The question asked was,

The initial concentration of $\ce{N2O5}$ in the first order reaction $$\ce{N2O5 -> 2NO2 + 1/2O2}$$ is $1.24\cdot 10^{-2}~\mathrm{mol\,L^{-1}}$ at 318K.

The concentration of $\ce{N2O5}$ decreases to $0.20\cdot 10^{-2}~\mathrm{mol\,L^{-1}}$ after 1 hour.
Calculate the rate constant at this temperature.

How i attempted to solve this is as follows:- Attempted Solution

The answer given, however, is $4.82 \times 10^{-4}\ \mathrm{s{-1}}$. What have I understood incorrectly?

  • 1
    $\begingroup$ You assumed incorrectly that the rate of the reaction does not change as the concentration of the reactant decreases. You need to solve this problem as a differential equation. $\endgroup$ Dec 30, 2016 at 12:09

2 Answers 2


This Khan Academy Video explains it very nicely.

So since this is a first order reaction $$\ce{ -\Delta N2O5/\Delta t = k[N2O5]}$$

This gives you the rate of reaction at the very start. A second later however, $\ce{[N2 O5]}$ decreases which will in turn change the rate of reaction. So lets apply calculus. We can write this as

$$\ce{ \frac{-d [N2O5]}{d t} = k[N2O5]}$$

Then we can integrate to find out the total change over a period of time.

$$\int_{[N2O5]_0}^{[N2O5]t} \frac{d[N2O5]}{[N2O5]}= \int_0^t-k dt$$

since $-k$ is constant, we can take it outside of the integral

$$\int_{[N2O5]_0}^{[N2O5]t} \frac{d[N2O5]}{[N2O5]}= -k\int_0^tdt$$

So then we get

$$ \ln([N2O5]_t) - \ln([N2O5]_0) =-kt$$

So putting in your values

$$\ln(0.20\cdot 10^{-2}) - \ln(1.24\cdot 10^{-2}) = -k\cdot 60\cdot 60$$

So I'm pretty sure the answer is $$5.068\cdot10^{-4}~\pu{s^{-1}}$$



For first order reaction . here $a,(a-x)$ are initial and final reactant concentration.

$k\to rate constant$

$t\to time$

Instantaneous rate is proportional to instantaneous concentration of the reactant . So, u can't take the value of conc. as ini conc. and $dx$ is small change in conc. so u can't take it as $C_{1}-C_{2}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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