Q. In a Ist order reaction the concentration of reactant decreases from $800\ mol/{dm}^{3}\$to$\ 50\ mol/{dm}^{3}\$in$\ 2 \times 10^{4}\ s.$ The rate constant of reaction in $S^{- 1}$ is
A) $2 \times 10^{4}$
B) $3.45 \times 10^{- 5}$
C) $1.386 \times 10^{- 4}$
D) $2 \times 10^{- 4}$
Explanation
(c)
First calculate number of half-lives with
$C_{n} = \frac{C_{0}}{2^{n}}$
Then calculate $t_{1/2}\ as\ t_{1/2} = \frac{total\ time\ taken}{number\ of\ half - lives}afterwards\$
$t_{1/2} = \frac{0.693}{k}$
$\ 50 = \frac{800}{2^{n}}$
$\ 2^{n} = \frac{800}{50} = 16 = 2^{4}$
$\ n = 4$
$\ T = n \times t_{1/2}$
$\ t_{1/2} = \frac{2 \times 10^{4}}{4} = 0.5 \times 10^{4}\ s$
$k = \frac{0.693}{0.5 \times 10^{4}} = 1.386 \times 10^{- 4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.