than concentrated phosphoric acid.
This does not imply that phosphoric acid will be your solvent of reaction; indeed you may even use polyphosphoric acid (CAS 8017-16-1) -- which is a water free, waxy solid -- to catalyse the reaction.  As alternatives to the above route, I suggest

using strongly acidic ion exchangers (like Amberlyst-15), provided you find a way to remove the water generated (Dean-Stark distillation, trapping in an other reaction).
If you change to anhydrous conditions, a Lewis Acid ($\ce{AlCl3}$ is a classic example of the stronger ones) may be used, too. Along this classical route, of course, you would need to generate the acyl chloride first (typically $\ce{SOCl2}$, or $\ce{PCl5}$), to transform the acid in the acyl chloride. And subsequently, you will perform the Friedel-Crafts acylation.  Perhaps $\ce{AlCl3}$ may serve in a one-pot reaction, too -- yet you would need a bit more.

What do you mean by "catch the water" ? Why can phosphoric acid catch the water but not HCl ?

Concentrated $\ce{H3PO4}$ may be up to 85 m% and is highly hygroscopic.  Concentrated $\ce{HCl}$ is up to 37 m%, lesser hygroscopic than the former (63 m% of water, seen from the other perspective).  In addition, instead of phosphoric acid as solution, you may use polyphosphoric acid (CAS 8017-16-1) which is a water free solid. You need both generation of the acyl cation, and removal of water (because water would react just again on the acyl cation) to drive the reaction forward.  Of course this does need not to imply to run the reaction "just in phosphoric acid" (as sole solvent).

Please edit to add this to your answer so that I can upvote it! Thanks

@blue And till some moments ago, I did not know that instead of the more general "intramolecular Friedel-Crafts reaction" this may be called a "Haworth reaction", too.  The gain here is (at least) for both of us.

On the wiki page they don't show any conditions required for Haworth reaction. Not even H3PO4. Strange! https://en.wikipedia.org/wiki/Friedel%E2%80%93Crafts_reaction#Haworth_reactions

@blue According to Organic Synthesis (http://orgsyn.org/demo.aspx?prep=CV4P0898), once the acyl chloride is formed, even a milder Lewis acid as $\ce{SnCl4}$ (rather than the harsher $\ce{AlCl3}$) may work.  (But $\ce{SnCl4}$ might be not efficient enough to promote opening of the lactone.)  An other related entry, a bit different, although with $\ce{AlCl3}$ is http://orgsyn.org/demo.aspx?prep=CV6P0744. Likely eased because it is an intramolecular reaction, and the aryl does not bear a deactivating substitutent.

Eaton's reagent - $\ce{P2O5}$ in Methanesulfonic Acid - is a good alternative.
JOC 62(1997) 3552
Conc. Nitric acid risks nitrating the aromatic ring. I have never seen conc. Nitric or conc. $\ce{HCl}$ used for this

What about HCl, HNO3....? You answered the question only partly.

Edited to include nitric and HCl


text:
Will an acyl chloride be attacked by water molecule to form carboxylic acid?

$\ce{H2O}$ molecules can act as a nucleophile. So it should attack the carbonyl carbon. However, I think that that when the $\ce{O-}$ "kicks back" it will once again remove the water molecule keeping the $\ce{Cl}$ intact as water is definitely a better leaving group. But in my book they showed a reaction intermediate, which involved an acyl chloride and they used water molecule as a nucleophile to convert acyl chloride to carboxylic acid. I'm confused as to how that is possible? Isn't water a better leaving group than $\ce{Cl-}$ ?

Don't really remember, but in general proton transfers are very fast. For mechanistic organic chemistry Clayden is a good place to start.

You need to lose a proton, then chloride becomes a better leaving group than hydroxide. In general, proton transfers are incredibly fast; in this scenario there is tons of water around to accept the proton, and the tetrahedral intermediate with a positively charged oxygen is fairly acidic anyway.

Mechanism from Clayden 2nd ed. (p 199).
In the absence of an external base (e.g. pyridine), water will act as the base itself here. Another way of approaching it is to note that hydrolysis of an acyl chloride will release one equivalent of HCl:
$$\ce{RCO\color{red}{Cl} + \color{red}{H}OH -> RCOOH + \color{red}{HCl}}$$
so in the absence of a second base, you will necessarily have to produce $\ce{H3O+}$.


text:
Why are thioethers not better leaving groups than ethers?

Why are thioethers ($\ce{RSR}$) not better leaving groups than ethers ($\ce{ROR}$)? I learnt that better bases are poorer leaving groups. Here, clearly   the poorer base is thioether, as the lone pair is on a larger atom.

Then you should unlearn it. Being a leaving group is only tangentially related to basicity, as you can see from this table.

What this chart doesn't cover (and would help you with your understanding) is what is meant by "leaving group ability?" Is it kinetics? Is it thermodynamics? You can't rationalize if you don't know what you're trying to rationalize.

@Zhe It's from Wikipedia....

@Mithoron It would be better if you explain why....

@blue Let's assume we are talking about thermodynamics. Then look at it by considering the reverse reaction. Is it easier to S-alkylate RSR or to O-alkylate ROR? Kinetics would be a little trickier to rationalise, as steric hindrance and other effects come into play.

Wikipedia is not a source. Go to the reference they are citing and figure out what it's actually measuring.


text:
What is litre atm?
"The slope of a given straight line graph with constant temperature is found to be 0.2 l atm at 5 atmospheric pressure." What is l atm here? It is the unit of what?

$\pu{L atm}$ is a unit of (pressure)(volume). As $\pu{atm}$ is a multiple of $\pu{N//m^2}$ and $\pu{L}$ is a multiple of $\pu{m^3}$, then $\pu{L atm}$ is a measure of joules, the unit of energy, which is $\pu{kg m^2//s^2}$ in base SI units.

you might also want to recall the gas law $pV=RT$ and the work done  (from thermodynamics) as $p\Delta V$; both have units of energy.

$\pu{L atm}$ is a unit of pressure ($P$) times volume ($V$). As $\pu{atm}$ is the multiple of $\pu{N m-2}$. $\pu L$ is the multiple of $\pu{m3}$, then $\pu{L atm}$ is a multiple of Joules, the measure of energy, which is $\pu{kg m2/s2}$ in base SI units.


text:
What are the applications of quantum field theory to chemistry?
I have a physics background and I learned how quantum field theory is applicable to condensed matter physics, and since some parts of condensed matter physics overlap with chemistry I was wondering if there are any direct applications of QFT to chemistry as well. Also what are the research areas in chemistry that use QFT, because I'm considering doing my graduate work in chemistry and it would be cool if I could use QFT in chemistry as well.

There's a subfield of chemistry called relativistic chemistry that deals with using relativistic corrections for energy levels. It's not quite directly QFT though.

Have a look at this article: "The Physics behind Chemistry and the Periodic Table"
Pekka Pyykkö Chem. Rev. 2012 112 (1), 371-384
(http://pubs.acs.org/doi/abs/10.1021/cr200042e)
I think it answers your question and has most of the references to work done in that particular field. For most questions in chemistry the non-relativistic Schrödinger Equation is good enough. Even for systems with an appreciable amount of relativistic effects, the Dirac Equation is often not solved directly. So, the number of applications where Quantum Electrodynamics is used is really limited.

Thanks this looks like it answers my question, I'll read the full article at my university library to get access later this week. +1

Quantum field theory is sometimes used in chemistry to derive relativistic corrections to non-relativistic quantum mechanical energies. And this is useful when computing energies of molecular matter with heavy atoms, due to electrons moving at high speeds near a massive nucleus.
In reality one does not use quantum field theory. One starts with a Schrodinger-like wavefunction expression
$$H \Psi = E \Psi$$
where the Hamiltonian operator is the sum of a kinetic part $K$ and a potential part $V$, and then one seeks for a potential, sometimes named an "effective potential", so that $V$ gives the same scattering amplitudes than quantum field theory. I.e. quantum field theory is used to verify that the chosen potential $V$ is physically admissible. Note that scattering equivalence does not imply dynamical equivalence. And sometimes incorrect potentials have been proposed in the belief that compatibility with quantum field theory was enough [1].
Basically there is no known expression for $V$ at arbitrary orders in powers of $(v/c)$. The problem is not only computational, it is technical, because quantum field theory cannot really describe bound states; the only rigorous states in quantum field theory are those of free particles. This limitation is fine for particle physics and the study of scattering events, but it is not for molecular sciences as chemistry.
An attempt to unify quantum field theory (specially QED) with the traditional many body methods used in chemistry is that from Lindgren and coworkers. Their research is summarized in book [2]. The point is that QED does not consider all electron-electron correlation effects, which limits its applicability to systems where electron correlation effects are small. Lindgren and coworkers have developed a so named "covariant-evolution-operator method" which they claim

forms a suitable basis for a combined QED–MBPT procedure. This leads to a perturbative procedure that is ultimately equivalent to an extension of the relativistically covariant Bethe–Salpeter equation, valid also in the multireference case and referred to as as the Bethe–
  Salpeter–Bloch equation.

Their approach, however, doesn't solve fundamental questions associated to  Bethe–Salpeter equations; they simply mention the difficulties and move away! Also, from a numerical point of view, renormalization of the radiative effects is only partial in their procedure, with hope to improve in future:

Schemes have been developed for this process but so far not
  been implemented in a QED-MBPT procedure. When the procedure is more developed, critical tests can be performed to find out to what extent the new effects will improve the agreement between theory and accurate experimental data.

The article from Pekka Pyykkö mentioned in another answer is freely available here. There are several objections to this article. For instance, in page 4 he claims that electron-electron interactions can be taken as instantaneous in the Coulomb gauge. This is not true. He seems to believe that electron-electron interactions are only given by the Coulomb term, when the Breit term also describes electron-electron interactions. Pyykkö then claims that for certain uses the CB potential has to be surrounded by projection operators. He omits to say that without those projection operators the standard CB potential gives invalid answers. Pyykkö  only surrounds the interaction part of the DCB Hamiltonian, but in reality the Dirac Hamiltonian $h_i$ (equation 2 in his paper) also has to be surrounded  like $Ph_iP$. That is why in other texts you will find projection operators around the full DCB Hamiltonian (check equation 2.113 in [2]) and not only about the CB term. There are similar objections to the rest of his paper, but the biggest objection, in my opinion, is that he never mentions that quantum field theory is not needed to explain the phenomena he covers.
REFERENCES
[1] https://arxiv.org/pdf/hep-ph/9706219.pdf
[2] http://www.springer.com/gp/book/9783319153858

All the time-dependent quantum chemistry methods I am aware of treat the fermions (electrons) quantum mechanically while treating electromagnetic radiation (bosons/laser-photons)  classically.  A complete QM description of the radiation+matter interaction seems like a natrual extention of quantum chemistry to QFT.  I think the the classical treatment of EM breaksdown at wavelengths that approach to size of the interacting QM-treated matter.  So, the QFT treatment would be needed to describe matter interacting with very high energy photons.


text:
What are the microstates of the triplet oxygen molecule?
I am trying to understand molecular term symbols of oxygen molecule. Everywhere I looked, it is claimed that the molecule has 6 microstates, three of them are singlet, and three of them are triplet. Now I can draw the diagrams for the singlet states with no problem, but I am struggling to find the tree triplet configurations. According to my microstate matrix there should be a triplet state with a total spin multiplicity of zero - how is this possible?

Are you able to find the term symbols? What are the degeneracies of the terms? Note that the triplet state has $S = 1$ which allows $M_S = 1, 0, -1$.

I can find the term symbols, I have uploaded a diagram here: http://imgur.com/a/quLxo . The singlet state and the triplet state are both triply degenerate, and the triplet state can have a total ang. mom. of 1, 0, and -1, exactly as you have pointed out. The question is: how is the triplet state having zero total ang. mom. possible? On the table I've linked, why is the triplet state in the middle of the matrix considered triplet? Why isn't it singlet?

$M_S$ doesn't refer to a total spin angular momentum, but rather to a projection of the spin angular momentum onto the z-axis. The triplet state still possesses nonzero angular momentum ($= \sqrt{S(S+1)}\hbar$) but the projection onto the axis ($= M_S\hbar$) can be zero. In fact, there has to be one component of the triplet state ($S = 1$) with a zero projection $(M_S = 0)$.

(By the way the symbol $M_L$ is inappropriate for a diatomic molecule; it should be $\Lambda$ as the projection of orbital angular momentum taken is not onto the z-axis but rather the internuclear axis. But that is a relatively minor point.)

In a $(\pi2p)^2$ configuration the orbital $\lambda$ and spin s quantum numbers have to be tabulated and combinations removed which break the Pauli principle. The total angular momentum quantum number is $\Lambda$ and spin $\Sigma$, for example the combination
$\lambda_1 = 1, \lambda_2 =1 , s_1=1/2, s_2 =-1/2$
generates total angular momentum $\Lambda = 2, \Sigma =0$ and this is part of the $^1\Delta$ state.
There are only 6 valid combinations leading to $^1\Delta, ^3\Sigma^-,^1\Sigma^+ $ states of which $^3\Sigma^-$ is the ground state.
A vector spin diagram shows how a triplet can have spin multiplicity of zero. There is a picture of vector spin for singlets and triplets in my answer to this question; How does spin flipping of triplet carbenes occur?


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How is Standard Gibbs free energy measured?
I have seen hole theory of thermodynamics but almost nothing about the experiments that measure $\Delta G^\circ$. I know how to measure it by electrodes, when the reaction is a redox one. But what with other reactions?
What are the experimental ways to determine $\Delta G^\circ$? Some examples would be helpful

You're probably asking about $\Delta G$, which is actually quite different from $G$.

https://chemistry.stackexchange.com/questions/4752/how-is-hydration-free-energy-of-a-solute-measured-experimentally https://chemistry.stackexchange.com/questions/42066/is-there-a-way-to-experimentally-measure-entropy?rq=1

Generally you measure the equilibrium constant $K$ and then convert it to $\Delta G^\circ$ (note that is change in standard free energy)
$ \Delta G^\circ = -RT \ln K $
In the case of redox reactions, this can be done with electrodes as you mention. For other reactions you usually use some spectroscopic technique (NMR, IRC, UV, etc) that can distinguish reactants from products, but calorimetry can also be used in certain cases.
Here is a video describing one example using UV.

If we pick a different standard state (e.g. 2 bar instead of 1 bar) will the change in standard free energy change?

If the number of particles change in the reaction, then yes. Otherwise, no.


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Why does the Rosenmund reduction stop at the aldehyde stage?

In the Rosenmund reduction, an acyl chloride can be reduced to an aldehyde by hydrogenation over a palladium catalyst poisoned with barium sulfate.
Why isn't the aldehyde product further reduced to an alcohol?

In general, the Rosenmund reduction isn't actually used very often, mainly since the catalyst is also very good at reducing other functional groups which may be present in the molecule (the catalyst is similar to that used for a Lindlar reduction).
To quote from Comprehensive Organic Name Reactions and Reagents:

[...] the Rosenmund reduction is an unreliable and irreproducible method for organic synthesis, and it is often in competition with other side reactions. For example, alcohols are often the by-products of the Rosenmund reduction.

How the reaction works/why we get an aldehyde
As with many reactions involving catalyst systems in which palladium has been absorbed onto something (the classic one being palladium on carbon/charcoal), the mechanism of the Rosenmund reduction isn't well defined, and indeed there are likely to be different mechanisms operating depending on the quality/specification of the catalyst used.
One possible pathway has been proposed, and is given below. The critical thing here is that the acyl chloride is first activated by oxidative addition into the acyl chloride – this activation allows the reduction to take place.

The product of the reaction is often therefore the aldehyde, which should be kinetically inert to further reduction (this has been discussed on chemistry.SE previously, and can be found here), though the alcohol is often obtained in small amounts.

Not sure if it's useful to point out, but: if you created a metal hydride first via oxidative addition, then you can do a $\beta$-migratory insertion into the carbonyl. Reductive elimination would provide a reduced carbonyl. But either the preliminary oxidative addition of hydrogen or the migratory insertion pathways have been shut down by poisoning.