text:
Calculating temperature change of helium due to Joule-Thomson effect
We are trying to fill a rocket helium tank that is 78 liters with helium, final pressure of $\pu{10500 psi}$. We plan to pressurize high pressure tanks to $\pu{20000 psi}$ using a compressing pump system, so that on launch day we don't have to wait for the 78 liter tank to fill slowly by the compressing system. We plan to fill the rocket tanks from the high pressure tanks in a cascade manner (the number and volume of the tanks may vary, depending on the solution of the JT equation(s).
The problem is that helium has a negative JT coefficient. It will heat up during the transfer from the high pressure tanks to the lower pressure tank if I understand it correctly.
So, the concern is how hot will it get. Will the O-rings/seals melt? I have some equations, but it seems to me there may be too many unknowns.
I've seen others make certain assumptions on this site, but its not clear to me how that works. Can anyone advise how you would proceed?

Can we please focus on a typical single tank transfer as an example.  Tank volumes, pressures before transfer, etc?

Good question. I'll have to give more details tomorrow after I get to my desk and determine what tank volume is. Pressure before is 20,000,desired final  pressure in the 78 liter receiver is 10,500. The supply tank size may vary, I'll provide an update tomorrow.

So, now that I'm back at my desk it's all coming back to me. The design so far is to use four high pressure storage tanks pressurized to 20,000 psi. Open one to fill the rocket tank as much as it can towards the 10,500 psi GHe. Then close that valve and open the next one, and cascade until the desired pressure is reached.

The 78 liters converts to 4730 cu. in. for the rocket tank. The pressure vessels are 890 cu. in.  P3=(P1V1+P2V2)/(V1+V2)

So initially with P1 @ 20,000psi, V1 at 890 cu. in and the rocket tank at 14.7 psi and 4760 cu. in. the settling pressure is 3162 psi. On the next tank we get to 7,108 psi. The third tank gets us to 10,130 psi. The fourth tank could get us all the way to 12,443 psi, but we'd stop when we reach the desired pressure.  The question is, how to calculate the heat gain due to the helium transfer?

I still don't follow what you are trying to do.  Here is my understanding:  You start out with an 890 cu in tank at 20000 psi.  You then use it to fill a tank initially at 14.7 psi to a pressure 4760 cu in to 3162 psi.  After that, I'm lost.  Have you found an enthalpy-pressure diagram for helium that you can use?

I see that the formatting allows a way to place an image on here but the instructions aren't clear, does the image have to be hosted on another site with a link or can it be included directly in the comment?

@ChesterMiller I appreciate your interest. Try sketching 4 pressure holding tanks side by side. Put an valve on each tanks input and out put. Connect the output of each tank to a manifold that connects them to the rocket ship tank. Fill each of the 4 tanks individually to 20,000 psi, one at a time. Close all inlet valves. Open Tank 1 outlet valve to the rocket, wait for the tank to fill to the settling pressure. Close Tank 1 outlet valve.

Open tank 2 outlet valve to the rocket, wait for the tank to fill to the settling pressure. Close Tank 2 outlet valve. Open tank 3 outlet valve to the rocket, wait for the tank to fill to the settling pressure. Close Tank 3 outlet valve. Open tank 4 outlet valve to the rocket, wait for the tank to fill to the settling pressure, or until we reach launch pressure. I hope that is clear enough. Now, Helium has a negative JT constant, it heats up when it goes from a high pressure (20,000 psi) to a lower pressure. How much heat does the helium gain at each valve opening to the rocket tank?

I was unable to fine an enthalpy-pressure diagram for helium. I did find an Enthalpy Entropy Chart for helium on page II-F-2 on the PDF at https://www.bnl.gov/magnets/staff/gupta/cryogenic-data-handbook/Section2.pdf that has lines for pressure and temperature, would that be useful?

The pressures don't go high enough in the charts.  And I get the picture of what the system is.  One final question:  the "settling" pressure:  this is the pressure at which you shut the valve, or is it the pressure at which the two tanks equilibrate in pressure?

The worst case scenario is when the helium is going through the valve from 20000 psi in the first tank to 14.7 psi (just as the rocket tank begins to fill), right?

I would agree that the worst case temperature increase would be going from the highest pressure to the lowest.  The "settling pressure" is what I think is where equilibrium occurs by using the P3=(P1V1+P2V2)/(V1+V2) calculation, it is theoretical.

In any event, see my answer below.

To do this analysis, one needs data up to 20000 psi (1400 bars) on either enthalpy vs temperature and pressure (a graph) or compressibility factor Z vs temperature and pressure.  The only graphs I have found of the former type go up to only 100 bars, which is a factor of 14 too low.  However, I have found data on the compressibility factor Z at temperatures up to 300 K and pressures up to 1000 bars:  https://cds.cern.ch/record/1444601/files/978-1-4419-9979-5_BookBackMatter.pdf
Although this is still a factor of 1.4 too low, it might provide some idea of the temperature rise that might be expected in the valve.
So here is how the data would be used.
The effect of pressure on enthalpy (per mole) of gas is given by $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P\tag{1}$$For a real gas, the equation of state in terms of the compressibility factor Z=Z(P,T) is given by$$PV=ZRT\tag{2}$$If we substitute Eqn. 2 into Eqn. 1, we obtain:  $$\left(\frac{\partial H}{\partial P}\right)_T=-\frac{RT^2}{P}\left(\frac{\partial Z}{\partial T}\right)_P\tag{3}$$Integrating Eqn. 3 between P=0 and arbitrary P at constant temperature yields the so-called Residual Enthalpy $H^R$:$$H^R(P,T)=-RT^2\int_0^P{\left(\frac{\partial Z}{\partial T}\right)_{P'}\frac{dP'}{P'}}=-RT^2\frac{\partial}{\partial T}\left(\int_0^P{(Z(T,P')-1)\frac{dP'}{P'}}\right)\tag{4}$$where P' is a dummy variable of integration.
If the final pressure coming out of the valve is low (so that the gas exiting the valve is in the ideal gas region), we can write:  $$\Delta H=-H^R+C_p\Delta T=0$$where, for a monoatomic gas like Helium, $C_p=\frac{5}{2}R$.  Therefore, $$\Delta T=-\frac{2}{5}T^2\frac{\partial}{\partial T}\left(\int_0^P{(Z(T,P')-1)\frac{dP'}{P'}}\right)\tag{5}$$
This expression would be evaluated using the data presented in the reference above.
If a reference can be found with data going out to 1400 bar, that would be even better.

Thank you @Chester Miller for your effort in putting all this together. If I understand this correctly, I would have to integrate the equation inside the parenthesis, take the derivative of that over ∂T, then multiply that by -2/5T squared, is that what you are telling me?

Yes.  You would use the data in the reference, and integrate numerically with respect to lnP, and differentiate numerically with respect to T.  You would also be extrapolating to fill in the blank between 1000 bars  and 1400 bars.

To calculate the Joule Thomson heating or cooling there are basically two approaches.
When considering a small pressure change, the temperature change can be approximated by multiplying the JT coefficient at the outlet conditions by the pressure change. The issue with this is that on beforehand you don't know what the outlet temperature is, so you need to assume the same as the inlet, which causes errors, as the JT coefficient is temperature dependent.
The better approach is to use a TP diagram with isenthalps. This can be used to accurately determine the final temperature. It is based on the fact that the expansion is isenthalpic, i.e. the enthalpy doesn't change. Hence, you plot your initial condition in the PT diagram, then follow (parallel to) the isenthalp to your final pressure, and simply read the final temperature from the diagram. You can use this approach with an actual diagram, or you could use a computer program, which would perform the underlying calculations for you. In that case you would need to perform a two-step calculation:

At the initial conditions (T1, P1) you calculate the enthalpy H
Use the final pressure (P2) and the calculated enthalpy to calculate the final temperature T2

Many software programs exist that can perform these calculations, such as NIST Refprop.
Alternatively, you could manually perform the above calculations using a simpler Equation of state, such as Peng-Robinson, for which you only need to know the following parameters of Helium for the calculation:

Critical temperature (5.19 K)
Critical pressure (227400 Pa)
Acentric factor (- 0.36)
Ideal gas Heat capacity (20.78603 J/molK)

This will be somewhat less accurate, and a lot more work, than the use of dedicated software. Fortunately you're looking at Helium, which is just about the most well behaved gas there is, so the inaccuracies using a simple Equation of state are likely to be fairly small.


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Which Oxygen atom in HCOOH (formic/methanoic acid) does Carbon donate its electrons to, to obtain a partial positive charge?
I was wondering if the Carbon atom in HCOOH (methanoic/formic acid) forms a positive partial charge by donating its electrons to both the Oxygen atoms, since they both possess a higher electronegativity value, or just one of them.
I was also curious about how the single bond and double bond between Carbon and the Oxygen atoms would affect the electron donation.

I'm guessing the OP was trying to render $\ce{HC(=O)OH}$.

Electronic effects are cumulative. For example, the central carbon atom in an orthoester like HC(OEt)3 is surrounded by 3 oxygen atoms which add up.
In the case of formic acid (which is what you described in your question), yes the carbon atom is donating its electron density to BOTH oxygen atoms. However, the oxygen atom involved in the C=O bond will attract more electrons, because the double-bond is shorter.

There is no straightforward answer to such a question, as it is not always obvious to indicate how one should count electrons being donated to other groups, or how to correctly partition electrons between nuclei.
In 1978, using ab initio calculations F.A. Momany has computed the Mulliken charges of the atoms of formic acid at the STO-3G level, and compared this to the use of an electrostatic potential. If one only looks at the Mulliken partial charges, this would tentatively answer the question.
The partial atomic charges of the C atom, the C=O oxygen and the OH oxygen are respectively  is 0.262, -0.294 and -0.256. So it seems that the C might indeed be donating to both of the oxygen atoms. But one should also consider that the H atom bound to the OH oxygen is also donating some of its electron density to the O: the partial charge of the OH H atom is computed at 0.214. Thus, overall, the C atom is only contributing by a net positive charge of 0.042 to the O atom of the OH group.

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Probability of the single molecule reaction to happen
I am trying to simulate the single particle second order reactions with known rate constants $k_1$ and $k_{-1}$ in Python. The point of this simulation is to fit the real experimental data to the model and see which theory is working the best.
I have found many different equations for the probability of reactants $\ce{A + B}$ going to $\ce{C}$ and vice versa. The good explanations I found here and here, but the equations are not the same.
So far, I have used those equations:
$$
\begin{align}
\ce{A &->[$k_1$] B} &\quad &p(\ce{A}) = p_a = k_1[\ce{A}]\mathrm{d}t \\
\ce{A + B &->[$k_1$] C} &\quad &p(\ce{A}) = k_1[\ce{A}][\ce{B}]\mathrm{d}t / (V ([\ce{A}_0] + [\ce{B}_0]))\\
\ce{A + A &->[$k_1$] B} &\quad &p(\ce{A}) = 2k_1[\ce{A}]\mathrm{d}t/(2V[\ce{A}_0])([\ce{A}_0] - 1)
\end{align}
$$
I don't expect those equations to be correct, but they work. I would be very grateful for some hints or suggestions!
EDIT: I will post the sample code I was using:
 #A - list with 1 for particle in A or 0 for particle in B, starts with all as 1,
 #e.g:
 A = [1,1,1,1,1,1,1,1]

    for iteration, element in enumerate(A):
        A_conc = len(A[A > 0]) # number of particles in state A

        p(A) = k1 * A_conc * dt      # probability of changing to A->B

        if random.random() > pa:     # checking if random event is favouring transition
            A[n] = 0                 #A reacts to B

This is the way I want to use those equations, but still, the question remains if the equations are correct.

The Gillespie method is the one to follow, see your first 'here'. But you cannot write down rate equations as such but have to calculate via an algorithm. (I cannot  post the method now but will  later if you have not found out by searching for his papers or book)

I would really appreciate it. I was using random number generator, and if the random number was greater than the probability, the molecule reacted. Anyway, I will wait for your reply!

When two species $A$ and $B$ react, $A + B \to product$, the rate of reaction is $k\mathrm{[A][B]}$ where $k$ is the rate constant and $\mathrm{[A]}$ and $k\mathrm{[B]}$ the concentrations at any time during the reaction, which clearly change with time. To work out what happens in each small time interval means calculating the change in the number of reactant molecules in a given volume. Doing this means calculating two things. The first is the time at which any of possibly several reactions will occur, and the second is to decide which reaction this will be.
In the reaction $A + B \to product$, there is only one type of reaction but there are very many schemes where there are two or more reactions; for example, the photo-dissociation of halocarbons with UV light and subsequent reactions to destroy ozone.
$$\mathrm{CFCl_3} \overset{h\nu}\to \mathrm{CHCl_2  +Cl}\qquad \mathrm{Cl+O_3} \to \mathrm{ClO+O} \qquad \mathrm{ClO+O_2} \to \mathrm{Cl+O_2} $$
Given such a reaction scheme, the first step is to decide what reaction will happen next given the current state of the reaction at time $t$. At the level of considering individual molecules, once the reaction has started, it does not follow sequentially as shown in the scheme. This only happens when a whole ensemble of molecules, typically as large as Avogadro's number is examined, and then normal kinetic equations apply because of the averaging over many events that occurs. In any given time interval when considering molecules one at a time, it is not possible to predict exactly which reaction will happen next. It is quite possible, for instance, that several $\mathrm{Cl+O_3}$ reactions may occur before either of the other two. The best that can be done is to guess the outcome, and repeat the calculation many times and this approach is now described.
Suppose that the reactions are labelled with a number $R$ that represents reaction 1 or 2 or 3 etc. The chance of a reaction of type $R$ occurring will be defined as
$$prob(T + dT, R) \equiv a_RdT$$
Specifically, this is the probability that a reaction of type $R$ occurs in time interval $T$ to $T + dT$ in the reaction volume, given the current state of all reactions at an earlier time $t$. The values of $a_R$ can be easily worked out, and this is shown later. What is now needed is the probability $P_0(T)$ that no reaction of any type occurs up to time $T$ past the present time $t$, multiplied by the probability that only a reaction of type $R$ occurs in the small time interval $T \to T + dT$, which is defined as $a_RdT$. The probability of reaction type $R$ happening is therefore
$$ P(R,T)= P_0(T)a_RdT$$
and shown schematically in the figure.

Calculation of reaction probability.

To calculate the probability of no reaction occurring, consider any small interval $d\tau$ after some arbitrary time $\tau$, making an interval $\tau \to \tau + d\tau$. The probability of no reaction is then
$$P_0(\tau+d\tau) =P_0(\tau)\left(1-\sum_{R=1}^n a_Rd\tau  \right)$$
where the summation is over all $n$ reaction types $R$. The summation is made because this gives the chance that any one of the $R$ reactions will occur, and therefore one minus this number is the chance that no reaction occurs. Rearranged, this equation gives
$$ \frac{P_0(\tau+d\tau)-P_0(\tau)}{d\tau}=-\sum_{R=1}^n a_R$$
and in the limit of a small time interval $d\tau$, it becomes the true derivative $\displaystyle \frac{dP_0}{d\tau}=-\sum_{R=1}^na_R$.
For clarity only, a notational change is now made so that
$$a_0=\sum_{R=1}^n a_R \tag{8}$$
and therefore $\displaystyle \frac{dP_0}{d\tau}=-a_0$.The chance of no reaction up to time $T$ is found by integrating, and is $\displaystyle P_0(T)=e^{-a_0T}$.
The final result for the probability of reaction R occurring at time $T$ is the chance of no reaction up to $T $ multiplied by the chance of reacting at $T$, which is $a_R$, giving
$$P(T,R) = a_Re^{-a_0T} \tag{9}$$
an expression that shows that these times are Poisson distributed. Equation 9 is a joint probability density function, as it is the chance that reaction occurs in the time interval and that it will be reaction $R$.
Two terms now remain undefined. The first is the individual $a_R$ and their sum $a_0$, and the second determines which reaction happens next. Calculating $a_R$ is very easy, as each $a_R$ is the product of the number of molecules present in a reaction of type $R$ multiplied with the rate constant per unit volume.
If there is just a single reaction $\mathrm{A} \overset{k_1}\to \mathrm{B}$ then $a_1=k_1N_A$ where $N_A$ is the number of $A$ molecules present at time $T$. If the reaction is $\mathrm{A+B} \overset{k_2} \to \mathrm{C}$ then $a_2=k_2N_aN_B$ and if both reactions are present in the overall reaction scheme which would then be
$$\mathrm{A} \overset{k_1}\to \mathrm{B} \qquad \mathrm{A+B} \overset{k_2} \to \mathrm{C} \qquad a_0=\sum a_R=a_1+a_2 \tag{10}$$
In the special case where a reaction combines two identical molecules $\mathrm{A+A} \overset{k_2} \to $ the number of indistinguishable combinations, i.e the number of distinct A-A pairs of the two species must be calculated and this makes $\displaystyle a = k_2\frac{N_A(N_A - 1)}{2!}$. Normally, in dealing with chemical kinetics, $N_A$ is so vast that $\displaystyle a = k_2\frac{N_A^2}{2}$, but in the Monte Carlo simulations, $N_A$ is small because individual molecules are dealt with, and the exact formula must be used. Thus, macroscopic, (kinetic), chemical equations are only valid when deviations from the Poisson distribution are negligible.

An aside
The term $\displaystyle a = k_2\frac{N_A(N_A - 1)}{2!}$ implies that these numbers are averages in the reaction volume at time $t$ and should be written as the average of the product, i.e. as $\displaystyle \frac{\langle N_a(t)(N_A(t)-1)\rangle }{2!}$. The question then arises as how to deal with this. Two approximations can be made; the first is to assume that $N_A$ is large compared to 1 and so $\langle N_A(t)(N_A(t)-1)\rangle \rightarrow \langle N_A^2(t)\rangle $, which seems to be a reasonably sensible approximation, the next step is more problematic and is $\langle N_A^2(t)\rangle \sim \langle N_A(t)\rangle^2$. This is a drastic step and assumes that the variance in the number of molecules is zero, i.e. that $N_A(t)=\langle N_A(t)\rangle$ which means that $N_A$ is a deterministic process, so that the random nature of the numbers of $N_A$ entering into the reaction volume is always the same, which is not what was assumed at the outset. (To see that the variance is zero use the definition $\sigma^2_{N_A} = \langle N_A^2 \rangle -\langle N_A \rangle^2$ where $\sigma$ is the standard deviation)
In any normal kinetic experiment the number of molecules is vast, $\langle N_A(t)\rangle \sim 10^{18}$ would not be untypical, so that the fluctuations in the  number of molecules is tiny and the approximation made above is valid.
$$ \frac{\sigma_{N_A}}{\langle N_A \rangle} \sim \frac{\sqrt{\langle N_A \rangle}}{\langle N_A \rangle} \sim \frac{1}{\sqrt{\langle N_A \rangle}}\sim 10^{-9}$$
Even if there are as few as 1000 molecules the relative fluctuation  is not too drastic but in experiments using small numbers of molecules, as in the case of many 'single molecule' experiments then clearly the rate equation approach needs to be modified.

The final part of the calculation determines which particular reaction step is going to occur and at what time this is done by choosing pairs of numbers that have the distribution $P(T, R)$ equation 9. Gillespie (1997, 2007) argues that $P(T, R)$ is equivalent to two distributions, $P_1$ and $P_2$;
$$P(T,R)= P_1\cdot P_2 = \left(a_0e^{-a_0 T} \right)\left( \frac{a_R}{a_0}\right)$$
and which can be determined by choosing two uniformly distributed random numbers, $r$, each in the range 0 to 1. The equation
$$T=-\frac{1}{a_0}\ln(r) \tag{11}$$
generates a random number $T$ according to the probability density function $\displaystyle P_1=a_0e^{-a_0 T}$.
Equation 11 produces times that are Poisson distributed, and the time $T$ is added to the current time of the reaction. The second distribution, $P_2$, is calculated by choosing a uniform random number according to the probability density function $a_R/a_0$. For example, the chance that reaction 1, (equation 10), occurs is $a_1/(a_1 + a_2)$ and for reaction 2 this chance is $1 - a_1/(a_1 + a_2) = a_2/(a_1 + a_2)$. Which reaction occurs is determined by choosing a random number between 0 and 1 and seeing which of the two regions it falls into. If it falls between 0 and $a_1/(a_1 + a_2)$, reaction 1 occurs; otherwise, reaction 2 occurs. If reaction 1 occurs, $\mathrm{A} \to \mathrm{B}$, then 1 is subtracted from the original number of $A$ molecules in each step of the reaction and 1 is added to the number of $B$ molecules.
If reaction 2 happens, $\mathrm{A+B} \overset{k_2} \to \mathrm{C}$ then 1 molecule is removed from each of the numbers for $A$ and $B$. How many molecules are added and subtracted, of course, depends on the stoicheiometry of each particular reaction.
In the case of a termolecular reaction $\mathrm{ A + B + C} \rightarrow $ when the probability of three molecules simultaneously colliding is calculated the probability the chance that this occurs in  an infinitesimal time interval $d t$ is proportional to $dt^2$ so becomes vanishingly small and thus does not occur with any physical reality. It is often convenient, however, to consider a termolecular scheme as a simplified approximation to a scheme such as $N_A+N_B \leftrightharpoons X^*; \; X^* + N_C \rightarrow$
The outline plan for the Gillespie algorithm is
(1) Define the initial amounts of each species.

Define the arrays of data points to hold results.   Set the maximum time. Set the time to zero; t = 0.

(2) Start a loop around the calculation until the time is up or no molecules are left.

Calculate $a_1,\;a_2,\;a_3$ etc. and $a_0 =a_1 +a_2 +a_3 +\cdots$ Calculate time $\displaystyle T = -\frac{1}{a_0}\ln(r)$
Increment time $t = t + T$
Calculate which reaction occurs: if a random number falls between 0 and $a_1/a_0$, reaction 1 occurs, and so forth for other reactions.
  Add or subtract numbers of molecules from the totals, depending on which reaction has occurred and its stoicheiometry.
  Store the results.

(3) Continue the loop until finished.

Refs
'Markov Processes. An Introduction for Physical Scientists' D. T. Gillespie, publ Academic Press 1992.
D. Gillespie  J. Phys. Chem. 81, p2340, 1977 eqn 29
The python code for a Lotka-Volterra predator-prey reaction is given below. I'm assuming that matplotlib and numpy, as np.blabla() etc are used.
def lotka():                # Lotka reaction Gillespie method
                            # x0 + Y = 2Y  (k1) ,  Y +  Z = 2Z (k2) , Z = Q(k3)

    #k1 =10.0   k2 = 0.01   k3 = 10.0
    #y0 =2000.0  z0 =2000.0
    ycount = np.zeros(bins,dtype=int)
    zcount = np.zeros(bins,dtype=int)
    dtime =  np.zeros(bins,dtype=float)
    for i in range(bins):
        dtime[i] = maxt*(i+1/2)/bins
        pass
    ny = y0
    nz = z0
    ycount[0] = y0
    zcount[0] = z0
    t = 0.0
    indx = 0
    while indx < bins :
        r1 = k1*ny
        r2 = k2*nz*ny
        r3 = k3*nz
        a0 = r1 + r2 + r3+1e-20
        t0 = -np.log( np.random.rand() )/a0
        t = t + t0
        indx = int(np.trunc( t*bins/maxt ))
        if indx < bins :
            r = np.random.rand()
            if  r > ( r1 + r2 )/a0  :
                nz = nz - 1
                zcount[indx] = nz          # react 3
            elif r > r1/a0 :
                nz = nz + 1
                ny = ny - 1
                zcount[indx] = nz
                ycount[indx] = ny          # react  2
            else:
                ny = ny + 1
                ycount[indx] = ny          # react 1
                pass
            pass
        pass         # events

    return zcount,ycount,dtime      # end lotka

    bins = 1000
    maxt = 200.0
    k1 = 0.5
    k2 = 0.001
    k3 = 0.5
    y0 = 400.0
    z0 = 100.0

    zcount,ycount,dtime = lotka()

    fig1= plt.figure(figsize=(10, 6))
    plt.rcParams.update({'font.size' : 16})  # set font size for plots

    plt.plot(dtime,zcount,color='red')
    plt.plot(dtime,ycount,color='blue')#'ro',markersize=2)

    plt.xlabel('time')
    plt.ylabel('population')
    plt.xlim([0,maxt])

    plt.tight_layout()

    plt.show()

Thanks this was very good explanation. Also thanks to your post I have done some more digging to understand the idea behind this algorithm (some parts of the code were not so clear) and found some good tutorial which might be beneficial for others: https://github.com/karinsasaki/gillespie-algorithm-python/blob/master/build_your_own_gillespie_solutions.ipynb

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text:
Is Beer-Lambert law also valid for fluorescence and luminescence?
When I measure optical density in a microplate reader the absorbance is proportional to the concentration and the height of the liquid. Is Beer-Lambert law also valid for fluorescence and luminescence?
Let's say I prepare a fluorescent solution and measure e.g. the signal of 100 μL and of 50 μL. Do I get the half signal for 50 μL?

Related (and is probably containing an answer to this question): https://chemistry.stackexchange.com/questions/38289/how-does-absorption-spectroscopy-work

I tried to read through the linked answer but unfortunately didn't understand it at all

No the Beer Lambert law  ($I_{trans}=I_0\exp(-\epsilon[C]L)$) does not apply as this shows that the transmitted light intensity decreases exponentially with concentration as it passes through a solution.
Fluorescence/luminescence is directly proportional to the concentration of molecules excited provided that the solution  is dilute so that dimers or excimers are not formed and that self-absorption of the fluorescence light by another part of the same solution does not occur.

Thanks for your answer. So beside the Beer's Law when I would measure 3 different volumes e.g. [25, 50, 75 uL] with the same fluorescence concentration will I receive a linearity, as more volume means more light or is that wrong? The reason why I am asking is about volume verification for pipetting devices. Optical density could do that too but the dynamic range isn't very high

Yes if you account for all porphyrin said. But even so you should integrate all the emitted light . Not viable at all. I would rather trick in a way that you always measures the same sample and reasons on dilution steps. This let you know if you are really picking up 25 50 75 ...


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Should I use MCPBA here?
I came across a question that asked for a method to convert propene to glycerol. I used the method in the pic, but I'm worried about two steps-
Step-2, This reaction is just made up by myself! I'm using a 'bulky' base to favour elimination. Is this reaction feasible? Will it occur without giving any byproducts?
Step-3, Can I use MCPBA to convert an alkene containing an alcohol group to its corresponding epoxide? I'm worried that MCPBA can protonate the alcohol to generate the carbocation (moreover, the allylic carbocation makes it even more probable!) and that will lead to byproducts.
That's all.
Any help will be appreciated.

Step 2 is a known procedure run industrially by passing propylene oxide over a catalyst example here https://patents.google.com/patent/WO2017087479A1/en

MCPBA is a pretty weak acid so I would not worry about the possibility of alcohol protonation. You could equally use the Sharpless procedure  conditions - it did win a Nobel prize  https://en.wikipedia.org/wiki/Sharpless_epoxidation

@Waylander Okay! So step 2 is reasonable reaction. Am I right?

@Waylander May I know the reason why MCPBA is a weak acid? Because it looks to me as a pretty strong acid because it has a carbonyl carbon withdraws electrons pretty well moreover it has a chloro group attached to benzene ring which enhances I- effect further.

@Waylander Wow! It's a very weak acid. I saw its Pka value 7.57. Would love to know the reason why it is so.


text:
Is there any difference between thermal energy at room temperature in fluids(water) vs air?
Context of the question: I was reading on hydrogen bonds, specifically how close they are to the thermal energy of the fluids. So I googled the value of thermal energy at room temperature, which was $\pu{2.5 kJ/mol}$. Which is a difference of $10$ order of magnitude (H bonds are $\pu{23 kJ/mol}$), which makes sense to me given the time taken to break ($1$ to $\pu{20 ps}$) and reform a bond ($\pu{0.1 ps}$). That is how the question I posed in the title arised.
Since sea water can retain heat for longer periods of time, and they release it more slowly as well – what is the difference between thermal energy at ambient room temperature for gas (atmosphere) vs fluids (water), and how does this change with volume?

This may provide some clarification: https://van.physics.illinois.edu/qa/listing.php?id=30379&t=heat-capacity-and-melting-points

First of all, just a minor thing, the thermal energy itself is always the same. You can't grab it, you can just measure it. What is different is the amount of energy "stored" in air or water, but its no different "energy".
The retainment of heat in the sea is due to the heat capacity.
https://en.wikipedia.org/wiki/Heat_capacity
This is a specific property of each molecule.
To increase the temperature of 1 g water by 1 K you need more Energy than for 1 g of air. That is the reason why the temperature near the sea is more moderate (absorbing heat in the summer and releasing heat in winter).
This is also a reason, why you can go into a sauna with 90 °C without burning yourself, but you can't grab into 90 °C hot water. This is also due to thermal conductivity but also due to the capacity. When your skin touches the 90 °C hot air, it is absorbing the heat providing you with cooler air around your skin. The same happens in water, but first, your skin has to absorb much more energy to cool down the water, second of all the surrounding water heats up the cooler water layer around your skin much quicker, than hot air does. That's why air is considered an isolation.

Ok, so there is no difference in thermal energy, but the stored energy changes. Water stores more energy. Got it.


text:
Natural orbital population analysis in Gaussian
Within (restricted) density functional theory and Hartree-Fock, respectively the Kohn-Sham orbitals and spin orbitals are the natural orbitals, with occupation numbers either 2 or 0.
As I understood, this also applies to spin-unrestricted calculations, is this correct?
Now, when doing UDFT or UHF with Gaussian and specifying pop=NaturalOrbitals, the natural orbitals are different from the KS/canonical orbitals and some occupations numbers are also different from 2 and 0. Unfortunately Gaussian manual provides no reference to how the natural orbitals analysis is done exactly and the literature is fairly confusing (spin-projection vs non-spin projection etc.).
Could anybody give me an overview or reference regarding what Gaussian exactly does when this pop=NaturalOrbitals keyword is turned on?

I really cannot find any documentation of what Gaussian actually does to produce the orbitals. You might have more luck asking that question on the CCL, as this mailing list is closer to the people who develop the code. However, judging from the definition in the Gold Book, natural orbitals, it seems to be a reasonably simple transformation.

Maybe this will help: https://doi.org/10.1103/PhysRev.97.1474

Thanks @Martin-マーチン, those references are very useful. As much as I have since understood, the KS/canonical orbitals of a broken-symmetry UHF solution are the "natural spin orbitals" BUT if one constructs the spinless first order density matrix and diagonalizes that, then the natural orbitals will be different from the KS orbitals.


text:
Would sodium explode in salt water?
I’ve seen the video etc of sodium being put into freshwater and the reaction of flames/small explosion but I wonder if the same reaction would occur if conducted in salt water such as the ocean.

https://chemistry.stackexchange.com/questions/2606/what-exactly-is-happening-when-sodium-potassium-or-other-alkali-metal-explodes

https://youtu.be/LmlAYnFF_s8?t=398

@uhoh: I believe https://youtu.be/e9AiK5zulf8 is a much more straightforward way to answer OP's question. :)

@Quuxplusone yes that certainly sounds like a coulomb explosion!

Yes, sodium metal is also going to react exothermically with salt water or any other aqueous solution as long as it comes in contact with water:
$$\ce{Na (s) + H2O -> Na+ (aq) + OH- (aq) + 0.5 H2 (g)}$$
eventually leading to explosion of hydrogen-oxygen mix forming near the water surface. Presence of sodium chloride in salt water isn't going to influence the reaction as it neither capable of shifting equilibrium to the left (gas formation), nor passivate sodium metal surface as it would do, for instance, some viscous oil.

According to this post, this is an incomplete explanation of the reason for explosion.

@Ruslan Of course it's incomplete because that's not what OP asked about. Thanks for providing the URL, now those who are also interested in how the detonation occurs, may quickly find the answer:)

Sodium catches fire even in humid atmosphere. Salt water will make no difference. It will react with water in solution of salt in water, and the heat evolved will be very high and will vapourize sodium and it will fly away and may cause injuries. Do not do this at home.

Yes it will, I tried it once, with 2 kgs straight into the Atlantic ocean. It'll sink, fizzle a little, pretending to be all over, and then roar up like a red hydra with molten sodium sputtering everywhere. It experiences as an explosion, though I am sure nitpickers will say it was only a very vigorous fire.
Rowing a boat that was on fire, with oars that were on fire, I claim the right of definition. It decidedly explodes.

I upvoted because 1) you did (and survived) the bucket list experiment every chemist wants to do, whether or not they would admit to it, and 2) the boat and oars were on fire! Bravo!

Shot-putting sodium from a rowboat: a new olympic sport!

@BuckThorn You have to have a NaK for it! ;-)

@EdV When the professor asks you to destroy what is surplus on the old chemicals storage in advance of the government inspection you do what you must ;). But I still defend my decision not to touch the 60 year old organic peroxides (they had crystals rimming the flask stoppers - a very gentle but firm message to not even consider touching the flasks. "Here be unknown and unstable depositions...")

Well noted! When one of our professors retired due to Alzheimer illness, his lab had several pounds of NaK to be safely eliminated. Our campus Environmental Health and Safety folks treated it like a leftover WW II bomb and “neutralized” it in a parking lot well away from main campus, i.e., where new hires get their parking. ;-) I wouldn’t have gone near the peroxides either. Years ago, we had multi-pounds of old potassium that was removed because of the peroxide possibility.


text:
Solubility and pH
I was reading about antacids. It stated that in earlier times $\ce{NaHCO3}$ and a mixture of aluminium and magnesium hydroxide were used as antacids. In excess $\ce{NaHCO3}$ can make the medium alkaline but as

metal hydroxides (say, magnesium hydroxide) are insoluble thus they do not increase the pH above neutrality.

I am unable to understand how solubility is related to pH. Please explain.

When magnesium hydroxide dissolves in water, it increases the hydroxide ion concentration in solution:
$$\ce{Mg(OH)2(s) <=> Mg^2+(aq) + 2 OH-(aq)}$$
When hydroxide reacts with the $\ce{H+}$ given off by acids, it gets consumed:
$$\ce{OH-(aq) + H+(aq) <=> H2O}$$
The reason magnesium hydroxide stops dissolving is that there is too high a concentration of magnesium and hydroxide ions in solution (the reverse reaction is as fast as the forward reaction, and the system is at saturation or, more generally speaking, at equilibrium). Now if the second reaction happens as well, this will lower the hydroxide concentration, allowing more magnesium hydroxide to dissolve.

This is related to LeChatelier's principle, to certain coupled reactions in Biochemistry, and to the common ion effect.

Your stomach pH is in the range of 1 to 3 (Wikipedia). If it goes into the very acidic rage and you develop heartburn, antacids can be used to raise the pH and soothe the irritation.
Antacids aren't used to raise the pH above 7, just to move it above 1 or 2. The amount of NaHCO3 you would ingest is so small that it would not do more than this, and even tho its pH in plain water is 8.5, it is quite harmless.
Magnesium hydroxide is only very slightly soluble, but it will raise the pH of plain water to about 9; still pretty harmless, but it will dissolve in strong acid and neutralize it. The amounts ingested are small enough not to raise the stomach pH much, and it won't hurt the esophagus when going down. Aluminum hydroxide is similar.
Very strong and very soluble alkalies would burn the esophagus going down, so are not used. Antacids don't need to bring the pH into the alkaline range, just to increase the pH enough to be less irritating. Commercial proton pump inhibitors can decrease stomach acid by up to 99% (equivalent to about 2 pH units). That's all that's needed.

The pH is the negative decadic logarithm of the H+ concentration in the solution.
If you have a salt (like NaHCO3) that can dissolve, it can form OH- or H+ during that process. Therefore the solution can become acidic or basic.
If a salt is not soluble, nothing will happen.


text:
Model of an ideal liquid in chemistry
In chemistry there are ideal models/concepts of ideal crystal, ideal gas and even ideal solution, but there is no model/concept of ideal liquid.
In physics (fluid dynamics, to be precise) there is a model of perfect fluid, a fluid without internal friction and, as a consequence, no viscosity, and thermal conductivity. The model covers macroscale physical properties and doesn't really touch what's happening on the "chemistry level".
Why chemistry lacks a proper model of ideal liquid when a seemingly more complex model of an ideal solution has been developed? Isn't there a need for one, and if so, is there a concept of ideal fluid that comes the closest to be an alternative of, let's say, an ideal gas law, but for liquids?

The nearest to an ideal fluid is a hard sphere fluid, but this is removed from the ideal gas or even solution concept of ideality in a critical way.
Ideal (also called perfect) gases are ideal because they lack intermolecular interactions. A statistical description that ignores the intermolecular potential suffices to describe an ideal gas. A first extension toward nonideality, following the vdW equation, describes particles in a gas as hard spheres (having an excluded volume).
The property of reduced compressibility encoded by a hard sphere potential also provides the simplest model for a liquid. The hard sphere potential suffices to describe a simple liquid, and it is about as ideal an abstraction of a liquid as you can get. However, while it shows a liquid-solid phase transition, it does not display a liquid-gas transition. In that sense it is not a universal model, it does not capture some key properties of liquids, in the same sense that the ideal gas law is not universal because it too does not address some simpler key properties of real gases (such as excluded volume). However, introduction of an additional attractive term as in the full vdW model, bringing us further away from ideality, allows us to observe a critical point below which a phase transition clearly occurs between liquid and gas. I should add that the hard sphere potential can be expressed in reduced units, making it universal within the scope of that limited model. It is a suitable first approximation for understanding some properties of nonpolar liquids, for instance condensed noble gases.
I would add that Raoult's law is another ideal model for a liquid, and consistent with the ideal gas concept (or rather Dalton's law). It is also a limiting law (or model), like the ideal gas, and applies to dilute solutions.

The vdW equation exhibits a critical point, and a liquid-gas transition below the critical point. It does not show a liquid-solid transition. See, for example, https://physics.stackexchange.com/questions/453490/critical-point-of-a-pure-substance/454219#454219

@JonCuster I meant the hard sphere fluid shows a liquid-solid transition, not the vdW gas. The hard sphere fluid can be considered a van der Waals substance without an attractive potential term.

Fair enough. But, I might claim the vdW model as an 'ideal' fluid. But we all know that ideality is never the case for our real problems (sigh).

@JonCuster Yes, the vdW fluid is an ideal fluid, no doubt about it. You can regard hard sphere fluids as a subset of vdW fluids, without an attractive potential term.


text:
Does diluting the titration flask change the molarity
I was solving the following problem on a test and I thought the answer was A, but the answer is actually D. Below is the problem and my reasoning:

Answer choice B is clearly incorrect because the number of drops of indicator largely does not affect the observed end point. Answer choice C is not mathematically valid, so it can also be ruled out.
Between A and D, I thought that A was correct because if some neutralized solution remains in the flask, then the apparent volume of NaOH in the flask is higher and the concentration of NaOH is lower than the actual value. Furthermore, I thought D was incorrect because spilling $\ce{NaOH}$ solution does not change the concentration of the remaining $\ce{NaOH}$.
Can someone please help me understand this question?

If your titration reached the end point and you add water, will that change the color of the indicator? If the concentration is all that matters, why even measure the volume in a titration?

@KarstenTheis What's your point? I'm not getting it

The way you know the relation between HCl and NaOH is the stoichiometry of the reaction given by the chemical equation. That means you know how many moles of HCl react with how many moles of NaOH. When you add water (or water and sodium chloride), you are not changing any amounts. In fact, you often add unspecified amounts of water in a titration before you start. Don't get me wrong, there are lots of other scenario where concentration is very important, such as reaction rates, equilibrium reactions, colligative properties, to name just a few.

@KarstenTheis But if you add water into the titration flask, doesn't that erroneously reduce the concentration of the NaOH in the flask?

It does decrease the concentration and increase the volume. You still need to add the same amount of HCl to neutralize it, though. As long as you know the volume of NaOH you put in, and get the amount $n_{OH}$ (how many moles) of hydroxide from the titration, you can calculate the initial concentration. Think about it, when you are titrating, the volume is no longer the original volume of NaOH solution because you add indicator solution before and HCl solution during the titration.

@KarstenTheis then why isn't answer choice A correct? Doesn't it say that leaving neutralized solution resulted in lower concentration?

The error resulted in a calculated concentration of NaOH that was too low because they made an error somewhere along the way. The concentration of the original NaOH solution is what is it, the calculation just gives the wrong answer (too low in this case).

@KarstenTheis Thank you. Can you please help me understand why D is wrong? I don't think that spilling a solution will change the remaining solution's concentration.

(D) is somewhat ambiguous. If you spill it some time before adding it to the buret, it doesn't matter. But if it's spill after you've measured the initial amount in the buret, then the effective volume that you think you've added is larger, leading to a smaller computed concentration.

The answer to the title question is obviously YES.
Regarding the test quiz:
It is simply that titration involves equivalents of substance, not their concentration. The amount of NaOH to be neutralised does not change if you dilute its solution.
All the careful V measurements (volumetric flask and pipetting) that you perform before titration are just to ensure that each trial does indeed contains the same amount of analyte.*
After that amount enters the titration flask, it doesn't matter if you dilute it. Dilution might be even required for practical reason.
*The grid with variable V of NaOH solutions makes little sense. In practice the given sample would have been brought to a standard V and finally divided into trials.

The concentration of the original NaOH solution is calculated as:
$$c_\ce{NaOH} = c_\ce{HCl} V_\ce{HCl} / V_\ce{NaOH}$$
with $c_\ce{HCl}$ and $V_\ce{HCl}$ the concetration and volume of HCl added during the titration, and $V_\ce{NaOH}$ the volume of NaOH titrated. If any of these values are off, the result is off.
For example, if the HCl has a different concentration than we thought, our result will be wrong. If half of the HCl solution goes on the bench instead of the Erlenmeyer, the result will be wrong. The same goes for spilling some of the NaOH solution after measuring the volume (volume in the Erlenmeyer will be different from the volume claimed to be in it).


text:
Determining the equilibrium constant
This is freely translated, I hope it's not too bad.

The reaction to dehydrate methylcyclohexane to toluene in the gaseous phase:

Formulate the reaction equilibrium's constant.

Now, that's not a problem - the problem is that I don't know the aggregate state. I know that $\ce{H2}$ is a gas and that other thing left of it seems to be toluene, which is also a gas according to the text above. So how should I now know what aggregate state the methylcyclohexane is in? I also assume everythings at standard conditions.
The solution is apparently (but it might be wrong, since it's not an official solution)

$$K_1 = a_\ce{C7H14}^{-1}\cdot a_\ce{C7H8}^1\cdot a_\ce{H2}^3 = \left(\frac{c_\ce{C7H14}}{c^\circ}\right)^{-1}\left(\frac{c_\ce{C7H8}}{c^\circ}\right)^{1}\left(\frac{p_\ce{H2}}{p^\circ}\right)^{3}$$
$$K_2=a_\ce{C7H14}^{1}\cdot a_\ce{C7H8}^{-1}\cdot a_\ce{H2}^{-3} = \left(\frac{c_\ce{C7H14}}{c_\ce{C7H8}}\right)^1\left(\frac{p_\ce{H2}}{p^\circ}\right)^3$$

whereas $K_1$ is right and $K_2$ is left.
I mean that's clear and everything but how do I know the aggregate state of $\ce{C7H14}$?

I think you read is as "(dehydrate Methylcyclohexane to Toluene) in the gaseous phase", so everything is in the gas phase.

ok - that's what I first did but then the the solution given here would be wrong. No?

For species in the gas phase you have a choice of describing their approximate activity either in terms of concentration or partial pressure. Your standard conditions are well-defined as $c^\circ$ for the organic molecules and $p^\circ$ for dihydrogen. Different choices are possible.

The standard for equilibrium constants of gas phase substances is to use $K_p$, so in this case $p^\circ$ would be more appropriate than $c^\circ$ for the organics as well as the hydrogen. If you use $c^\circ$, you should use it for all gas phase components and indicate your $K$ as $K_c$. Using different standards for gas phase components in the same reaction is unorthodox.
And if you use your equilibrium constant for calculation of $\Delta G$ for example, the simple form of the equation $\Delta G^\circ = -RT\ln K$ requires that the $K$ be $K_p$ for gas phase reactions. Using $K_c$ will give an incorrect result.

The thing is I am very unsure about the aggregate state of all components, so it's hard to formulate $K_{1,2}$. The text might indicate that all components are a gas but then it might just say the product is a gas. Was wondering if I somehow should be able to know.

I see how you could read it to mean that only toluene is in the gas phase, but in that case, the phase of the starting material would be separately indicated. The problem really only makes sense if it means that all components are in the gas phase.


text:
Compare stability of the given carbocation

Give the order of stability of the given carbocations.
Usually stability is determined by studying effects taking place in the carbocation. All are resonating and there is no diff in inductive effect.So I want to know how the molecular weight might affect stability

Give us your thoughts on this or the question will be closed

Usually stability is determined by studying effects taking place in the carbocation. All are resonating and there is no diff in inductive effect.So I want to know how the molecular weight might affect stability

Edit your question to include this comment

Ok i did it...now its a valid question right?

I think so, I have voted to reopen

The hyperconjugative effects are not identical because the C-H/D/T bonds are of slightly different bond strengths due to differing differing zero point energies.

The question is pretty easy though the INDUCTIVE and MESOMARIC effects are same but there is a diffrence in the HYPERCONJUGATIVE Effect as in HYPERCONJUGATIVE Effect the bond between $C-H$ breaks to provide stability thus any change in bond energy of the bond to be broken changes the strength of effect.
Now as the bond strength of $C-T$ bond if highest hence the HYPERCONJUGATIVE Effect of $\ce{CT3}$ decreases and hence the carbocation is least stable.And $\ce{CH3}$ is most stable
You can read more about Hyperconjugation in the following Site:
https://chemistry.tutorvista.com/inorganic-chemistry/hyperconjugation.html


text:
Calculating the amount of a product at the reaction equilibrium
$\ce{Cr2O3}$ can be reduced to the elements chromium with high temperature and carbon by this two reactions:
$$
\begin{align}
\ce{Cr2O3(s) + 3C(s) &-> 2Cr(s) + 3 CO(g)} \label{rxn:1}\tag{1}\\
\ce{Cr2O3(s) + \frac{3}{2}C(s) &-> 2Cr(s) + \frac{3}{2}CO2(g)} \label{rxn:2}\tag{2}
\end{align}
$$
The following image shows the equilibrium constants.

Using the image, we can get $K_1$, $\Delta_rG^\circ_1$, $\Delta_rH^\circ_1$, $\Delta_rS^\circ_1$ of reaction $\eqref{rxn:1}$ as well as $K_2$ for $T = 1500\,\mbox{K}$.
Assume all of them are given. Now:
In a closed container with $V = 1.00\,\mbox{dm}^3$ inner volume, we put $5.00\,\mbox{g}$ of $\ce{Cr2O3}$ and $20.0\,\mbox{g}$ carbon. For $T = 1500\,\mbox{K}$ we have the reaction equilibrium $\eqref{rxn:1}$ and $\eqref{rxn:2}$.
Problem: What mass (in g) reduced chromium do we have in the equilibrium?
Question: Now I don't know how to solve the problem. I can do everything above like determining the Gibbs-Energy or the equilibrium constants, but what are my though now to get the amount of reduced chromium?
First, I don't get exactly what it means if they say "in the equilibrium". Does that refer to $\eqref{rxn:1}$ or $\eqref{rxn:2}$ or both?
Further, how do I actually calculate it?

You already have the equilibrium constants, so you don’t need to mess with the thermodynamics.

The equilibrium constant values determined at 1500K tell you something about the partial pressures of the two gases in the final state.

Yeah they do. But how can I use that to get some statement about the chromium here?

If you have the partial pressures, then you have the number of moles.

I think I might have seen my mistake. So I basically use $pV=nRT$ to get the moles of $CO_2$ and $CO$. I further know how much moles in total I put into the reaction. Now I can take the difference and get how much $Cr$ I got, right?

Sounds good.  Go to it

yeah, I'll do some calculations - still a bit unsure about the relationship of (1) and (2) though - thanks! You could make a simple answer so I could accept it.


text:
Can I have a homogeneous 2 molal solution of copper(II) sulfate at 20 °C?
I'm doing a high school chem project where I am making a calibration curve of copper(II) sulfate for spectrophotometry in which I make solutions with five different concentrations. The most concentrated one being what I though was 2 molal.
But I noticed that in that solution I could see specs of crystals floating around still after stirring for a while, and it did not want to dissolve completely.
It says on Wikipedia that copper(II) sulfate has a solubility of 1.26 molal (20 °C) in water. Does that mean that I can't have a solution with a concentration of 2 molal at that temperature?

Yes, it does...

Yes, that's what it means. You could try to make a super-saturated solution, but that would mess up your experiment because some solute will eventually precipitate.
You don't want to have too high a concentration anyway because you might be out of the range of the instrument. It makes sense to have concentrations that the instrument can detect (not too low) and that don't max out the instrument (very concentrated solutions won't let any light through at the wavelength of interest).
You might also think about what the range of concentrations is for your unknowns. The calibration curve should cover that range nicely - calibration standards outside of that range don't contribute to your results much.


text:
Determining the concentration after half-life period
I was given the reaction

In diluted solution and with neutral pH, the reaction reacts according to a reaction velocity law of pseudo-1. order:
$$-\frac{dc_{DNPA}}{dt}=k_{obs},\quad k_{obs}=k\cdot c_{H_2O}$$
At $t=0$ we have $96\mu M$ DNPA.
How big is the concentration of DNPA and DNP after three half-time periods ($3\cdot t_{1/2}$)?
Of course, for DNPA we would just have $1/8$th of $96\mu M$ which is $12\mu M$. But what about DNP?
For me it's clear that it correlates directly with the concentration of DNPA since we have a heavy diluted solution, so the concentration of the water can be assumed as constant.
Further I know that $v_c(t)=\frac{1}{\nu_i}\frac{dd_i(t)}{dt}$ whereas $\nu_i$ are the stoichiometric coefficients. So basically, the concentration od DNPA, DNP and HOAc are the same since all have a stoichiometric coefficient of 1. Right? So they all change the same. Meaning: if DNPA changes to $1/8$-th, so does DNP and HOAc.
I am still a bit unsure, but can I now conclude, that $c_{DNP}(3t_{1/2})=12\mu M$?

I'm afraid you are mistaken. The DNPA essentially turns into DNP in the course of the reaction, so the answer is $84\ \mu M$, assuming the initial DNP concentration is zero. Formally, the stoichometric coefficient of DNPA is -1, which illustrates nicely that the sum of the amounts of DNPA and DNP is constant in this system.

S0 $c_{DNPA}(3t_{1/2})=12\mu M$ and $c_{DNP}(3t_{1/2})=84\mu M$? I'f so, do I also have $c_{HOA_c}(3t_{1/2})=84\mu M$? And of course you are right about the coefficient.

@xotix Full agreement.

Thanks! I actually just saw that I missed a little bit of theory here and of course everything makes sense! THanks.


text:
What fluorescent materials absorb visible light and emit UVC light?
What fluorescent materials, if any, exhibit anti-stokes shift such that they absorb visible light and emit ultraviolet light in the range of 250-285 nm? I'm interested in particular in materials that are efficient in their conversion and not difficult to acquire.

Almost by definition, upconversion is not efficient...

@JonCuster I think I know what you are referring to, that it takes multiple lower-energy photons to produce one higher-energy photon. I would like to use "efficient" in the sense that the input energy primarily gets converted to ultraviolet light in the desired range.

Yeah, and that is the problem. For two photon processes you need lots of intensity and even then the efficiency is lousy. Or you are asking for multiple sequential absorption through states with really long lifetimes. Also really unlikely. It just isn’t happening...

If you want to convert visible light into UV, you have two options.

Two-photon excited fluorescence. The process is inherently wasteful, but the field is studied due to applications in microscopy. See example paper on fluorophores https://pubs.acs.org/doi/abs/10.1021/ol017150e . Also, there are commercially available dyes.
Optical frequency multiplier. The process is fairly efficient, but I think it requires laser as the light source. Also, AFAIK it doesn't involve excitation of a molecule, but is an arcane optical device exploiting non-linear optical properties of some materials. Most laser diodes in violet and UV range use frequency multipliers to up-convert light from more convenient red and green diodes.

Naturally I think that the second option is closer to what you want, but in this case you should look for commercial products, that might be available as "laser frequency doubler/tripler".  Good luck.

Two photon absorption is a nice way of doing this but you will need a short pulse laser to make it work well.  Laser frequency doublers etc are widely used in many laser systems and use non-centrosymetric crystals such as potassium dihydrogen phosphate (KDP). You can also generate continuum spectra with short (sub picosecond) laser pulses focussed into, say, water or sapphire, but these are not what as asked about.

It is not possible in a single molecule as energy conservation will forbid it. There is not enough thermal energy, $k_BT \approx 210 $ wavenumbers, visible to uv thousands of wavenumbers.  It is sort of possible using triplet-triplet annihilation ($\ce{T + T \to S^* + S}$ ) but this needs significant initial excitation and long lived triplets, which means de-oxygenated solution, and produces the same singlets as were excited initially. There are also inorganic up-conversion materials, for example some lanthanides, such as ytterbium based ones will convert near ir light to visible light.


text:
Conformer generation for radicals
Is there an automatic way to generate conformers for radical like systems (spin multiplicity is doublet)? My system for the study is a substituted fullerene with a missing H atom from the functional group.
I have tried using different conformer generators for this purpose (Baloon, RDKit, Obabel, Fullmonte, CCDC's Conformer_generator) but none of them successfully yielded out any conformers. I do not know how to run MD of such a system for which there is no general force-filed available. I have also tried to go through the AMBER tutorial but was unsuccessful: http://ambermd.org/tutorials/basic/tutorial4b/
I would really appreciate if anyone can suggest something that actually works for the system in consideration.
Thanks in advance.

Are you sure that a non-reactive MD calculation would be meaningful for a radical system?

I recommend gfn2-xtb from the Grimme group: https://www.chemie.uni-bonn.de/pctc/mulliken-center/software/xtb/xtb I have not tried the confscript yet, but the siman option works reasonable well for my systems.

RDKit should work. Can you post the code somewhere (e.g. https://gist.github.com)?  This code should also work for you https://github.com/jensengroup/get_conformations

Just saw this - can you post the compound? I'm surprised that OB (and others) don't see "conformers" and it's possible that your structure falls beyond the usual rules. I'll post more later.

tldr: Manually or script dihedral angle changes, then use a GFN or another DFTB method to find any/all conformers
Without seeing the compound, it's really hard to know the problem. Most programs, including those you mentioned, have a set of rules to identify "rotatable bonds" and suggest likely dihedral angles.
As you might expect, such rules are derived for neutral, ground-state organic molecules.
There are a few options:

As suggested by Jan Jensen's comments above, you can use RDKit, Open Babel or another tool to drive your own dihedral angle search.
As suggested by Martin's comments above, there are some tools in Grimme's xtb  program to sample conformations. These will use the GFN1 or GFN2 approximate density functional (pretty much DFTB + dispersion correction) to evaluate energies.

I'd probably suggest a combination of both. From what I understand of Grimme's conformer search scripts, it calculates the vibrational modes of the molecule and samples geometries along vibrational modes. This is a good idea, although in some molecules it may not sample all minima (e.g., if you have many conformations in a large molecule, you might never "reach" the whole potential energy surface from any given initial geometry).
So you can change the dihedral angles using your program of choice, generate a few initial structures, then use xtb to optimize the geometries and confscript to sample a few more geometries.
I feel pretty confident that in small to medium geometry molecules, this will do a good job.
A few notes:

If you're sampling a radical, you definitely want to use a good DFT or quantum method to evaluate the energies. I doubt any current force field will do what you want.
We haven't finished our evaluation, but right now, GFN2 is performing better than any current method on conformer energetics for neutral organic molecules. We haven't checked radicals, but I would trust it more than any other current "fast" method.
If you don't have the xtb package, the method and conformer sampling scripts are now available as part of ORCA 4.1 too. There's a description of this kind of conformer sampling in the manual when discussing prediction of NMR spectra.

Your question is a great illustration of why my group is currently working on a "conformer toolbox" - an NSF-funded project to create a set of easy-to-use software for generating many kinds of conformers. I'll come back and revise when we have some better tools for problems like this.


text:
Does dissolving something in water change the specific heat of the solution?
I’m a little confused on this topic. If I have a solid that I dissolve in water and that undergoes an exothermic reaction, can I assume that the specific heat $c$ of the solution instantaneously after the dissolution occurs is equal to $\pu{4.18 J g-1 °C-1}$? Does dissolving something in water change the specific heat of the solution? Or can I just say that the specific heat of the solution (solid plus water) equals the specific heat of water?
My first instinct was to think that yes, dissolving a solid in water does change the specific heat of the solution. If this is true, then I thought I might determine the overall specific heat of the solution by taking the specific heat of water and multiplying it by the percentage of the solution that is water, then adding that to the specific heat of the solid times the percent of the solution that is ions from that solid.
I’ve overthought this quite a bit, and I’ve tried to find an explanation on the internet. I’m pretty sure I’m thinking about this incorrectly, and I apologize if this is a bad question. Any clarity you could provide would be much appreciated.

It depends on the solute concentration. If the solution is dilute in terms of the concentration of solid you added, then the heat capacity will be dominated by that of the solvent, or, the heat capacity of the solution will be little altered by addition of a minute amount of solute relative to the value for the pure solvent.
To borrow an expression from a different answer, we can write
$$c_p=x_\mathrm{solv}c_{p,\mathrm{solv}}+(1−x_\mathrm{solv})c_{p,\mathrm{solute}}$$
where $c_p$ is the (molar) heat capacity of the mixture, $c_{p,\mathrm{solv}}$ and $c_{p,\mathrm{solute}}$ are the (molar) heat capacities of solvent and solute, respectively, and $x_\mathrm{solv}$ is the mole fraction of solvent in the mixture. Since for dilute solution we can often make the approximation $x_\mathrm{solv}\approx 1$, the second term on the right is approximately zero and can be ignored. Then
$$c_p\approx c_{p,\mathrm{solv}}$$
So for many examples, in particular those involving dilute solutions, you can safely ignore the effect of the added solute and assume that the heat capacity of the mixture is equivalent to that of the pure solvent.
To give a slightly more quantitative example, consider the molar heat capacity of solutions of sodium chloride in water, data from the Dortmund Data Bank:

A 1 molal solution corresponds to an NaCl mole fraction of 0.0177. At this concentration of NaCl the heat capacity is ~1.4% smaller than that of pure water. An error of this size is often acceptable.

If you were adding two liquids ideally, that is, assuming that the mixing occurs without any change in enthalpy - which occurs for chemically similar substances, then yes, the heat capacity of the mixture is a weighed average of the heat capacity of the components ($A$, $B$):
$$c_p = x_Ac_{pA}+(1-x_A)c_{pB}$$
However, when we are talking about dissolving solids, the situation is a little different. That is because infinite-dilution properties of a solute that derives from a solid source do not mimic the properties of the (crystalline) solid. For example, table salt ($\ce{NaCl}$) has a certain density as a powder, but this value has nothing to do with the density it acquires in solution once the ions dissociate and live freely in solution. The same goes for heat capacity, which is also a thermodynamic property.
So the answer is yes, the heat capacity of the solution does change by adding a solute and, in fact, it depends on the amount that is added. Only experimentally can you measure the exact function of concentration that defines the heat capacity of a particular solution. From a molecular perspective, it will be highly dependent on how the phenomenon of solvation develops.
As always, you can always look up the scientific literature to get the data of the thermodynamic functions of certain solutions including heat capacity.


text:
Kolbe's electrolysis of acetylenedicarboxylic acid
I have been wondering what this reaction would lead to?
The ethyne diradical can't dimerize because triple bonded cyclic structures don't form until 8 carbons. And even if it polymerizes I find no way to end it's chain. This could probably lead to the carbyne.
Is there any literature on this out there?

I'm not completely up on Kolbe mechanism, but why do you think a diradical would form?  Presumably it could only form stepwise, with a mono-radical forming first.  This would have to last long enough, not reacting with anything else, and be as prone to homolysis as the original non-radical carboxylate ion, to form.  Not clear to me that it could.  What is the Kolbe product of even a "regular" dicarboxylate like succinic acid?

I think you are confused about the 8 carbon ring. While (E)-cyclooctene is stable. cyclodecyne is the smallest isolable cycloalkyne. In your query I believe your only source of carbon is the "diradical of acetylene". The polymerization would give a linear polyacetylene that would have to be extremely long to bite its own tail, if it could wind it. As to termination, the C-H bond dissociation energy of acetylene is 133 kcal/mol. The radical is very reactive and can abstract hydrogen from  the walls of the vessel.


text:
dSsur, dS and dStotal relation
By definition,

dS=dqrev/T.

Based on my understanding, the subscript rev means that the dq is heat exchanged in a reversible pathway from a state i to another state f, regardless of the actual pathway by which the process proceed. I am always curious why ΔS is defined only for reversible process, it would be good if someone can explain it, nonetheless it is not our main concern here-let's just accept the definition first.
So, similarly,

dSsur=dqrev,sur/Tsur......(1)

From what I have read, V of surrounding can be regarded as constant, so

dqrev,sur=dUsur

and since dUsur is a state function,

dUsur=dqsur

and therefore dqrev,sur=dqsur implying

dSsur=dqsur/Tsur......(2)

(a) If V is constant, q=ΔU, so q is a state function at constant V?
(b) Let's stop at equation (1). Suppose we would not like to simplify it further, and since the equation (1) is an intermediate result it must be true, we have

dS=dqrev/T and
dSsur=dqrev,sur/Tsur

I suppose dS without any subscript is dS of system. By First Law,
dqrev= -dqrev,sur
and then if T=Tsur and the chemical reaction has not reached equilibrium (is this scenario possible?),

dS= -dSsur

By summing up all entropy changes for system and surrounding, we have
ΔStotal=0 for non-equilibrium state? But as I know ΔStotal=0 if and only if surrounding is in equilibrium with system. What's my misconception here?
(c) How can dqsur,irrev=dqsur,rev? Does not make sense for me physically, without looking at the equation with dUsur

(a) is true. (b) whenever you write $dS=dq/T$ you imply reversibility (equilibrium). For any process this is generally assumed for the surroundings (we assume it to have properties consistent with this), but not (generally) for the system. (c) Again, we generally assume that the surroundings can give or accept heat reversibly.

(a) is not true in general. A battery loses internal energy when it is discharged even if it does not transfer any heat (i.e. perfect battery that does not warm up when discharged, or perfect insulation around the real battery).

@KarstenTheis True, the assumption I made is that there is only PV work.

In thermodynamics, the surroundings are typically treated as an ideal constant temperature reservoir for which $\Delta S$ is always equal to $Q_{surr}/T_{surr}$, irrespective of whether the process that the combined system and surroundings experiences is reversible.  As such, the surroundings is assumed to have an infinite heat capacity and thermal conductivity, and to not change significantly in temperature (its temperature is always $T_{surr}$), no matter how much heat it receives or rejects.  So it always presents the temperature $T_{surr}$ to the system at its interface with the system.  And, all the entropy generated in any process is confined to the system.
In evaluating the entropy change for system and surroundings, the system and the surroundings must first be separated from one another, and then each must be separately subjected to a reversible path.  These separate reversible paths do not have to match up in any way whatsoever.  For the surroundings, the final state is that which was obtained with the irreversible path when $-Q_{irrev}$ was transferred from the system. So the change in entropy for the surroundings is $-Q_{irrev}/T_{surr}$.  And the change in entropy for the combination of system and surroundings is then given by:
$$\Delta S=\Delta S_{sys}-\frac{Q_{irrev}}{T_{surr}}=\left[\int{\frac{dQ_{rev}}{T}}\right]-\frac{Q_{irrev}}{T_{surr}}$$

For your last equation, you have simplified my equation (1) into equation (2). If I do not want to simplify equation (1) [this intermediate result must be true], and I plug the equation (1) in place of your delta S for surrounding, I will get both q are reversible, and by First Law the two must add up to zero and that gives delta S always=0. It was just like you put rev q for system although the process is irrev, can't I do that on delta S for surrounding?

No.  It is impossible to devise a reversible process for the combination of system and surroundings that takes them both simultaneously between the same two thermodynamic end states as those obtained with an irreversible process.  To get the entropy change between the end states, you must first separate the system from surroundings, and then subject each of them individually to a different reversible path.  The reversible heats and works for these individual paths do not have to match.  For more on this, see https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/

(a) It is correct to say that $q$ is equal to a state function under certain conditions. Your example of constant V making $q=\Delta U$ is one. Another specific case is at constant pressure, $q=\Delta H$. However, the fact that $q$ is equal in value to a state function in these specific cases does not mean that it is itself a state function. It still does not meet the definition of a state function. By enforcing the condition of constant V or constant P, you are selecting a path for the process, so it is no longer path independent.
(b) In order for a process to be truly reversible in the thermodynamic sense, it must be at equilibrium at all times. That is because $\Delta S_{universe}$ is only 0 at equilibrium. By this definition, though, no reversible process can happen at all, since we can never be away from equilibrium conditions. To keep the math easy and get around this issue, we introduce the idea of "microscopic reversibility", meaning that we move the system along a path via very very small deviations from equilibrium, and equilibrium is restored before we make the next increment change.
For example, if we are adding heat to a system, we create a nearly reversible process by starting with the system and surroundings at the same temp (at equilibrium) and then changing the temp of the surroundings by a very small amount and waiting for the system to come up to that temp before changing the temp again. This is not truly a reversible process, but as the increment approaches zero (and the time for the process approaches infinity), the limit is the reversible process.
Going back to your question, if the chemical reaction is not at equilibrium, then the process by which it gets to equilibrium is necessarily irreversible. Thus, the assumption that $q_{rev,sur}=-q_{rev,sys}$ is not valid. The surroundings are not in the same state that they would be for a reversible process that took the system from $i$ to $f$. Thus the two $q_{rev}$ values are not the same magnitude, which is question (c).
Finally, I don't think it's correct to say that $\Delta S$ is defined only for reversible processes. It's a state function, so we don't need to assume reversible to determine it if we have all the other information necessary. The only definition of $\Delta S$ that invokes a reversible process is $\Delta S = \frac{q_{rev}}{T}$, in which it's the heat transfer that must be from a reversible process, not the $\Delta S$. That $\Delta S$ would apply to an irreversible process if the final state is the same as for the reversible process with heat transfer equal to $q_{rev}$.
One last comment - you stated that V of surroundings can be assumed constant. That is not always true. We often consider cases where the surroundings are a local volume that is considered to be isolated from the universe for all practical purposes. For example, we could have a reaction occurring in a balloon inside a sealed container, and if the reaction resulted in gas production that increased the volume of the balloon, the volume of the interior of the box would correspondingly decrease. If we were interested in the energetics of the reaction, we would treat the interior of the box as the surroundings and the inside of the balloon as the system.
[Really last comment - a picky bit of nomenclature - since $q$ represents a transfer of heat, $dq$ is not the appropriate term in your equations. For example, $\Delta S =\frac{q}{T}$, not $\frac{dq}{T}$.]

$q = \Delta U$ is true if there is no work. Constant volume ensures that there is no pV work, but you also have to specify there is no non-pV work such as mechanical work, electrical work, magnetic work or gravitational work Wikipedia

Good point. I was only considering PV work.

For (b), if the chemical reaction is at equilibrium, qrev,sur=−qrev,sys; if the chemical reaction is not at equilibrium, the equation does not hold, yet by definition we still have to put qrev,sys for dS of system, but for dS of surrounding we will put the actual q released/accepted by the system (surrounding=actual system) through the actual irreversible pathway, so the two will not add up to zero, is that what you meant?

Essentially yes. The actual heat transfer to the surroundings is in most cases going to be the same as what would be transferred in a reversible process to get to the final state of the surroundings, as you describe. This is based on the typical case of an irreversible process where energy gets transferred as heat (irreversible) rather than work (reversible), which is why the maximum possible magnitude of work is in the reversible process.


text:
Does the starting concentration in a isochoric container depend on the pressure?
Given is the following reaction of carbonic acid and carbon dioxide:
$$\ce{RCOOH (g) → RH (g) + CO2 (g)}$$
The reaction rate of this decarboxylation was measured in a container with constant volume at $T = \pu{277 °C}$. The following starting pressures $p(0)$ and half-time periods $t_{1/2}$ were measured whereas there was only carbon acid at the beginning.
$$
\begin{array}{lrrrrr}
\hline
p(0)/\pu{mbar} & 20   & 40   & 100  & 150  & 200  \\
t_{1/2}/s      & 95.7 & 47.8 & 19.2 & 12.9 & 9.58 \\
\hline
\end{array}
$$
Problem: Why does this data imply that the concentration of the acid doesn't follow a kinematic of the first order?
Answer: A kinetic of first order would have the same half-time period for different starting concentrations. This isn't given here. So this isn't a kinetic of the first order.
Question: It's clear that a kinetic of first order doesn't depend on the starting concentration. What I don't get is: how do I see that we actually have different starting concentrations? Why should the starting concentration depend on the starting pressure?
My though: We have
$$ c = \frac{n}{V}; \qquad pV = nRT; \qquad V = \text{const}$$
so we get
$$V = \frac{nRT}{p} = \text{const}$$
and from that we get
$$c = \frac{n}{V} = \frac{np}{nRT} = \frac{p}{RT} = \text{const} \quad \to \quad c = \text{const}$$
So apparently, we have the same concentration for different pressures. So how should I see from that data, that the the starting concentration has changed?
I also assumed that we always use the same amount of moles and just change the pressure externally. Maybe they though we change the amount of moles we put in and that changes the pressure, which we then measured?
Edit: I just noticed that sometimes people give the amount of moles of a gas in bars (which is still kind of strange to me), so basically in the units of pressure. I guess that's what's going on here?

Regarding the mol-bar thing you asked at the end, in reality it's neither and you have to use activities instead. It just so happened that it's convenient to use bars for the partial pressures of gases as they need to be related to the standard state of pressure (1 bar). Maybe read this topic of refer to your PC textbook for more details.

I don't understand how you conclude that the concentration c is constant. You seem to head in the right direction but then you miss the key point, that c is proportional to p, and as the table indicates, p is changed!


text:
Solid state and Bragg's Equation
A frst order reflection from (111) plane is observed for LiX with 2theta =24.6 degree (X-ray of wavelength 1.54 angstrom). Assuming LiX to be a cubic crystal system, calculate the length of the side of the unit cell in angstrom.
This is a question of JAM 2014, here, the length is considered as a=2a for ionic system in a solution provided by Arihant's book. Why is it so? Is it correct?

That's what happens when you are reading a book written by a troll. Indeed, bluntly saying $a=2a$ is a great way of leaving your readers confused and dumbstruck.

$a = 2a$ is a mistake of some kind.

From Cubic crystal system on Wikipedia:

In the rock-salt or sodium chloride (halite) structure, each of the two atom types forms a separate face-centered cubic lattice, with the two lattices interpenetrating so as to form a 3D checkerboard pattern.

If LiX crystallizes like NaCl, then the unit cell length is twice the distance of the shortest Li - X distance. The magnitude of the answer given in the book looks like it could be an interatomic distance, but the calculation does not make any sense to me (it does say $a = 2a$, as @Ivan_Neretin already pointed out in the comments. Also, when a calculation drops the units midway and sticks them back in the result, it demonstrates a certain nonchalance that does not instill confidence in the source.
The question asks about the unit cell parameter a (in a cubic system, this is all you have to specify to describe the dimensions of the cell because the other two lengths are the same and the angles are all right angles). If that's all you want, the internal structure of the unit cell has no bearing on the calculation (there might be systematic absences in a particular space group, but the question says that the 111 reflection was observed, so there is no issue there either). Why not just plug in the angle, get d, and calculate the unit cell length from that?

Yeah sir... Thanks for clearing the confusion . I'll do it that way.

Your book must have wrong solution

This would be more helpful if you rotated the image and walked us through it. If someone writes a = 2a, I think there needs to be more discussion.


text:
How do dibromopyrazolones undergo base-promoted decomposition?

I was not sure but seeing the base as NaOH I used the fact that it will take most acidic hydrogen that is N-H and electron pair will come over it and it will resonate from two sides. I could make upto this intermediate but next I am not really sure the ring will break up and give the product as option D.

Although a mechanism was not reported in Carpino's paper of 1958, there is a more concise mechanism than the one offered by @Soumik Das. There is no need to dissemble structure 1 to a tricarbonyl compound. Vinylogous elimination of the elements of HBr from dibromide 1 leads to species 2. [Alternatively, attack of hydroxide at the carbonyl of 1 would also suffice.] Attack of hydroxide at the carbonyl group of 2 produces 3, which via a Grob-like fragmentation affords the propargylic carboxylate 4 of the acid 5. ADDENDUM: The ring opening of 3 should not be construed as concerted. Just saving space.

Carpino, L. A., J. Am. Chem. Soc., 1958, 80, 599.

In the first question, the formation of the product D) can be justified in the following way. First the reagent $\ce{OH-/H2O}$ will basically perform alkaline hydrolysis of the starting compound to form 2,3-diketocarboxylate ion. But by the hydrolysis, it will also release hydrazine ($\ce{H2N-NH2}$) in basic medium, which will ultimately perform Wolff-Kishner reduction on the $2,3$-diketone part of the intermidiate compound. It is interestingly observed that 1,2-diketones under Wolff-Kishner conditions always produce alkynes.
The product formation can be summarised as follows,

Also the mechanism of Wolff-Kishner reduction of $\alpha - \beta$ diketone compound is similar to the following image (Some intermidiate steps are skipped to make it concise).

The release of nitrogen gas is the main reason for driving this kind of elimination reaction forward.


text:
Ring contraction when cyclohexene oxide is treated with methyl Grignard

My book writes that:

Recall that epoxides rearrange with Lewis acids in a pinacol-like fashion, and that Grignard reagents in THF exist in the following equilibrium:

I approached this question by reasoning that the Grignard reagent would lead to opening of the epoxide:

but the answer given in the book is instead

How is the five-membered ring formed?

I think in this case the question is asking you to think of the Grignard rearranging to give a Lewis acid that induces a rearrangement of the epoxide

I'm calling BS on the provided answer. The semi-pinacol rearrangement seems reasonable, but that means you're going to end up with an aldehyde in the presence of methyl Grignard. I have trouble believing that it will stay an aldehyde. In some sense, the problem suggests that as well. Why else would one need to oxidize with PCC at all?

Perhaps the carbonyl remains bound to the LA-like Mg species which deactivates it for methyl Grignard attack - but I do rather agree with you.  Odd choice of reduction conditions too. Has someone database access to find this reaction?

@Waylander I don't like the binding argument because such a bound species would putatively be present during the additional of methyl Grignard to an aldehyde. On the other hand, if the Grignard were used in catalytic amounts (not explicitly specified in reaction description), then it would be possible not to obtain the product with methyl addition, except sacrificially to generate the true active Mg catalyst.

@Waylander slightly belated, but the accepted answer is correct: the product formed is 1-cyclopentylethan-1-ol. Oxidation of this gives cyclopentyl methyl ketone. The reactions (in Et2O, not THF) are in French journals from the 1920s.

@orthocresol  , i have understood the mechanism of ring contraction.But why is there selectivity between ring contraction and traditional base catalised ring opening of epoxide ? Is this an isolated case or does it occur for acyclic epoxides also

@Chakravarthy Kalyan: Acyclic epoxides rearrange with Lewis acids [DOI:10.1021/jacs.5b10419] Cuprates, with or without a Lewis acid, react in the "normal" SN2 fashion. That is, without rearrangement. J. F. Normant, et al., Tetrahedron, (1986), 42, 5607.

Which book is it? @Sarah jane

Sara Jane: Lest you miss the point of the stimulating discussion between Waylander and Zhe, I would like to sum up their thoughts with a diagram. Your first clue should have been the opening line: "Recall that epoxides rearrange with Lewis acid...". It is common that catalytic Lewis acids rearrange 2,3-dialkyl oxiranes [epoxides] to ketones. In the case of cyclohexane epoxide (1), rearrangement to ring-contracted cyclopentane carboxaldehyde (4) occurs. In the illustration I have used CH3MgBr as the Lewis acid. Alternatively, one might use the other components of the Schlenk equilibrium, i.e., (CH3)2Mg or MgBr2. The rearrangement is shown as two separate steps (2 and 3) although it is likely to be a concerted process.
The aldehyde 4 in the presence of the Grignard reagent will undergo addition. Given sufficient Grignard reagent, addition to the carbonyl group will occur to afford the secondary alcohol 6. The only ways that the aldehyde 4 can be isolated is if there is insufficient methyl magnesium bromide or that the Grignard reagent acts as a base and forms the unreactive enolate of the aldehyde. The aldehyde 4 would be produced upon aqueous workup. While Grignard reagents with β-hydrogens, e.g., isopropyl and ethyl Grignard reagents, are capable of reduction of aldehydes and ketones, methyl Grignard reagents are incapable of this reaction. Therefore, the primary alcohol 9 may be assumed to be a misprint in place of secondary alcohol 6.
The oxidation of alcohol 6 to ketone 7 is straightforward. The final step is the formation of the alkene 8 (most likely of the (E)-configuration) via the McMurry coupling.


text:
OF2 vs O2F2 oxidising strength
My teacher said that OF2 is the third most powerful oxidising agent after F2 and atomic O, but I think O2F2 should be even more oxidising, since it less stable than OF2. I just read on quora that O2F2 is so unstable that it readily oxidises anything that comes in contact even when it is frozen. Also, in case of hydrides of p-block we say that reducing power increases down the group as the tendency to donate H increases due to decreasing thermal stability, so on similar lines, I think O2F2 should be a more powerful oxidant.
Or is it because in OF2, the oxidation state of O atom is +2 (compared to +1 in O2F2), which it wants to get rid of and thus gets reduced, oxidising the other species readily. But if this is true then why is it more stable as we know that stability is inversely proportional to reactivity? Or are the concepts of stability and oxidising power completely unrelated?
I was not able to find the electrode potentials of these species, so how could they be even compared reliably?

Your right. O2F2 has no std. el. potential - it decomposes in very low temp. but even in low temp. it's still stronger.

related https://chemistry.stackexchange.com/questions/108262/compound-having-highest-standard-reduction-potential https://chemistry.stackexchange.com/questions/102502/what-is-the-strongest-oxidising-agent

also https://chemistry.stackexchange.com/questions/65079/relative-o-o-bond-lengths-in-o2x2-molecules-x-h-f-cl https://chemistry.stackexchange.com/questions/4624/please-explain-why-dioxygen-difluoride-is-so-dangerous

and https://en.wikipedia.org/wiki/Oxygen_fluoride https://en.wikipedia.org/wiki/Krypton_difluoride

In one of the answers, it says that OF2 produced on the decomposition of FOOF is more oxidising only due to the additional O2 produced with it. And that during oxidation by O2F2, the actual job is done by OF2 itself. Does that mean it is actually inappropriate to compare the two since they are essentially the same during the actual reaction?  And technically FOOF is the better oxidant since it actually is OF2 plus O2?

No this info seems to be incorrect - it decomposes into O2 and F2. And OF2 is very weak in comparison - it only slowly hydrolyses in water when FOOF sets ice on fire.

<Of the products formed by the self-decomposition of FOOF, oxygen gas is a pretty strong oxidizer, but it's the least of our worries here. The other product, oxygen difluoride, is almost as strong an oxidizer as dioxygen difluoride, which is only stronger because it's oxygen difluoride plus more oxygen. OF2, , like its FOOF parent, reacts very strongly and exothermically with almost anything, especially water. It will literally burn ice, rapidly;>                     This is the text from an answer in the 3rd link you shared. This seems to directly contradict your opinion.

It's not an opinion and I put link because these are related posts not necessarily correct. I you'd read the Wikipedia art. then you'd see info with citation to actual paper "(...) decomposes to O2 and F2 close to its normal boiling point of about 216 K.https://pubs.acs.org/doi/abs/10.1021/cr60226a003

https://en.wikipedia.org/wiki/Dioxygen_difluoride https://en.wikipedia.org/wiki/Oxygen_difluoride


text:
Simple soda–lime decarboxylation
This is my homework question I tried it hard the question is as follow:

I attempted it as follow:

According to my understanding as acid base reaction are fastest the carbanion will gain a $\ce{H+}$. From the water formed and will the product as I made, but the answer is completely different:

Please guide me where I am wrong.

You are wrong in thinking that the anion of the chloroalkene will simply pick up a proton. These reactions are run at a high temperature ao there is plenty of energy around for forming unstable species

You have a leaving group beta to anion. You would have no doubt about what will happen if it were on an alkane.

You are correct in the first step - the decarboxylation gives the anion of the chloroalkene.
It is important to remember that such soda lime reactions are run at high temperature and CaO is a powerful drying agent, so there will be no water present to quench the anion. What happens next would be obvious if you were dealing with an alkane; there is a leaving group beta to an anion so there would be an elimination with loss of Cl-. This is what happens; Cl- leaves to give the triple bond (again remember that this is a high temperature reaction so plenty of energy for forming unstable species). This is clearly unstable within a [6] ring so it dimerises by a 2+2 process to give the product drawn.


text:
Name of this glassware?
Could you tell me the name of this glassware? I found it in the laboratory with a label that says "container of stoichiometry", but I have not found more information about it, especially about the uses that it could give.
Thank you!

Could you please describe what the glass tube inserted through the large screw cap looks like inside (or where does it lead)? It's hard to see on the photo due to the glare.

To follow-up on @andselisk's comment, can you just remove it and take a separate picture of it, please?

Looks somewhat like this: https://chemglass.com/distillation-receiving-vessels-cylindrical-chemrxnhub

It looks like a vessel for sublimation. The tube going in the top would be filled with cold liquid making a tiny cold finger. The crude mixture is put in the cylindrical cup at the bottom. A vacuum is pulled from the bottom to lower the pressure. The side arm is for pressure release. After the material sublimates and deposits on the cold finger, it can be removed with the pure material on it in order to easily recover the material.

Usually, the sublimation apparatus doesn't have an adapter at the bottom for two different input/ouputs...

Yes, but having a three-way valve there would allow you to break the vacuum in the line while keeping a vacuum in the vessel if you don't have a valve in your vacuum line to do that. Just a possibility.

That's a good point, Andrew. I just would have expected such a joint to be at the top.

With regard to the label reference to stoichiometry, it could have been used to sublimate $\ce{I2}$ for use in iodometric titrations?

Perhaps it might also be connected to a nitrogen / argon line if a cycling vacuum / nitrogen distribution line isn't in place. Surely it is a cold finger.


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Theory of symmetry breaking in water?
As we all know $\ce{H2O}$ is bent ($C_\mathrm{2v}$) in its ground state equilibrium structure, rather than linear ($D_\mathrm{\infty h}$), which can be readily understood e.g. from the MO diagram of bent versus linear case:

The question, that puzzles me here, is if that apparent case of "symmetry breaking" can be described by a Jahn-Teller or Renner-Teller effect? Jahn-Teller seemingly is ruled out since the "high symmetric" from is linear, which is excluded from the Jahn-Teller theorem while the Renner-Teller effect is apparently a subtle ro-vibronic effect.
Since the water ground state is a closed shell singlet (even in the linear case), my take on it would be, that it must be a second-order JT-effect of coupling between ground and excited states by virtue of the bending distortion. But I am not sure since I never heard or read about water as a second-order JT-example.
Can anyone comment on that?

Check out Albright et al. Orbital Interactions in Chemistry 2ed pp 131-9

@orthocresol: So it means: Perfectly normal second order JT, if I understand correctly?

I think so, but will refrain from giving a more certain answer, I only briefly flicked through.

I don't think so. If I remember correctly, the ground state of linear water is a sigma state, so that excludes it from any Jahn-Teller or Renner-Teller effects. Similar phenomena, (such as 2nd order JT or pseudo-JT) require the presence of low-lying excited states, which is not the case in linear water. The first excited state is several eV higher in energy, not enough for a significant vibronic effect.

I don't think so. Strictly speaking there is no kind of structural distortion between a higher and a lower symmetry structure at all that cannot be described by perturbation theory in chemistry. It may require high order PT like 3,4,5th and so on, however. In the end it may depend only on how you define the JT effect.


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Why is Fluoride anion less stable than iodide anion?
Down the halogen group , electropositive character increases , then why is it that Fluoride ion is the least stable ?
The counter arguement is that fluoride ion has four lone pairs and a small size which makes it least stable.Down the group , since atomic size increases the extra electron is stabilized.
In all such stability trends there are multiple counteracting factors , how to decide which one to give more weightage ?

Actually, the answer is a bit more nuanced. If we observe the pKb values (i.e. tendency to form HX from X-, where X is a B-L base) it is pretty clear that F- would form HF very quickly.
The explanation is that I- would have a much larger area to balance out the negative charge compared to F-, and the anion would be more stable as a result.