Question

For a first-order reaction, the ratio between the time taken to complete $\frac{3}{4}$th of the reaction and time taken to complete half of the reaction is

• A. 2
• B. 3
• C. 1.5
• D. 2.5

Hint:

First-order: $t_{1/2} = \frac{\ln 2}{k}$. $t_{3/4} = \frac{\ln 4}{k}$.

Answer 1

The Correct Option is A

For a first-order reaction, the integrated rate law is: $$ \ln[A]_t = \ln[A]_0 - kt $$ where $[A]_t$ is the concentration at time $t$, $[A]_0$ is the initial concentration, and $k$ is the rate constant. For completion of half of the reaction, $[A]_t = \frac{1}{2}[A]_0$, and the time taken is $t_{1/2}$. $$ \ln\frac{[A]_0}{2} = \ln[A]_0 - kt_{1/2} $$ $$ kt_{1/2} = \ln 2 $$ For completion of $\frac{3}{4}$th of the reaction, $[A]_t = \frac{1}{4}[A]_0$, and the time taken is $t_{3/4}$. $$ \ln\frac{[A]_0}{4} = \ln[A]_0 - kt_{3/4} $$ $$ kt_{3/4} = \ln 4 = 2\ln 2 $$ Therefore, $$ \frac{t_{3/4}}{t_{1/2}} = \frac{2\ln 2}{\ln 2} = 2 $$