Question

$75$% of a first order reaction was completed in $32$ min, then when was $50$% of the reaction completed?

• A. $24$ min
• B. $16$ min
• C. $8$ min
• D. $48$ min

Answer 1

Answer: (B)

$16$ min

$\ln \left(\frac{A_{0/4}}{A_o}\right)=kt\frac{2\ln \left(2\right)}{k}=t_{1/4}$

Now, $t_{1/2}=\frac{\ln \left(2\right)}{k}$ (known)

Thus, $t_{1/2}=\frac{t_{1/4}}{2}=\frac{32}{2}=16min$