Question

For a second order reaction,
$2A\to \text{Products}$, a plot of $\text{log}{t}_{\frac{1}{2}}$ vs $\text{log a}$ (where 'a' is the initial concentration) will give an intercept equal to which of the following?

• A. $\frac{1}{k}$
• B. $\text{log}\left(\frac{1}{2k}\right)$
• C. $\text{log}\left(\frac{1}{2k}\right)$
• D. $-\text{log k}$

Answer 1

The correct option is D $-\text{log k}$

For a second order reaction,
$2A\to \text{Products}$
$k=\frac{1}{t}[\frac{1}{\left[A\right]}-\frac{1}{[A{]}_0}]$
when, $t=t_{\frac{1}{2}},\left[A\right]=\frac{[A{]}_0}{2}$
Hence, $t_{\frac{1}{2}}=\frac{1}{k}[\frac{2}{[A{]}_0}-\frac{1}{[A{]}_0}]=\frac{1}{k[A{]}_0}$
$\therefore \text{log}{t}_{\frac{1}{2}}=-\text{log k - log}\left[A_0\right]=-\text{log k - log a}$
Plot of $\text{log}{t}_{\frac{1}{2}}$ vs $\text{log a}$ will be linear with intercept $=-\text{log k}$