Consider the reaction carried out at $\mathrm{T}(\mathrm{K})$ $\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g})…
Consider the reaction carried out at $\mathrm{T}(\mathrm{K})$
$\mathrm{A}(\mathrm{g})+\mathrm{B}(\mathrm{g}) \rightarrow \mathrm{C}(\mathrm{g})$
The rate law for this reaction is $r=k[A]^1[B]^2$. The concentration of $\mathrm{A}$ in experiment 2 and rate in experiment 3 shown as $\mathrm{x}$ and $\mathrm{z}$ in the table, $\mathrm{x}$ and $\mathrm{z}$ are respectively
\begin{array}{|l|l|l|l|}
\hline \begin{array}{l}
Experi- \\
ment
\end{array} & \frac{[\mathrm{A}]}{\mathrm{mol} \mathrm{L}^{-1}} & \frac{[\mathrm{B}]}{\mathrm{mol} \mathrm{L}^{-1}} & \begin{array}{l}
Initial rate \\
\left(\mathbf{m o l ~ L}^{-\mathbf{1}} \mathbf{~ s}^{-\mathbf{1}}\right)
\end{array} \\
\hline 1 & 0.05 & 0.05 & \mathrm{R} \\
\hline 2 & \mathrm{x} & 0.05 & 2 \mathrm{R} \\
\hline 3 & 0.20 & 0.10 & \mathrm{z} \\
\hline
\end{array}
$x=0.10 z=8 R$
$x=0.05 z=4 R$
$x=0.10 z=16 R$
1) $x=0.20 \mathrm{z}=16 R$
Solution
According to the rate law:-
When $[\mathrm{A}]_2=4[\mathrm{~A}]_1$ and $[\mathrm{B}]_2=2[\mathrm{~B}]_1$
$\begin{aligned}
\text { rate } & =\mathrm{r}=\mathrm{z}=\mathrm{k}[4 \mathrm{~A}][2 \mathrm{~B}]^2=\mathrm{k} \times 4 \times 4 \times[\mathrm{A}]_1[\mathrm{~B}]_1 \\
& =16 \mathrm{R} .
\end{aligned}$
And when $[\mathrm{B}]$ is kept constant, and $[\mathrm{A}]_2=\mathrm{x}$
$\begin{aligned}
\Rightarrow \text { rate } & =r_2=2 \mathrm{R}=2 \mathrm{k}(0.05)(0.05)^2 \\
& =\mathrm{k}(0.10)(0.05)^2 \\
& =\mathrm{k}(\mathrm{x})(0.05)^2
\end{aligned}$
Thus, $x=0.10$.