Question

The concentration of a reactant $X$ decreases from $0.1\, M$ to $0.005\, M$ in $40$ minutes. If the reaction follows $1$ order kinetics, the rate of the reaction when the concentration of $X$ is $0.01\, M$ will be

• A. $1.73 \times 10^{-4}M\, min^{-1}$
• B. $3.47 \times 10^{-4}M\, min^{-1}$
• C. $3.47 \times 10^{-5}M\, min^{-1}$
• D. $7.5 \times 10^{-4}\, M\, min ^{-1}$

Answer 1

Answer: (D)

$k=\frac{2.303}{t} \log \frac{A_{0}}{A}=\frac{2.303}{40} \log \frac{0.1}{0.005}$

$=\frac{2.303}{40} \log 20=0.075$

Rate of reaction when concentration of $X$ is $0.01 M$ will be

$0.075 \times 0.01=7.5 \times 10^{-4} M \min ^{-1}$