Question

Rate constant of a first order reaction is $1.15\times 10^{-5}se{c}^{-1.}$ The percentage of initial concentration remained after $1$ hour is

• A. $89.15\%$
• B. $95.96\%$
• C. $55.86\%$
• D. $83.45\%$

Answer 1

Answer: (B)

$1.15\times 10^{-5}=\frac{2.303}{1\times 60\times 60}log\frac{a}{a-x}$
(time $1hour=1\times 60\times 60seconds)$
$log\frac{a}{a-x}=\frac{1.15\times 10^{-5}\times 1\times 60\times 60}{2.303}=0.0179$
$\frac{a}{a-x}=1.042$
$\frac{a-x}{a}=\frac{1}{1.042}=0.9596$
Percentage of reactant remained = $0.9596\times 100=95.96\%$