Question

For $2\mathrm{A}+\mathrm{B}\to \mathrm{C}.$ Find the rate law.

$\left\{\mathrm{A}\right\}$	$\left\{\mathrm{B}\right\}$	$\left\{\mathrm{C}\right\}$
$0.05$	$0.05$	$0.045$
$0.10$	$0.05$	$0.09$
$0.20$	$0.10$	$0.72$

• A. $\mathrm{R}=\mathrm{K}\left[\mathrm{A}\right]\left[\mathrm{B}\right]$
• B. $\mathrm{R}=\mathrm{K}\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^2$
• C. $\mathrm{R}=\mathrm{K}{\left[\mathrm{A}\right]}^2\left[\mathrm{B}\right]$
• D. $\mathrm{R}=\mathrm{K}{\left[\mathrm{A}\right]}^2{\left[\mathrm{B}\right]}^2$

Answer 1

The correct option is B

$\mathrm{R}=\mathrm{K}\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^2$

Answer: (B)

Explanation for correct option:

$$\begin{gathered} \mathrm{r}=\mathrm{K}{\left[{\mathrm{A}}_1\right]}^{\mathrm{x}}{\left[\mathrm{B}\right]}^{\mathrm{y}} \\ 0.045=\mathrm{K}{(0.05)}^{\mathrm{x}}{(0.0.5)}^{\mathrm{y}}\dots \dots \dots \dots \left(1\right) \\ 0.090=\mathrm{K}{(0.010)}^{\mathrm{x}}{(0.05)}^{\mathrm{y}}\dots \dots \dots \dots \left(2\right) \\ 0.72=\mathrm{K}{(0.20)}^{\mathrm{x}}{(0.10)}^{\mathrm{y}}\dots \dots \dots \dots \dots \dots \left(3\right) \\ \mathrm{From}\left(1\right)÷\left(2\right), \\ \frac{0.045}{0.090}={\left(\frac{0.05}{0.10}\right)}^{\mathrm{x}}\Rightarrow \mathrm{x}=1 \\ \mathrm{From}\left(2\right)÷\left(3\right), \\ \frac{0.090}{0.720}={\left(\frac{0.10}{0.20}\right)}^1={\left(\frac{0.05}{0.10}\right)}^{\mathrm{y}}\Rightarrow \mathrm{y}=2 \\ \mathrm{r}=\mathrm{K}\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^2 \end{gathered}$$

Hence option (B) is the correct option.