Question

If the rate constant of a reaction is $2.303\times {10}^{-4}{s}^{-1}$, then the time required to reduce 1.6 M of the reactant to 0.4 M is:
(log 2 = 0.3)

• A. 50 min
• B. 20 min
• C. 150 min
• D. 100 min

Answer 1

The correct option is D 100 min

• The reaction is first order as the unit of rate constant is $s^{-1}$.
• For first order reaction,

$t=\frac{2.303}{k}log\frac{[R{]}_0}{\left[R\right]}$
$=\frac{2.303}{2.303\times {10}^{-4}}log\frac{1.6}{0.4}=\frac{2.303}{2.303\times {10}^{-4}}log4$
$=\frac{2.303}{2.303\times {10}^{-4}}\times 2\times 0.3=6000s$

$t=100min$