The first order integrated rate equation is

For first order,
\(
\text { rate }=\frac{d[R]}{d t}=k[R]
\)
or
\(
\frac{d[R]}{[R]}=k d t
\)
On integrating Eq. (i)
\(
\begin{array}{l}
\int \frac{d[R]}{[R]}=k \int d t \\
\ln [R]=-k t+C
\end{array}
\)
\(\ldots\)
\(
\text { At } t=0,[R]=\left[R_{0}\right]
\)
[where, \(R=\) final concentration, ie, \(a-x\) and
\(R_{0}\) is the initial concentration, \(\left.i e, a .\right]\)
\(
\ln \left[R_{0}\right]=C
\)
On putting the value of \(C\) in Eq. (ii), we get
\(
\begin{array}{r}
\ln [R]=-k t+\ln \left[R_{0}\right] \\
-k t=\ln [R]-\ln \left[R_{0}\right] \\
k t=\ln \left[R_{0}\right]-\ln [R]
\end{array}
\)
or
\(
\begin{aligned}
k &=\frac{1}{t} \ln \frac{\left[R_{0}\right]}{[R]} \\
or k&=\frac{1}{t} \ln \frac{a}{a-x}
\end{aligned}
\)