Question

For the reaction, $ A+B \rightarrow p=-\frac{d\left[A\right]}{dt}=-\frac{d\left[B\right]}{dt}=k\left[A\right]\left[B\right] $ and $ kt=\frac{1}{\left[A\right]_{0}-\left[B\right]_{0}} \ln \frac{\left[A\right]\left[B\right]_{0}}{\left[B\right]\left[A\right]_{0}} $ when, $ \left[A\right]_{0} \ne \left[B\right]_{0} $
If, $ \left[A\right]_{0}=\left[B\right]_{0} $ then the integrated rate law will be

• A. $kt = \ln \frac{\left[A\right]}{\left[B\right]} $
• B. $ \frac{1}{\left[B\right]}=\frac{1}{\left[A\right]_{0}}+kt $
• C. $ \frac{1}{\left[A\right]}=\frac{1}{\left[B\right]_{0}}+kt $
• D. $ \frac{1}{\left[A\right]}=\frac{1}{\left[A\right]_{0}}+kt $ or $ \frac{1}{\left[B\right]}=\frac{1}{\left[B\right]_{0}}+kt $

Answer 1

Answer: (D)

For a second order reaction,
$A+B \to P$
If $[A]_{0}=[B]_{0}$, then integrated rate law will be calculated as follows
Then, $-\frac{d \left[A\right]}{dt}=-\frac{d \left[B\right]}{d t}=k\left[A\right]\left[B\right]=k\left[A\right]^{2}$
$-\frac{d\left[A\right]}{dt}=k\left[A\right]^{2}$
$\frac{d\left[A\right]}{\left[A\right]^{2}}=-kdt$
Integrating between $ t=0, t=t$
$\int\limits_{\left[A\right]_0}^{\left[A\right]} \frac{d\left[A\right]}{\left[A\right]^{2}}=-k \int\limits_{0}^{t}dt $
On integrating the equation,
$\frac{1}{\left[A\right]_{1}}=\frac{1}{\left[A\right]_{0}}+kt$
Since, $\left[A\right]_{0}=\left[B\right]_{0}$ above equation can also be written as
$\frac{1}{\left[B\right]_{t}}=\frac{1}{\left[B\right]_{0}}+kt$