Question

Two first order reactions have half-lives in the ratio 8 :1, calculate the ratio of time intervals $T_1:T_2$. The time $T_1$ and $T_2$ are the time period for ${\left(\frac{1}{4}\right)}^{th}$ and ${\left(\frac{3}{4}\right)}^{th}$ completion respectively:

• A. $1:0.0301$
• B. $0.125:.0.602$
• C. $1:0.602$
• D. none of these

Answer 1

The correct option is A $1:0.0301$

For first order reaction,

$k_1=\frac{ln2}{t_{1/2}}$

Given $\frac{(t_{1/2}{)}_1}{(t_{1/2}{)}_2}=\frac{k_2}{k_1}=\frac{8}{1}$...(i)

$T_1$=time interval for 1/4th completion of reaction (1):

$N=N_0{e}^{-kt}$

$\frac{3}{4}=e^{-k_1{T}_1}$

$\Rightarrow k_1{T}_1=ln\left(\frac{4}{3}\right)$

$T_2$=time period of 3/4th completion of reaction (2):

$\frac{1}{4}=e^{-k_2{T}_2}$=> $k_2{T}_2=ln\left(4\right)$

$\frac{T_1}{T_2}=\left(\frac{k_2}{k_1}\right).\frac{\frac{ln4}{3}}{ln4}=\frac{8\times ln\frac{4}{3}}{ln4}=1:0.0301$