Question

In a 2nd order reaction, when the concentration of both the reactants is equal, the reaction is completed 20% in 500 sec. The time it would take in seconds for the reaction to go to 60% completion is _______.

Answer 1

For a second order reaction $k_2=\frac{1}{t}\times \frac{x}{a(a-x)}$

When the reaction is 20% complete, x is 20% of the initial concentration a that is 20% of x

$\frac{20a}{100}=0.2a$

$k_2=\frac{1}{500}\times \frac{0.2a}{a(a-0.2a)}=\frac{1}{2000a}$

Now suppose it takes $t_1$ seconds for the reaction to go for $60\%$ completion.

Then

$x=\frac{60a}{100}=0.6a$

Thus $k_2=\frac{1}{t_1}\times \frac{0.6a}{a(a-0.6a)}$

$\frac{1}{2000a}=\frac{1}{t_1}\times \frac{3}{2a}$

$t_1=3000seconds$