Chemistry - Chemical Kinetics Question with Solution | TestHub

For first order, $$\text{ rate }=\frac{d[R]}{dt}=k[R]$$ or $$\frac{d[R]}{[R]}=kdt$$ On integrating Eq. (i) $$\begin{array}{l}\int \frac{d[R]}{[R]}=k\int dt\\ \ln [R]=-kt+C\end{array}$$ $\dots$ $$\text{ At }t=0,[R]=\left[R_0\right]$$ [where, $R=$ final concentration, ie, $a-x$ and $R_0$ is the initial concentration, $ie,a.]$ $$\ln \left[R_0\right]=C$$ On putting the value of $C$ in Eq. (ii), we get $$\begin{array}{r}\ln [R]=-kt+\ln \left[R_0\right]\\ -kt=\ln [R]-\ln \left[R_0\right]\\ kt=\ln \left[R_0\right]-\ln [R]\end{array}$$ or $$\begin{array}{rl}k& =\frac{1}{t}\ln \frac{\left[R_0\right]}{[R]}\\ ork& =\frac{1}{t}\ln \frac{a}{a-x}\end{array}$$