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)

$k=\frac{2.303}{t}\log \frac{R_0}{R_0-x}$

$k=\frac{2.303}{30}\log \frac{100R_0}{25R_0}$

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

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

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

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

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

$=59.8\approx 60min$