The reaction of $A_{2}$ and $B_{2}$ follows the equation…

The reaction of $A_{2}$ and $B_{2}$ follows the equation
$\mathrm{A}_{2}(\mathrm{~g})+\mathrm{B}_{2}(\mathrm{~g}) ightarrow 2 \mathrm{AB}(\mathrm{g})$
The following data were observed
$\begin{array}{|c|c|c|}
\hline\left[\mathrm{A}_{2}ight]_{0} & {\left[\mathrm{~B}_{2}ight]_{0}} & \text{Initial rate of appearance of } \mathrm{AB}(\mathrm{g})\left(\mathrm{in} \mathrm{Ms}^{-1}ight) \\
\hline 0.10 & 0.10 & 2.5 \times 10^{-4} \\
\hline 0.20 & 0.10 & 5 \times 10^{-4} \\
\hline 0.20 & 0.20 & 10 \times 10^{-4} \\
\hline
\end{array}$
The value of rate constant for the above reaction is:
  1. $2.5 \times 10^{-4}$
  2. $2.5 \times 10^{-2}$
  3. $1.25 \times 10^{-2}$
  4. None of these

Solution

Order w.r.t. $\mathrm{A}=1$; order w.r.t. $\mathrm{B}=$ 1
Rate $=\frac{1}{2} \frac{\mathrm{d}(\mathrm{AB})}{\mathrm{dt}}=\mathrm{k}_{\mathrm{r}}[\mathrm{A}][\mathrm{B}]$
$\frac{1}{2} \times\left(2.5 \times 10^{4}ight)=\mathrm{k}_{\mathrm{r}}(0.1)(0.1)$
$\mathrm{k}_{\mathrm{r}}=1.25 \times 10^{-2}$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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