Question

For a second order reaction, $2A⟶Products$, a plot of log $t_{\frac{1}{2}}$ vs log a (where 'a' is the Initial concentration) will give an intercept equal to:

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

Answer 1

The correct option is C log $\left(\frac{1}{k}\right)$

For a second order Reaction, $2A⟶products-\frac{d\left[A\right]}{dt}=k{A}^{2}k=\frac{1}{t}[\frac{1}{a-x}-\frac{1}{a}]$
When, $t=t_{1/2},x=\frac{a}{2}$
Hence, $t=t_{1/2}=\frac{1}{k}(\frac{2}{a}-\frac{1}{a})=\frac{1}{ka}$
$\therefore log{t}_{1/2}=-logk-loga=-logk-loga$
This is of the form $y=mx+c$
Plot of log $t_{1/2}$ vs log a will be linear with intercept = -log k