The decomposition reaction $\ce{2N2O5(g) ->[\Delta] 2N2O4(g) + O2(g)}$ is started in a closed cylinder under isothermal conditions at an initial pressure of $\pu{1 atm}$. After $\pu{Y \times 10^{3} s}$, the pressure inside the cylinder is found to be $\pu{1.45 atm}$. If the rate constant of the reaction is $\pu{5 \times 10^{-4} s-1}$, assuming ideal gas behavior, the value of Y is?

I used the equation given here for a first order reaction, $[A]=[A]_oe^{-kt}$

Using $k = \pu{5 \times 10^{-4} s-1}$, I am getting 4.606 but the answer given is 2.3.

Why is this wrong?

The question is from JEE Advanced 2019 paper 2.

  • 1
    $\begingroup$ You can't directly use this formula.. the co-efficients of rate are different.. I would suggest that you read through the proof of this property and then try to solve it from the first principles. Concepts matter.. $\endgroup$ Jul 14 '20 at 15:50
  • $\begingroup$ @Maurice I have done this same question before(and got the same answer).. So, I would assume that the OP has blindly copied the formula and not gone through the derivation which makes a difference in this case (as you may notice). $\endgroup$ Jul 14 '20 at 16:02
  • 1
    $\begingroup$ If the pressure has increased from 1 atm to 1.45 atm, it means that the number of moles of gas was n in the beginning, and 1.45 n at the end. It means that 2·0.45 · n mole of N2O5 has been transformed into 2·0.45·n mole of N2O4 + 0.45·n mol of O2. So [A]/[Ao] = 0.1/1 = 0.1. Do try to finish this calculation. $\endgroup$
    – Maurice
    Jul 14 '20 at 16:08
  • 5
    $\begingroup$ You should your work to get help here. Just copy and paste a portion of a web page won't help. Also, do not post a question as an image. $\endgroup$ Jul 14 '20 at 16:14
  • $\begingroup$ @Safdar: A great job on editing! There was a typo on equilibrium constant what I fixed. Also, make sure to do complete editing such as including appropriate tags and editing the title according to question. The title should be attractive to the keywords used by future searchers. Other than that, excellent job. $\endgroup$ Jul 14 '20 at 16:56

The basic law of chemical kinetics states for $\ce{aR -> bP}$, that

$$A = kR = -\frac{1}{a}\frac{\mathrm dR}{\mathrm dt} = \frac{1}{b}\frac{\mathrm dp}{\mathrm dt}$$

(where $A$ is activity and $k$ is called rate constant1)

Note that, in our case, the stoichiometric coefficient of reactant (i.e., $\ce{N2O5}$) is 2, therefore the differential rate equation becomes,

$$-\frac{1}{2}\frac{\mathrm dR}{\mathrm dt} = kR$$


$$-\frac{\mathrm dR}{\mathrm dt} = (2k)R$$

On integrating and re-arranging, we get,

$$R = R_0 \mathrm e^{(2k)t}$$

which means, $$t = \frac{2.303}{2k}\log_{10}{\frac{R_0}{R}}$$

This factor of $\frac{1}{2}$ was what you missed, I think.

1: As per my first equation, it's clear how rate constant is specific to a certain reaction. The following is an extract from Concentration and Rates (Differential Rate Laws) (under the heading "Reaction Orders"):

The proportionality constant (k) is called the rate constant, and its value is characteristic of the reaction and the reaction conditions. A given reaction has a particular rate constant value under a given set of conditions, such as temperature, pressure, and solvent.

  • $\begingroup$ I don't follow, even if we change the coefficient of the equations by which I mean we multiply throughout with 2, still the reaction will happen in the same time right. What I mean if we the concentration drops by a factor of ten in ten minutes it would drop by ten in ten minutes whatever the stoichiometric coefficient are. $\endgroup$
    – Wood
    Jul 15 '20 at 3:39
  • $\begingroup$ Think of it this way, when the co-efficient is 1, every second a certain number of molecules react (this is an oversimplification), now if the co-efficient becomes 2, then that would mean that the number of molecules used up per second doubles. and so the time taken to reach a specific concentration halves. $\endgroup$ Jul 15 '20 at 3:53
  • $\begingroup$ @Wood: I've corrected some parts of my answer. Read the first bracket carefully. Further, I agree that time is independent of concentration or pressure, but it is dependent on rate constant, which is what I've edited. Thanks for pointing out! $\endgroup$ Jul 15 '20 at 4:49

The other answer is correct, but I'd like to try out a different way of explaining it.

  • We know that at fixed volume and temperature, the pressure is proportional to the number of particles. From the formula given, we get 45% more particles when 90% of the original $\ce{N2O5}$ is broken down, since it takes two to make one $\ce{O2}$.

  • The units of our rate constant include no concentration, so we know the reaction is first order, meaning it progresses at the same rate without regard to pressure. It means rarely, something just "goes wrong" with a molecule of $\ce{N2O5}$, at random, with no obvious cause.

  • Our rate constant can be rewritten as $\pu{\frac{1}{2000} s-1}$. The $\pu{2000 s}$ is the mean lifetime - like a half-life but the time for the overall amount of a starting reactant to be reduced to not 1/2 but to $1/e ≈ 1/2.718 ≈ 0.368$ of what it was. (The reason for the odd number is that some of the remaining sample will survive multiple mean lifetimes, and when you average it all out, the mean lifetime is the average time any one molecule will survive)

  • We want the amount of $\ce{N2O5}$ remaining to be $10\% = 1/10$, so we have to figure out an exponent $n$ so $(1/e)^n = 1/10$. In other words, we need $\ln(10) ≈ 2.30$.

  • At this point I'm just about ready to make the same mistake -- I get $2.30 \times \pu{2000 s}$. But ... we need to remember that 2 molecules of $\ce{N2O5}$ are used per reaction, so the rate of its depletion is twice the rate of the reaction. So we divide by 2 to get $2.30 \times \pu{1000 s}$.

The computation is similar to half-life computations with radioisotopes, but in those we never have a stoichiometric coefficient to worry about because there's no way that each atom of, say, $\ce{^{32}P}$ can reliably "take out" another such atom when it decays. Notice this time is also different from the time it would take 45% of $\ce{N2O5}$ to react if it did not affect a second molecule. Even though it is not rate limiting, using up the second molecule of $\ce{N2O5}$ in each reaction reduces the amount available to continue reacting.

  • $\begingroup$ by e-life, did you mean mean-life ? $\endgroup$ Jul 15 '20 at 3:55
  • $\begingroup$ The 'e-life' is called the lifetime $\tau=1/k$ $\endgroup$
    – porphyrin
    Jul 15 '20 at 7:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.