Question

For a first order reaction, $A \to P, t_{1/2}$ (half life) is 10 days. The time required for $\frac{1^{th}}{4}$ conversion of A (in days) is :
(ln 2=0.693, ln 3=1.1)

• A. 5
• B. 3.2
• C. 4.1
• D. 2.5

Answer 1

Answer: (C)

$A \rightarrow P$

For first order$t_{1/2}=\frac{In 2}{k}=10$ days

For $t_{1/4}=\frac{In 43}{k}=t$

$\Rightarrow \frac{In2}{2In2-In3}=\frac{10}{t}$

$t_{1/4}=\frac{\left(2\times0.693-1.1\right)}{0.693}\times10=4.1$