Question

A reactant decomposes with a half-life of 29.5 s when its initial concentration is 0.229 M. When the initial concentration is 0.639 M, this same reactant decomposes with a half-life of 82.3 s. What is the value and unit of the rate constant for this reaction?

Answer 1

Answer :

The order of reaction is, 0 (zero order reaction).

The value of rate constant is, $0.00388Ms^{-1}$

Explanation :

Half life : It is defined as the time in which the concentration of a reactant is reduced to half of its original value.

The general expression of half-life for nth order is:

$t_{1/2}\propto \frac{1}{[A_o]^{n-1}}$

or,

$\frac{t_{1/2}_1}{t_{1/2}_2}=\frac{[A_2]^{n-1}}{[A_1]^{n-1}}$

or,

$n=\left(\frac{\log\frac{(t_{1/2})_1}{(t_{1/2})_2}}{\log\frac{(A)_2}{(A)_1}}\right )+1$           .............(1)

where,

$t_{1/2}$ = half-life of the reaction

n = order of reaction

[A] = concentration

As we are given:

Initial concentration of A = 0.229 M

Final concentration of A = 0.639 M

Initial half-life of the reaction = 29.5 s

Final half-life of the reaction = 82.3 s

Now put all the given values in the above formula 1, we get:

$n=\left (\frac{\log \frac{29.5}{82.3}}{\log\frac{0.639}{0.229}}\right )+1$

$n=0.000196\approx 0$

Thus, the order of reaction is, 0 (zero order reaction).

Now we have to determine the rate constant.

To calculate the rate constant for zero order the expression will be:

$t_{1/2}=\frac{[A_o]}{2k}$

When,

$t_{1/2}$ = 29.5 s

$[A_o]$ = 0.229 M

$29.5s=\frac{0.229M}{2k}$

$k=0.00388Ms^{-1}$

Thus, the value of rate constant is, $0.00388Ms^{-1}$