What source of error in an acid-base titration could lead to a higher reading of the acid concentration? Methanoic acid ($\ce{HCOOH}$) is found in stinging nettles. In an analysis of the concentration of methanoic acid in a solution extracted from nettles, the solution was titrated with a standard solution of potassium hydroxide. Aliquots of the methanoic acid were transferred to a conical flask and titrated against the potassium hydroxide solution in a burette.
Which one of the following would cause the calculated concentration of methanoic acid to be higher than it actually is?
A Rinsing the conical flask with deionised water
B Rinsing the burette with deionised water
C Rinsing the pipette with deionised water
D Rinsing the conical flask with potassium hydroxide solution

For this question, im not sure if A and B are correct answers. The book only has B.
For A: if you rinse the conical flask, would this mean that there would be a greater volume in the flask after the titration than expected? Therefore when you have a lower concentration?
For B: if you rinse the burette with deionised water, this would dilute the $\ce{KOH}$, therefore causing a greater titre required. This means a greater mole of $\ce{KOH}$ and hence acid. Would this then lead to a greater concentration?
So B is the only correct answer, is my reasoning correct?

A Excess water in the conical flask will have no influence whatsoever* (in first order approximation). Since you are determining the the amount of substance of methanoic acid with the titration, the volume is irrelevant. The volume you use to determine the concentration is from when you take the aliquot with the pipette.
This option is therefore incorrect.

B You are correct, the excess water left in the burette will dilute you potassium hydroxide solution, causing you to use more of it, and therefore also reading a higher amount of substance of methanoic acid than would be correct. Obviously this leads to a higher concentration.
This is a correct option.

C Rinsing the pipette with water will cause you analysis solution to be diluted. Therefore the calculated concentration would be lower than the actual concentration.
This option is therefore incorrect.

D Rinsing the conical flask with potassium hydroxide solution will lead to having some of the methanioc acid already neutralised. You will use less of your analysis solution, therefore calculating a lower concentration.
This option is also incorrect.

In order to achieve the highest accuracy, you should

rinse the burette with the potassium hydroxide solution you use to analyse.
rinse the conical flask with plenty of deionised water (to remove any residues).
rinse the pipette with the solution which is to be analysed.

* If it is too much water, then the volume significantly changes. This will cause the indicator substance to react ever slightly differently, which might lead to you not seeing the equivalence point early enough. This could lead to reading a higher concentration.
However, you eyes are a larger source of error if it is a few drops of water.


Is activation energy temperature-independent? I know that activation energy for a reaction is the extra energy given to the reactants to reach the threshold energy so that they can collide and react. But then, why is it said that the activation energy does not depend on the temperature?

If the temperature is higher, the reactant molecules have a higher kinetic energy, and thus the energy difference between the threshold energy and the energy of the molecules will decrease, that is, the activation energy will decrease.

There is a simple (some might say simplistic) way to get an intuition about this and it involves thinking at the molecular or atomic level rather than about the bulk properties of the reaction (the thermodynamic view).

For a simple reaction where molecule A has to bang into molecule B to create molecule C, the reaction will only happen if the amount of energy involved in the collision is large enough. The amount of energy involved depends on the nature of the reaction which is determined by the electronic structures of A and B. Within limits, those electronic structures don't vary with temperature and a given reaction will only occur when there is enough energy in the collision for the reaction to happen.

What changes with temperature isn't the energy required for the reaction to occur, it is the number of molecules that have enough energy to overcome the barrier. At higher temperatures more molecules have that energy, but the amount of energy required doesn't change. The amount of energy require to initiate the reaction is roughly equivalent to the activation energy. It can be calculated by looking at the way the overall rate of the reaction changes with temperature. So the rate of reaction changes with temperature but not the barrier to the reaction happening.

This view is a little simplified but it gives the right intuition about the problem.


Will room-temperature water in a sealed flask evaporate? Some liquid water is present inside a sealed flask at room temperature. Is water evaporating, and is the system at equilibrium? First of all, can water evaporate in a sealed flask at room temperature? If so, is the reaction happening a reversible reaction? Any help would be appreciated. Thanks.

Even at room temperature, some of the water molecules in the liquid will have sufficient energy to 'escape' and become vapour. Because the flask is sealed however, some of the water vapour molecules will return to the liquid phase.

Eventually, an equilibrium will be reached when the rate at which water molecules which escape the liquid is equal to the rate at which water molecules in the vapour enter the liquid. At this point, the pressure reached by the water vapour is known as the vapour pressure of water (at the given temperature).

The 'reaction' is not completely reversible since some heat is lost to the surrounding environment through the wall of the flask.


Why is the chemical formula for vinegar CH3COOH what it is? I am quite new to chemistry, but I would like to think that I have mastered the basics. This would include naming and writing out chemical formulas. However, once I had seen the formula for vinegar ($\ce{CH3COOH}$), I was bewildered.

It would have made sense to me that the formula would be more, say, "compacted" into just $\ce{C2H4O2}$. What is the reasoning behind this odd way of writing such formulas and when should I use it (it seems like this could be a huge gap in my knowledge)

CH3COOH is an accepted and common form of writing the structure of acetic acid (commonly known as vinegar when diluted in water). You are right in saying the formal notation should be used - which is C2H4O2 - since it gives the reader the types of atoms and their quantities, and these two characteristics define the exact composition of the molecule, and there's nothing wrong with that.

However, the notation CH3COOH takes you one step further and tells you something about the actual structure of the molecule, since it groups together the atoms that comprise a functional group, and a very important one at that. Grouping together COOH is an accepted and convenient form of communicating to the reader: "there is a carboxylic group here", or "this molecule is actually a carboxylic acid", which is something you'd like to know when dealing with a certain substance.

To summarize, C2H4O2 can be the molecular structure for acetic acid or formate (formic ester, as mentioned in another answer). In order to avoid confusion and, more importantly, indicate that there is a carboxylic group in the molecule, it is more convenient and straightforward to note it in the formula than letting readers figure it out themselves.


determine density of oxygen gas using modified ideal gas equation I am trying to determine the density of a gas (oxygen). As the gas is a compressible fluid a modified version of the ideal gas equation is required. I have been given the following equation and I am struggling to interpret it. I would like advice on the unit conversions and whether this is the correct equation at all.

$$\rho=\frac{(P+1.01325).10^5.MW}{z.8314.(T+273)}$$

where:

$\rho$ = density (Kg/m3)

P = pressure (bar(g))

MW = molecular weight (in kg/kmol)

z = compressibility factor (dimensionless)

T = temperature  (deg. C)

I think that the 1.01325 converts the bar.g. to bar.a. then to pascals using the 100,000 multiplier?

Has the universal gas constant $8.314 J/(k.mol)$ been converted to $J/(K.kmol)$ ?

What an awkward, ugly equation that is! Mithoron is correct about the traditional, elegant and understandable form.

Yes, (P+1.01325)*10^5 converts gauge pressure in atmospheres to absolute pressure in Pascals.

z is compressibility - look up the Nelson-Obert generalized compressibility charts.

As for the 8314, the units here should be Pa*m^3/kmol-K. Dividing it by the molecular weight in kg/kmol will give you the specific gas constant for oxygen in Pa-m^3/kg-K. That's already been done in this equation.

Look up compressibility factor on wikipedia then find the Nelson Obert charts online, also critical data for oxygen, and use them to determine whether your oxygen is ideal in the P, T range you are dealing with (z factor between 1.01 and 0.99).  The specific volume of the gas depends on z as well as on P and T, but you can neglect the z correction if it's within a percent or two of the ideal value of 1.

Responding to your subsequent comment:

P, T and z determine the gas's specific volume in m^3/kg, which along with mass flow rate (kg/s) and pipe cross-sectional area (m^2) determine flow velocity (m/s), which along with pipe roughness will determine friction factor, which along with gas viscosity, pipe length and the presence of fittings like elbows and open valves, will determine pressure drop.


In a galvanic cell where the two electrodes are in the same electrolyte solution, why do reduction and oxidation occur separately? In a certain book, I was presented with the following solution:

An iron nail is attached by a piece of wire to a magnesium ribbon, and the iron and magnesium are placed into the same container, with dilute sulfuric acid.

My understanding is that due to the difference in reduction potential of the iron and magnesium, the magnesium ribbon will oxidise in preference to the iron, so the Mg metal will oxidise and lose its electrons to form Mg2+ ions, and the electrons will travel to the iron electrode and react with the H+ ions in the electrolyte solution, and H2 gas will be evolved at the iron electrode.

My question, is why must oxidation and reduction occur at the two different electrodes? If the magnesium isn't connected to the iron, it will just react with the H+ ions there, and the same overall redox reaction will occur. Why is it, that if there is a wire attached from the magnesium to the other electrode, the electrons from the oxidation of magnesium would prefer to follow the path of the wire, rather than just react with the H+ ions already present at the magnesium strip?

Magnesium blocks are connected to iron structures to serve as cathodic protectors. A key element is that the magnesium should react slowly enough (i.e., selectively) to last for a long time (years), while providing enough electron flow to keep the iron negative in the midst of all its electron acceptors (oxidizers). If the iron were in a severely corrosive situation and you needed more current, and you think of using sodium as a sacrificial anode to provide a faster reaction, it wouldn't be practical for many reasons.

The sodium situation is where the electrons wouldn't bother traveling to the iron cathode; things are happening too fast because of the energetics of sodium reacting with water (forget the acid!). This situation does not permit the selectivity to form separate anodes and cathodes.

Dry cells are constructed of zinc anodes and carbon cathodes; impurities in the zinc surface can short-circuit the cell: the zinc becomes both anode and cathode, and is consumed inefficiently. This condition was ameliorated by amalgamating the zinc surface with a tiny bit of mercury. Mercury has a high hydrogen overpotential (i.e., it takes a higher potential to force enough protons to hydrogen atoms and then to molecules and then to bubbles which rise up.) The mercury prevented the initiation of local cathodes, and did not prevent the migration of zinc to its surface and thru it. So the selectivity was improved by the mercury amalgamation, allowing the zinc to be a purer anode. (The next step was to improve it further so as to eliminate the mercury!)

A single iron nail can have electrochemical corrosion all over its surface with anodes and cathodes that switch around and/or are so small that you can't distinguish them. You see only a rusty nail. One way to separate the anodes and cathodes and to make them visible is to put the nail in an agar gel with a little NaCl, phenolphthalein and potassium ferrocyanide. The nail begins to corrode at a random spot. This first spot will lose Fe++ ions to the gel, where they will react with ferrocyanide ion to turn blue. Electrons from the reaction will flood the iron nail; somewhere else, O2 will be reduced to hydroxide ions, which turn the phenolphthalein red. Typically, the head and point of the nail will become anodes (because they are more highly stressed, therefore a little more anodic) while the shaft will be cathodic. Pieces of rebar were more random in their reactivities.

So, oxidation and reduction can occur at the same electrode; it's just more illuminating (and useful) to see them separated instead of all mixed up.


Does "easily hydrated" imply highly soluble in water too? Does "easily hydrated" imply highly soluble in water too?

Small ions like $\ce{Li^+}$ and $\ce{Be^{2+}}$ are said to be easily hydrated but does it also imply they are highly soluble?

I looked up some books and they say that compounds of Lithium like $\ce{Li_2O}$
and $\ce{LiF}$ are sparingly soluble.Why?

Note that you are considering two different things: easily hydrated cations and soluble compounds. Solubility is complicated, and compounds containing the same cations may well differ in solubility-- lithium oxide reacts violently with water to form lithium hydroxide, for example, whereas lithium fluoride is only sparingly soluble.

What causes the difference between the two compounds? For one, the $\text{O}^{2-}$ anion is an extremely strong base, and readily reacts with water to form $2\,\text{OH}^-$; likely the high enthalpy of reaction provides a thermodynamic driving force for the dissolution of lithium oxide. If we then consider lithium fluoride, no large driving force exists to favor dissolution. While $\text{F}^-$ may react with water to form $\text{HF}$ and $\text{OH}^-$, $\text{F}^-$ is only weakly basic and such a reaction would not be as exothermic as that with lithium oxide.

We must then consider smaller driving forces that were drowned out when considering lithium oxide. Even though $\text{Li}^+$ and $\text{F}^-$ are both small, highly charged species with large charge densities that will interact favorably (with regards to enthalpy) with the polar water molecules in solution, their charge densities will also cause water molecules to cluster around the ions, forming hydration shells and hence decreasing entropy; this is thermodynamically unfavorable. Furthermore, the electropositivity of $\text{Li}^+$ and the electronegativity of $\text{F}^-$, combined with their small ionic radii, lead to a large lattice energy for $\text{LiF}$, further disfavoring dissociation. These effects all contribute to the sparing solubility of $\text{LiF}$.

Related question.


How do electrolytes work? I just want to know how to electrons travel from one end to the other.

The concept I have in my mind is that the ions functions as "buses" for the electrons then send them to the other side, but doesn't that make them neutral molecules? (in which case they wouldn't be attracted to the + side.)

In compact matter, electrons "trickle" from a source to one atom, forcing off electrons which move to the next atom, ionizing each in turn. The same holds true in the opposite direction, where electrons are removed and the corresponding "hole", or positive ion, is propagated towards the negative electrode. Individual electrons and ions move slowly at the drift velocity, but electromotive force propagates much faster. This is similar to the motion of the particles in the toy, "Newton's Cradle".

In an ionized gas, electrons and ions may move much faster, actually crossing the full volume of the plasma (though electrons move faster, due to their lesser mass for the same charge).


Can we prolong life? This is rather a bunch of questions that I decided to post on Chemistry StackExchange since I thought the chemists would have the most knowledge about the chemical processes of life.

So recently I have came across stem cells and a lot of revolutionary medical products that can rebuild tissue and even organs.

And then I thought that can we use these products to transplant our organs and in a sense, rebuild our own body. I have even came across making a heart and lungs as well.

If so is there any thing that stands between having a longer life?

What else do we need to have in order to prolong and in a sense immortalize ourselves.

I know that it would be impossible to do this for everyone since our planet cannot sustain more life, but theoretically is it possible to do this?

What are you views into this?

Ageing is a disease.  Just as treatments have been found for other diseases, ageing can be slowed or reversed as well.

Not that long ago, scientists noticed that at the end of each DNA strand there were thousands of nucleotides.  Nucleotides are used for genetic coding.  That is to say, certain sequences of nucleotides instruct the body to synthesize  certain proteins when the genetic code in the DNA is being read.  DNA is really like a massive software program.

In any case, when scientists noticed all of this code at the end of a DNA strand, at first they couldn't figure out what is was for, so they thought it must be gibberish.  Later they noticed that everytime a cell divided (e.g. everytime the DNA was read) a piece of this gibberish was lopped off.  Eventually it became understood that this "gibberish" was the basis of aging.

Each lopped off piece was termed a telomere.  Each telomere contains the same nucleotide sequence (code).  There are about 15,000 telomere nucleotides at the end of the DNA strands in an embryo.  This has been shortened to about 10,000 when you are born and when it reaches a length around 5,000, the body dies.  It turns out that the telomeres play a role in stabilizing DNA strands during the copying process, when their length gets down to around 5,000 nucleotides the telomere length becomes to short to provide adequate stabilization and errors (genetic mutations) occur in the copying process and/or the cells become senescent (become inactive, go to sleep).  Errors or sleep, either way, the body stops functioning efficiently.

This telomere shortening occurs during cell division in all human cells except for sperm and egg cells.  Sperm and egg cells are immortal.  There is an enzyme called telomerase, it is expressed only in the sperm and egg cells.  After these cells divide a telomere is lopped off, but telomerase re-adds a telomere to the DNA in sperm and egg cells.  Hence, their telomere length never shortens - they live forever.

All cells contain the code to make telomerase, but that code  (or signaling) is turned off in all cells except for sperm and egg cells.  A specific protein combines with a certain portion of DNA and that is all that is required to prevent telomerase expression in a cell.

If scientists could find a way to prevent that (repressor) protein from complexing with DNA, then telomerase expression could be turned on and that cell would then be immortal.

As you can imagine, this is a big area of research.  Here is a link to a company that has been active in this area for a long time, and here is a link to a basic tutorial on the subject of telomeres.  Poke around the web site, the videos and other tutorials are extremely interesting.  Even aspects such as the social consequences of curing aging are discussed.

Enough progress has been made that you can buy products in the over-the-counter market that reduce telomere shortening, but they're very expensive at the present time.


Prepare 1 liter of 28% NH3(aq) solution from 1 gal of %10 NH3(aq) I have one gallon of commercial, surfactant free, ammonium hydroxide solution that I would like to use to prepare a concentrated solution.

I have two options as I see it, and I am still getting into the chemistry side (amateur chemistry), but am pushing ahead and want to do this even without the understanding. I am hoping the solutions will provide adequate explanation as to what is being done, and why, which will help me learn to solve problems like this in the future. Chemistry newbie, here.

Solubility at room temperature is to be assumed.

I see that I have two options.

1) Prepare a receiving flask with cold distilled water (likely in an ice bath) and then prepare a distillation flask with the %10 ammonium hydroxide solution. Slightly heat the distillation flask to the point where the ammonia gas begins to off-gas, which is funneled into the receiving flask and dissolve.

2) Prepare a receiving flask and distillation flask both with an amount of the 10% ammonium hydroxide solution. As above, gently heat the distillation flask to off-gas the ammonia into the receiving flask where it should dissolve, and increase the concentration beyond 10%.

Since I would like a 1 liter solution containing 28% NH3(aq), what volume of cold distilled water should be placed in the receiving flask, which I ask because I assume the solvation of NH3 into water will cause the volume to increase. If I placed 1 liter of water into the flask initially, then this would result in more than 1 liter of solution. This also depends on the amount of ammonia that needs to be dissolved to reach my desired concentration, which varies from option #1 to #2.

So what I want to know is,

Q1) How to determine the volume of a solution given a specific concentration of NH3(aq). This will allow me to account for the increase in volume when adding ammonia to an existing solution, in order to reach a desired concentration.

Q2) How do I know when I have reach %28 NH3(aq) concentration?

Q3) A procedure leading from 10% to 28%, and any advice on more technically correct approaches to this problem. Is it a problem that should be considering molarity, etc.?

Q4) Any quick background info that I can look up to help me understand this problem would be welcome!

Bottom line: Prepare a 1 liter solution of 28% NH3(aq) from 1 gallon of 10% NH3(aq), and please provide an explanation.

Thank you for any help. I am a software programmer, and have delved into he world of chemistry as a hobby, and am getting confused by terms like molarity, normality, and both of those versus concentration, etc. It's maddening to say the least.

Disclaimer: ammonia is extremely hazardous including flammable in some vapor concentrations.

  How do I know when I have reach %28 NH4OH concentration?

You can measure the specific gravity and compare to the table on page 2 of
this document:

Next, it is important to understand that if you heat/boil/distill the 10% solution the vapor coming off is not 100% ammonia, it is a mixture of ammonia and water, but with increased ammonia concentration.

To predict the percent ammonia in the vapor phase, you need to look at the dew point line in a table like Fig. 1.2 here, but that one is for 16 bars or Fig. 2 of Ganesh et al. Journal of Mechanical Engineering Research Vol. 3. (1), pp. 25-39 that shows many pressures including 1 bar.

As the original solution is heated, the system progresses toward the upper left in Fig. 1.2/Fig.2, the liquid phase along the bubble point line and the vapor phase along the dew point line.  The concentration of ammonia in both phases progressively decreases.  Eventually, the concentration of ammonia in the vapor phase drops below the concentration of ammonia in the original solution.

From Fig. 2 of Ganesh et al. it can be seen that at 1 bar starting from 10 weight% ammonia liquid phase, the initial vapor phase concentration would be close to 28% ammonia.

This shows that even if you added no water to the condensate, there is no way you can get 1 liter of 28% ammonia from 1 gallon of 10% ammonia in one simple distillation step.

Only through fractional distillation could you get that much 28% ammonia from that little 10% ammonia.

Also, practically speaking, it will be difficult to avoid lossing ammonia vapors to the environment and safety needs to be considered.


Does enthalpy equal heat when PV work is done? The derivation that enthalpy equals heat at constant pressure goes like:

$$\begin{align}
H &= U + P_{\mathrm{int}}V \\
\Delta H &= \Delta U + \Delta (P_{\mathrm{int}}V)
\end{align}$$

If $P_{\mathrm{int}}$ is constant, then

$$\Delta H = \Delta U + P_{\mathrm{int}}\Delta V$$

We also have

$$q = \Delta U - w$$

If external pressure is constant, $w = -P_{\mathrm{ext}}\Delta V$. Then

$$q = \Delta U + P_{\mathrm{ext}}\Delta V$$

So in order for $\Delta H$ to equal $q$, we need $P_{\mathrm{int}} = P_{\mathrm{ext}} = \mathrm{constant}$. This is true when the system is open to a constant external pressure and no PV work is done. But when there's PV work done (and the temperature is still constant), $P_{\mathrm{int}}$ clearly changes, so does $\Delta H$ still equal $q$?

I shall start off by mentioning that your question is not so much about p-V work being done, but rather about the pressure not being constant. It is perfectly possible for p-V work to be done when the pressure is constant, and indeed under these circumstances, the equation $\Delta H = q$ is still obeyed. However, this would necessitate a change in temperature.

Well, you can always try to come up with a counterexample.

Your hypothesis would be that if pressure is not constant and p-V work is done, then $\Delta H = q$. So, let's try calculating those two quantities for a reversible isothermal expansion of $1\mathrm{~mol}$ of an ideal gas, from an initial state of $(p_1, V_1)$ to a final state of $(p_2, V_2)$. Let's just arbitrarily set $V_2 = 2V_1$. Of course, since this is an ideal gas we are talking about, $p_2$ will be equal to $p_1/2$. And since this is a reversible process, $p_\mathrm{int} = p_\mathrm{ext}$ throughout, so let's drop the int and ext subscripts.

Let's calculate the work done (on the system) first.

$$\begin{align}
w &= -\int \! p\,\mathrm{d}V \\
&= -\int \! \frac{nRT}{V}\,\mathrm{d}V \\
&= nRT \ln{\left(\frac{V_1}{V_2}\right)} \\
&= -RT \ln 2
\end{align}$$

Now the internal energy change of the system has to be zero, since this is a closed system with an ideal gas and therefore $U = U(T)$. $\Delta T = 0$ therefore implies $\Delta U = 0$. So:

$$\begin{align}
q &= \Delta U - w \\
&= RT \ln 2
\end{align}$$

However, the enthalpy $H$ is also only a function of temperature: $H = H(T)$. Therefore,

$$\Delta H = 0 \neq q$$

Hypothesis: disproved!


Can I make a buffer from two different weak acids or bases? Can I make a buffer using two different acids/bases? For example, adding 1 mole of acetic acid and 1 mole of ammonia?

The only information I can ever find is that I can make a buffer using only an acid (or base) and its conjugate base (or conjugate acid). So, carbonic acid and carbonate, $\ce{H2CO3}$ and $\ce{HCO3-}$.

Why do I need a conjugate acid-base pair and not two random weak acid-base pairs to make a buffer?

Yes, having multiple weak acids and/or bases having different pKa values would give buffering capacity in more than one pH range, or over a greater pK range.

There is a term "universal buffer" which refers to a buffer having multiple pKas spaced thoughout the pH ~2-12 range to give some buffering capacity over the enter pH range.

As explained in A Neutral Buffered Standard for Hydrogen Ion Work and Accurate
Titrations Which Can be Prepared in One Minute J. Am. Chem. Soc., 1932, 54, pp 1911–1912, ammonium aceate has the property that if it is added to pure water it produces a buffered solution of pH 7.00, accurate to a hundreth of a pH unit.


Is a hydrogen bond considered to be a van der Waals force? Is a hydrogen bond considered to be a Van der Waals force?

According to the IUPAC gold book a van der Waals force is:

  The attractive or repulsive forces between molecular entities (or between groups within the same molecular entity) other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules. The term includes: dipole–dipole, dipole-induced dipole and London (instantaneous induced dipole-induced dipole) forces.

Hydrogen bonding is a type of dipole-dipole interaction, so it would fit the definition of a van der Waals force.

The way I think of it is: van der Waals forces are anything that make a gas non-ideal, since that's how they were originally discovered and defined.


Why is the sum of two inexact differentials exact? From the first law of thermodynamics, $\mathrm{d}U = đQ + đW$, where $\mathrm{d}$ represents an exact differential and $đ$ an inexact differential. Exact differentials correspond to state functions and are path-independent.
How is it possible, and what does it imply, when we say that the sum of two inexact differentials is an exact differential? Is the sum of two path-dependent functions always a path-independent function? Also, what is the difference between a partial derivative and an inexact derivative / differential, apart from mathematical implications?

There's not much to do, as you nearly spotted the difference by yourself: $U$ is a state function, thus is path-independent. In other terms, one can estimate it at any moment only with an initial and a final value (and express it in terms of its variables only). But with $Q$ and $W$, it's different: these are path functions (or process functions) consequently depending on the path. Hence, their differential is not something one might express depending only of its variables. (these are not "real" functions for mathematicians...)

Is the sum of two path-dependent functions a path-independent function?

The sum of inexact differentials is an exact differential in that particular case. In other cases, it depends on the units involved: for instance, with entropy, we have $\mathrm{d}S = \delta Q_\mathrm{rev} / T$. So, no typical case here!

what is the difference between a partial derivative and an inexact derivative?

These are absolutely not the same! A partial derivative is a differentiation of a multi-variable function regarding one of its variables only, while what we have here is an inexact differential, thus not a derivative as 1) no variable is specified here, and 2) an inexact differential can be of a function depending on only 1 variable.


Is it possible to dope Ca ions in an NaCl crystal to produce a vacancy defect? I was just reading about defects and doping in solids and read about the defect in  ionic crystal ($\ce{NaCl}$). There were examples of $\ce{SrCl2}$. Could this also occur with $\ce{CaCl2}$?

A peek at the NaCl - CaCl2 phase diagram (say at FactSage) will show you that NaCl can hold quite a reasonable fraction of CaCl2 in the NaCl crystal. In contrast, CaCl2 has very limited to no solubility of NaCl.


Which factor should I take into priority to determine reactivity of of Halides? Such as in the attached question I was taking into account the stability of the carbocations (by the number of resonating structures) formed after removal of Chlorine from and concluded to be the order I>II>III>IV>V the answer states the order to be I>III>IV>V>II

Can conclude what factors they are priority to. Help needed

If you need to consider the stability of the carbocations obtained after chloride leaves, we can, as Klaus commented already, use Hückels rule, which more or less states that fully $\ce{sp^2}$ (i.e. usually $\ce{sp^2}$, as long as there is atleast one full p-orbital) hyberdized  compounds with $4n+2$ electrons are particularly stable or aromatic. The same type of compounds that have $4n$ are particularly unstable, anti-aromatic. Of course a carbocation has an empty p-orbital, so compunds I and III will be aromatic and compound II anti-aromatic. Compound I is fully benzenoid and also less strained than compound III and will hence be more stable, so I > III. Then follows the allylic carbocation IV which is obviously more stable than its non-allylic counterpart V. And last but not least the anti-aromatic compound II.

In conclusion I > III > IV > V > II, as the answer indicated.


What does "totally reduced" mean? My textbook mentions that:

  Iron is unable to be totally reduced... the most common non-zero oxidation states of iron is (o for oxidation, r for reduction) are: +2r, +3, +4o, +5o, +6o.

I thought it maybe meant going to $\ce{Fe^4+}$ (as the most reduced I commonly see is $\ce{Fe^3+}$), but then upon googling there is also $\ce{Fe^5+}$ ions, so how far does it have to go until it is totally reduced (if this is even what it meant by totally reduced)? Edit: Sorry, got reduction and oxidation mixed up (thanks for pointing this out DVSA). However, my question still stands, though replacing the above with more to e.g.  $\ce{Fe^4-}$ than  $\ce{Fe^3-}$  .

The question, for context, is below (I have already typed up what I think are the relevant parts above). FYI the answer is D, but I want to know what B means.

Reduction means that electrons are added, you are talking about oxidation. I haven't heared the term "fully reduced" in this context before, but I guess it means a full shell. In case of iron this would mean a oxidation state of -10, while as far as I know the highest we can get is -4.

And don't confuse oxidation states with charge of ions.

Well, now that I see the actual question I think there's nothing to really understand here. Just a wrong, made up answer.


In the carbonate anion, which atoms gain the two electrons? In the carbonate ion ($\ce{CO3^2-}$), is it the oxygen or the carbon that has gained the two additional electrons?

One can draw Lewis structures to see where the lone pairs of electrons may be.

Here you can see that you can draw the ion like this, with one of the oxygen atoms having a double bond with the carbon, and the other two having a single bond. This means that the double bonded oxygen atom has no charge. The single bonded oxygen atoms have a net charge of $-1$ each as they have gained a lone pair of electrons.
However this picture does not tell the whole story. In reality, the negative charges are not localised to individual oxygen atoms in the ion. We can draw a total of three structures (called canonical forms):

These can be drawn because in actual fact there is delocalisation of electrons between the carbon and each oxygen atom. The true structure is not changing/oscillating between these three forms, but is actually called a resonance hybrid of all of the canonical forms. This is shown below. Therefore you can see that the negative charges are not localised to single oxygen atoms, but are instead delocalised across the entire ion.

If you want to read more about the carbonate ion and its structure: https://en.wikipedia.org/wiki/Carbonate#Structure_and_bonding


Activity of different positions in phenol with respect to electrophilic aromatic substitution Which of the statements below is correct?

  a) The ortho and para positions are more active than the meta position.

or

  b) The ortho and para positions are activated while the meta position is deactivated.

As the figure below illustrates, substituents on aromatic rings that are electron donating like the methyl and t-butyl groups (and also the hydroxy group [e.g. phenol]) increase the reaction rate at all positions (ortho, meta and para) relative to benzene.

On the other hand, electron withdrawing substituents (the chlorine and ester carbonyl examples) decrease the reaction rate at all positions relative to benzene.

So for phenol, or any other electron donating substituent, while the meta position does react more slowly than the ortho and para positions, it still  reacts faster than benzene.  Therefore, it would be incorrect to say that the meta position is deactivated. Only when the rate at a given position is less than the rate for a single position in benzene could one say that the position is deactivated.

Statement "a" in your question is correct.

(image source)

For future reference a note on terminology. The above figure tells us that nitration of toluene proceeds 24.5 times faster than the nitration of benzene; that is to say that the relative rate for nitration of toluene is 24.5 times greater than benzene.  The meta position in toluene reacts 3 times faster than a single position in benzene, this is referred to as the partial rate factor.


When to use internal energy and enthalpy in balances I am confused as to when you use ΔU and when to use ΔH, I don’t really understand the difference. Is it that ΔU is used for closed systems whilst ΔH is used for open systems?

Is it that ΔU is used for closed systems whilst ΔH is used for open systems?

That's not quite correct.

Consider the case of a simple system: a volume of gas. The problem is that when processes take place, the gas may expand or contract. Since the gas pushes back on the external pressure, some amount of work may be done the system ($W = p\Delta V$). The internal energy change $\Delta U$ does not automatically factor in this work.

Enthalpy, on the other hand, is a quantity which is defined ($U + pV$) such that it does account for the pressure-volume work of the system, so $\Delta H$ is the change the energy of the system and the pressure-volme work done by the system.

So, the summary is that if you have a constant volume process (where pressure volume work is by definition zero), use $U$. If you have a constant pressure system, use $H$.

Even the free energies are this way as well. There's Helmholtz free energy ($A$ or $F$) for constant volume processes and Gibbs free energy ($G$) for constant pressure processes.

In chemistry, most processes are done in open air (unless you're using a bomb calorimeter), in which case all processes are simply subject to atmospheric pressure, so these processes would all be best described using enthalpy and Gibbs free energy.


Is the rate of electrolysis the same throughout as long as voltage is same In an electrolytic cell, where the electrolysis of dilute sulfuric acid is taking place, would the rate of electrolysis be constant always, even at the beginning?

By rate of electrolysis, I mean the rate of gas production or the rate of charge delivery. And would this rate be established as soon as I turn on the voltage, or does it slowly increase to a constant level?

The rate of electrolysis (decomposition of water) is defined by the current that passes. If you apply a constant voltage, factors arise as suggested in the comment by MaxW.

At constant current, the rate of hydrogen (and oxygen) evolution would be somewhat different initially because on a microscopic level, hydrogen atoms have to cover the surface to some extent to form H2; H atoms may reduce some oxides on the electrode surface, and H2 has to agglomerate to form bubbles (and dissolve in water to saturate it with H2). And then the bubbles have to rise to the surface.


where can I find experimental compressibility factor values? I am interested in experimental compressibility factors for various gases, such as methane and carbon monoxide. Not only for standard pressures and room temperature, but in a range from 1 to a few 1000 bars and various temperatures. Where can I get them? I've done a quick google search, but they seem to be well hidden.

EDIT: Perhaps a better question is where can I find second, third, fourth, fifth... virial coefficient for these 2 gases? I am interested at 2 temperatures: 200 K and 287 K. I found some so far, but not at 287 K.

The NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) is a program that can calculate the thermodynamic and transport properties of some important fluids and their mixtures. It was developed by the National Institute of Standards and Technology (NIST) and is commercially available as NIST Standard Reference Database 23 here.

For example, the following selected results have been obtained with REFPROP Version 9.0. The number of digits is set to the default value (5 digits), which are not necessarily significant.

Methane
Temperature: 200.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99369
2nd Virial Coef.: −0.0065173 m³/kg
3rd Virial Coef.: 0.000015219 (m³/kg)²

Methane
Temperature: 287.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99802
2nd Virial Coef.: −0.0029520 m³/kg
3rd Virial Coef.: 0.000010232 (m³/kg)²

Carbon monoxide
Temperature: 200.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99739
2nd Virial Coef.: −0.0015501 m³/kg
3rd Virial Coef.: 0.0000035524 (m³/kg)²

Carbon monoxide
Temperature: 287.00 K
Pressure: 1.0000 bar
Comp. Factor: 0.99952
2nd Virial Coef.: −0.00040865 m³/kg
3rd Virial Coef.: 0.0000027781 (m³/kg)²

REFPROP can also be used to plot the results, for example for carbon monoxide:


What type of chemical reaction is it when sodium bicarbonate is added to citric acid? I can't tell what type of chemical reaction it is? Is it a combustion chemical reaction? Or Displacement or decomposition or something else?

The first step is usually called an acid-base reaction or a proton transfer. Any acid of similar strength to citric or acetic (vinegar) acids will react with bicarbonate this way, and it is very fast in water solution at room temperature. The products are carbonic acid and the sodium salt of the acid used--for example, sodium citrate.

Carbonic acid is energetically unstable relative to carbon dioxide and water, to which it is rapidly converted via what most chemists would call an elimination reaction. I used to think that this occurs instantly, but carbonic acid is a species that actually exists in solution and can be observed and studied.

While there are many named organic reactions, I  agree with Ivan that there is no one universally agreed upon way to name them, so others may wish to weigh in with their opinions about this reaction.


Do external electronic shells influence internal ones? There is a well-known electron shielding effect: negatively-charged internal electronic shell screens external shells from positive nuclei, thus increasing their radii.

Thus, 2S orbital radius of Li atom (ground state) should be greater than radius of excited Li2+ cation with single 2S electron.

But is there a reverse influence: from external electronic shells to internal ones? For example, are radii of 1S orbital in Li and Li+ equal, or not?

I am sure there is not a big difference anyway, but this is theoretically interesting. Probably it can be proved by comparision of internal shell ionisation energies of Li and Li+ but I am not sure where to look for appropriate experimental data.

I have some intuitive considerations, while they did not lead me to certain answer:

Concept of Faraday cage looks to be irrelevant, because electron
shells are not literally material conductive shells, but they are
filled space regions.
Shells interpenetrate, thus external shell
can a bit force out internal shell from nucleus by increasing
electron density in space  adjacent to nucleus.
Classically,
internal one of two concentric  elastic charged spheres should
shrink. If we imagine big spheres, their small areas should interact
repulsively as parallel plates. But considered spheres are actually
not "big".

Not only do "outer" electrons impart some shielding to "inner" ones, but an analytical technique is based on that fact.

In X-ray photoelectron spectroscopy, binding energies of electrons just below the outer shell are measured.  These binding energies are typically hundreds of electron volts, roughly 280-300 eV for carbon as a typical example.  An important feature is that this binding energy is measurably not fixed for a given element, rather is varies according to how the atom is chemically bound.  As an example, the 1s electrons in carbon lie at about 285 eV binding energy in hydrocarbons, but shifted by as much as 10 eV by bonding to fluorine (see here), which is more electronegative and draws electrons away from the carbon 2s and 2p  shells.  Similarly, the 1s electrons in oxygen range from about 529 to 534 eV in various oxides, with the higher values tending to occur in nonmetal oxides (see here).


How can one calculate the pH of a solution? So, I have 1 litre of acetate buffer, and 0.1 mol of oxonium ions/$\ce{H3O+}$ is added. The task is to calculate the pH of this solution.

What I was thinking:

$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{c(\ce{Ac-})}{c(\ce{HAc})} = \mathrm{p}K_\mathrm{a} + \log\frac{n(\ce{Ac-})}{n(\ce{HAc})}$$
$$\mathrm{pH} = 4.74 + \log\frac{n(\ce{Ac-}) - 0.1\ \mathrm{mol}}{n(\ce{HAc}) + 0.1\ \mathrm{mol}}$$

But neither $c(\ce{Ac-})$ nor $c(\ce{HAc})$ (which should, in theorey, be equal) is given. The only thing I know (or I think I know) is this: $$\frac{c(\ce{Ac-})}{c(\ce{HAc})} = \frac{1}{1} = 1$$

So, is it possible to calculate the pH from that? Am I missing something?

I'm tring to write the methodology
If anything is wrong please point it out to me i'll be happy to correct!

To find $\small\ce{pH} $ of a solution you can:

Find the conjugated base of your acid  (for example $\small\ce{HCOOH}$ becomes $\small\ce{HCOO-}$ (correct me if i'm wrong here)
Write the equation of the solution
Now you want to prepare an aqueous solution.
Find the concentration of $\small\ce{[CH_3COOH]}$
you use the formula to determine the concentration in solution of acid

$$
   \ce{[CH_3COOH]\ =\ C\ -\ [CH3COO- ]}
   $$

($\ce{C}$ is the concentration of brought acid in $\ce{mol/l}$)
Then you write that:

$$\qquad\ce{[H+ ]\ =\ [CH3COO- ]}$$
as both dissociate from the same acid as

$$\ce{CH3COOH\ ->\ CH3COO- \ +\ H+}$$
the $\small\ce{pH}$ concentration is given by:
$$\ce{pH\ =\ -\log(\small[H_3O+ ])}$$

$pH$ here you are! (without any unit)


Is "family" synonymous with "group"? Is using the term "family" strictly correct when we mean a group of the periodic table or is it just in general use (and should perhaps be discouraged in favour of using "group" exclusively)?  At my highschool, other teachers refer to the transition metals as a family; is this a common practice?

Family is an informal term. Only a group and a period is an accepted terminology for the periodic table by the International Union of Pure and Applied Chemistry, a widely respected authority on chemical nomenclature and terminology.

The Wiki page of periodic table should be modified because a dubious and layman type reference (www.shmoop.com) for group (= also known as family) is being cited.


Why does the ionic product of water remain constant after addition of non-neutral solute? In my textbook, it is given that the ionic product of water $K_\mathrm{w}$ remains constant even when a non-neutral solute such as an acid is added to it.

$$K_\mathrm{w} = \ce{[H3O+][OH-]}$$

When a strong acid is added to water, the concentration of $\ce{[H3O+]}$ increases. Yet the $K_\mathrm{w}$ must remain constant. But how? How does the concentration of $\ce{[OH-]}$ decrease?

$K_w$ is the equilibrium constant of this reaction:
$$\ce{2H2 O <=> H3O+ +OH-}.$$
So, it's constant at a fixed temperature. It equals $10^{-14}$ at $25~^\circ\mathrm{C}$. At this temperature $\ce{[H3O+]=[OH^{-}]}=10^{-7}~\mathrm{mol/L}$.

If you add acid to water, the above reaction equilibrium is  displaced in order to counteract the change in the concentration of ion hydronium, i.e. the reaction is displaced to the left. This means that  ions hydroxide will react with the excess of ions hydronium to give water. As a result: the concentration of ion hydroxide will be less than $10^{-7}~\mathrm{mol/L}$, and the concentration of ion hydronium will be more than  $ 10^{-7}~\mathrm{mol/L}$. But their product is constant. $\ce{[H3O+] \cdot[OH^{-}]}=10^{-14}$.


Separation of Methanol-water in a plate distillation column I conducted an experiment to separate methanol and water. I am trying to determine the mole fraction of the distillate collected but having a little trouble with the calculation.

So i had a flow rate of 55ml/min. When i collected my sample i got a specific density of 0.815 and  a volume of 334ml. I tried to work out the mass but i cant seem to use specific density as its a ratio so has no units.

Would i just assume the specific density is the density and set units to g/cm3 so i can get the mass of the distillate , the concentration and then determine the mole fraction from there or does that not make sense
?

Yes, you can do that.

Specific gravity is considered as unitless ratio of density related to water density.

Density as a specific mass is mass per volume, in $\pu{ g/cm^3}$ or $\pu{ kg/m^3}$

Regardless of country specific quantity naming convention:

Numerically, for 3 valid decimal digits,considering value uncertainty, the unitless quantity and quantity in $\pu{g/cm^3}$ are about interchangeable at $\pu{20 ^{°}C}$, as water density is near $\pu{0.9982 g/cm^3}$


Tetrahedron is minimum energy conformation for atom with 4 electron domains in VSEPR theory Background

A student asked me to prove that the regular tetrahedron is the minimum energy geometry available to describe the locations of the electron domains in three dimensions – the prediction of valence shell electron pair repulsion (VSEPR) theory for a central atom with 4 mutually repelling electron domains. With the tetrahedron demonstrated, it could then be shown using vector multiplication that the angles between all substituent atoms is $109.5^{\circ}$.

Some ideas

I can think of a few ways to start this proof:

As mentioned above, we could define a regular tetrahedron and work backwards to see if the distance between all vertices is equal and maximized. Start with the vertices: A(0, 0, 0), B(1, 1, 0), C(1, 0, 1) and D(0, 1, 1), and a central point O($\frac{1}{2}$, $\frac{1}{2}$, $\frac{1}{2}$). With these, we can get vectors that point to each vertex. Call them $\vec{r_1}$ through $\vec{r_4}$. We remember that we can calculate the angle between any two vectors using the geometric and algebraic definitions of the dot product. I won't bore you with the details of this calculation but if anyone wishes to see it, I would be happy to share it. If you do it you will see that all the angles are equal and satisfy: $\cos^{-1}\left(-\frac{1}{3}\right) = \theta_i \approx 109.47122^{\circ}$. From here we would need to calculate the six distances bewteen the four vertices to show that the distances are all equal.
Using multivariable calculus, we could treat this as an optimization problem subject to the constraint matrix, $g$; the sum square distance function, $f$; and use the optimization formalism described by Lagrange. This results in a system of 13 unknowns – the 4 points with three coordinates each plus $\lambda$ – and 13 equations – the four constraint equations plus the nine partial derivative equations from the Lagrange method:

$$g = 1 = \left\{\begin{array}{c}g_1(x_1,y_1,z_1) = {x_1}^2 + {y_1}^2 + {z_1}^2\\
g_2(x_2,y_2,z_2) = {x_2}^2 + {y_2}^2 + {z_2}^2\\
g_3(x_3,y_3,z_3) = {x_3}^2 + {y_3}^2 + {z_3}^2\\
g_2(x_4,y_4,z_4) = {x_4}^2 + {y_4}^2 + {z_4}^2\\
\end{array}\right.$$
$$f = \mathop{\sum_{i=1}^{3}\sum_{j=1}^{3}}_{i<j}\left[\left({x_i}-{x_j}\right)^2+\left({y_i}-{y_j}\right)^2+\left({z_i}-{x_j}\right)^2\right] $$
$$\nabla f=-\lambda \nabla g$$

An approach similar to the one outlined above could be used with spherical coordinates instead. This version reduces the set of variables from 13 to 8 (the four sets of two angles for each point $\theta$'s and $\phi$'s; note the spherical radius could be set to 1 or any other bond distance, $a$, so is not a variable). These could be further reduced using symmetry arguments. For example, if we take the first vertex at the point $(0, 0, 1)$, we know based on symmetry that if we look down the z axis at the xy-plane, the angle between the three other vectors must have an angle of $120^{\circ}$ with respect to one another. Let one of the other vertices have the point $(x, 0, -\sqrt{1 - x^2})$ – satisfying the radius of 1 unit constraint we have imposed. Now based on symmetry, we know that the other two points must be $(x\cos(\frac{2\pi}{3}),x\sin(\frac{2\pi}{3}),-\sqrt{1-x^2})$ and $(x\cos(\frac{4\pi}{3}),x\sin(\frac{4\pi}{3}),-\sqrt{1-x^2})$ using spherical symmetry. From here, we can write a formula for the total sum square distance between all the points and set the derivative with respect to $x$ to zero to find the maximum sum square distance. Notice we have reduced the problem with these symmetry arguments from a multivariable calculus problem to a calculus BC problem. Alternatively, we could use some other guess and check method to solve for $x$. Note, we should find the same answer using this method as the others above (and below).

Sketch of a Simpler Solution
There is one remaining way that is simpler than all of the above, and it has to do with circumscribing a tetrahedron with a cube.
I will leave it to the reader to work on this most beautiful version. The properties of the cube ensure that the distances are maximized, and some more geometry will show that the inscribed shape is indeed a regular tetrahedron. For a hint, check out this excellent answer to this question on the math stack exchange site.


What are the standard math prerequisites for physical chemistry? I'm currently a chemistry major and I know that physical chemistry requires lots of calculus.  Which mathematics courses should I take so that I am well prepared for my physical chemistry courses? Specifically, I need to take courses in thermodynamics, quantum chemistry and statistical thermodynamics and beyond, as I want to specialize in physical chemistry.

Is it true that the more mathematics courses that one takes, the easier physical chemistry becomes? I'm currently taking a multivariate calculus course now, and I wonder which is the next level of math courses I should take.

I plan to take as many mathematics courses as possible. I'm thinking of taking ordinary differential equations after i finish multivariate calculus. What additional courses beyond ordinary differential equations should I take? My school offers courses in complex analysis, real analysis, and partial differential equations. Will they be useful?

In my experience, the most helpful advanced math courses you can take for Physical Chemistry are Differential Equations and Linear Algebra. Multivariate Calculus is helpful, but not as helpful as Diff. Eq. would be, especially in the beginning. Linear Algebra is very helpful for quantum chemistry, Diff. Eq. is also helpful for quantum, but more so for statistical mechanics. In the comments you mentioned several different Diff. Eq. options - as @permeakra said, taking all of them is a good idea, but ODEs and PDEs will probably be the most useful.


Reading CGTF basis from Basis Set Exchange site I'm making a simple electronic structure code as an exercise, and I want to confirm that my understanding about basis set data from BSE is correct.
The CGTF of H in aug-cc-pVTZ basis is given as:
H    S
 33.8700000              0.0060680
  5.0950000              0.0453080
  1.1590000              0.2028220

H    S
  0.3258000              1.0000000

H    S
  0.1027000              1.0000000

H    S
  0.0252600              1.0000000

H    P
  1.4070000              1.0000000

H    P
  0.3880000              1.0000000

H    P
  0.1020000              1.0000000

H    D
  1.0570000              1.0000000

H    D
  0.2470000              1.0000000

END
S-type GTF is simple, but I have questions about the number of other high-l GTFs.
P-type GTFs
I think the values
H    P
  1.4070000              1.0000000

mean that $\chi = (1.00000) g_{p}$. Then Does that means there are three p-type function ($x$, $y$, $z$)?
D, F-type GTFs
If that so, in the upper case. How D, and F-type GTFs? Specifically, if we consider D-type GTFs, are there six gaussians ($x^{2}$, $y^{2}$, $z^{2}$, $xy$, $yz$, $xz$) or five gaussians following the atomic convention ($xy$, $xz$, $yz$, $x^{2}-y^{2}$, $3z^{2}-r^{2}$)?

Yes it is correct that for $p$-type basis functions that you have all three $p$-functions, i.e. $p_x$, $p_y$, $p_z$.

For basis functions of larger angular momentum, you are also correct that there is two "different" conventions. The cartisean coordinate and the spherical coordinate convention.

In many electronic-structure programs both conventions are used, since the integrals are calculated in cartisean coordinates and then afterwards transformed to spherical coordinates, i.e. as you stated for $d$-type functions going from 6 basis functions to 5 basis functions.

In regard to make you own electronic-structure code, I would recommend keeping everything in cartisean coordinates, because this will make your life simpler. (Keep in mind that if you compare you energy with from a Hartree-Fock calculation with other programs you will have a slightly lower energy, because you have 1 more $d$-type basis function).

For a simple electronic-structure program a very nice guide for molecular integrals can be found here, http://joshuagoings.com/2017/04/28/integrals/


Do solid insoluble ions react with soluble ions in aqueous solution? Do solid and insoluble ions (like aluminium phosphate) react with soluble ions in aqueous solution (like potassium sulfate)?  When would a solid ion react with an ion in aqueous solution in general?

What you seem to be asking about is a two phase system, a solid phase in equilibrium with ions in solution. For such a system there are two reactions occurring at the same rate. Thus some of the solid is constantly dissolving and some of the ions from solution depositing at the same rate.
There is a named process, Oswald ripening, that describes the interaction in such a solution. The overall effect is that smaller particles are more likely to dissolve so the particle size tends to grow as the process continues since larger particles have a smaller surface area to volume ratio.


Current in a Concentration Cell I would like to know if there is a simple function that gives the current as a function of time in a concentration cell. The current is measured in the wire connecting the two terminals of the cell. The two half cells are isolated i.e. not connected by a salt bridge and the electrolyte is a liquid.

I am looking at a very simple case without involving the factors that affect the current and the internal resistance of the cell.

From my workings, I've got $\displaystyle I=C\ln\left( \frac { A-q }{ B+q }  \right) $ or more specifically

$\displaystyle \frac { dq }{ dt } =\frac { RT }{ nrF }\ln\left( \frac { VnF{ C }_\pu{ cathode }-q }{ VnF{ C }_\pu{ anode }+q }  \right) $

where $R$ represents the gas constant

$T$ represents the absolute temperature

$n$ represents the number of moles of electrons exchanged

$r$ represents the resistance of the wire

$F$ represents the Faraday's constant

$V$ represents the Volume of the solution (assuming both electrolytes' to be same)

$C$ represents the initial concentration of the electrolyte in each half cell.

Now I've tried integrating this function using WolframAlpha and then differentiating it to obtain the current explicitly as a function of time. Seems this function is too complicated for Wolfram to evaluate.(I've used the expressions with A,B,C)

Is my derived expression correct?

Also, shouldn't a simple dynamic process have a simpler equation?

Thanks.

Yes, your derived expression looks correct to me.

As to the closed form for the integral, WolframAlpha is most likely unable to find one because it is simply not there among the elementary and special functions. This has nothing to do with the equation being simple. Of course it is simple; nature "solves" it quite easily. It is just that the nature's definition of "simple" does not necessarily coincide with ours. Helium atom is simple, so good luck trying to solve the Schrodinger equation for it. Would you argue an atom is not actually that simple? Well, your concentration cell is made up of quite a few atoms.

So it goes.


Is there a danger in electrolyzing NaCl? To make $\ce{NaOH}$ one could electrolyze a solution of $\ce{NaCl}$ in water ($\ce{H2O}$) It would go like this:

$$\ce{2H2O_{(l)} + 2Cl-_{(aq)} + 2Na+_{(aq)} -> H2_{(g)} + Cl2_{(g)} + Na+_{(aq)} + 2OH-_{(aq)}}$$

After calculating a bit, I found that even just a few minutes of electrolyzing this at $10\ \mathrm{A}$ could be deadly.

Can anybody confirm this?

If the cell is mixed, you would make bleach, sodium hypochlorite and some free chlorine (and probably chew up your anode). Since the solution is not very basic, hypochlorite would probably disproportionate to chlorite, and that possibly to chlorate.

One mole of electrons is one faraday, about 96 500 coulombs. One ampere is one coulomb per second. Five minutes passes $(5\ \mathrm{min})(60\ \mathrm{s/min})(10\ \mathrm{C/s}) = 3000\ \mathrm{C}$ or $0.031\ \mathrm{F}$ in a reaction that requires two moles input for each mole of chlorine. $0.01554\ \mathrm{mol}$ chlorine or $350\ \mathrm{ml}$ at STP, tops. It wouldn’t be good for you.


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

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

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

No doubt it is aromatic.

The $4n+2$ rule is intended for a single cycle of π-electrons, which may possibly include multiple rings but can't have any interior atoms or pendant π- bonds (which would not fit in a cycle). Here, as Kartik indicates in a comment, there are evidently two such cycles and you have to consider them separately. One cycle has six π-electrons and the other has ten. You have a molecule (or more accurately, a molecular ion) with two aromatic cycles.


What's considered a "higher" level of theory in quantum mechanics calculations? Comparing wavefunction-based methods and density-based methods for quantum mechanical calculations, how is "rank" determined?

For example, in wavefunction methods, CCSD(T) is considered a higher level than MP2, which is a higher level than HF.

What about if I'm comparing MP2 to B3LYP? Or if I add dispersion
corrections to the DFT method to use something like B3LYP-D3BJ?
What if I just want to compare different DFT functionals? Is hybrid higher than pure? Is B3LYP higher than PBE0?

For single reference wavefunction based methods, the comparison is relatively straightforward within method "families". This is due to the nature of how these methods are formulated, usually involving a very long series that has been terminated at some point to make the calculation feasible.

You start with Hartree-Fock, which ignores electron correlation.
After that you can add perturbative corrections of increasing order (this will be the MP2, MP3, MP4, MPn family), do something called configuration interaction, truncated at increasing levels of excitation (CIS, CISD, CISDT,...,FCI), or something called the coupled cluster methods, again truncated at some level of excitation (CCSD, CCSDT, CCSDTQ,...,FCI).

In these cases the most straightforward interpretation of "higher level" would mean a less aggressive truncation of the series that includes more terms. In case of CI and CC, this almost always results in more accurate results, and both converge to the same limit, where all possible terms are included and correlation effects are exactly calculated. (of course there are errors from other approximations) The MPn perturbation series has no such strong guarantees, and this is one of the reasons why pretty much noone uses anything beyond MP4 these days. (even MP4 is seldom used)

For DFT you have something called Jacob's ladder, but things are generally more murky, as most DFT functionals have many empirical (fitted) parameters, and there is no straightforward way of comparing them.

In general, the only way to compare them, is to take a big database of either experimental or calculated (via CCSD(T) or similar high level method) data, and compare the average/maximum errors.


Organic solvent for cleaning autosampler syringe I work in a laboratory where gas chromatography is used extensively. We handle phthalates. What are the possible wash solvent for autosampler syringe? I've been using acetone.

Methanol is less expensive. Don’t know how well it would work in your particular situation. We do semi volatiles (EPA 624) .  Methanol works fine for the GC autosampler syringe.


Relative basicity of fluorine and oxygen in silyl ether deprotection Why does the $\ce{O}$ take the $\ce{H+}$ instead of the $\ce{F}$?

What I took into consideration:

$\ce{O}$ pulls more $e^-$ from atoms near it: $\ce{O}$ is more basic
$\ce{F}$ has one more $e^-$ pair than $\ce{O}$: $\ce{F}$ is more basic
$\ce{F}$ is more electronegative than $\ce{O}$   : $\ce{O}$ is more basic(?)
$\ce{O}$ has more substituent group making it sterically hindered: doesn't matter(?)

Is it okay to think this way? What's the most important factor?

Because oxygen has a lower electronegativity than fluorine, oxygen is less stable with a negative charge (and more stable with a positive charge). This is why $\ce{OH^-}$ is more basic than $\ce{F^-}$ and why $\ce{HF}$ in water is an acid (hydrofluoric acid), while $\ce{H2O}$ in water is neutral.


Why does whipped cream use nitrous oxide instead of nitrogen gas? It seems that nitrous oxide $(\ce{N2O})$ is frequently used to create whipped cream. But why can't just regular nitrogen gas $(\ce{N2})$ be used instead?

There are two ways to efficiently make an aerosol product:

Use a gas that liquifies under the pressure inside the can. For example, butane lighters. Nitrogen is one of the "fixed gases", meaning it's a gas under most conditions (but take a look at the temperatures and pressures needed for liquid nitrogen—it's not going to ever be found in consumer products).

or

Use a gas that is highly soluble in the liquid (carrier) and that will "substantially" vaporize when the higher pressure inside the can is reduced to atmospheric.

The US government restricts the pressures that can be used in aerosol cans (and requires 100% quality control testing—when's the last time you heard of an aerosol can exploding? [although it does happen]). If you cut up a can, you'll notice that it's pretty flimsy. The higher the pressure (and a gas that has been dissolved in a liquid doesn't exert much pressure), the more expensive it will be to build the container (aerosol can).

Thus, the fixed gases are almost never used, except in some medical products. Why? Because they just aren't soluble enough to help move the liquid (or, in other cases, solid) out of the can and also to disperse it into a very fine mist. The customer wants basically one thing when using an aerosol: uniform, consistent spray from first to last drop. Using a very soluble gas helps, and using one with a boiling point near room temperature also helps. But the laws of thermodynamics say that the temperature of the can will drop as you spray out its contents. This may dramatically interfere with the liquid-to-gas phase change, while solubility is less sensitive to temperature. The trick is to make a product (and I've made some) that sprays out consistently and also doesn't leave so much left behind in the can that the customer feels ripped off.


Properties of azeotropes Consider two liquids with vapour pressures in pure form being ${P^o}_A$ and ${P^o}_B$. ${P^o}_B >{P^o}_A$ which implies liquid B is more volatile. If the liquids formed an ideal solution, then a straight line (drawn with red in the diagram) will be obtained for the vapour pressure of the solution. However, if they formed a non ideal solution with positive deviation, the red curve would have been obtained.

Now, we get a maxima on that curve. At that composition, the solution would be called a minimum boiling point azeotrope. I understand why minimum boiling point. It's because of the maximum vapour pressure but what I don't understand at all are these points:

The azeotrope boils at a constant temperature.
The composition of the azeotrope remains fixed while boiling.
The azeotropic mixture cannot be separated by fractional
distillation.

These points are really not explicit from the graph. I am not able to figure out the reasons for those points. Can someone please provide a mathematical/ intuitive view of azeotropes relating to these 3 properties?

Also, "in an azeotrope or constant temperature boiling mixture, the vapour has the same composition as the liquid." How can we confidently claim that the point of maxima is the point where the vapour has same composition as liquid?

I will first enumerate the four points that need to be addressed.

The vapor and liquid phases have the same composition in an azeotrope.
The azeotrope boils at a constant temperature.
The composition of the azeotrope remains fixed while boiling.
The azeotropic mixture cannot be separated by fractional distillation.

I will assume the definition of an azeotrope to be a solution whose composition is such that it exists at a local extremum of a Pxy or Txy plot. I prefer to instead take 1. as a definition, in which case my original definition follows as a property of an azeotrope, but both approaches are equivalent.

"Proof" of 1., by  contradiction. Overlay the corresponding vapor pressure curve as a function of $y_A$ to make your plot a Pxy plot for species $A$. This new vapor pressure curve must touch the original curve at the azeotropic composition $x^*=y^*=z$, for otherwise a tie line would connect two vapor phases at some $P$, which is physically unreasonable.

Proof of 2. Refer to the Pxy plot from 1. The dew point and bubble point coincide at the azeotrope, so a phase transition happens only at one value of $P$ and hence only at one value of $T$.

Proof of 3. Refer to the Pxy plot from 1. The dew point and bubble point coincide at the azeotrope, and so must the compositions of each phase.

Proof of 4. Fractional distillation is based on the composition of the vapor and liquid phases being different. This is precluded by 3. in an azeotrope.


How do you calculate the norm of a function describing a molecular orbital? I'm a bit confused by a past paper question I came across for one of my courses, shown below.

  The wavefunctions corresponding to the lowest, $\psi_{1}$, and second-lowest, $\psi_{2}$, energy molecular orbitals in 1,3-butadiene are:
  $$\psi_{1}=0.372\phi_{1} + 0.602\phi_{2}+0.602\phi_{3} +0.372\phi_{4}$$
  $$\psi_{2}=0.602\phi_{1} + 0.372\phi_{2}-0.372\phi_{3} -0.602\phi_{4}$$
  where $\phi_{i}$ is the $p_{z}$ orbital on atom $i$. Calculate the norm of $\psi_{1}$, and the overlap between $\psi_{1}$ and $\psi_{2}$. Is the result what you would expect? Explain.

Neither calculating norm nor the overlap between 2 orbitals in the way the question asks came up.
I assume by norm it means the normalisationg factor, but that makes no sense to me since it's already normalised. I assume the overlap should be 0 since they should be orthonormal, but don't know how to show that. I'd really appreciate some help!
Sadly the lecturer for the course has contracted Covid-19 and so of course is not available to answer questions.

I'm unsure if you're meant to solve this via the Hückel molecular orbital theory (see Chemistry LibreTexts), however, if you need to calculate the norm of a wave function $\Psi$, this is $|\Psi |$. Given what we know in your post about both $\Psi_{1, 2}$ then
\begin{align}
  |\Psi_1|^2
    &= \langle\Psi_1 | \Psi_1\rangle \\
    &= 2\cdot(0.602^2)+2 \cdot (0.372^2) \\
    &= 1.001576 \\
  \to |\Psi_1| &= 1.000788 > 1\\
\end{align}
I write that this is greater than one, as, on closer inspection,  is down to rounding from the coefficients of each $p_z$ orbital.

The repeat process can be found for the overlap part, which I am interpreting as $\langle\Psi_1 | \Psi_2\rangle$.


Formation of trifluoride anion I wanted to understand mechanism of the formation of $\ce{F3-}$ anion through the molecular orbital theory.

UPDATE - What I really want to know is that when $\ce{F3-}$ is formed from $\ce{F2-}$ and $\ce{F}$, the unpaired electron of $\ce{F}$ enters the anti-bonding orbital $\sigma^*_{\mathrm{2s}}$ then shouldn't the $\ce{F2-}$ ion break up, as all the bonding and anti-bonding orbitals would be filled up, resulting in a bond order of 0?

I'm not too sure, but let me try. In the case of the F3- ion, the effective bond order is 1 (see attached images), not 2. In other words, the only effective bonding MO holding the entire thing together is a 3-centre 2-electron bond (think linear H3+ and FHF-).
I presume you know how the F2 molecule MO diagram looks like, so I'll just explain what F3- looks like. Using the D2h character table and forming our reducible representations, we get the following:

Note s-p mixing occurs in the Ag and B1u orbitals, as shown:

The MO diagram for F3- would look disastrously messy, so I'll simply show how the molecular orbitals look like and how they're ordered.

The first thing you'll probably notice is how the central fluorine isn't allowed to s-p mix anymore; although the symmetry remains D2h the now-presence of a central atom means the s and p orbitals of the central fluorine are no longer of the same symmetry, and no longer interact. That, in addition to the fact that the sole bond that is effectively holding the ion together is a 3-centre 2-electron bond, makes answering your question rather difficult.
I think a way to consider it would be in terms of the formation of a Lewis adduct; in other words, the pz electrons in the F- ion interact with the empty LUMO of the F2 molecule while all the lower-energy orbitals form bonding/antibonding sets with the lower-energy F2 MOs, forming a new HOMO and LUMO from the "HOMO" of F- and LUMO of F2.
Hope this helped.


Can a reaction with positive Gibbs free energy change ever occur? If you have an endothermic reaction with a negative entropy change, is it still possible to induce the reaction in some way despite the fact that the Gibbs free energy change is positive or all temperatures?

Yes, $\Delta G = -RT\ln K$.

If a reaction is endothermic with a negative entropy change, $\Delta G$ is positive.

That $\Delta G$ is positive only means that $K<1$.

$\Delta G$ would need to approach infinity for $K$ to approach zero.

Therefore, at equilibrium, there will always be some products in principle no matter how unfavorable the reaction.

As a practical matter, the reaction would need to be only slightly endothermic and having only a slight negative entropy change to get a significant amount of product.

However, if the small amount of product is continuously removed from the system, as for example by precipitation from a liquid phase, or gas bubbling out of a liquid phase, or by liquid-liquid extraction, one can keep the reaction going forward.


Is molecular weight of proteins additive? I have an elementary question regarding calculation of molecular weight of a complex of two proteins. Essentially, I have two proteins binding as

$$\ce{A + B <=> AB}$$

I have the concentration of [A] and [B] in ng/ml, which I can convert to nM using its molecular weight (available in kDa). I have only concentration of [AB] in ng/ml, but I am not able to convert that into nM. Is the assumption that molecular weight of [AB] is equal to sum of molecular weights of [A] and [B] correct assumption?

I am not able to see the intuition behind this logic, and if someone can help me here, I would be very thankful.

The molecular mass in Dalton, amu or whichever unit you prefer is equivalent to the molar mass $M$ in $\mathrm{g/mol}$ that chemists prefer to use for this discussion. Both are strictly additive. The molar mass of a larger complex is the sum of the molar masses of all its components.

This is because of how the unit is defined. Focussing on molecular mass, we could see ot as mass of a molecule. The mass of a molecule is put together by adding the masses of the atoms that this molecule is made up of. If that molecule now forms larger aggregates, the mass of the resulting thing (still a single ‘thing’ albeit larger) is the sum of the masses of its constituents. And so on.

You just need to beware in case you have aggregates that are not strictly $\ce{AB}$. Consider two proteins that form an $\ce{AB2}$ complex: In this case $M(\ce{AB2}) = M(\ce{A}) + 2 \cdot M(\ce{B})$, because we have two $\ce{B}$ in the final complex.


Why is the nomenclature glucose 6-phosphate, not 6-phosphateglucose? What exact nomenclature rule define the naming of glucose 6-phosphate?

Reading this closely, technically, the compound should be 6-phosphonoglucose but glucose 6- phosphate is acceptable usage, probably derived from considering it as a salt analogue rather the ester that it is.


Heat of formation method For enthalpy calculations why is it that in some cases the "heat of formation method" is used? I know that $ΔH_f$ represents the heat required to form the compound at standard temperature and pressure ($25^oC$, $1$ atm) but why is it used?

For example, in $Ex.1$

  NO(g, $300^oC$):  $H$ = $\int (Cp)_{NO}dT$
  where $T_1 = 25^oC$ and $T_2 = 300^oC$

The enthalpy of nitrogen monoxide is simply the integral from the initial temperature to the final temperature.

However, in $Ex.2$

  $CH_4$($150^oC$):  $H$ = ($ΔH_f)_{CH_4}$ +$\int (Cp)_{CH4}dT$
  where $T_1 = 25^oC$ and $T_2 = 150^oC$

The enthalpy of methane is the heat of formation + the sensible heat from the initial temperature to the final temperature.

What tells me when I should add the heat of formation?

You're confusing different enthalpies (don't worry, it seems to be rather common). Let's stick with NO as an example. The standard enthalpy of formation, $\Delta_fH^\circ$, is defined as the enthalpy of the reaction $\ce{0.5N2 + 0.5O2 = NO}$ at any given temperature. That is, if you need $\Delta_fH^\circ(298\textrm{ K})$, or $\Delta H^\circ_1$, your reaction is $$\ce{0.5N2(gas, 298) + 0.5O2(gas, 298) = NO(gas, 298)}, (1)$$ and for $\Delta_fH^\circ (T)$, or $\Delta H^\circ_2$, provided that the gases in question do not liquify at $T$, the formation reaction is $$\ce{0.5N2(gas, T) + 0.5O2(gas, T) = NO(gas, T)}. (2)$$

Remember that $\Delta_fH^\circ$ for the elements in their stable states is zero at any temperature.

Now let's say that we have to calculate $\Delta_fH^\circ(T)$ using the known value of $\Delta_fH^\circ(298\textrm{ K})$. For that we need to know the heats of three other processes: $$\ce{N2(gas, 298) = N2(gas, T)}, (3)$$ $$\ce{O2(gas, 298) = O2(gas, T)}, (4)$$ $$\ce{NO(gas, 298) = NO(gas, T)}. (5)$$

The enthalpies of $(3)$, $(4)$ and $(5)$ are basically the enthalpies of heating from $298$ to $T$. They are often called "enthalpy increments" and are defined as $\Delta_{298}^TH^\circ=\int_{298}^T C_p\,dT$ . Note that the enthalpy increments can be written between any two temperatures, not just from 298 K.

It is clear that when we add $(1)$ and $(5)$, and subtract half the $(3)$ and $(4)$, we'll get $(2)$. The same with enthalpies: $\Delta_fH^\circ(T) = \Delta_fH^\circ(298\textrm{ K}) + \int_{298}^T \Delta C_p\,dT$, where $\Delta C_p$ is the difference between $C_p$ for products and reactants of reaction $(2)$ with stoichiometric corresponding coefficients (1.0 and 0.5 in our case) taken into account.

Returning to your examples: in the first one, you are asked to find the enthalpy increment for $\ce{NO}$, which is just the amount of heat required to, well, heat it up. In the second example, you are asked to find the enthalpy of formation of $\ce{CH4}$, and I hope that my example above shows you the way of doing it. Also, if you copied the formulae from the textbook right, there's $\Delta$ missing from under the integral (should be $\int \Delta C_p(\ce{CH4})\,dT$).

A rule of thumb: when in doubt about the enthalpies of any process, write down the process thoroughly, with temperatures and phases of all reactants and products. This way it becomes much clearer. It is also useful to remember that in the real world there is no such thing as "The Enthalpy": what one has to deal with is always some change, some $\Delta$.


Reaction of 2,3-dichlorobutanoate with a decarboxylating agent followed by hydrochloric acid and then aq. potassium hydroxide Question

$$\ce{CH3-CH(Cl)-CH(Cl)-COO- ->[-CO2]A->[HCl]B->[aq KOH]C}$$ Product C is:
a) $\ce{CH3-CH2-CHO}$
b) $\ce{CH3-CH(OH)-CH2-CH2-OH}$
c) $\ce{CH3-CHO}$
d)$\ce{CH3-C(=O)-CH3}$

Answer

 The answer given is (a)

My attempt

Where did I go wrong?

Considering the follow up doubt, why geminal and not vicinal?

the carbocation formed is more stable because of $\ce{Cl}$ +M.


What is the Ka of OH- and Kb of H3O+? What is the $K_\mathrm{a}$ of $\ce{OH^-}$ and $K_\mathrm{b}$ of $\ce{H_3O^+}$? Have these constants been determined?

In water both equilibrium constants will be extremely small; the concentrations of $\ce{O^2-}$ or $\ce{H4O^2+}$ ions in water are essentially zero.

According to Cotton and Wilkinson's Advanced Inorganic Chemistry, for the hydrolysis of solid oxide ions you have

$$\ce{O^2- (s) + H2O (l) -> 2 OH- (aq)} \qquad \qquad K_b > 10^{22}$$

so you can never have any significant concentration of oxide ions in aqueous solution. As DavePhD points out, $K_\mathrm{a}$ for the hydroxide ion is less than $K_\mathrm{w}/K_\mathrm{b} = 10^{-36}$.

Now, $\ce{H3O+}$ may have some capacity to act as a base in media like $\ce{HF:SbF5}$ (see this paper on the role of $\ce{H4O^2+}$ in isotopic exchange reactions between hydronium ions). $\ce{H4O^2+}$ can exist in sulfolane solution, too (see this paper).

If $\ce{H4O^2+}$ exists in water at all, it's probably going to actually be two protons bridged and hydrated by a lot of waters, rather than an actual $\ce{H4O^2+}$ ion. It will have a much shorter lifetime than even a hydronium ion. The molecule can exist in theory, though, and its electronic structure has been studied for at least half a century (see Rosenfeld's 1964 paper including SCF calculations on $\ce{H4O^2+}$).


Accessible statement on the intuition and usefulness of group theory to chemistry I am writing a lesson plan for high school students who are studying group theory,
and I am having difficulty in relating the subject matter to the real world.
I have quite a few examples using the Rubik's cube, but I would like to use
some statement that might provide some intuition to students of how group theory
is useful for chemists.
I understand that group theory is used with spectroscopy, but I am not sure how to
relate this back to the theory.

I have seen some good reasonable non rigorous attempts to relate group theory to
the insolubility of quintic polynomials, but I think it would great to have a second
more concrete example, such as chemistry.

I understand this may be a big thing to ask, and would be happy with just a broad brush strokes answer without being too technical or trivial.

Individual molecules rely on point group symmetry - that is, the symmetry about a point. Crystal structures rely on space group symmetry including the translational symmetry in a crystal.

Many molecules, particularly important ones, are highly symmetric. The different symmetry operations (reflections, rotations, etc.) form a finite group.

There are a number of interesting group theory results, but a simple statement comes down to two things:

Understanding Orbitals, Bonding, Vibrations..: Using point groups, any function of the molecule corresponds to a linear combination of irreducible representations. This means I can quickly understand vibrations of a molecule, spectroscopy (as you mentioned) and orbital diagrams and bonding using just pencil and paper.
Faster Calculations: Using symmetry, you can greatly reduce the amount of time required for quantum calculations. For one, you only need to consider the symmetry-unique atoms and bonds. For another, you can show that certain types of integrals and calculations will always be zero and can be completely ignored. This leads to immense speedups.

For example, chemists label orbitals with symmetry labels based on the point group. Using the direct product we can realize that certain interactions will never occur.

Similarly, vibrations have symmetry labels, and we can determine what types of spectroscopy will interact with those vibrations, or whether they are "silent."


Calculation of overall rate constant of a parallel reaction Consider a parallel reaction

$\ce{A->B}$ (rate constant for this reaction is $k_1$ and it follows first order kinectics),

$\ce{2A->C}$ (rate constant for this reaction is $k_2$ and it follows first order kinetics).

Then we can write
$$\mathrm dA/\mathrm dt = \mathrm dA_1/\mathrm dt + \mathrm dA_2/\mathrm dt$$
(where $\mathrm dA_1$ is the amount of $\ce A$ going through the first path and $\mathrm dA_2$ is the amoubt of $\ce A$ going through 2nd path),

$$\mathrm dA/\mathrm dt = k_1[\ce A] + 2k_2[\ce A]$$

Then to calculate the $k$ where $k$ is the rate constant for the overall reaction we write

$$k[\ce A] = k_1[\ce A] + 2k_2[\ce A]\tag i$$

And so $k$ comes out to be $k_1+2k_2$.

My doubt is that to write $-\mathrm d[\ce A]/\mathrm dt = k[\ce A]$ did we assume that in the overall reaction the coefficient of $\ce A$ in the overall reaction to be 1.

Taking a general case
$$\ce{$a$A->products}$$
we can write
$$-\mathrm d[\ce A]/\mathrm dt=ak[\ce A]$$

But according to the equation $\text{(i)}$ we took $a=1$. Why?

If suppose $\ce A$ is a radioactive substance and we have to calculate the mean life of $\ce A$ then we require to know the overall rate constant, so foe that what should be the value of $a$?

The key to answering these kinds of riddles is to think about units.  Rates of reaction have units of $\mathrm{\frac{mol}{L\cdot s}}$, but that' isn't fully specified.  You need to know moles of what.

In your case, the confusion is stemming from confusion of what the "moles" in $\mathrm{\frac{mol}{L\cdot s}}$ refers to.

The answer can either be moles of "reaction", or moles of A.  The other key thing to realize is that you can write "the" reaction however you want.  The value of the "rate constant" $k$ depends on how you define the rate.  If for the overall reaction you write:

$$\ce{3A -> \varnothing}$$

...and if the "rate" you are calculating is the rate of reaction, then there are three moles of A lost per mole of reaction, so

$${dA \over dt}=-3k_{\alpha}\ce{[A]}$$

If instead you decide "the" reaction is

$$\ce{A -> \varnothing}$$

then you can rate the rate of disappearance of A as

$${dA \over dt}=-k_{\beta}\ce{[A]}$$

The choice is up to you and it is arbitrary.  You just have to keep track of the moles of what the rate laws you write refer to.

  Then to calculate the k where k is the rate constant for the overall reaction we write [[Equation (i)]]

You need to think carefully about what you meant by the words "the overall reaction".  When you write equation 1, you are adding together rates of loss of A, i.e. the units are $\mathrm{\frac{mol~of~A}{L\cdot s}}$.  The only reaction you can write that has the same rate is one where the coefficient of A is 1.

"The" half-life for radioactive elements usually refers to units where units are $\mathrm{\frac{mol~of~element}{s}}$.  The fact that the unit is $\mathrm{\frac{mol~of~element}{s}}$ and not $\mathrm{\frac{mol~of~reaction}{s}}$ means that if we wanted to write the corresponding chemical reaction, the stoichiometric coefficient had better be 1.$%edit$


Should titration stop at the intermediate color or a drop past the intermediate Using Bromcresol green as an indicator, should one stop when the blue color changes to green or to yellow for more accurate results.

Shouldn't one aim to get a green because its far closer to the end point than a yellow could be. For example, if the solution turned yellow and an additional drop was added, it would be indiscernible.

Neither of your options is correct.

Ideally an acid base indicator would be colorless at one pH and colored after the transition. Such an indicator is phenolphthalein. So when using phenolphthalein it is indeed impossible to detect when all the phenolphthalein is colored. So you titrate to a point where there is the faintest pink color.

So for bromcresol green you should either stop when the solution starts to turn from blue to blue green (ph about 4.6), or as the solution turns from green-yellow to yellow (ph about 5). It would very difficult to judge the deepest green (at about 4.8) accurately.

So which transition you would choose would depend on balancing between which side of pH 4.8 that you wanted to end on, and which transition was easier to detect visually.

You can look at a list of some acid/base indicators at Wikipedia. There are few that go between colorless to colored. Most are color-1 to color-2.

I also point out that you question contains a fallacy.

    Should titration stop at the intermediate color or a drop past the intermediate

The burette column is marked "to contain" an amount of fluid along its length. So you can let a partial drop form, then touch the tip of the burette to the flask and get a partial drop of titrant. You don't have to titrate in "quantized" drops.


What is the timescale of disulfide bond formation? What is the timescale of disulfide bond formation in a given protein? What influences it? Finally, how does one actually measure it?

From what I understand, the reduction and oxidation of bonds happens on the order of 10-8s. However, it appears that the timescale of disulfide bond formation in macromolecules scales with the size of the macromolecule.

I agree that the actual redox process is very fast. The rate limiting issue with proteins and other macromolecues involves the reactive groups finding each other. Thus, from the time of translation to the time of disulfide formation, the peptide chains need to fold properly. Perhaps other subunits need to be assembled and attached. Once everything is in the right place, the disulfide bond should form quickly provided there is an appropriate oxidant present.

You could measure the rate of the entire process by attaching reporters to positions on the protein - say a fluorophore and a quencher. When the process is complete, the quencher and the fluorophore are close in space, and fluorescence stops. This approach will measure the overall rate of the entire process. I am not sure if you can separate out the different components, e.g. specifically the S-S bond forming. I can't point to a specific paper where this is strategy is used, since I am away from my usual computer that likes my journal subscriptions. Hopefully I will be able to update my answer.


What happens during cleaning silverware? I know two distinct recipes for cleaning silver:

Put them into a pot with some club soda and a piece of aluminium foil and pour over a hot water.
Gently clean silverware with a fine chalk powder (obtained from a pharmacy).

I'd be happy to know:

What happens chemically during these procedure - why does it work, and what are the byproducts of these reactions?
How toxic (if at all) are the byproducts?
How much silver is lost in such a procedure?
Are these procedures "idempotent"? I mean, if I apply one of these procedures to an already cleaned item, will it consume some more silver, or will it remain intact? In other words, do the reactions consume just tarnish or do they consume pure silver too?

Put them into a pot with some club soda and a piece of aluminium foil and pour over a hot water.

I am familiar with this process. I am not familiar with the other. Perhaps someone else can provide an answer for it.

  What happens chemically during these procedure - why does it work, and what are the byproducts of these reactions?

This process cleans silver by a redox process. Aluminum is a more active metal than silver. This means that aluminum metal, Al(s), will reduce silver salts (salts of Ag+). The most common type of tarnish on silver is silver oxide, Ag2O.
The reaction is:

$$\ce{3Ag2O(s) + 2Al(s) -> 3Ag(s) + Al2O3(s)}$$

The club soda - containing sodium bicarbonate - plays two roles. First it acts as a supporting electrolyte so that electrons can be conducted through the solution. Second, the carbonic acid in it dissolves the passivation layer of aluminum oxide that protects the aluminum foil. This produces aluminum carbonate, which is unstable in water and decomposes into amphoteric aluminum hydroxide, which is soluble through the formation of a number of complex ions.

$$\ce{Al2O3(s) + 3H2CO3(aq) -> Al2(CO3)2(s) + 3H2O(l)-> 2Al(OH)3 (aq) + 3CO2(g)}$$

The net reaction is then:

$$\ce{3Ag2O(s) + 2Al(s) +3H2CO3(aq) -> 3Ag(s) + 2Al(OH)3(aq) + 3CO2(g)}$$

The byproducts are carbon dioxide and aluminum hydroxide.

How toxic (if all) are the byproducts?

Not very. Aluminum salts have very low health impact on you or the environment. Additionally, you produce only very small amounts of aluminum salts this way.

How much silver is lost in such a procedure?

None (Almost). The process coverts solid silver oxide into solid silver metal, both of which are negligibly soluble in club soda (or any other aqueous solution). The $K_{sp}$ of AgOH (which is what the surface of Ag2 looks like when exposed to water) is $1.52\times10^{-8}$. All of the action happens at the surface of both metals. The only common silver salts of appreciable solubility in aqueous solutions are silver nitrate and silver sulfate.

Is this procedure "idempotent"?

Yes. This procedure converts silver oxide into metallic silver. It has no effect on metallic silver: metallic silver does not react with most acids; it cannot be reduced farther by aluminum metal; and it is not soluble in water or most aqueous solutions.


Is the decomposition reaction of nitrogen monoxide to dinitrogen and dioxygen possible? Is the reaction $\ce{2NO -> N2 + O2}$ possible? Of course with some kind of catalyst plus high temperature. How can I find the answer to such questions and where can I find more information about this reaction?

You should look up enthalpy of formation and enthalpy of reaction in your textbook. Also have a look at (Gibbs) Free energy change on reaction. This will tell you  if the reaction is going to be spontaneous, as it is in this case.
The thermodynamic energies never tell you how fast the reaction will be as these relate only to the starting (reactant) and ending (product) energies. The activation energy, the barrier between reactants and product is unknown to thermodynamics.
Secondly, a catalyst will not change the thermodynamics of the reaction, the starting and ending states are the same, but it speeds up a reaction by allowing the reaction to proceed by an alternative pathway with a smaller activation barrier.


Kinetics Vs Thermodynamics : Apparent contradiction in the definition of Enthalpy in the two disciplines My book gives the following curve:

It gives the following relation :$$ΔH= E_{\mathrm{activation,forward}}-E_{\mathrm{activation,backward}} \tag{1}$$
But I suspect that $ E_{\mathrm{activation,forward}}-E_{\mathrm{activation,backward}}$ corresponds to change in Internal Energy of the system ΔU.
or $$ΔU= E_{\mathrm{activation,forward}}-E_{\mathrm{activation,backward}} \tag{2}$$
However,at a constant pressure, heat of reaction or Enthalpy change $$ΔH = ΔU + PΔV \tag{3}$$
Statement (3) contradicts statements (1) and (2) or is valid only for $ΔV=0$

This source contains the same graph : https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Book%3A_Introductory_Chemistry_(CK-12)/18%3A_Kinetics/18.04%3A_Potential_Energy_Diagrams

What is going on here?
In case, you feel that the graph is given only to cover a particular kind of reaction, please provide relevant comments.

The reason might be that while drawing the reaction energy profile, we forget to mention what energy we are mentioning in the Y-axis. The following conventions are generally used:

If reaction conditions are constant NVT, energy in Y-axis should represent internal energy.
If reaction conditions are constant NPT, energy in Y-axis should represent enthalpy.
If reaction conditions are constant $\mu$VT, energy in Y-axis should represent Helmholtz free energy.
If reaction conditions are constant $\mu$PT, energy in Y-axis should represent Gibbs free energy.
If you are looking at single molecule, the energy will be the total energy of the molecule (kinetic energy + potential energy).

In case of reactions, the last three are generally used.
Hence, the "Energy" Y-axis changes based on reaction conditions. In the question, it seems you might have got mixed up somehow.

Summary of the terms used:

N: No of molecules
$\mu$: Chemical potential
V: Volume
P: Pressure
T: Temperature


Why would the following reaction not proceed through SN2 reaction? I made a guess that the OH- group cannot do a backside attack on the carbon because the C double bond with O and the 5 carbon ring may act as steric hindrance and prevent it from forming a bond but I'm not really sure.

The substrate in this reaction is an Acyl Chloride, which is highly susceptible to nucleophilic acyl substitution, more so than other carboxylic acid derivatives, so SN2 attack (as mentioned in the comments) is highly improbable. The mechanism for this nucleophilic acyl substitution is as follows:

With equilibrium constant $K_c >> 1$ favouring the products on the right.