Question

g The decomposition reaction of A to B is a first-order reaction with a half-life of 2.42×103 seconds: A → 2B If the initial concentration of A is 0.163 M, how many minutes will it take for the concentration of A to be 66.8% of the initial concentration?

Answer 1

Answer:

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.

Explanation:

The equation used to calculate the constant for first order kinetics:

$t_{1/2}=\frac{0.693}{k}}$ .....(1)

Rate law expression for first order kinetics is given by the equation:

$t=\frac{2.303}{k}\log\frac{[A_o]}{[A]}$ ......(2)

where,  

k = rate constant

$t_{1/2}$ =Half life of the reaction = $2.42\times 10^3 s$

t = time taken for decay process = ?

$[A_o]$ = initial amount of the reactant = 0.163 M

[A] = amount left after time t =  66.8% of $[A_o]$

[A]=$\frac{66.8}{100}\times 0.163 M=0.108884 M$

$k=\frac{0.693}{2.42\times 10^3 s}$

$t=\frac{2.303}{\frac{0.693}{2.42\times 10^3 s}}\log\frac{0.163 M}{0.108884 M}$

t = 1,409.19 s

1 minute = 60 sec

$t=\frac{1,409.19 }{60} min=23.49 min$

In 23.49 minutes the concentration of A to be 66.8% of the initial concentration.