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)

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),
$\Rightarrow \frac{2.303}{20} \log \frac{100}{90}=\frac{2.303}{t} \log \frac{100}{81}$
$\Rightarrow \frac{1}{20} \times 0.04575=\frac{1}{t} \times 0.09151$
$\therefore t=40\, min .$