Question

For the reaction, $A+B\to p=-\frac{d\left[A\right]}{dt}=-\frac{d\left[B\right]}{dt}=k\left[A\right]\left[B\right]$ 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, $-\frac{d\left[A\right]}{dt}=-\frac{d\left[B\right]}{dt}=k\left[A\right]\left[B\right]=k{\left[A\right]}^2$
$-\frac{d\left[A\right]}{dt}=k{\left[A\right]}^2$
$\frac{d\left[A\right]}{{\left[A\right]}^2}=-kdt$
Integrating between $t=0,t=t$
$\underset{{\left[A\right]}_0}{\overset{\left[A\right]}{\int }}\frac{d\left[A\right]}{{\left[A\right]}^2}=-k\underset{0}{\overset{t}{\int }}dt$
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
$\frac{1}{{\left[B\right]}_t}=\frac{1}{{\left[B\right]}_0}+kt$