Question

A compound decomposes by a first-order process. If 26 of the compound decomposes in 60 minutes, the half-life of the compound is ________minutes.

(A) 26
(B) 31
(C) 18
(D) 5
(E) 138

Answer 1

Answer:

(E) 138

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given:

26 % is decomposed which means that 0.26 of $[A_0]$ is decomposed. So,

$\frac {[A_t]}{[A_0]}$ = 1 - 0.26 = 0.74

t = 60 min

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.74=e^{-k\times 60}$

Taking natural log both sides, we get that:-

$ln\ 0.74=-k\times 60$

k = 0.005018 min⁻¹

Also,  Half life expression for first order:-

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,  

$t_{1/2}=\frac {ln\ 2}{0.005018}\ min=138\ min$

The correct option is:- (E) 138