Question

Compound $A$ and $B$ react to form $C$ and $D$ in a reaction that was found to be second-order over all and second-order in $A$. The rate constant -at ${30}^{0}C$ is $0.622$ L ${mol}^{-1}{min}^{-1}$. What is the half-life of A when $4.10\times {10}^{-2}$ M of A is mixed with excess $B$?

• A. $40$ min
• B. $39.21$ min
• C. $28.59$ min
• D. None of these

Answer 1

The correct option is B $39.21$ min

$A+B⟶C+D$

rate$=k{\left[A\right]}^2$ (given)

$=0.622{[4.10\times {10}^{-2}]}^2$

$=0.001$ is the rate of reaction initially

Half-life$=t_{1/2}=\frac{1}{K\left[A\right]}=\frac{1}{0.622\times [4.1\times {10}^{-2}]}=0.3921\times {10}^2=39.21minutes.$