Question

Given that for a reaction of $n^{th}$ order, the integrated rate equation is: $k=\frac{1}{t(n-1)}\left[\frac{1}{C^{n-1}}-\frac{1}{C_0^{n-1}}\right]$ , 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. $t_{\frac{3}{4}}=t_{\frac{1}{2}}[2^{n+1}+1]$

Answer 1

Answer: (A)

$t_{1/2}=\frac{1}{k(n-1)}\left[\frac{2^{n-1}}{C_0^{n-1}}-\frac{1}{C_0^{n-1}}\right]=\frac{1}{k(n-1)}\left[\frac{\left(2^{n-1}-1\right)}{C_0^{n-1}}\right]\to (1)$
$t_{3/4}=\frac{1}{k(n-1)}\left[\frac{\left(2^{2n-2}-1\right)}{C_0^{n-1}}\right]\to (2)$
$\frac{t_{3/4}}{t_{1/2}}=2^{n-1}+1\Rightarrow t_{3/4}=t_{\frac{1}{2}}\left(2^{n-1}+1\right)$