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

The half-life for a first-order reaction is constant and independent of the concentration of the reactant. Since only 26% decomposes in 60 minutes, the half-life must be greater than 60 minutes. Therefore, the correct answer is (E) 138 minutes.

Step-by-step explanation:

The student's question involves determining the half-life of a compound that decomposes by a first-order process. Given that 26% of the compound decomposes in 60 minutes, we understand that this is not a complete half-life since a half-life would be the time it takes for 50% of the compound to decompose.

In the case of a first-order reaction, the half-life is constant and does not depend on the concentration of the reactant. Therefore, even without the exact rate constant, knowing that after 60 minutes only 26% has decomposed tells us that the half-life must be greater than 60 minutes. Since none of the provided answers (A, B, C, D) are greater than 60, the correct answer must be (E) 138 minutes, assuming that the half-life is indeed not dependent on the concentration for this reaction.

Answer 2

(E) 138

Step-by-step 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* t)`

`0.74=e^(-k* 60)`

Taking natural log both sides, we get that:-

`ln\ 0.74=-k* 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