Question

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

Answer 1

For second order reaction, $\frac{1}{[A{]}_t}=\frac{1}{[A{]}_o}+kt$

where $[A{]}_t$ = concentration of reactant left after time t

$[A{]}_o$ = initial concentration at time t = 0

The value of rate constant k:

$\frac{1}{[A{]}_{10min}}=\frac{1}{[A{]}_o}+kt$

$\frac{1}{0.04M}=\frac{1}{0.08M}+k\left(10\right)$

$k=1.25M^{-1}mi{n}^{-1}$

Time at which concentration of reactant becomes 0.01 M:

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

$\frac{1}{0.01M}=\frac{1}{0.08M}+\left(1.25M^{-1}mi{n}^{-1}\right)t$

$t=70$ min