Question

The time taken for $10\%$ completion of a first order reaction is $20$ minutes. Then for $19\%$ completion the reaction will take

• A. $40 \,min$
• B. $60 \,min$
• C. $30\, min$
• D. $50\, min$

Answer 1

Answer: (A)

$40 \,min$

Answer 2

When the reaction is $10 \%$ completed, $k=\frac{2.303}{20} \log \frac{100}{100-10}\ldots$(i) When the reaction is $19 \%$ completed, $k=\frac{2.303}{t} \log \frac{100}{100-19}\ldots$(ii) From Eq (i) and (ii), $\Rightarrow \frac{2.303}{20} \log \frac{100}{90}=\frac{2.303}{t} \log \frac{100}{81}$ $\Rightarrow \frac{1}{20} \times 0.04575=\frac{1}{t} \times 0.09151$ $\therefore t=40\, min .$