Question

A reaction with respect to X is zero order till the concentration is reduced to half of the initial concentration. Then the reaction becomes first order with respect to it. If the value of the rate constant for the zero order and first order are equal to k (in magnitude), then find the time taken to reduce the concentration of X to $\frac{1}{16}th$ of its original concentration X${}_0$:

• A. $\frac{1}{2k}+\frac{3ln2}{k}$
• B. $\frac{x_0}{2k}+\frac{3ln2}{k}$
• C. $\frac{x_0}{2k}+\frac{4ln2}{k}$
• D. $\frac{1}{2k}+\frac{4ln2}{k}$

Answer 1

Answer: (B)

$\frac{x_0}{2k}+\frac{3ln2}{k}$

When, the reaction is zero-order, the concentration changes from $X_0$ to $\frac{X_0}{2}$. The rate constant is k.

This requires on half life, $t_{\left(1/2\right)}=\frac{X_0}{2k}$

Now the reaction becomes first order.

$\frac{X_0}{2}\underset{\frac{ln2}{k}}{\to }\frac{X_0}{4}\underset{\frac{ln2}{k}}{\to }\frac{X_0}{8}\underset{\frac{ln2}{k}}{\to }\frac{X_0}{16}$

Total time required to reduce the concntration from $X_0$ to $\frac{X_0}{16}$ is $\frac{X_0}{2k}+\frac{3ln2}{k}$.