A first order reaction takes 10min to complete 40% of the reaction. Another first order reaction requires 15min to complete 60% of the reaction. Calculate the ratio of the rate constants of the two reactions.

Explanation

For a first order reaction , `k=(2.303)/(t)log.(a)/(a-x)`
Let, the rate constants of the two first order reactions be `k_(1)and k_(2)` respectively.
For the first reaction ,when `t=10"min",x=axx(40)/(100)=0.4a`
`therefore(a-x)=(a-0.4a)=0.6a`
For the second reaction ,when t = 15 min,
`x=axx(60)/(100)=0.6a`
`therefore(a-x)=(a-0.6a)=0.4a`
`thereforek_(1)=(2.303)/(10)log.(a)/(0.6a)=0.05109" min"^(-1)`
and `thereforek_(2)=(2.303)/(15)log.(a)/(0.4a)=0.06109" min"^(-1)`
`therefore` The ratio of rate constants, `(k_(1))/(k_(2))=(0.05109)/(0.06109)=0.8363`