The kinetic datas for the reaction: $2 \mathrm{~A}+\mathrm{B}_{2} ightarrow 2 \mathrm{AB}$ are as given…
$2 \mathrm{~A}+\mathrm{B}_{2} ightarrow 2 \mathrm{AB}$ are as given below:
\begin{array}{|l|l|l|}
\hline[\mathrm{A}] \mathrm{mol} \mathrm{L}^{-1} & {\left[\mathrm{~B}_{2}ight] \mathrm{mol} \mathrm{L}^{-1}} & Rate \mathrm{mol} \mathrm{L}^{-1} \mathrm{~min}^{-1} \\
\hline 0.5 & 1.0 & 2.5 \times 10^{-3} \\
\hline 1.0 & 1.0 & 5.0 \times 10^{-3} \\
\hline 0.5 & 2.0 & 1 \times 10^{-2} \\
\hline
\end{array}
Hence the order of reaction w.r.t. $\mathrm{A}$ and $\mathrm{B}_{2}$ are, respectively,
- 1 and 2
- 2 and 1
- 1 and 1
- 2 and 2
Solution
$5 \times 10^{-3}=\mathrm{K}[1.0]^{\mathrm{a}}[1.0]^{\mathrm{b}}$...(2)
$1 \times 10^{-2}=\mathrm{K}[0.5]^{\mathrm{a}}[2.0]^{\mathrm{b}}$...(3)
Dividing equation (1) and (2)
$\frac{1}{2}=\left[\frac{1}{2}ight]^{\alpha}$
hence $\mathrm{a}=1$
Dividing equation (1) and (3)
$\frac{2.5 \times 10^{-3}}{1 \times 10^{-2}}=\left(\frac{1.0}{2.0}ight)^{\beta}$
$\frac{1}{4}=\left(\frac{1}{2}ight)^{\beta}$
$b=2$ /
Asked in: JEE-TOPICTESTS-CHEMISTRY