Question

$75 \%$ of a first order reaction is completed in $ 30$ minutes, what is the time required for $93.75 \%$ of the reaction in minutes?

• A. 45
• B. 120
• C. 90
• D. 60

Answer 1

Answer: (D)

$k=\frac{2.303}{t} \log \frac{R_{0}}{R_{0}-x}$

$k=\frac{2.303}{30} \log \frac{100 R_{0}}{25 R_{0}}$

$k =\frac{2.303}{30} \times 0.602$

For $93.75 \%$ reaction, $t$ is given by

$t=\frac{2.303}{k} \log \frac{a}{a-x}$

$t =\frac{2.303 \times 30}{2.303 \times 0.602} \log \frac{100}{100-93.75}$

$t=\frac{30}{0.602} \log 16=\frac{30 \times 1.20}{0.602}$

$=59.8 \approx 60\, min$