Question

For a first-order reaction $X\to$ product, the concentration of $X$ changes from $0.2$ M to $0.05$ M in $40$ minutes, find the rate of reaction of $X$ when its concentration is $0.02$ M.

• A. $1.73\times {10}^{-4}M{min}^{-1}$
• B. $6.93\times {10}^{-4}M{min}^{-1}$
• C. $1.73\times {10}^{-5}M{min}^{-1}$
• D. $3.47\times {10}^{-5}M{min}^{-1}$

Answer 1

Answer: (B)

$6.93\times {10}^{-4}M{min}^{-1}$

For a first order reaction we use:

$ln\left(\frac{A_o}{A}\right)=kt$ (Integrated Rate Law)

Using $A_o=0.2M$ $A=0.05M$ we get

$k=\frac{2ln\left(2\right)}{40}mi{n}^{-1}$

Rate of the reaction, $-\frac{dA}{dt}=k\left[A\right]=\frac{2ln\left(2\right)}{40}mi{n}^{-1}\times 0.02M$

$=6.93\times {10}^{-4}Mmi{n}^{-1}$
Option B is correct.