Question

In a first-order decomposition reaction, 50.0% of a compound decomposes in 10.5 min.

Answer 1

In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol $t_{1/2}$.

(a) For first order reaction, rate constant and half life time are related to each other as follows:

$k=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{10.5 min}=0.066 min^{-1}$

Thus, rate constant of the reaction is $0.066 min^{-1}$.

(b) Rate equation for first order reaction is as follows:

$k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}$

now, 75% of the compound is decomposed, if initial concentration $[A_{0} ]$ is 100 then concentration at time t $[A_{t} ]$ will be 100-75=25.

Putting the values,

$0.066 min^{-1}=\frac{2.303}{t}log\frac{100}{25}=\frac{2.303}{t}(0.6020)$

On rearranging,

$t=\frac{2.303\times 0.6020}{0.066 min^{-1}}=21 min$

Thus, time required for 75% decomposition is 21 min.