Question

Two first order reaction have half-lives in the ratio $8:1$. Calculate the ratio of time intervals $t_1$ and $t_2$ are the time period for the ${\left(\frac{1}{4}\right)}^{th}$ and ${\left(\frac{3}{4}\right)}^{th}$ completion.

• A. $1:0.301$
• B. $0.125:0.602$
• C. $1:0.602$
• D. none of these

Answer 1

The correct option is C $1:0.602$

As their half-lives in the ratio $8:1$ so their rate constant ratio is $1:8$ so ratio of time taken is $t_1/t_2=\frac{\left(2.303/k_1\right)log\left(1/\right(1-1/4\left)\right)}{\left(2.303/k_2\right)log\left(1/\right(1-3/4\left)\right)}=\frac{8\times log\left(4/3\right)}{log4}=1:0.602$