Question

Given that for a reaction of order n, the integrated form of the 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 $t_{3/4}$ and $t_{1/2}$ where $t_{3/4}$ is the time required for C to become 1/4 $C_0$?

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

Answer 1

The correct option is A $t_{3/4}=t_{1/2}[2^{n-1}+1]$

$k=\frac{1}{t(n-1)}\left[\frac{1}{C^{n-1}}-\frac{1}{C_0^{n-1}}\right]t_{1/2}=\frac{1}{k(n-1)}\left[\frac{2^{n-1}}{C{o}^{n-1}}-\frac{1}{C{o}^{n-1}}\right]...\left(1\right)t_{3/4}=\frac{1}{k(n-1)}\left[\frac{4^{n-1}}{C{o}^{n-1}}-\frac{1}{C{o}^{n-1}}\right]...\left(2\right)equ.\left(2\right)÷equ.\left(1\right)\frac{t_{3/4}}{t_{1/2}}=\left[\frac{4^{n-1}-1}{2^{n-1}-1}\right]=\frac{2^{2(n-1)}-1}{(2^{n-1}-1)}=\frac{(2^{n-1}+1)(2^{n-1}-1)}{(2^{n-1}-1)}{t}_{3/4}=t_{1/2}[2^{n-1}+1]$