Question

A first order reaction is $75\%$ completed in $100minutes$. How long time will it take for its $87.5\%$ completion?

• A. $125min$
• B. $150min$
• C. $175min$
• D. $200min$

Answer 1

The correct option is B $150min$

For first order reaction,
$t=\frac{2.303}{k}lo{g}_{10}\frac{R_0}{R}\begin{array}{c}\to \text{initial concentration}\\ \to \text{left out concentration}\end{array}$
$75\%$ reaction completed in $100min$ than rate constant is $\therefore R_0=100;R=100-75=25$
$k=\frac{2.303}{t}lo{g}_{10}\frac{100}{25}$
$k=\frac{2.303}{100}lo{g}_{10}\frac{100}{25}$
$k=\frac{2.303}{100}lo{g}_{10}4$
Time taken for completion of 87.5% of the reaction; $(\therefore R_0=100;R=100-87.5=12.5)$
$t=\frac{2.303}{k}lo{g}_{10}\frac{100}{12.5}$
On substituting the value of $k$ in equation we get,
$t=\frac{2.303lo{g}_{10}8}{\frac{2.303}{100}lo{g}_{10}4}=\frac{100\times lo{g}_{10}8}{lo{g}_{10}4}$
$=100\times 1.5$
$=150min$