Question

For a first-order reaction, $A⟶$ Products, the concentration of $A$ changes from $0.1M$ to $0.025M$ in $40$ minutes. The rate of reaction when the concentration of $A$ is $0.01M$, is:

• A. $1.73\times {10}^{-3}M/min$
• B. $3.47\times {10}^{-4}M/min$
• C. $3.47\times {10}^{-5}M/min$
• D. $1.73\times {10}^{-4}M/min$

Answer 1

Answer: (B)

$3.47\times {10}^{-4}M/min$

$t_{1/4}=\frac{2\ln \left(2\right)}{k}k=\frac{2\ln \left(2\right)}{t_{1/4}}$
$=\frac{2\ln \left(2\right)}{40}mi{n}^{-1}\therefore v=k\left[A\right]$

$=\frac{2\ln \left(2\right)}{t_{1/4}}\times 0.01$

$=\frac{2\ln \left(2\right)}{40}\times 0.01$

$=3.47\times {10}^{-4}Mmi{n}^{-1}$