I was solving this question from this book and I was not able to get any of the options.

The gaseous reaction $\ce{n_1 A_{(g)} -> n_2 B_{(g)}}$ is first order with respect to ‘A’. The rate constant of reaction is ‘k’. The reaction is studied at a constant pressure and temperature. Initially, the moles of ‘A’ was ‘a’ and ‘B’ was not present. How many moles of ‘A’ are present at time t?

A) $\mathrm{a.e^{-kt}}$

B) $\mathrm{a.e^{−n_1kt}}$

C) $\mathrm{a.e^{−n_2kt}}$

D) $\mathrm{a(1− e^{−n_1kt})}$



When I looked into the solutions, it seemed, for me that the solution is wrong.

Solution given in the book

$$\mathrm{r_{rxn} = \frac{-1}{n_1} . \frac{dn_a}{dt} = k.n_A}$$ $$\mathrm{\implies n_A = n^o_A.e^{-n_1kt}= a.e^{−n_1kt} }$$

My approach to the question

The rate of a chemical reaction is defined as $$\mathrm{ r= \frac{1}{\mu_i}\frac{dC}{dt}}$$ where $\mathrm{dC}$ is the infinitesimal change in concentration of reactant or product in infinitesimal interval of time $\mathrm{dt}$ , ${\mu_i}$ is the stoichiometric coefficient of reactant or product.

When volume of system, $\mathrm{V}$ is constant, $$\mathrm{\frac{dC}{dt}=\frac{1}{V}\frac{dn}{dt}}$$

But if volume of the system changes then $$\mathrm{\frac{dC}{dt}=\frac{d(n/V)}{dt}}$$ which is equivalent to $$\mathrm{\frac{dC}{dt}=\frac{1}{V}.\frac{dn}{dt} - \frac{n}{V^2}.\frac{dV}{dt}}$$ and proceeded further using the ideal gas equation.

So, is the solution provided in the book correct or my approach?

  • $\begingroup$ Your reasoning does make sense. But how can we guess what you call "the solution provided by the book" ? $\endgroup$
    – Maurice
    Commented Jun 3, 2021 at 14:53
  • $\begingroup$ @Maurice look for the image of the solution. He has inserted that $\endgroup$ Commented Jun 3, 2021 at 19:32
  • $\begingroup$ But I don't get why you feel the solution is incorrect. What part did you feel wrong, please explain? $\endgroup$ Commented Jun 3, 2021 at 19:35
  • $\begingroup$ @Jayadithya This may help. chemistry.stackexchange.com/questions/43558/… $\endgroup$
    – Rishi
    Commented Jun 4, 2021 at 3:21
  • $\begingroup$ @Jayadithya The first question’s solution looks wrong to me. $\endgroup$
    – Rishi
    Commented Jun 4, 2021 at 3:32

3 Answers 3


The answer and the solution given in the book are wrong

I have searched and found the following in Chemical Kinetics by Laidler which talks about the volume change during reaction. It states:

....Equation (1.11) gives a general definition of the rate of reaction, but Eqs. (1.13)-(1.14) apply only if the volume remains constant during reaction. To remove the restriction of constant volume, we differentiate the relationship $$\mathrm{n_B = [B]V}$$ for a species B and obtain $$\mathrm{dn_B = V.d[B] + [B]dV \tag{1.15}}$$ Insertion into Eq. (1.11) gives $$\mathrm{r = \frac{1}{\mu_B}.\frac{d[B]}{dt} + \frac{[B]}{\mu_B . V}.\frac{dV}{dt} \tag{1.16} }$$ This becomes equivalent to Eq.(1.13) when volume remains constant. The final terms in Eq.(1.16) is a correction for the change in concentration brought by change in volume. Even if no reaction were occuring an increase in volume decreases in concentration of the substances present. The rate of this decrease is equal to $\mathrm{\frac{[B]}{V}.\frac{dV}{dt}}$ and therefore addition of the term $\mathrm{\frac{[B]}{\mu_B . V}\frac{dV}{dt}}$ is required to make $\mathrm{r=0}$ for this case of no reaction.....

Using the ideal gas equation and the first order rate law $$n_{\mathrm{1}}\ce{A} \ce{->} n_{\mathrm{2}}\ce{B}$$ \begin{array}{|c|c|} \hline \text{Time} &\text{Number of moles of A }& \text{Number of moles of B} \\ \hline \text{t=0} & a & 0\\ \text{t=t} & a-n_1x& n_2x \\ \text{t= infinite} & 0 &\frac{n_2a}{n_1} \\ \hline \end{array} Using the ideal gas equation $pV = nRT$ at the time 't' when $n_1x$ moles of $\ce{A}$ are consumed, $$pV_t = (a-n_1x+n_2x)RT$$ $$\implies V_t = \frac{(a-n_1x+n_2x)RT}{p}$$ Using the first order rate law, $$\mathrm{\frac{{d(n_A/V_t)}}{dt} = k\frac{n_A}{V_t}}$$ Substituting $\mathrm{n_A} = a-n_1x$ and $V_t = \frac{(a-n_1x+n_2x)RT}{P}$ in the above rate law and solving the differential equation, the result is $$\mathrm{\frac{akt}{V_0} = x - \frac{n_2x}{n_1} - \frac{a}{n_1}.ln(\frac{a-n_1x}{a})}$$


  1. Chemical Kinetics, 3e by Keith J. Laidler, Pearson publications.

  2. Atkins Physical Chemistry 11e Volume 3 Molecular Thermodynamics and Kinetics.


When the reactants and products are both gaseous and at same pressure and temperature you might want to prefer using volume as a unit of measurement instead of concentrations as it simplifies the equations tremendously. Using volume as a unit of measurement is possible because the moles of each gas $\ce A$ and $\ce B$ will be proportional to their volumes.

This way instead of dealing with three variables you now are concerned with only two. And we all can agree that solving a differential equation even in two variables can sometimes be tricky lest it be in three.

Thus for the reaction,


The rate law can be written as, $$-\frac{1}{n_1}\frac{\mathrm dV_{\ce A}}{\mathrm dt}=kV_{\ce A}$$

Which simplifies to,

$${(V_{\ce A})}_t={(V_{\ce A})}_0e^{-n_1kt}$$

Now the same expression can also be written in the form of moles as the volume was proportional to the number of moles.

$${(n_{\ce A})}_t=ae^{-n_1kt}$$

Which is same as the answer provided by your book.

Coming to your solution:

Your solution is also technically correct. But you are overlooking the fact that the moles of a gas and it's volume are not independent variables. They are interdependent on each other and thus creating a differential equation containing two interdependent variables does no good.

Thus before differentiating the $V/n$ quantity limit it to only one variable. If you will eliminate $n$ than it will be equivalent to the approach used in this answer and if you eliminate $V$ than it will be equivalent to the solution provided by your book.

  • 1
    $\begingroup$ “Using volume as a unit of measurement is possible because the moles of each gas A and B will be proportional to their volumes.” Could you explain this in more detail in your answer because I am getting something else, through ideal gas equation;$$\mathrm{P_AV_A=n_ART}$$ Here note that Volume of gas A will be equal to that of the ‘volume-changing container’ say V then; $$\mathrm{\frac{P_AV}{n_A} = constant}$$ Now using partial pressure equation for gas A and simplifying; $$\mathrm{\frac{P (n_A)V}{(n_A+n_B)n_A}=constant}$$ where P is the total pressure $$\mathrm{\frac{V}{n_A+n_B}=constant}$$ $\endgroup$
    – Rishi
    Commented Jun 4, 2021 at 11:08
  • 1
    $\begingroup$ This is an intriguing problem. Atkins 5th edn. shows a problem where azomethane (g) decomposes to ethane (g) and N2 (g) at 600 K. It gives partial pressure values for azomethane vs time, and asks to show that it is a first order reaction. The solution says "the partial pressure of a gas is proportional to its concentration", and it proceeds to show that by plotting $ln(p_{azomethane})$ vs $t$, one gets a straight line, hence the reaction is indeed first order. This would seem to support NisargBhavsar's answer, as partial volumes and partial pressures are the same. $\endgroup$ Commented Jun 21, 2021 at 20:19
  • 1
    $\begingroup$ And it is true: it is easy to show that $p_A = C_ART$. So if one works literally, from $\frac {dC_A}{dt} = - kn_1C_A$, it follows that $\frac {dp_A}{dt} = - kn_1p_A$, and one does not have to worry about changing volumes at all. Perhaps the confusion comes from the fact that, in a reaction where 2 mols of gas form from 1 mol of gas, the partial pressure is not linearly related to the 'conversion' ($1- \frac {n_A}{n_{A,0}}$), as one normally assumes in this kind of problems. Still, only one answer can be correct, unless they are both saying the same in different forms. $\endgroup$ Commented Jun 21, 2021 at 20:34
  • 1
    $\begingroup$ And in fact NisargBhavsar's answer makes sense mechanistically. Saying that azomethane decomposes by a first order mechanism means that each molecule of azomethane reacts independently from its own concentration. It does not need to 'collide' with other molecules to react. So even though its decomposition causes an increase of the total volume, that does not change the relative rate. In fact I suspect that even in a solution, if you start a 1st order reaction and then keep adding solvent, you will still get a constant $\frac 1 C \frac{dC}{dt}$. To be verified, but I am relatively confident. $\endgroup$ Commented Jun 21, 2021 at 20:56
  • 1
    $\begingroup$ @user6376297 I might be stupid here but in your first comment, “as partial volumes and partial pressures are the same” can you prove this, mathematically? I don’t see how this is true. $\endgroup$
    – Rishi
    Commented Jun 22, 2021 at 4:28

As I mentioned, I agree with Nisarg Bhavsar's conclusion.

For the justification, I think a more formal one can be obtained starting from this Wikipedia article on mass balance:


In this case the system is indeed closed, so the mass (not concentration) balance is:

$r_A \cdot V = \frac {dn_A}{dt}$

The reaction rate $r_A$ by convention is moles of A produced minus transformed, per unit time, per unit volume. [That's why its product by the instant reaction volume $V$ gives indeed the change in moles per unit time].

It is stated that the reaction is first order in $A$.
This implies the following:

$r_A = -k \cdot n_1 \cdot C_A$

The stoichiometric coefficient $n_1$ must be added because, again by convention, the rate constant $k$ is given for the 'standard' case where $n_1 = 1$.

Substituting in the mass balance:

$-k \cdot n_1 \cdot C_A \cdot V = \frac {dn_A}{dt}$

As $C_A \cdot V = n_A$, this becomes:

$-k \cdot n_1 \cdot n_A = \frac {dn_A}{dt}$

which leads to the same conclusion as the one from the book and from Nisarg Bhavsar, i.e.:

$n_A(t) = n_{A,0} \cdot e^{-k \cdot n_1 \cdot t}$

The conversion or advancement is:

$y_A(t) = \frac {n_{A,0}-n_A(t)}{n_{A,0}} = 1 - e^{-k \cdot n_1 \cdot t}$

This is in line with the idea that the conversion of a first order reaction does not depend on the initial concentration.

In practice, if one wanted to follow this reaction experimentally, they could sample it, taking only a very small volume of gas out, and measure the concentration of $A$ in the sample, e.g. by gas chromatography.
Knowing what the total volume of the reaction is at the sampling time, one could calculate the total number of moles of $A$, to fit the above integrated equation.

Could one use 'partial volumes' instead of number of moles?

Yes. Given that it is stated that this reaction is occurring at constant (total) pressure and temperature, and the mixture is ideal:

$V = n \cdot \frac { R \cdot T} {P} = n \cdot constant$

$V_A = n_A \cdot constant$

So one can substitute $n_A$ by $V_A$, and the equation still holds.

Why did Atkins' example about azomethane use partial pressure?

$\ce{CH3-N=N-CH3(g) \rightarrow CH3-CH3(g) + N2(g)}$

If one starts with pure azomethane, this is clearly a reaction (in the gas phase, at 600 K) where the total number of moles of gas are doubled when the reaction is complete.

Atkins states that the reaction is first order in azomethane, and the system is closed.
So the same equation as above applies (this time with stoichiometric coefficient $1$):

$-k \cdot n_A = \frac {dn_A}{dt}$

Atkins uses the following integrated law:

$\ln {\frac p {p_0}} = - k \cdot t$

where $p$ is the partial pressure of azomethane (not sure how it was determined experimentally).

By showing that the $\ln p$ vs $t$ data can be fitted to a straight line, it concludes that the reaction was first order.

However, this requires that the reaction be at constant volume (not mentioned by Atkins, but in fact mentioned in other books, like Silbey, Alberty - Physical Chemistry).

Atkins' argument starts from the assumption that:

$\ln {\frac {[A]} {[A]_0}} = - k \cdot t$

and then it says "the partial pressure of a gas is equivalent to its concentration", thus substituting $[A]$ by $p_A$.

It is true that for an ideal gas mixture:

$p_A = x_A \cdot P = \frac {n_A} {n} \cdot P$

and, as:

$P = n \cdot \frac {R \cdot T} V$

$\implies p_A = n_A \cdot \frac {R \cdot T} V = C_A \cdot {R \cdot T}$

So, as long as the temperature is constant, it is indeed legitimate to substitute $C_A = [A]$ by $p_A$, in a ratio where the constant term ${R \cdot T}$ cancels out.

But this is missing the point.
The real question is: where did the integrated rate law come from in the first place?
Going back to the mass balance:

$-k \cdot C_A \cdot V = \frac {dn_A}{dt}$

As far as I can tell, the only way for this to become the integrated rate law used by Atkins is by assuming constant volume. Then, dividing by $V$ on both sides and taking $V$ into the differential:

$-k \cdot C_A = \frac {d \frac{n_A} V}{dt} = \frac {dC_A}{dt}$

which, integrated, is indeed:

$\ln {\frac {C_A} {C_{A,0}}} = \ln {\frac {[A]} {[A]_0}} = - k \cdot t$

If, like in the example from the original post, the total pressure, not the total volume, were constant, one could not derive this equation.
The one based on the number of moles of course would still be valid, but then:

$n_A(t) = n_{A,0} \cdot e^{-k \cdot t}$

$p_A = \frac {n_A} {n} \cdot P$

$n = n_A + n_{ethane} + n_{N_2} = n_A + 2 \cdot (n_{A,0} - n_A) = 2 \cdot n_{A,0} - n_A$

$p_A = \frac {n_A} {2 \cdot n_{A,0} - n_A} \cdot P$

$n_A = \frac {2 \cdot n_{A,0} \cdot p_A}{p_A + P}$

$\frac {2 \cdot n_{A,0} \cdot p_A(t)}{p_A(t) + P} = n_{A,0} \cdot e^{-k \cdot t}$

$p_A(t) = P \cdot \frac {e^{-k \cdot t}} {2 - e^{-k \cdot t}}$

I tested this out numerically and it seems to work, but then again, I am not a physical chemist, so if anyone sees any errors in my logic or if I am using circular arguments, please do correct me.

On the other hand, if Nisarg Bhavsar and I are correct, maybe someone should warn Jayadithya that the accepted answer is not correct (is it?).


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