Question

When the concentration of A in a reaction was changed from 1.60M to 0.80 M , the half life increased from 2.0 min to 4.0 min at 25C. Calculate the the order of the reaction and the rate constant .

Answer 1

Answer :

The order of reaction is, 2 (second order reaction).

The value of rate constant is, $0.31M^{-1}min^{-1}$

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 = 1.60 M

Final concentration of A = 0.80 M

Initial half-life of the reaction = 2.0 min

Final half-life of the reaction = 4.0 min

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

$n=\left (\frac{\log \frac{2.0}{4.0}}{\log\frac{0.80}{1.60}}\right )+1$

$n=2$

Thus, the order of reaction is, 2 (second order reaction).

Now we have to determine the rate constant.

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

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

When,

$t_{1/2}$ = 2.0 min

$[A_o]$ = 1.60 M

$2.0min=\frac{1}{k\times 1.60M}$

$k=0.31M^{-1}min^{-1}$

Thus, the value of rate constant is, $0.31M^{-1}min^{-1}$