Question

For a first order reaction, the time required for completion of 90% reaction is ‘x’ times the half life of the reaction. The value of ‘x’ is (Given: In 10 = 2.303 and log 2 = 0.3010)

• A. 1.12
• B. 2.43
• C. 3.32
• D. 33.31

Answer 1

Answer: (C)

3.32

Answer 2

A → Products

For a first order reaction,

$t_{1/2} = \frac{ln2}{k} = \frac{0.693 }{ k}$

Time for $90\%$ conversion,

$t_{ 90\%} = \frac{1}{k} \;In \;\frac{100}{10 }= \frac{ln10}{k} = \frac{2.303}{k}$

$t_{90\%} = \frac{2.303}{0.693} \;t_{1/2} = 3.32 \;t_{1/2}$