Question

Given that for a reaction of $n^{th}$ order, the integrated rate equation is: $k=\frac{1}{t \left(\right. n - 1 \left.\right)}\left[\frac{1}{C^{n - 1}} - \frac{1}{C_{0}^{n - 1}}\right]$ , where $C_{0}$ and $C$ are the values of the reactant concentration at the start and after time $t$ . What is the relationship between $t_{\frac{3}{4}}$ and $t_{\frac{1}{2}}$ ?
( $t_{\frac{3}{4}}$ is the time required for $C$ to become $\frac{1}{4}C_{0}$ )

• A. $t_{\frac{3}{4}}=t_{\frac{1}{2}}\left[2^{n - 1} + 1\right]$
• B. $t_{\frac{3}{4}}=t_{\frac{1}{2}}\left[2^{n - 1} - 1\right]$
• C. $t_{\frac{3}{4}}=t_{\frac{1}{2}}\left[2^{n + 1} - 1\right]$
• D. $t_{\frac{3}{4}}=t_{\frac{1}{2}}\left[\right.2^{n + 1}+1\left]\right.$

Answer 1

Answer: (A)

$t _{1 / 2}=\frac{1}{ k ( n -1)}\left[\frac{2^{ n -1}}{ C _{0}^{ n -1}}-\frac{1}{ C _{0}^{ n -1}}\right]=\frac{1}{ k ( n -1)}\left[\frac{\left(2^{ n -1}-1\right)}{ C _{0}^{ n -1}}\right] \rightarrow(1)$
$t _{3 / 4}=\frac{1}{ k ( n -1)}\left[\frac{\left(2^{2 n -2}-1\right)}{ C _{0}^{ n -1}}\right] \rightarrow(2)$
$\frac{ t _{3 / 4}}{ t _{1 / 2}}=2^{ n -1}+1 \Rightarrow t _{3 / 4}= t _{\frac{1}{2}}\left(2^{ n -1}+1\right)$