Question

Rate constant of a first order reaction is $1.15 \times 10^{-5} sec^{-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)

$1.15\times10^{-5} =\frac{2.303}{1\times60\times60}log \frac{a}{a-x}$
(time $1 hour = 1 \times 60 \times 60 seconds) $
$log \frac{a}{a-x }=\frac{1. 15 \times10^{-5}\times 1 \times60 \times60}{2.303} =0.0179$
$\frac{a}{a-x} =1.042$
$\frac{a-x}{a} = \frac{1}{1.042} =0.9596$
Percentage of reactant remained = $0.9596 \times 100 = 95.96\% $