Question

For the second order reaction, if the concentration of reactant changes from 0.08 to 0.04 m in 10 minutes. calculate the time at which concentration of reactant becomes 0.01 m

Answer 1

using second order integrated equation

Answer 2

Answer: After 70 minutes concentration of reactant becomes 0.01 M.

Solution:

For second order reaction:

$\frac{1}{[A]_t}=\frac{1}{[A^o]}+kt$

$[A]_t=\text{concentration left at time t},[A^o]=\text{Initial concentration}$

k= rate constant with$M^{-1}min^{-1}$

1) The value of k:

$\frac{1}{[A]_{10 min}}=\frac{1}{[A^o]}+kt=\frac{1}{0.04M}=\frac{1}{0.08M}+k(10)$

$k=1.25M^{-1}min^{-1}$

2) Time at which concentration of reactant becomes 0.01 m:

$\frac{1}{[A]_t}=\frac{1}{[A^o]}+kt=\frac{1}{[0.01M]_t}=\frac{1}{0.08M}+(1.25M^{-1}min^{-1})t$

$t=70 min$

After 70 minutes concentration of reactant becomes 0.01 M.