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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?

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    $\begingroup$ 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. $\endgroup$ – Chet Miller Dec 30 '16 at 12:09
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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 $$5.068\cdot10^{-4}~\pu{s^{-1}}$$

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$k=\frac{2.303}{t}*\log{\frac{a}{a-x}}$

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 $C_{1}-C_{2}$

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