Question

For the second-order reaction, if the concentration of a reactant changes from $0.08M$ to $0.04M$ in $10$ minutes. Calculate the time at which concentration of reactant becomes $0.01M$ :-

• A. $20$ min
• B. $30$ min
• C. $50$ min
• D. $70$ min

Answer 1

Answer: (D)

$70$ min

$\begin{array}{c}\frac{d\left[A\right]}{dt}=k\left[A^2\right]\\ {\int }^{}\frac{d\left[A\right]}{\left[A^2\right]}={\int }^{}kdt\\ Integrating,\\ {\left[-\frac{1}{\left[A\right]}\right]}_{A_0}^A=kt\\ \frac{1}{A_0}-\frac{1}{A}=k\\ \frac{1}{0.08}-\frac{1}{0.04}=k\times 10\\ \frac{12.5}{10}=k\\ 1.25=k\\ \frac{1}{0.08}-\frac{1}{0.01}=1.25\times t\\ \frac{87.5}{1.25}=t\\ t=70min\end{array}$