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:- 
The answer given, however, is $4.82 \times 10^{-4}\ \mathrm{s{-1}}$. What have I understood incorrectly?
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 $$5.068\cdot10^{-4}~\pu{s^{-1}}$$
$k=\frac{2.303}{t}*\log{\frac{a}{a-x}}$
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 $C_{1}-C_{2}$
