Q. A follows first order reaction , A$\rightarrow \$product Concentration of A, change from 0.1 M to 0.025 M in 40 min. find the rate of reaction of A when concentration of A is 0.01 M.
A) 3.47$\times 10^{- 4}M\ \min^{- 1}\$
B) 3.47$\times 10^{- 5}M\ \min^{- 1}\$
C) 1.73$\times 10^{- 4}M\ \min^{- 1}\$
D) 1.73$\times 10^{- 5}M\ \min^{- 1}\$
Explanation
(a)
A$\rightarrow product\$(first order reaction )
For first order reaction ,
Rate constant (k)=$\frac{2.303}{t}\log_{10}\frac{{\lbrack A\rbrack}_{0}}{{\lbrack A\rbrack}_{t}}$
At t= 40min,
$= \frac{2.303}{40}\log_{10}\frac{0.1}{0.025} = \frac{2.303}{40}\log_{10}4$
$\frac{2.303}{40}{\times 2log}_{10}2 = \frac{2.303}{40} \times 2 \times 0.3010$
=$0.0347\min^{- 1}\$
At concentration of $A = 0.01\ M = \lbrack A\rbrack$
Rate $\frac{dx}{dt}k\lbrack A\rbrack$
$\frac{dx}{dt} = 0.0347 \times 0.01$
$= 3.47 \times 10^{- 4}mol\ L^{- 1\ }\min^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.