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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!!


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pH of 0.5M K2CO3
$\ce{H2CO3}$ ionises as
\begin{align}
\ce{H2CO3 + H2O &<=> H3O+ + HCO3-}  & K_\mathrm a &= \pu{4.0*10^{-7}}\\
\ce{HCO3- + H2O &<=> H3O+ + CO3^2-} & K_\mathrm a &= \pu{5*10^{-11}}\\
\end{align}
The question is to find out the $\mathrm{pH}$ of $\pu{0.5 M}$  $\ce{K2CO3}$ solution.
Since $\ce{K2CO3}$ is a salt of weak acid and strong base hence its pH is given by
$$\mathrm{pH} = \frac{\mathrm pK_\mathrm w + \mathrm pK_\mathrm a + \log c}{2}.$$
I am unsure of which $\mathrm pK_\mathrm a$ I should consider. Any help shall be appreciated.

Consider obtaining $K_{\mathrm{b}}$ and solving equilibrium from the perspective of base.

@Zhe we are given $K_a$ and not $K_b$. I think there will be anionic hydrolysis of the salt.

$\mathrm{K}{b}$ of the conjugate base is $\mathrm{K}{a}$ of the acid.

@Zhe i know that $K_b$ of conjugate base is $K_a$ of acid but couldnot infer anything from this. can you please elaborate. Thanks.

$$\ce{CO3^{2-} + H2O <=> OH- + HCO3-},\ K_{\mathrm{b}} = ?$$

The second equilibria has little effect on your calculation, so you would be fine ignoring it, but if you want to be completely accurate, see here: https://chemistry.stackexchange.com/a/50894/15410

Write down the actual (equilibrium) reaction that develops between carbonate ion and water when one proton is exchanged (only full-fledged strong acids or bases might exchange a second proton to any significant extent).  Which conjugate acid/base appears?  $\ce {H2CO3/HCO3-}$ or $\ce {HCO3-/CO3^{2-}}$?  When you see which one it is, use that dissociation constant.

The  carbonate ion is the Conjugate base of the weak acid
$\ce{HCO_3^-}\ (K_{a(HCO_3^-)}={5\times10^{-11}})$, so this solution will alkaline. Given the concentration of this solution ,the pH should be sufficiently high to preclude the formation of any significant amount of $\ce{H_2CO_3}$ , so the solution of this problem as a solution of a monoprotic weak base:
$\ce{CO_3^{-2} + H_2O <=> HCO_3^-  + OH^-}$
$$\ce{K_{b(CO_3^{-2})}}=\frac{[OH^-][HCO_3^-]}{[CO_3^{-2}]} =\frac{K_w}{K_{a(HCO_3^-)}}=\frac{10^{-14}}{5\times 10^{-11}}=\ce{2\times10^{-4}}$$
Neglecting the $\ce{OH^-}$ produced by the autoprotolysis of water, it is valid to make the usual assumption that $\ce{[OH^-]}={[HCO_3^-]}$,and little of the carbonate goes to bicarbonate,so$\ce{[CO_3^{-2}]}={[CO_3^{-2}]_0}=0.5$ ,so thus
  $$\dfrac{[OH^-]^2}{0.5}=K_{b(CO_3^{-2})}= \ce{2\times10^{-4}}$$
$\ce{[OH^-]}={10^{-2}}$ Which corresponds to $\ce{pOH = 2}$ or $\ce{pH = 12}$

Re: The equilibrium expression must be solved as a quadratic yes and no. It depends on the precision of the answer. Little of the carbonate goes to bicarbonate, so to two significant figures you really don't need the quadratic solution. $$\sqrt{(2\times10^{-4})\times 0.5} = 0.010$$ so that solution is off by about 1 part in 50. Seems good enough.

@MaxW I edit my answer your comment makes the calculation easy

Making the calculations easy is the point. In reality I'd be surprised if the "true" value was within 5% of the calculation. Also the "exact" solution, neglecting activities, is a cubic equation. So even the quadratic is an approximation.

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Electroplating an iPhone
I thought it would be cool to electroplate the rear case of an iPhone 6 or newer since it's made out of aluminum. It's been a long time since I learned about the topic, so I looked online to research this, but hardly found anything about the topic other than some aftermarket companies that sell expensive gold electroplated iPhones. Is this a feasible DIY idea? If so, are there certain metals that would electroplate to iPhone aluminum easier than others?

Electroplating aluminum is tricky because of the adherent oxide coating, preventing plating, and because, once the coating is removed, Al reacts so quickly with aqueous solutions that the coating is not adherent.  It can be done, but may require use of cyanides, not for the amateur!

You could also check out anodizing for added coolness.

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Relation between acidity and reactivity?
Up till high school, I've been taught that reactivity and acidity of a substance in a relation that if a substance is acidic and if it's acidity is higher (comparatively to some other substance) than its reactivity will also be higher. Examples were mostly given of $\ce{HCl}$ and $\ce{H2SO4}$ stating that since $\ce{H2SO4}$ provides more hydrogen ions when ionized hence its reactivity is much higher than $\ce{HCl}$.
However today I stumbled onto a contradiction of this theory. It was stated in a website that the acidity of phenol is much greater than an average alcohol due to a greater stability of conjugate base.
Some links ahead it was also stated that the reaction of acyl chloride is less vigorous with phenol as compared to other alcohols due to reactivity of $\ce{-OH}$ group being stabilized by delocalization.
My question is, why does the theory "Acidity is proportional to reactivity" not applies for reaction of acyl chlorides with phenols and alcohols?

The issue is that reactivity is not an inherent property of a compound. It can be reactive towards specific types of reactions and completely inert to others. Acyl chloride reacting with phenol doesn't rely on phenol behaving as an acid, the type of reaction for which you would describe it as more reactive.


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How to make a primary alcohol from an alkane?
Alcohols can be made from alkyl halides, which can be made from alkanes. However $3° > 2° > 1° > 0°$ for alkyl halides synthesis from alkanes. What approach could make an alkane with primary and secondary carbons into only primary alcohols?

I'm pretty sure you'd be on course to win a Nobel prize if you can do that selectively with good yields. This is what the C-H activation chemists are looking at all the time, and it's hard.

I have suggested a possible mechanism for free radical reaction of alkanes to produce alcohols, give it a read :) https://chemistry.stackexchange.com/questions/72902/alcohols-by-free-radical-substitution

@PrittBalagopal Seems like an experiment is needed.

Recently a research paper[1] has been published which claims to effectively convert n-alkanes into n-alcohols.
It proposes a quadruple relay catalysis for selective synthesis of n-alcohols from n-alkanes via the following reaction pathway:

The reaction seems to get yields higher than 75%. The paper itself goes into much detail about each catalyst and its role.
The only catch here is that as this is a hydromethylation, you need to start with an alkane which has one carbon less in its chain than the required alcohol.

Reference:
[1]: Tang, X.; Gan, L.; Zhang, X.; Huang, Z. N-Alkanes to n-Alcohols: Formal Primary C─H Bond Hydroxymethylation via Quadruple Relay Catalysis. Sci. Adv. 2020, 6 (47), eabc6688.

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Structure and bonding in ClO2
To draw the structure of $\ce{ClO2}$, I tried to find the hybridization of the central atom i.e $\ce{Cl}$. I figured out that there would be two $\ce{Cl - O}$ double bonds, which means $\ce{Cl}$ would have 4 bonding electrons. But this assigns 3 electrons as nonbonding residing on the $\ce{Cl}$ atom. So, $\ce{ClO2}$ is an odd electron species.
Then, how do I find its hybridization and draw its structure?
Can I not just count the number of σ bonds and the number of lone pairs as usual, i.e, considering the 3 non bonding electrons as 1 lone pair, and find the number of hybrid orbitals and thus, find its hybridization? This gives the answer of $\ce{ClO2}$ being $\ce{sp^2}$ hybridized and it's geometry being trigonal planar. Is this the correct answer? Am I right in my determination of bonding and nonbonding electrons of the atom or is there something special in this compound $\ce{ClO2}$?

The $\ce{Cl-O}$ bond is a 3 electron bond.

@Yashas, what is a 3 electron bond? Never heard of that. How is it formed? Can you please explain it.

related https://chemistry.stackexchange.com/questions/39001/what-to-do-when-there-is-an-odd-number-of-valence-electrons-left-on-the-the-cent

The answers to all these questions are on the wikipedia page. But you're mostly right. I would call the structure bent, not trigonal planar. That is, it's in the $C_{2v}$ point group.

@jheindel, I have mentioned that it's geometry is trigonal planar i.e including the configuration of lone pair. Indeed, it's shape would be bent.


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Calculating volume % in a mixture of gases at equilibrium

In a mixture of $\ce{N2}$ and $\ce{H2}$ initially In a mole ratio of $1:3$ at $\ce{30 atm}$ and $\pu{300^\circ C}$, the percentage of ammonia by volume under the equilibrium is $17.8$. I have to calculate the equilibrium constant $K_\mathrm p$ of the mixture for the reaction
  $$\ce{N2(g) + 3H2(g)<=>2NH3(g)}$$

When I read the question, the first thing that comes to me is Dalton's law of partial pressure and (I think) that according to that law we can imagine the components of gaseous mixture occupying a volume ($V$) each equal to the total volume of container. Thus, that is why we simply write the partial pressure of any gas as ($P_iV=n_iRT$) and sum all the partial pressure to get total pressure.
Coming back to original question, according to the previous logic, each gas in the container at equilibrium is occupying same volume ($V$) equal to the total volume of the container. Then the term volume % doesn't make any sense because each gas is occupying the whole volume of the container (under ideal gas assumptions) so volume% of each gas $= 100\%$.
So what does the question mean by "Volume %" (for gases)? Is there any flaw in the above logic?
P.S: I would have easily solved the question if mole % was given as each gas has different moles so no problem there.

I know it doesn't make sense, but, when the term volume percent is used, it means mole percent.

@Chester Miller if we mean mole percent then why do we even bother to use the term volume percent ?? I just can't think of any logic for doing so .. it feels so Bizarre to me to think that any serious chemistry teacher or book writers will use such terms so lightly and then don't even mention that they actually meant something else .. Making situations troublesome for weak students like me ( Another Reason got added to my list of why I don't like chemistry )

Anyway thank-you for pointing out a small thing which was troubling me for quite some time..

I can't answer your questions because it never made sense to me either.  In atmospheric science, they talk about "volume mixing ratio."  It took me a while to realize that this is the same thing as mole fraction.  Go figure!

Perhaps "volume %" is meant to distinguish from "weight %" (or mass %), and the question simply assumes that you will remember that the volume of one mole of gas is the same as the volume of a mole of another gas (because we are not even thinking of solids or liquids).


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Why copper wires submerged in salt (or fresh) water oxidize much faster while they conduct electrical current?
On a lot of flooded vehicles (salt/fresh water), I've noticed that wire that carries more current is always the one with most destruction (at the connector/splice), in comparison with lower current carrying wire. Wire turns black on the outer layer, accumulating green/white oxidation(?) on top of it.
How does electricity affect the rate of oxidation?

Hello and welcome to Chemistry.SE. If you have any questions about the site you can take the short [tour] or visit the [help]. Best of luck with finding a good answer to your interesting question.

Electricity in a wire submerged in water effectively turns it into an electrochemical cell, and then it is no surprise that the anode gets oxidized pretty quickly. A modest potential of a few volts would suffice to oxidize any metal, even gold. This works for AC as well (a wire oxidizes during the positive half-wave and then does nothing during the other half-wave).


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Why are heavy metal ions such as those of Hg and Cd toxic to animals?
Why are heavy metal ions such as those of Hg and Cd toxic to animals?
While studying green chemistry I came to know about the Minamata disease caused by Hg and other disorders due to the accumulation of Hg and Cd so why at all are heavy metals toxic is the reason chemical or biological?

Part of the reason is that these metals compete with other metals that are important for the body, or they alter the structure of a protein. For example, Hg will bind to certain proteins that contain selenium. This in turn will cause the protein to stop working properly. Another example is the NMDA receptor in neurons. Hg disrupts NMDA receptors because it is biologically similar to Ca (which are the metals that the NMDA receptors are designed for).

I think this question would get better answers at biology SE.

If I'm not mistaken, I think binding to sulfur is a bigger issue than selenium, but yes. As to why Hg(II) and Cd(II) bind to sulfur or selenium well, you may want to look into hard-soft acid-base theory.


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Are allotropes of sulfur two different phases?
Monoclinic sulfur and rhombic sulfur are two allotropes of sulfur. In phase equilibria, part of the system that is having distinct boundary, mechanically separable and different chemical or physical properties that other part of the system is considered as a phase. Now I want to ask, do chemical properties of Allotropes differ and why so that we can consider them different phases?

Solid allotropes of pretty much anything are different phases. They have different crystal structures, and you can't have a single phase with two different crystal structures inside it.
As for the other criteria, some of them are sufficient but not necessary, and some are difficult to check. The distinct boundary must be there, but that's a knowledge that comes from them being two phases and not vice versa. Physical properties are not all that different, though you can tell one allotrope from another by density. (We used a solution of $\ce{BaI2}$ to this end.)
Chemical properties are even more similar since the molecules are the same. I doubt that a purely chemical test can be devised. But then again, do we really need it?


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Comparing electronegativities of aluminium and gallium
If I were to compare the electronegativities of Al and Ga, shouldn't I be saying that the electronegativity of elements (in general) decreases down the group, I say this by thinking of the position of fluorine amongst the halides and about the metallic character trends (which increases down the group) so Al should be more electronegative. However Wikipedia's data page says otherwise. For aluminium it's 1.61 whereas for gallium it's 1.81 (I'm using the Pauling Scale). I then calculated $Z_{eff}$ values for both of them using Slater's rule which indicated a greater effective nuclear charge for gallium, so is this due to the poor shielding by d-orbitals?

The classical explanation is poor shielding by 3d orbitals, see also https://chemistry.stackexchange.com/a/49598/16683


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What exactly is a solubility product, and what is its use?
What exactly is a solubility product, and what is its use?
I've come across this term while reading, and don't have access to anyone who can help me understand this concept.

If we take small particles of a sparingly soluble salt, say AgCl, then one or two particles may dissolve in water and a saturated solution is obtained pretty soon. Upon adding extra AgCl to the solution precipitation of AgCl starts. At this stage, you'll find an equilibrium is established between the dissolved and the undissolved AgCl.
$$\ce{AgCl(s) <=> Ag+ (aq.) + Cl- (aq.)}$$
And for this equilibrium if you're to find the equilibrium constant, let's call it  $K_{sp}$, you get:
$$ \ce{K_{sp} = [Ag+] \cdot [Cl-]} $$
(Both the concentrations are the ones taken at equilibrium.)
This $K_{sp}$ is called solubility product. It's useful to predict whether or not more solute can be dissolved in your solution or not. We do this by comparing it to something which is called the ionic product.

I think that a sentence about the meaning of 'saturated solution ' can be useful

The solubility product is an equilibrium constant for the dissolution of a sparingly soluble but highly ionised electrolyte. It does not apply in principle to strong soluble electrolytes or to compounds that exist in an unionised state in solution, such as $\ce{HgCl2}$.
In a  reaction such as
$$\ce{AgCl <=>Ag^+ + Cl^-}$$
which must have reached equilibrium, which means that  some solid is present together with the solution (and so the solution is saturated), the solubility product is
$$\ce{[Ag^+][Cl^-]=}K_s $$
In the general case $\ce{A_nB_M <=> nA + mB }$ then $\ce{[A]^n[B]^m=}K_s$
(The thermodynamic derivation of $K_s$ depends on using activities which may be replaced by concentration when the concentration is low as it is in the case of sparingly soluble / insoluble salts)
The solubility product illustrates a quantitative application of Le Chatelier’s principle. If, for example, the concentration of $\ce{Cl^-}$ is increased by adding sodium chloride some AgCl  is precipitated until equilibrium is re-established, i.e. $\ce{[Ag^+]}$ is reduced until $\ce{[Ag^+][Cl^-]=$K_s$} $. Similarly if $\ce{[Ag^+]}$ in solution is reduced by forming a soluble complex, such as by adding ammonia, then more $\ce{Cl-}$  will be produced as more solid AgCl dissolves until equilibrium is established.
The solubility product $K_s$ can be used to calculate the concentration of an ion in solution and hence its total amount in a solution. In the case where there is a common ion it can be used to determine how solubility varies with common ion concentration. Thus it can be used to answer questions such as 'How much $\ce{Ag^+}$ is there in 100 ml of a solution of 0.1 M potassium chromate  saturated with silver chromate'.


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Redox Titrations Half & Full Equations
I was presented with the following problem and I am finding it difficult to present the half equations for the answers.

I managed to produce the half-equation for the Iodine redox.
$\ce{2I- -> I2 + e-}$
I then proceeded to the $\ce{CrO4^{2-}}$ equation.
$\ce{CrO4^{2-} + 3e- -> Cr3+}$
This is because the oxidation number of the Cr in $\ce{CrO4^{2-}}$ is $+6$ and it is reduced to $+3$
However, the answer is given as:
$\ce{CrO4^{2-} + 4H2O + 3e- → Cr3+ + 8OH-}$
I am struggling to get from the equation I made to the one presented in the answer for the $\ce{CrO4^{2-}}$ redox.
Can someone explain why there are $\ce{4H2O}$ on the left hand side and $\ce{8OH-}$ on the left hand side of the equation?
Please help!
Many thanks.

The addition of OH- and water to balance basic solutions is a very common procedure for Redox reactions. https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Balancing_Redox_reactions


text:
Using acidic detergents
If detergents were acidic rather than basic, can we still achieve a clean surface? What would be more effective? Strong acid or a weak acid?
How would the result of cleaning be different if an acid of pH 4, for example, is used?
Is it possible to achieve the same result with an acid and a base, or even if they are both present in a detergent?

I'm leaving this question open. Not everything that contains more than one question mark is too broad, people.

Many commercial detergents are composed of pH-neutral surfactants like alkyl polyethylene oxides plus alkaline additives like sodium silicate or carbonate or borax, which help dissolve greasy materials better than acids.
There are some acidic detergents, like citric acid cleaners for stainless steel and other metals and toilet bowl cleaners. The acidic function is paramount, and the reduced surface tension is secondary, useful in providing better wetting.
It is instructive to consider what cleaning really means, and the individual steps that go on. First, water is the primary cleaning agent. It is the universal solvent, but not quite good enough or fast enough. So we add surface tension lowering agents - big improvement: oily materials are wetted and can be suspended better as the first step in removal. Alkaline additives then combine with polar surfaces to further displace oily materials, improving cleaning action. Acids don't help with oily materials much but can remove some contaminants better.
Dispersants for solids, like mud or clay, are typically anionic polymers, usually alkaline because the anion contains groups that will adsorb on polar surfaces and disperse the particles by electrostatic repulsion, assisting their removal.
There are classes of surfactants where the oleophilic group is anionic, like sulfonates or phosphonates, and these can be acidic if the cation is hydrogen or alkaline if an alkali metal, or nearly neutral if the cations are mixed. Another class of surfactants, called cationic, has the oleophilic group attached to nitrogen, and is typically somewhat acidic because the anion is the anion of a strong acid like HCl.
So you could optimize your cleaner by selecting a pH of water that cleans best; then add a specific surfactant to improve removal of your specific dirt - or the reverse. And consider interactions and whatever other additives might help.


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Work done by a gas in thermodynamics

A given mass of gas expands from state A to state B by three paths 1, 2, 3 as shown in the figure below.

If $W_1,$ $W_2$ and $W_3$ are respectively, be the work done by the gas along three paths, then
(A) $W_1 > W_2 > W_3$
(B) $W_1 < W_2 < W_3$
(C) $W_1 = W_2 = W_3$
(D) $W_1 < W_2 : W_1 < W_3$

The work done is $W = -P \,\mathrm dV$ that is the negative of the area under the graph; since Area 3 > Area 2 > Area 1, thus, $W_1 > W_2 > W_3.$
But the answer given is option (B). Why is this so?

The question is a little confusing. Clearly the third path does the most work because the area under the curve is the largest. But if you compare them using the sign convention, it would look the smallest. I would say to always compared works by their magnitude rather than their sign, which really just tells us about direction. As a comparison if you wanted to know who was going faster, someone moving at -100 or -1 m/s, you wouldn't say the one moving at 1 m/s was faster.

Can you say that since Work is a scalar we compare scalars by their magnitude? And by the way thank you for that beautiful analogy.

Yes, work is a scalar. From a math perspective this is because we can represent it as the dot product of two vectors (displacement and force). The sign is just for purposes of keeping track of whether the energy is moving to system or surroundings.

Sign conventions can make 'work' confusing, but the question here asks for the work done by the gas. When a gas expands, it's doing work (it's losing energy), so it's actually $W = + P \, \mathrm dV$, inverting all the inequality signs. So option B) is correct.

But isn't it the standard convention in chemistry that work done by a gas means the system would lose energy hence -PdV? In physics it is +PdV.

The question asks for the work done by the gas. It's the only explanation I can think of, but maybe some other genius can shed more light on it.

@TheLostGuardian0, work adopts a sign convention. work done by and work done on are unambiguous, and do not.

In the same exam they had asked a previous question where they had given a P-V graph and asked the work done by the gas and the answer was -(area under graph). This was why my doubt arised.

The work done by the gas on the surroundings is the integral of +PdV.  The work done by the surroundings on the gas is the integral of -PdV.  So B is correct.

Work done by the gas and on the gas are two different things.
In your question, the volume is increasing and the pressure is decreasing.
Let's consider some things before making an attempt to answer your question,
For simplification, consider the adiabatic process where $Q=0$.
If the external agent does positive work, the volume always decreases.
But in your question, as the volume is increasing, the external agent isn't the one doing the positive work, its actually the gas because only the gas can expand the system and still do positive work.
In the first law of thermodynamics,
$\mathrm dU=Q+W$
$\mathrm dU=Q-p\,\mathrm dV$,
The term $W$ represents the work done by the external agent.
If the external agent does positive work, volume decreases (i.e, $\mathrm dV$ is negative => $W$ is positive) and the internal energy increases and vice versa.
So, the work done by the gas becomes $-(-p\,\mathrm dV)$, because for a reversible process the work done by external agent and the gas are equal in magnitude and opposite in sign.
Now, as $\mathrm dV$ is positive, work done by the gas in also positive.
As you can see, magnitude wise $W_3$ is the greatest and now we also know that the net positive work is done by the gas because the final volume is greater.
This should answer your question.


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What does "gamma" in gamma-oryzanol mean?
The structure of gamma-oryzanol is:

What does the "gamma" in its name mean?

No designations such as alpha-oryzanol, beta-oryzanol, etc. appear to be in use, and according to this WebMD entry, as well as this PubChem document, another name for gamma-oryzanol is simply "oryzanol".
There seems to be some ambiguity as to just what oryzanol, or gamma-oryzanol refers to. According to Chemical Book:

Gamma Oryzanol is a mixture of 6 substances derived from rice bran oil, including sterols and ferulic acid, called Cycloartanyl Ferulate or Triterpene alcohol ferulate.

But then other sources, including this US National Institutes of Health publication, consider gamma-oryzanol composed of four components:

There are four main components of γ-oryzanols, namely, cycloartenyl ferulate, 24-methylene cycloartenyl ferulate, campesteryl ferulate, and sitosteryl ferulate [original publication:  Lerma-García M. J., Herrero-Martínez J. M., Simó-Alfonso E. F., Mendonça C. R. B., Ramis-Ramos G. Composition, industrial processing and applications of rice bran γ-oryzanol. Food Chemistry. 2009;115(2):389–404. doi: 10.1016/j.foodchem.2009.01.063

Further confusing the issue is that the PubChem article referenced above gives a single structure for gamma oryzanol (the same as that given in the OP), but does not include any physical properties that one would expect for a single pure compound (i.e. no melting point, solubility, vapor pressure data) although it does give the molecular weight. Again, PubChem lists "oryzanol" as a synonym and does not have any other oryzanol-named compound in it's database.
Summary TL/DR:
The gamma in "gamma-oryzanol" doesn't refer to a specific structure or isomer and there is no corresponding compounds named alpha-oryzanol, beta-oryzanol, etc. Rather, the term gamma-oryzanol is also simply called "oryzanol" and it refers to a combination of several different moieties as described above.

Is it possible that this name is related to the type or location of the ring?

That's sure the first thing you'd think, but I don't see where there's a significant functional group "gamma" to another. Also, you'd expect the term in that context to differentiate from another conformation like an alpha or beta position and there is just nothing in any chemical database I can find referring to anything but "gamma" or simply oryzanol. Maybe there is some historical significance to the naming, but I'm not sure what that would be. As far as modern nomenclature, I think of PubChem as being a pretty authoritative source.

No problem. If you found the answer useful, the way we say thanks here is by accepting the answer (with the check box next to the answer).  Cheers...

Some works directly mention the alpha and beta designated substances – that were (incl. gamma) rather fractions obtained during first historic isolations:[1]

(…) Oryzanol occurs in the unsaponifiable fraction of rice bran oil and is so named because it was first discovered in rice bran oil (Orysae Sativa L.) (Kaneko and Tsuchiya, 1954)[2] and contained a hydroxyl group. Oryzanol was originally considered a single compound but later was determined to be a mixture of ferulic acids esterified with normal sterols or triterpene alcohols, called α-, β- and γ-oryzanol, of which γ-oryzanol has been the most commonly mentioned. The sterol components of γ-oryzanol are primarily campesterol and sitosterol, and the triterpene alcohol components are cycloartenol and 24-methylene cycloartanol (…)

Other text sound like alpha a beta fractions were not oryzanol at all:[3]

(…) During their studies, a number of lipid fractions were isolated from rice bran oil; the third (gamma) fraction isolated was oryzanol[2]. Its chemical composition was later identified as a mixture of
  ferulic acid esters of various plant sterols (…)

(Hard to tell, the article[2] is in Japanese.)
As airhuff already said in his answer, γ-oryzanol (or just oryzanol) is a mixture of various sterol ferulates, which now usually are listed by their chemical names, however a couple of notable compendia[4,5] list the two major[6] formerly isolated constituents under names oryzanol A and C (no B, since it turned out to be a mixture of A and C[7]).

(I) oryzanol A, CAS 21238-33-5, cycloartenyl ferulate,
9,19-cyclo-5α-lanost-24-en-3β-yl (2E)-3-(4-hydroxy-3-methoxyphenyl)prop-2-enoate
(II) oryzanol C, CAS 469-36-3, 24-methylenecycloartanyl ferulate,
24-methylidene-9,19-cyclo-5α-lanostan-3β-yl (2E)-3-(4-hydroxy-3-methoxyphenyl)prop-2-enoate

Your structure happens to be oryzanol A.

References:

Huang, C. J. Potential Functionality and Digestibility of Oryzanol as Determined Using in Vitro Cell Culture Models, Dissertation thesis, Louisiana State University and Agricultural and Mechanical College, 2003.
Kaneko, R.; Tsuchiya, T. New Compound in Rice Bran and Germ Oils. J. Chem. Soc. Jpn. 1954, 57, 526–529.
Wheeler, K. B.; Garleb, K. A. Gamma Oryzanol—Plant Sterol Supplementation: Metabolic, Endocrine, and Physiologic Effects. International Journal of Sport Nutrition 1991, 1 (2), 170–177.
O’Neil, M. J.; Heckelman, P.; Koch, C.; Roman, K. The Merck Index, 14th ed.; 2006.
Hill, R. A.; Makin, H.; Kirk, D.; Murphy, G. Dictionary of Steroids; CRC Press, 1991.
Xu, Z.; Godber, J. S. Purification and Identification of Components of γ-Oryzanol in Rice Bran Oil. J. Agric. Food Chem. 1999, 47 (7), 2724–2728.
Shimizu, M.; Ohta, G. Studies on the Constituents of Rice Bran Oil. V. Reexamination of Oryzanol-B. Chem. Pharm. Bull. 1960, 8 (2), 108–111.


text:
Is chromium the shiniest metal?
I have seen many pictures of chromium, and it's pretty lustrous as I must say. Here are some examples. Is it the most lustrous metal known?
Edit: By lustre, I mean suppose if we have a mirror smoothly polished made out of a material, then I consider lustre as the percentage of reflection of light within the wavelength of visible light (400 - 700 nm). Then according to this scheme, can we consider chromium as the most reflective for visible spectrum?

All metals are lustrous when polished. It is just that many of them fade quickly.

Your mirror is in silver and this is pretty lustrous :)

Could you better define 'lustrous' in a technical fashion that can be measured? Otherwise, this is opinion based.

A better question would be "Why is chrome mostly used for electoplating?"

@Jon Custer, I have made an edit, defining my perception of lustre

With your definition, one has to go with untarnished silver as the most reflective across the visible. However, to keep it that way (in, say a spectrometer) requires purging with an inert gas such as nitrogen or argon.

Aluminum used for telescope mirrors.

You likely refer to remission, especially if you would collect spectroscopic data with an integration sphere at hand.  More generally, you refer to reflectance, which depends on the material, the selected wavelength, and the roughness of the top surface -- assuming the selected wavelength is not able to penetrate all across the reflective layer.

(source)
Metals and non-metals (e.g. polymer lamp housings) may be covered by a layer of chromium, and then may be polished to obtain a very smooth surface; in contrast to other metals such a surface remains "bright", even when exposed to air.
Beside the silver mirror already known from sugar chemistry, mercury / almagam based mirrors were quite common till about 1800.  The liquid nature of mercury renders this material still useful in the niche of liquid mirror telescopes.
Glancing briefly over the (n,k) values of metals provided here provides access to calculate the reflectance $R$ as the materials in question
$$ R = \frac{(n-1)^2 + k^2}{(n+1)^2 + k^2}$$
Matching the experience to see much more often silver top-coated mirrors in the optics lab, than of other materials, silver is of higher reflectance than chromium:

(The value at 10 µm equally provided by the referenc intentionally was omitted by mine.)
So why spokes are chromium plated, and not with silver? I speculate the much higher hardness of chromium (Mohs 8.5) than the one of silver (Mohs 2.5) contributes to the reasons why the former is used, and not the later.  It remains a smooth surface.  Maybe electrochemical reasons (chromium likely already is part of the steel / alloy the rims are made of; rather than silver) and the ease Ag would be tarnished by the environment (thinking about $\ce{Ag2S}$) are other reasons, beside obvious cost in material and processing.

Thank you for your wonderful information, but I'm not still clarified whether Chromium is the most lustrous or not. You have mentioned they are used as plating, but perhaps simply lustre wouldn't have been the reason, maybe cost-effectiveness as well?

@PrittBalagopal It is almost "second nature" to look for a silver coated mirror if one wants to use a material of high reflectance across a wide span in the range of visible and invisible light (e.g. the IR reflective coatings of Dewars).


text:
Is the ethyl cation really more stable than benzylic and allylic carbocations?

Source: Concepts of Organic chemistry by O.P. Tandon, page no. 235

My book appears to state that the ethyl cation (a primary carbocation) is more stable than both allyl and benzyl carbocations.
I knew that cation stability depends on the following factors: resonance > hyperconjugation > +inductive effect.
Can someone please clarify whether the ranking shown above is true?

This seems completely wrong and like a terrible textbook. 3 alpha-H AND +I effect? The alpha-Hs are the ones providing the +I effect so it's the same thing and there's no way that all three contribute significantly at the same time.

Clearly the order given in the book is wrong. A lone pair containing centre adjacent to electrophilic carbon atom gives the carbocation extraordinary stabilization.

I might be completely wrong here, but doesn't $\to\text{Stability}$ indicate that the stability increases from left to right? Without more context and explanation, I can see how this would be very confusing though.

I believe this to be a simple typo in the book. In the original image you'll see that in the surrounding schemes $<$ and $\to\text{Stability}$ are given. Only in this line the relations are reversed, while the stability indicator remains the same.
However, since we're at it, we can at least try to put some values to the task. I'll adopt the same scheme used in the comparison of the t-butyl cation and the benzyl cation, or the radicals. I calculated the isodesmic reactions of the form in $\eqref{isodesmic}$ at the DF-B97D3/def2-TZVPP level of theory.
$$\ce{ R+ + CH4 -> RH + H3C+ }\tag{1}\label{isodesmic}$$
According to this we'll find the following order in decreasing stability: t-butyl, benzyl, allyl, (ethyl). Note that the ethyl cation is a non-classical cation, basically a proton coordinating to the π-bond of ethene; see also: Which carbocation is more stable, the ethyl- or 1-propyl-carbocation?.
\begin{array}{llr}
  \ce{R+}  &  \ce{RH}  &  \Delta G / \pu{kJ mol-1}\\\hline
  \ce{H3C+}           &  \ce{CH4}         &      0.0  \\
  \ce{[H2C=CH2]H+}    &  \ce{H3C-CH3}     &   -197.9  \\
  \ce{H2C=CH-CH2+}    &  \ce{H3C-CH=CH2}  &   -259.3  \\
  \ce{H5C6-CH2+}      &  \ce{H2C-C6H5}    &   -343.0  \\
  \ce{(H3C)3C+}       &  \ce{HC(CH3)3}    &   -370.9 \\\hline
\end{array}
(Side note: I was unable to remove a small imaginary mode from the isopropyl carbocation. That has almost no influence though.)

(I won't attach geometries or absolute energies this time, because that would exceed the character limit.)

If I interpreted this right, tertiary carbocations are more stable than benzyl carbocations? And to think that all this time I was taught the contrary, +1!

@paracetamol You are doing an extrapolation from two points. Apart from this, the other question was already about that issue. (https://chemistry.stackexchange.com/q/74943/4945)

I hope I'm not being silly now, but should the sign of $\Delta G$ in the tables all be positive (or equivalently, should the direction of equation $(1)$ be reversed)? (Same goes for the radicals post.)

As per Jerry March in Advanced Organic Chemistry, the stability order of the cations is benzyl > allyl > t-butyl as expected due to resonance. The ethyl cation does not even exist because it is so unstable that when it is tried to be made, the t-butyl cation is obtained instead. Same with methane and propane.
A special case is the cation with cyclopropyl substituents wich is still more stable than the benzylic cation because the bended p orbitals stabilize the cation.


text:
Conversion of Kc into Kw
In deriving the expression for $K_\mathrm w$, we often start with the equilibrium equation of water
$$\ce{H2O <=> H+ + OH-}$$
The $ \rm K_c$ expression is quoted as $$K_\mathrm c = \frac{[\ce{H+}][\ce{OH-}]}{[\ce{H2O}]} $$ followed on by multiplying the $[\ce{H2O}]$ by $K_\mathrm c$.
The problem I face is I have been taught that the pure liquids aren't included in the expression of $K_\mathrm c$ since they have a concentration of 1. So why are we including in the above example?

It's really a faulty derivation. This page gives a good explanation why.

"The problem I face is I have been taught that the pure liquids aren't included in the expression of $\mathrm{K}_{c}$ since they have a concentration of 1."
Liquids have a constant concentration (really activity), so in general, it is not helpful to include them in the equilibrium expression. However, that does not mean that the value is 1. Pure water has a concentration of roughly $55.5\ \mathrm{M}$. You can absorb this concentration into the equilibrium constant, since it is a constant. Therefore,
$$\mathrm{K}_{w} = \ce{[H+][OH-]} = K_{\mathrm{c}}\ce{[H2O]}$$
EDIT:
Just for the record, note that the autoionization of water is generally written:
$$\ce{H2O + H2O <=> H3O+ + OH-}$$
since we don't have naked protons in water.
This changes the equilibrium constant slightly:
$$\mathrm{K} = \frac{\ce{[H3O+][OH-]}}{\ce{[H2O]}^{2}}$$
but $\mathrm{K}_{w}$ is still the same.


text:
What should be the correct sequence of reagents to be used in this reaction?

The correct sequence of reagents for the following conversion will be

(1) $\ce{CH3MgBr, H+/CH3OH, [Ag(NH3)2]+OH-}$
(2) $\ce{CH3MgBr, [Ag(NH3)2]+OH-, H+/CH3OH}$
(3) $\ce{[Ag(NH3)2]+OH-, CH3MgBr, H+/CH3OH}$
(4) $\ce{[Ag(NH3)2]+OH-, H+/CH3OH, CH3MgBr}$

I think the correct option is (3), however, the key says it is (4).
The first step is definitely the oxidation of the $\ce{-CHO}$ group to $\ce{-COOH}$ using Tollen's reagent. But I think that the second step should be to use Grignard reagent to form tertiary alcohol and then acidify it. Because if we first acidify and then use Grignard reagent, the $\ce{CH3MgBr}$ will react with the leftover acid and the $\ce{CH3OH}$ due to its mildly acidic proton.
Also, I am rather doubtful that the $\ce{CH3MgBr}$ will even react with the acid formed in the first step since it will be in the form of $\ce{R-COO^-Ag+}$, which is not electrophilic.
So what is the correct answer? And how would the $\ce{CH3MgBr}$ react with the acid?
Thank you.

In case you are wondering, this question appeared in the JEE Main exam on 2nd April 2017.

That is not an acidification, it is an esterification. In any case, the steps are separate, it does not mean you leave the MeOH there and then add the Grignard (which is a complete waste of time for the reasons you mentioned!). You will somehow extract the organic product, re-dissolve it in a different (aprotic) solvent, then only add the Grignard. That is what is implied by the separate steps (2) H+, MeOH (3) CH3MgBr.

@orthocresol Oh! Duh. You are right. You should add that as an answer! The ester will also react more effectively with CH3MgBr, so their answer is correct.

I'll try to answer if I get the time, but no guarantees

@orthocresol if we follow the steps mentioned in option 4, then we will obtain a compound like this https://i.stack.imgur.com/jNewl.jpg . Final compound will only be obtained after hydrolysis, so why is option 4 correct?


text:
Is Mg(OH)2 more insoluble than MgCO3?
In removing temporary hardness by boiling, $\ce{Mg(HCO3)}$ is converted to $\ce{Mg(OH)2}$ but not $\ce{Mg(CO3)}$. The answer that I found in our coursebook was that it has a higher solubility product, but if its solubility product is higher, then it should be more soluble than $\ce{Mg(CO3)}$. Can you tell me the correct reason?

I finally found the answer.
In temporary hardness, we have to remove $\ce{Mg^2+}$ ions by precipitating it. In $\ce{Mg(OH)2}$, $\ce{Mg^2+}$ ions in water are present in the concentration that is cube root of the solubility product, but in $\ce{Mg(CO3)}$ it is the square root of the solubility product. So, despite $\mathrm{K_{sp}}$ of $\ce{Mg(OH)2}$ being higher than $\ce{Mg(CO3)}$, $\ce{Mg^2+}$ ions in $\ce{Mg(OH)2}$ are comparatively less than $\ce{Mg^2+}$ ions in $\ce{Mg(CO3)}$.

Yeah but you are comparing 2 different solubility products so you cannot claim on the basis of that that MgCO3 has more solubility and Mg(OH)2 has less solubility.

The solubility data for $MgCO_3$ and $Mg(OH)_2$ are these: for $MgCO_3$ 0.01 g/100 mL cold H2O; for $Mg(OH)_2$ 0.0009 g/100 mL H2O @ 18C, 0.004 g/100 mL H2O @ 100C (CRC Handbook).
The published Ksp data are all over the place: for MgCO3, I have found $10^{-5}$, $6.8× 10^{-6}$, and $ 3.5 × 10^{-8} $(twice). For Mg(OH)2, in the same set of 4 documents, I have $ 1.8 ×10{-11}$ (twice), $1.6 × 10^{-12} $ and $5.6 × 10^{-12}$
Calculating Ksp from the solubility data gives: for Mg(OH)2 @ 18C, $1.5 × 10^{-11}$ and @ 100C, $1.31 × 10^{-9}$. For MgCO3, Ksp calculates to  $1.4 × 10^{-6}$.
From the solubility data, Mg(OH)2 is by far less soluble than MgCO3. From the Ksp data, any way you look at them, the solubility product of Mg(OH)2 is smaller than for MgCO3, indicating a lower solubility. So, no conflict.
There is a complexity in comparing a solubility product of MgCO3 (which multiplies 2 concentrations) vs Mg(OH)2, which multiplies 3 concentrations. Not to mention the fact that solubilities and Ksp are given for hydrates of MgCO3, which I did not deal with. And you might consider that it is not we who are removing $Mg^{+2}$ ions from the temporary hard water, but that heating removes CO2 from the water, leading to production of Mg(OH)2 in solution and then precipitation. The loss of CO2 minimizes the possible precipitation of MgCO3, especially since it is more soluble at high temperatures, where the transformation is occurring.


text:
Difference between Side Chains and Functional Groups
From what I've read, a functional group seems to be a specific case of a side chain, but this group must be directly responsible for some of the chemical characteristics of the molecule. However, in some cases, methyl groups (such as R-CH3) are listed as functional groups as well. For example, I have the following question:

The correct answer is listed as C - but isn't the methyl a site of reactivity as well?

So the example you present is of a poor question. It's a classic case of a "best answer" MCQ that requires you pick the answer that is most accurate, rather than an answer that is unambiguously right. (I hate these.) Alkanes are listed in most textbooks and guides to organic chemistry as a "functional group." The reality is that it's hard to do chemistry on an alkane in most circumstances, so it's probably the least "functional" of the functional groups.
That said a side chain is an imprecise term. It's most consistently used in amino acid chemistry to refer to the differentiating moiety. But it's really used to indicate any moiety that hangs off of a common structure. What constitutes the side chain versus the primary structure is a function of context. A large polymeric structure can have side chains that consist of aromatic rings, while with a smaller structure consisting of a single benzene ring with some alkanes hanging off of it, the alkanes will be considered the "side chain." There's no substantive difference between a side chain and a functional group except for emphasis in a discussion.
So in brief: Functional groups are the things you learned about like amines, amides, carboxylic acids, etc. Side chains are parts of a molecule that are considered to be "hanging off" the "main" part of the molecule, and are highly dependent on the context of the discussion. Sometimes these side chains are single functional groups, and sometimes they're much bigger structures. It really depends. But functional groups are always functional groups. Though in general, chemists tend to "forget" alkanes as functional. But as I indicated earlier, the question isn't all that great, so don't feel too bad if it confused you.


text:
Is coupled cluster variational for two electrons?
I know that coupled cluster (CC) is not variational for the general case. However, if we only have two electrons with one nucleus, CCSD should be exact for this system like full configuration interaction (CI). Since full CI is variational, can we argue that CCSD is also variational for this special case?

No. The reason for this is not to be found in the excitations, but in the evaluation of the method, i.e. the working equations.
$$%Introducing some shortcuts
\require{cancel}
\newcommand{\op}[1]{\mathrm{#1}}  %\op{H}
\newcommand{\bracket}[2]{\left\langle#1\middle|#2\right\rangle}
\newcommand{\bra}[1]{\left\langle#1\right|}
\newcommand{\ket}[1]{\left|#1\right\rangle}
\newcommand{\order}[1]{^{(#1)}}         %E_n\order{1}
\newcommand{\overlap}[3]{\mathcal{#1}_{#2}\order{#3}} %\overlap{S}{m}{1}
\newcommand{\integral}[3]{\mathcal{#1}_{#2,#3}} %\integral{V}{i}{j}
$$
The variational principle
In short: a trial wave function always has a higher energy than the exact wave function.
Assume we know the exact solutions to the Schrödinger equation, which form a complete basis and there is one wave function $\Psi_0$ which leads to the lowest energy $E_0$.
\begin{align}
\op{H}\Psi_i &= E_i\Psi_i &
\{i \in \mathbb{N}, i\geq 0\}\tag{1}\label{schrodinger}
\end{align}
We choose the solutions to be orthonormal (boundary conditions).
$$\bracket{\Psi_i}{\Psi_j} = \delta_{ij} =
\begin{cases}1, & i=j \\ 0, & i\neq j \end{cases}
\tag2\label{orthonorm}$$
The trial wave function $\Phi$ can be expressed as a linear combination of the complete set of wave functions.
$$\Phi = \sum_i^\infty a_i\Psi_i\tag3\label{expansion}$$
The expectation energy of the trial wave function is given through
$$W = \frac{\bra{\Phi}\op{H}\ket{\Phi}}{\bracket{\Phi}{\Phi}}.
\tag4\label{expen}$$
With $\eqref{expansion}$, $\eqref{orthonorm}$, and $\eqref{schrodinger}$ we can expand and simplify.
\begin{align}
W &=
  \frac{
    \sum_i^\infty\sum_j^\infty a_i a_j \bra{\Psi_i}\op{H}\ket{\Psi_j}
  }{
    \sum_i^\infty\sum_j^\infty a_i a_j \bracket{\Psi_i}{\Psi_j}
  }\\
  &=
  \frac{
    \sum_i^\infty\sum_j^\infty a_i a_j E_j\bracket{\Psi_i}{\Psi_j}
  }{
    \sum_i^\infty\sum_j^\infty a_i a_j \bracket{\Psi_i}{\Psi_j}
  }\\
  &=
  \frac{
    \sum_i^\infty a_i^2 E_i\bracket{\Psi_i}{\Psi_j} +
    \sum_{j\neq i}^\infty a_i a_j E_j\bracket{\Psi_i}{\Psi_j}
  }{
    \sum_i^\infty a_i^2 \bracket{\Psi_i}{\Psi_j} +
    \sum_{j\neq i}^\infty a_i a_j \bracket{\Psi_i}{\Psi_j}
  }\\
  &=
  \frac{
    \sum_i^\infty a_i^2 E_i
  }{
    \sum_i^\infty a_i^2
  } & | -E_0\\
W - E_0 &=
  \frac{
    \sum_i^\infty a_i^2 (E_i - E_0)
  }{
    \sum_i^\infty a_i^2
  } \\
W - E_0 &=
  \frac{
    \sum_{i>0}^\infty a_i^2 (E_i - E_0)
  }{
    \sum_{i>0}^\infty a_i^2
  } + \frac{a_0^2 (E_0 - E_0)}{a_0^2}\tag{4'}\label{notnegative}\\
W - E_0 &\geq 0\tag5\label{varprin}
\end{align}
In $\eqref{notnegative}$ the last term is zero. Since $E_i > E_0$ (by definition) and $a_i^2$ is always greater or equal zero, the expectation value of the trial wave function must always have a higher energy than the exact solution of the Schrödinger equation, i.e. $\eqref{varprin}$.
Configuration Interaction
The CI wave function is set up as a linear combination of determinants from a reference calculation. This reference calculation can in principle be any other method the only requirement is that it also forms a complete (orthonormal) basis.
When Hartree-Fock is chosen, we know that we obtain a complete set and that this method itself is variational. The CI trial wave function can then be expressed as a linear combination of solutions of the HF wave function.
$$\Psi^\mathrm{CI} = \sum_i^\infty a_i\Phi_i =
a_0\Phi^\mathrm{HF} + \sum_{i>0}^\infty a_i\Phi_i$$
An alternative expression of the above is using an "excitation operator" $\op{T}$ to generate the higher energy determinants from the reference.
$$\Psi^\mathrm{CI} = (1 + \op{T})\Phi^\mathrm{HF}$$
The energy of the CI trial wave function can be evaluated with the Lagrange multiplier method, which requires, that the wave function is normalised (adding zero).
$$L = \bra{\Psi^\mathrm{CI}}\op{H}\ket{\Psi^\mathrm{CI}}
     - \lambda\left[\bracket{\Psi^\mathrm{CI}}{\Psi^\mathrm{CI}}-1\right]$$
Since the boundary conditions did not change, the variational principle still holds.
Coupled Cluster
Because of the inherent complexity of the CC ansatz, one key element to evaluate to CC energy is missing: the requirement of the normalised CC trial wave function.
We generate the trial wave function with an exponential approach (rather than a linear) expanded into a Taylor series.
\begin{align}
\Psi^\mathrm{CC}
&= \mathrm{e}^{\op{T}}\Phi^\mathrm{HF}\\
&= \left[\op{1} + \op{T}_1 + \left(\op{T}_2 + \frac12\op{T}_1^2\right)
   + \left(\op{T}_3 + \op{T}_2\op{T}_1 + \frac16\op{T}_1^3\right)
   + \cdots \right] \Phi^\mathrm{HF}\\
&= \Phi^\mathrm{HF} + \left[\op{T}_1 + \left(\op{T}_2 + \frac12\op{T}_1^2\right)
   + \left(\op{T}_3 + \op{T}_2\op{T}_1 + \frac16\op{T}_1^3\right)
   + \cdots \right] \Phi^\mathrm{HF}\\
\end{align}
Or to shorten the journey
$$\Psi^\mathrm{CC} = \Phi^\mathrm{HF} + \sum_i^\infty t_i \Phi_i.$$
One can easily see what kind of mess it would become requiring the wave function to be normalised $\bracket{\Psi^\mathrm{CC}}{\Psi^\mathrm{CC}}=1$, so the Lagrange ansatz won't work. Instead we choose the wave function to be orthonormal to the reference, which in the Hartree-Fock (complete orthonormal basis) takes away a lot of work.
\begin{align}
\bracket{\Phi^\mathrm{HF}}{\Psi^\mathrm{CC}}&= 1&
\implies\quad \bracket{\Phi^\mathrm{HF}}{\Phi_i} &= 0
\end{align}
We can write the CC-Schrödinger equation as
$$\op{H}\mathrm{e}^{\op{T}}\Phi^\mathrm{HF} =
  E^\mathrm{CC}\mathrm{e}^{\op{T}}\Phi^\mathrm{HF}$$
and evaluate the energy as
\begin{align}
\bra{\Phi^\mathrm{HF}}\op{H}\mathrm{e}^{\op{T}}\ket{\Phi^\mathrm{HF}}
  &= E^\mathrm{CC}
     \bracket{\Phi^\mathrm{HF}}{\mathrm{e}^{\op{T}}\Phi^\mathrm{HF}}\\
E^\mathrm{CC}
  &= \bra{\Phi^\mathrm{HF}}\op{H}\mathrm{e}^{\op{T}}\ket{\Phi^\mathrm{HF}}\\
E^\mathrm{CC}
  &= E^\mathrm{HF} +
  \bra{\Phi^\mathrm{HF}}\op{H}
    \ket{\left[\op{T}_1 + \left(\op{T}_2 +
     \frac12\op{T}_1^2\right)
     + \left(\op{T}_3 + \op{T}_2\op{T}_1 + \frac16\op{T}_1^3\right)
     + \cdots \right]\Phi^\mathrm{HF}}.\\
\end{align}
From this we can see that the evaluation of the expectation value of the energy is significantly different from what is required for the variational principle $\eqref{expen}$.
Without going further into details, the dropping of the requirement of a normalised trial wave function leads to not obeying the variational principle.
Additional considerations
It is often said that Full-CC is equivalent to Full-CI, which is a convenient half truth. It is extremely important that this gets more true with a better description of the reference wave function, i.e. with larger basis sets. Full-CI is exact at the complete basis set limit (for the electronic Schrödinger equation). This is not necessarily true for Full-CC.
Since we are in computational chemistry hardly ever concerned about absolute energies, the non-variational nature of the method is by far outperformed by the "infinite" approach generating the "excited" determinants. Relative energies don't suffer significantly (compared with the size consistency and extensivity) under the non-variational approach.

When you say "the Lagrange ansatz won't work here", do you mean you can't use a Lagrangian ansatz for CC methods in general? (Because that would be kinda wrong)

@Fl.pf. Yes of course you can use Lagrange, as you can use that for any kind of minimum search with given boundary conditions. I might have to be more precise in my wording, but I need to think about that. I assume in variational CC you'd use the Lagrange evaluation in the analogous way to CI.

Yeah ;), $\mathcal{L} = E_0 + \langle HF| \hat{\phi}^T|HF\rangle + \sum_{\mu} \epsilon_{\mu} t_{\mu} \bar{t}{\mu} + \sum{\mu} \bar{t}_\mu \langle\mu | \hat{\phi}^T | HF \rangle$ . Nice answer by the way!

The comments on complete basis set are irrelevant and wrong.  Full CC and Full CI are equivalent at any basis set.  CCSD is equivalent to CISD and exact for two electrons.  You can observe this empirically or work out the math.

@JeffHammond I have written what I have written. I certainly can't and as a result won't do what you have suggested in your comment, but I'll read your answer providing this proof. I'm certainly willing to add a bounty it you need an incentive.

To be clear, I have verified this empirically.

@JeffHammond this probably makes for an adequate answer, too. The offer with the bounty still stands.

You want a mathematical proof that CCSD generates the same excitation manifold as CISD relative to the Hartree-Fock vacuum state?

My previous answer was downvoted (-3!), but I maintain that in the particular case where there are only two electrons, CCSD is equivalent to CISD and Full-CI.
If the discussion was about asking if for a one-electron system Coupled Cluster gives an upper bound to the energy, everyody would agree that it is true. I agree with all that was said in the top answers for the general case, but with two electrons things are a bit special since there are no triples and quadruples.
1) The CCSD equations can be rewritten in such a way to be solved with an intermediate Hamiltonian in the space of singly and doubly excited determinants.[1] In the special case where there are 2 electrons the space is the FCI space, and  the equations of FCI, CISD and CCSD are strictly equivalent : the dressing matrix $\Delta$ is zero so the effective Hamiltonian is equal to the Hamiltonian.
2) I don't agree with the argument that the fact that the energy is computed by projection implies that the energy is non-variational. What makes the result usually approximate is the fact that the wave function is not an eigenfunction of the Hamiltonian. But when it is, computing the energy by projection is exact. In the special case where we have 2 electrons, the CISD is the Full-CI. So if one projects the equations on the singles and doubles, the Full space is considered and the FCI wave function is obtained (up to a normalization factor). In the FCI, the wave function is an eigenfunction of the Hamiltonian, so computing the energy by projection gives exactly the FCI energy (in a less numerically stable manner in practice).
3) $\Psi_{\rm CCSD} =  \exp^{T_1+T_2} \Phi_{\rm HF} = (1 + (T_1 + T_2) + \frac{1}{2}(T_1+T_2)^2 + \dots) \Phi_{\rm HF}$
With only 2 electrons,
$\Psi_{\rm CCSD} = (1 + T_1 + T_2 + \frac{1}{2}T_1^2) \Phi_{\rm HF}$
and $\Psi_{\rm FCI} =  (1+T_1+T_2)$
Both $T_1^2\Phi_{\rm HF}$ and $T_2 \Phi_{\rm HF}$ span the same determinant space, so finding the $\Psi_{\rm CCSD}$ and the $\Psi_{\rm FCI}$ which minimize the energy will give the same solution in the basis of determinants.
All the additional constraints of Coupled Cluster are on the amplitudes of the Triples and Quadruples, which are absent here.

[1] Eigenvalue problem formulation of coupled-cluster expansions through intermediate Hamiltonians, I Nebot-Gil, J Sanchez-Marin, J.L Heully, J.P Malrieu, D Maynau, Chemical Physics Letters, Volume 234, Issues 1–3, 3 March 1995, Pages 45-49, https://doi.org/10.1016/0009-2614(95)00026-Z
Note: In the paper the energies they obtain are slightly different from standard CCSD, but this is due to an approximation on the $T_1\times T_2$ terms, which don't exist with only two electrons.

In the most general case, CC can be understood simply as a prescription for a trial wave function (ansatz) that uses the exponential excitation operators. This ansatz can be then optimized variationally, and this is variational CC.[1] It is much more expensive than the common CC that uses another approximation (besides the ansatz), which linearizes the CC amplitude equations in a certain way. In most cases, this is a good approximation.
"Variational" means that a method has some ansatz for the wave function which is optimized (varied) to get the minimal energy. This is then guaranteed to be greater or equal to the true energy by the variational principle. In this sense, the common CCSD is not variational even for two-electron systems: there is no energy that would be optimized in the procedure.
Perhaps you could understand "variational" as a method that is in some way guaranteed to always give an energy greater or equal to the true energy. In general, the non-variational CC is not such a method. But for two-electron systems, the CCSD energy is exact, so technically, it is variational in this other derived sense. But as it is exact, which is a stronger statement, I see little sense in stressing the "variational" aspect. Also, I know of no method that would be variational in this second sense, but not in the first sense.
Reference

Troy Van Voorhis and Martin Head-Gordon. J. Chem. Phys. 2000, 113, 8873.

"But for two-electron systems, CCSD is exact, ..." This is not true. It considers all possible "excitations", but the wave function is not normalised, hence it does not obey (all) the boundary conditions of a physically sensible wave function.

@Martin, correct, I have modified the sentence to reflect this. Practically though, this is not a problem because the wave function can be always normalized a posteriori.

There is a well-known formula for converting T amplitudes in CC to CI amplitudes.  I'd like to see @Martin-マーチン show a case where CCSD produces different CI amplitudes than a CISD.

Yes,
in the particular case of 2 electrons, CCSD is strictly equivalent to CISD since there can't be any excitation higher that singles and doubles. CISD is the Full-CI of a 2-electron system, so CCSD is variational for 2 electrons.

This answer is 100% correct, so I do not know why there's 3 (or more) downvotes.

Here is a partial bibliography of papers that refer to CCSD as exact for two electrons, meaning equivalent to FCI, or CISD, for the same basis set.

https://doi.org/10.1016/0009-2614(93)85124-7: "The QCISD and CCSD transition energies are size consistent, and exact for two-electron systems"
https://doi.org/10.1063/1.5128795: "CCSD is exact for 2-electron systems."
https://doi.org/10.1063/1.5094788: "An important feature of these methods is that they are exact for systems with a number of electrons smaller or equal to the highest excitation level (i.e., CCSD is exact for two-electron systems)."
https://doi.org/10.1021/acs.jctc.9b00674: "In contrast, it is clear that CCSD (despite being exact
for an ensemble of isolated two electron pairs)..."
https://doi.org/10.1016/0009-2614(93)E1398-Z: "This CPMET-C based excited state method is size-consistent, but unlike CCSD and QCISD, is not exact for separated pairs of electrons."
https://doi.org/10.3389/fmats.2019.00123: "For two-electron systems, 2-fold excited Slater determinants can be generated at most and therefore CCSD theory becomes exact. Higher orders of coupled cluster theory become exact for systems with the corresponding number of electrons."

It is left as an exercise to the reader if a method that is equivalent to FCI is variational.


text:
What is the need for dipole moment if electronegativity already exists?
Say we have an HCl molecule. The difference in electronegativities of the H and Cl atoms tells us how polar the HCl molecule is. The dipole moment is also a measure of the degree of polarity in molecules.
So why do we need the "dipole moment" concept if we can see how polar a molecule is by looking at the electronegativities of the atoms in the molecule? Is it because in complex molecules it is easier to measure how polar a molecule is, using the dipole moment formula?

Hello and welcome to Chemistry.SE. If you have any questions regarding how the site works, please feel free to take the short [tour] or visit the [help]. Best of luck with your question!

Well, electronegativity difference tells us like almost nothing.

Dipole moment is a measure. Electronegativity is but a vague hint.

As @Ivan Neretin writes, the dipole moment is the experimental measurement so comes first , so you have your ideas the wrong way round.

Electronegativities can only qualitatively predict the nature of a bond in a molecule. For quantitative requirements,the concept of dipole moment is required.


text:
What is the pH of a solution that results of a mixture of two others with known pH?
Solution A has a pH of 1, soultion B has a pH of 6. If I mix the same volume V of each solution into one, what will be the pH of the resulting solution?
I'm guessing here as no more information has been provided: It is a mixture of a strong acid with a weaker acid. So I'm assuming that the final concentration of protons will be related with the equilibrium displacement of the weaker acid's dissociation. The full dissociation of the stronger acid forces the equilibrium of the dissociation of the weaker to be displaced to the left, as in there will be no proton contribution from the weak acid. So the pH of the resulting solution will be computed from the concentration of protons from solution A being diluted in the new volume, 2V.
Am I in the right direction? I feel that calculating the raw amount of protons contributed by each solution and then computing the new concentration will be too obvious for a method.
Anything that would guide me into this pondering will be very much appreciated.

In the absence of any information as to what acids these are, the only real answer you can give is probably the "too obvious" one.


text:
HCl and chrome plated plastic
I was trying to use $\ce{HCl}$ (muriatic acid) to remove chrome plating from plastic chrome plated shower parts (shower head holder) in order to make chromium chloride.  I understand that in chrome plating, the order of plating is generally, copper, then nickel, then chrome.  I also know that chrome and nickel dissolve in $\ce{HCl}$, copper doesn't.  Or isn't supposed to very quickly at least.  I let the parts set for several days to a week. and the parts were white, so everything was removed.
The chrome came off, but much off it came off in little flakes.  Apparently the nickel is removed also. But also the copper.  I filtered the solution to remove the black residue and save the chrome flakes and put them in a beaker of fresh $\ce{HCl}$.
Doesn't seem to do much. Why?


text:
2-Bromo-1-chloro-4-nitrobenzene reacts with sodium ethoxide
What are the major products?
I assumed that their is only one that is possible, which would be replacing the chloro group because the resonance structure supports that one. The bromo group doesn't seem to be able to be replaced. Would adding the ethoxide group, allow for the bromo group to replaced?

The reaction over here is a $\text{S}_{\text{N}}(\text{Ar})$ reaction. Normally, nucleophilic substitutions do not occur, but in presence of strong nucleophiles and electron withdrawing groups, it is possible.
Now, coming to your reaction, let's see what really happens:
Consider the resonation structures of 4-chloro-3-bromonitrobenzene:

You can see that the resonance structures produce a carbocation at the chlorine attached carbon, but not at the bromine attached one. Nucleophiles always tend to attack electrophilic centers, so it is the chlorine that will get replaced.
Continuing with the reaction, we will get:

So, the product would be 3-bromo-4-methoxynitrobenzene.

Formal charges on the nitro group are missing.

@orthocresol I am aware of that, I just made it simpler.

@PrittBalagopal If it's incorrect, it's not simpler.

@jerepierre I have added formal charges :)

@orthocresol I have added formal charges :)

This reaction is difficult to occur because it is a nucleophilic substitution to a benzene ring but in the presence of a strong electron withdrawing group this reaction moves ahead.
The point here to be noted is that -OEt is a strong nucleophile and attacks that carbon position where there is higher positive charge and thereby replacing the halogen out there.
And thereby we obtain the product....
Hope this was helpful


text:
Equation of enthalpy
This question arises because: by giving classes in thermodynamics, I have observed that students are often confused between the different definitions (or applications) of the enthalpy concept.
The enthalpy expression is obtained as follows:
From the first law of thermodynamics:
\begin{align*}
 U=Q+W
\end{align*}
Recalling the definition of work:
\begin{align*}
    W=-PV
\end{align*}
Variations added:
\begin{align*}
 \delta U=Q - \delta (PV)
\end{align*}
Its obtained that:
\begin{align*}
 \delta U&=Q-(V\delta P+P\delta V)
\end{align*}
Constant volumen:
\begin{align*}
 \delta U=Q-V\delta P
\end{align*}
Constant pressure:
\begin{align*}
 \delta U=Q-P\delta V
\end{align*}
Variations enlarged:
\begin{align*}
 (U_f-U_i)&=Q-P(V_f-V_i)\\
 (U_f-U_i)&=Q-(PV_f-PV_i)\\
         Q&=(U_f-U_i)+(PV_f-PV_i)\\
         Q&=(U_f+PV_f)-(U_i+PV_i)\\
         Q&=H_f-H_i
\end{align*}
Therefore, the definition of enthalpy is:
$$
    H = U - W
$$
Is this properly proposed?

https://en.m.wikipedia.org/wiki/Thermodynamic_square I had not seen this until just recently, but I think it can be useful.

Did you say that you teach thermodynamics to students?

Hi @ChesterMiller, yes, introductory, I do it from the kinetic molecular theory, the aim is to have that mentioned list to all the students with internet access.

It seems to me that some of the equations are incorrect.

@ChesterMiller please, feel free to edit the post

I don't know where to begin.  It's almost all incorrect.  Also, I don't want to be the bad guy.

@ChesterMiller, it is ok, the only thing that I think is not to helpful is to define enthalpy from legendre transformations or its equivalent: the maxwell relations. I prefer to develop the concept from the definitions of pressure and temperature from the kinetic molecular theory, then gas laws, the definitions of work and heat and finally the definition of enthalpy, which is shown in the post and your help would helpful to all people interested on this concept and its applications

and it would be great, because I see how I am doing wrong with my students, so, please @ChesterMiller , go ahead

Definition of internal energy:  U is a function of state, representing the total kinetic and potential energy of the molecules.  U = U(T,V)
First Law of Thermodynamics:$$\Delta U=\delta Q+\delta W$$where the symbol $\Delta$ is used to represent the change in a (path-independent) function of state (like U) between an initial and final thermodynamic equilibrium state of a closed system and the symbol $\delta$ is used to represent the change in a parameter that depends on the process path between and initial and final thermodynamic equilibrium state of a closed system. $\delta Q$ is the heat added to the system over the path, and $\delta W$ is the work done on the system over the path.
Relationship for the Work:$$\delta W=-\int{P_{ext}dV}$$where, for both for reversible and irreversible process paths, $P_{ext}$ is the force per unit area exerted by the gas on the piston face, and, by Newton's 3rd law, the force exerted by the piston face on the gas.  For an irreversible process path, the pressure typically varies with spatial location within the cylinder, so that the average gas pressure does not match $P_{ext}$ at the piston face.  For a reversible process, the gas pressure is uniform within the cylinder, so $P_{ext}=P$ where P is the gas pressure calculated from the equation of state of the gas (such as the ideal gas law), based on the number of moles in the cylinder, the gas pressure in the cylinder, and the gas temperature in the cylinder.
Combining the First Law with the Relationship for Work:$$\Delta U=\delta Q-\int{P_{ext}dV}\tag{rev and irrev processes}$$
$$\Delta U=\delta Q-\int{PdV}\tag{rev processes}$$
Constant Volume ($V_i=V_f)$:$$\delta W=0$$
$$\Delta U=\delta Q$$
Constant Pressure ($P_{ext}=P_i=P_f=P$):$$\delta W=-P_{ext}\Delta V=-P\Delta V$$
$$\Delta U=Q-P\Delta V$$
So,$$\Delta U+P\Delta V=\delta Q$$But, since P is constant,
$$\Delta U+\Delta (PV)=\delta Q$$
The definition of enthalpy is $$H\equiv U+PV$$
Therefore, for a constant pressure process (a process in which $P_{ext}$ over the entire process path is equal to the equilibrium pressures in both the initial and final equilibrium states of the system) $$\Delta H=\delta Q$$

Thank you so much Chester, now I see clearer how to introduce the enthalpy  concept :)

I believe that motivating the definition of enthalpy via the Legendre transform is indeed useful, and would like to attempt to do that here.
We start with the differential form of the first law, $\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V$. We see that $S$ and $V$ are the natural variables for $U$, in that the differential $\mathrm{d}U$ takes a simple "natural" form. In addition, if we're considering constant-volume processes, then $\mathrm{d}V = 0$, and $\mathrm{d}U = T\,\mathrm{d}S = \delta q_\text{rev}$, which is even simpler to work with.
Many processes, however, occur at constant pressure rather than at constant volume, and we'd like to work with a thermodynamic variable that undergoes the same simplification as does $U$ at constant $V$. This can be done by the Legendre transform*: $$\mathrm{d}H \equiv \mathrm{d}(U+PV) = \mathrm{d}U + \mathrm{d}(PV) = T\,\mathrm{d}S - p\,\mathrm{d}V + \mathrm{d}(PV) = T\,\mathrm{d}S + V\,\mathrm{d}P.$$ Adding the differential $\mathrm{d}(PV)$ on either side of the first law leads to a new thermodynamic variable $H$ that is a natural function of $S$ and $P$, instead of $S$ and $V$. We call $H$ the enthalpy. It easily  follows that, for a constant-pressure process, $\mathrm{d}H = T\,\mathrm{d}S = \delta q_\text{rev}$.

*One must be careful about adding arbitrary differential elements and calling it a Legendre transform. To be precise, the reason that adding the differential $\mathrm{d}(PV)$ works is because $P(V)$ is monotonic; equivalently, the inverse function $V(P)$ exists; equivalently, a value of $P$ defines a single value of $V$ and vice versa. The Legendre transform works only between variables satisfying such a criteria, for otherwise information is lost during the transformation.
Incidentally, that $P(V)$ is monotonic comes from a variational treatment of the second law. The definition of enthalpy thus relies on the second law, even though it is commonly introduced after solely the first law.


text:
Is the amount of substance of acid and base always equal at the equivalence point?
Does the amount of substance of acid always have to equal the amount of substance of base at the equivalence point? I know it "works" for monoprotic acids and monobasic bases, but what if we had a diprotic acid and monobasic base?

It sometimes depends on the indicator you are using for the titration. Phenolphthalein gives its equivalence point at a pH of 10-12, which is beyond the pH of 7.

Please use the correct term amount of substance for number of moles, which would be saying number of kilograms when referring to the mass.

We speak in this case of multiple equivalence points.
As a concrete example, if we are titrating $\ce{H2SO4}$ with $\ce{NaOH}$, then the first equivalence point occurs when all the $\ce{H2SO4}$ has reacted to form $\ce{HSO4-}$, and the second equivalence point when all the $\ce{HSO4-}$ has further reacted to form $\ce{SO4^2-}$. If you started with 1 mole of acid in solution, then adding 1 mole of base gets you to the first equivalence point, and another mole (2 moles total) gets you to the second equivalence point. In the same vein, one might consider triprotic acids, which have three equivalence points, and so on.
Incidentally, an interesting case to think about might be that of a titration between a diprotic acid and a dibasic base.

So what you're saying is that they require the same number of moles to reach the 1st, then 2nd, etc. equivalence points (if it applies) but in total it would require more moles (two times more) of NaOH to titrate H2SO4

@ChemStudent, yes, that's correct.

Oh ok that makes a lot more sense now. Thank you for responding! Have an awesome day!!!


text:
How does vapour pressure affect cooking?
Why should we cover a frying pan with plate while cooking or use pressure cooker? Will it save cooking gas?

Cover and pressure cooker are used for totally different reasons. Also, I recall there is a whole SE site for cooking.

afaik a pressure cooker is a very useful tool to mess up your stove extremely fast

Water in the food your trying to cook has a certain boiling point. This boiling point is elevated with increase in pressure (since under high pressure, vapors tend to condense back). To cook the food properly, you'd want the water to stay as liquid as much as possible, even under high temperatures, due to the fact that liquid water has a high specific heat capacity, storing lots of heat energy for a given temperature and can transmit heat to food better. Under pressure, boiling point of water is increased, so that it still remains liquid. This helps cook the food faster than without pressure.

I think that you are talking about pressurized pots.
If this is the case I think looking at the ideal gas equation will give us a hint on what happens. $$pV = nRT$$ Where $p$ is the pressure of the gas, $V$ is the volume of the gas, $n$ is the number of moles of the gas, and $R$ is the gas constant.
In the pressurized pot the volume is assumed to be constant, and this is a valid assumption as the amount of water that is vaporized is small and could be neglected in order to get a more qualitative result, if however we want a more detailed result, then we wouldn't neglect any thing and we would use a real gas equation.
As we increase the temperature, the number of moles of the vapor increases and they both increase the pressure of the water vapor. The increased pressure is the key for the result that we will get shortly.
If we looked at the phase diagram (specifically at the liquid phase) we see that as we see that as we increase the pressure the boiling point of water (or liquid) increases and this allows to get to higher temperatures that is not achievable within normal pressure.

In short, heating pressure pot increases the pressure above the water which increases the boiling point of water and allows reaching of higher temperatures which speeds up cooking.


text:
Making artificial sweat
I'm currently working on a chemistry investigation wherein I need to dilute a compound with artificial sweat. I have looked at various forums, including this, which gave answers, but is very outdated.
I have also checked on ISOs standards, but it seems that I cannot get the information I want - ISO 105-E04 - without paying a huge sum.
Do any of you know of a recipe for artificial sweat that I can easily conjure in a lab, that's also backed up by various sources?
Thanks
P.S. I cannot just use water + salt, as that would be too simple and cause too many uncertainties for my experiment.

Sweat is known to contain water, urea, and certain salts like NaCl. Try visiting Wikipedia page about sweat and get the exact composition.

ISO 105-E04 describes two solutions, an alkaline solution buffered to pH 8 and an acid solution buffered to pH 5.5. In addition to the phosphate buffer, both solutions contain sodium chloride and L-histidine. Is that what you need?

@Loong Yes. Unfortunately, it doesn't give the exact proportions of chemicals required to be mixed together unless you pay. In the link I put in my post, someone gave the composition of sweat from another ISO standard, but sadly it was from 2004.

@PrittBalagopal Wikipedia offers the composition, but it doesn't state the concentrations of lactic acid or urea. However, it does go in a lot of depth for the less noticeable components, like the different metals.

The international standard ISO 105-E01:2013 Textiles – Tests for colour fastness – Part E01: Colour fastness to water includes specifications for two different solutions:
Alkaline solution, freshly prepared, using grade 3 water complying with ISO 3696, containing, per litre:

0.5 g of ʟ-histidine monohydrochloride monohydrate ($\ce{C6H9O2N3.HCl.H2O}$);
5 g of sodium chloride ($\ce{NaCl}$);
and either
5 g of disodium hydrogen orthophosphate dodecahydrate ($\ce{Na2HPO4.12H2O}$)
or
2.5 g of disodium hydrogen orthophosphate dihydrate ($\ce{Na2HPO4.2H2O}$).

The solution is brought to pH 8 (± 0.2) with 0.1 mol/l sodium hydroxide solution.
Acid solution, freshly prepared, using grade 3 water complying with ISO 3696, containing, per litre:

0.5 g of ʟ-histidine monohydrochloride monohydrate ($\ce{C6H9O2N3.HCl.H2O}$);
5 g of sodium chloride ($\ce{NaCl}$);
2.2 g of sodium dihydrogen orthophosphate dihydrate ($\ce{NaH2PO4.2H2O}$).

The solution is brought to pH 5.5 (± 0.2) with 0.1 mol/l sodium hydroxide solution.


text:
Finding out the value of rate constant on titration
The question goes as follows
Consider the reaction $$\ce{X(aq) \longrightarrow Y(aq) + Z(aq)}$$
The above reaction is titrated with a reagent $A$ and all of $X, Y, Z$ reacts with $A$ with the ratio of 'n' factors of $X, Y, Z=1:2:3$
here n factor refers to change in oxidation number of substance per mole of the substance on titration with $A$
The following data was obtained.

The question is to find out the value of rate constant considering the titration reaction to be first order.
The value of rate constant is given by $\frac{0. 693}{t_{1/2}}$ So I have to find out half life of the reaction . I am facing trouble finding out since the rate of voume of A required depends on the amount of substance present at that instant. Any help shall be highly appreciated. Thanks.

From a mole balance, we find $C = n_x(t) + n_y(t) + n_z(t) = n_x(t) + 2n_y(t)$, where the second equality follows from the stoichiometry of the reaction. In particular, our boundary conditions are $C = n_x(0) = 2n_y(\infty).$
We assume that the reaction is first-order, so $n_x(t) = n_x(0)\exp(-kt) = C\exp(-kt).$ The two measurements give us the set of equations
\begin{align*}
80D &= n_x(5) + 5n_y(5) &&= C\exp(-5k) + \frac{5}{2}C(1-\exp(-5k)) \\
100D &= 5n_y(\infty) &&= \frac{5}{2}C,
\end{align*}
where $D$ is an undetermined proportionality constant for the titration, and the coefficient of 5 in $n_y$ accounts for the fact that 2 equivalents of $A$ react with $n_y$ and 3 equivalents with $n_z$. Dividing one equation by the other produces the final result $$\frac{4}{5} = \frac{2}{5}\exp(-5k)+1-\exp(-5k),$$ which can be solved for $k$ and hence $t_{1/2}$.

This is quite a tricky problem, and I'm not sure I have it all. The amounts of $X,Y, Z$ at time $t$ are $X=X_0\exp(-kt), ~Y=X_0(1-\exp(-kt)), ~ Z=X_0(1-\exp(-kt))$ and the measurement is  volume A. At infinite time $X_\infty=0$ and so using the ratios given $2Y_\infty+3Z_\infty=5\cdot 100$ is equivalent to $2X_0$ (by adding up $X,Y,Z$ at $t=\infty$) and at $t=5, $ $X_5+2Y_5+3Z_5=6\cdot80$ is equivalent to $X_0(2-\exp(-5k))$ so you should have all that you need if you now take ratios as $(500/480)=2X_0/(X_0(2-\exp(-5k)))$

Thanks for your answer. But I didn't get how you equated the number of equivalents of $Y$ and $Z$ with volume of reagent used. Can you please explain.

I'll try, I assumed that the ratios 1:2:3 are of volume. Then at infinity 100 ml is made of  5 parts of which 2 are Y and 3 Z , and at 5 sec there are 6 parts, 1 X, 2Y and 3Z.

no the ratios are n factors.for example the n factor of $\ce{K2Cr2O7}$ is $6$ according to the reaction $$\ce{Cr2O7^{2-} + 14H+ +  6e- => 2Cr^{3+} + 7H2O}$$ because 6 $e-$ are involved.

I assumed it was but then I don't see how you can answer the question as you don't know how much A is used for each of X, Y and Z . Thus I had to assume that if n=3 for Z then 3 times as much A is used than for X.


text:
Is converting SMARTS to SMILES a "lossless" operation?
The following three different SMARTS strings represent the same structure - at least when depicting it (e.g. at Uni Hamburg Smarts viewer).
SMARTS:
[#6]-1=[#6]-[#6](-[#6]-[#6](-[#6]-1)-[#6])=[#8]
[#6]-1-[#6]=[#6]-[#6](-[#6]-[#6]-1-[#6])=[#8]
[#6]-1-[#6](-[#6]=[#6]-[#6]-[#6]-1-[#6])=[#8]

Also, the chemical structure represented in SMILES leads to only one string, at least only one I can think of.
SMILES:
CC1CC=CC(=O)C1

Now, I do know the differences between SMARTS and SMILES, and I do know what the symbols in SMARTS mean. Still, I tried several softwares to find any differences in the use between SMARTS and SMILES representation for that specific case, and couldn't find any.
So I wonder if I could use SMILES in that case, or if there's any danger of missing information, for example when searching for substructures?

SMARTS is deliberately designed to be a superset of SMILES. That is, any valid SMILES depiction should also be a valid SMARTS query, one that will retrieve the very structure that the SMILES string depicts.
However, as a query language, SMARTS can be more general than SMILES is. For example, CC as a SMILES string depicts a single compound: ethane. As a SMARTS query, though, CC will match ethane, but will also match propane, acetic acid, cyclohexane, vancomycin, etc.
There's also SMARTS strings which are not valid SMILES strings. You list several in your question: [#6]-1=[#6]-[#6](-[#6]-[#6](-[#6]-1)-[#6])=[#8] is not a valid SMILES representation, unless your SMILES parser is being particularly generous. Even if it is, [#6]-1=[#6]-[#6](-[#6]-[#6](-[#6]-1)-[#6])=[#8,#7] is also a valid SMARTS query which should match your molecule, but not even a generous SMILES parser is likely to accept it.
Despite being similar, SMARTS and SMILES are intended for fundamentally different things - the purpose of SMILES is to represent particular compounds, whereas SMARTS represents a query against a range of possible molecules, or an abstract description of a set of possible molecules. As such, they're not inter-convertible, even if the strings are literally identical.
For your particular case, yes, CC1CC=CC(=O)C1 is both a valid SMILES and a valid SMARTS, but as a SMARTS query, it represents not just 5-methyl-2-cyclohexen-1-one, but also 5-propyl-2-cyclohexen-1-one and 3-hydroxy-5-butyl-6-amino-2-cyclohexen-1-one, as well as many others, all of which contain that substructure. The SMARTS viewer you link doesn't depict this explicitly, because it's implicit in the use of SMARTS that it's a substructure pattern for a broader class of compounds.

Thilo asked a similar question on the rdkit-discuss mailing list, where Andrew Dalke chimed in with this response, which he gave me permission to post here.  The answer uses the python-based rdkit library to give examples of converting between SMILES and SMARTS and other tasks.
From Andrew Dalke:
On Apr 19, 2017, at 12:03, Thilo Bauer  wrote:

Is converting SMARTS to SMILES a "lossless" operation, or does one loose
  information on doing so?

It is obviously not lossless if you include terms that cannot be represented in SMILES.
>>> from rdkit import Chem
>>> Chem.MolToSmiles(Chem.MolFromSmarts("[C,N]"))
'C'

or which don't make sense as a molecule:
>>> Chem.MolToSmiles(Chem.MolFromSmarts("c"))
'c'
>>> Chem.MolFromSmiles("c")
[23:02:24] non-ring atom 0 marked aromatic

It also loses some information which could be represented in SMILES:
>>> Chem.MolToSmiles(Chem.MolFromSmarts("[NH4+]"))
'N'
>>> Chem.MolToSmiles(Chem.MolFromSmarts("C[N+]1(C)CCCCC1"))
'CN1(C)CCCCC1'
>>> Chem.MolToSmiles(Chem.MolFromSmarts("[12C]"), isomericSmiles=True)
'C'

Do be careful if you want to handle aromatic atoms and bonds:
>>> Chem.MolToSmiles(Chem.MolFromSmarts("[#6]:1:[#6]:[#6]:[#6]:[#6]:[#6]:1"))
'C1:C:C:C:C:C:1'
>>> Chem.MolToSmiles(Chem.MolFromSmarts("c=1-c=c-c=c-c=1"))
'c1=c-c=c-c=c-1'

Background:
  I've got three different SMARTS strings representing the same structure
  - at least when depicting it. Also all three strings result in the exact
  same SMILES (see code and output below).

It looks like you want SMARTS canonicalization.
In general this is hard, because SMARTS can include boolean expressions and recursive SMARTS.
If you limit yourself to patterns like '[#6]-1=[#6]-[#6]...', with only atomic numbers and single/double/triple bonds, then I think RDKit will do what you want.
[[CF: Andrew Dalke also had important answer-level commentary on R.M.'s answer, which I copy below.]]

From chemistry stack exchange, an answer contributed by user R.M.:
SMARTS is deliberately designed to be a superset of SMILES. That is, any valid SMILES depiction should also be a valid SMARTS query, one that will retrieve the very structure that the SMILES string depicts.

Except, that last clause isn't true. Try matching tritium against itself.
>>> from rdkit import Chem
>>> mol = Chem.MolFromSmiles("[3H]")
>>> pat = Chem.MolFromSmarts("[3H]")
>>> mol.HasSubstructMatch(pat)
False

For hydrogens you must use '#1', because H in SMARTS means something different.
>>> pat2 = Chem.MolFromSmarts("[3#1]")
>>> mol.HasSubstructMatch(pat2)
True

SMILES input under Daylight and most other toolkits gets normalized to the chemistry model, including aromaticity perception:
>>> mol = Chem.MolFromSmiles("C1=CC=CC=C1")
>>> pat = Chem.MolFromSmarts("C1=CC=CC=C1")
>>> mol.HasSubstructMatch(pat)
False
>>> pat2 = Chem.MolFromSmarts("c1ccccc1")
>>> mol.HasSubstructMatch(pat2)
True

RDKit also does a small amount of additional normalization, or 'sanitization' to use the RDKit term. For example, it will convert "neutral 5 coordinate Ns with double bonds to Os to the zwitterionic form" (see GraphMol/MolOps.cpp):
>>> s = "CN(=O)=O"
>>> mol = Chem.MolFromSmiles(s)
>>> pat = Chem.MolFromSmarts(s)
>>> mol.HasSubstructMatch(pat)
False
>>> Chem.MolToSmiles(mol)
'C[N+](=O)[O-]'

I believe that the output SMILES from a toolkit, assuming that the SMILES doesn't have an explicit hydrogen, can be used a SMARTS which will match the molecule made from that same SMILES, by that same toolkit.

Thank you very much for the detailed answer and for posting the mailing list result here! The examples provided here clarify a lot.

In this case, the SMARTS and SMILES represent the same (sub)-structure, in particular because the SMARTS pattern explicitly defines the single atom type and connectivity for all atoms. So to answer your question, yes, this case is a lossless conversion.
In contrast, if the O could be an O or an N, the SMARTS would not catch that. And this both your SMILES and SMARTS would miss N=C1C=CCC(C)C1. However, you can write a SMARTS to match both N=C1C=CCC(C)C1 and O=C1C=CCC(C)C1, but a SMILES will not.
Also, note that the SMILES could be written in many (equivalent) ways, e.g., O=C1C=CCC(C)C1