Question

Which of the following is/are correct for second order reaction?

• A. $t_{1/2}$ is inversely proportional to initial concentration
• B. $k=\frac{1}{t}[\frac{1}{(a-x)}-\frac{1}{a}]$
• C. $k=\frac{2.303}{t}\log \left(\frac{a}{a-x}\right)$
• D. $k=\frac{1}{2t}[\frac{1}{{(a-x)}^2}-\frac{1}{a^2}]$

Answer 1

Answer: (A), (B)

The correct options are
A $t_{1/2}$ is inversely proportional to initial concentration
B $k=\frac{1}{t}[\frac{1}{(a-x)}-\frac{1}{a}]$

For a second order reaction,

$-\frac{d\left[A\right]}{dt}=k[A{]}^2$

${\int }_a^{a-x}-\frac{d\left[A\right]}{[A{]}^2}=k{\int }_0^{t}dt$

$\frac{1}{a-x}-\frac{1}{a}=kt$

$k=\frac{1}{t}[\frac{1}{(a-x)}-\frac{1}{a}]$

For $t_{1/2},a-x=\frac{a}{2}$

$t_{1/2}=\frac{1}{k}[\frac{2}{a}-\frac{1}{a}]=\frac{1}{ka}$ i.e. $t_{1/2}\alpha \frac{1}{a}$