Question

$99\%$ of a first order reaction was completed in $32$ min, when will $99.9\%$ of the reaction complete?

• A. $50$ min
• B. $46$ min
• C. $49$ min
• D. $48$ min

Answer 1

Answer: (D)

$48$ min

For first order reaction

$t=\frac{2.303}{K}\log \frac{\left[R_0\right]}{\left[R\right]}$

$\left[R_0\right]=100$

$\left[R\right]=100-99=1$

$K=\frac{2.303}{32}\log \frac{100}{1}=\frac{2.303}{16}{s}^{-1}$

For $99.9\%$ complete

$\left[R_0\right]=100$

$\left[R\right]=100-99.9=0.1$

$t=\frac{2.303}{2.303}\times 16\times \log \frac{100}{0.1}=16\times 3=48s$