Question

In a first-order decomposition reaction. 20.8% of a compound decomposes in 7.8 min. How long (in min) does it take for 88.2% of the compound to decompose?

Answer 1

t = 71.47 min

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:

20.8 % is decomposed which means that 0.208 of
`[A_0]` is decomposed. So,

`\frac {[A_t]}{[A_0]}` = 1 - 0.208 = 0.792

t = 7.8 min

`\frac {[A_t]}{[A_0]}=e^(-k* t)`

`0.792=e^(-k* 7.8)`

k = 0.0299 min⁻¹

Also,

Given:

88.2 % is decomposed which means that 0.882 of
`[A_0]` is decomposed. So,

`\frac {[A_t]}{[A_0]}` = 1 - 0.882 = 0.118

t = ?

`\frac {[A_t]}{[A_0]}=e^(-k* t)`

`0.118=e^(-0.0299* t)`

t = 71.47 min