Question

A second order reaction where $a=b$ is $20$% completed in $500$ seconds. How long will the reaction take to be $60$% complete?

Answer 1

Apply,
$k=\frac{x}{ta(a-x)}=\frac{20}{500[100\times (100-20)]}=\frac{1}{200000}{mol}^{-1}L{s}^{-1}$
$t=\frac{60}{k\times 100\times 40}=\frac{60\times 200000}{100\times 40}=3000$ seconds