Question

During the kinetic study of the reaction, $2\mathrm{A}+\mathrm{B}$ give $\mathrm{C}+\mathrm{D}$ following results were obtained.

Run	$\left[\mathrm{A}\right]$ in $\mathrm{M}$	$\left[\mathrm{B}\right]$ in $\mathrm{M}$	Initial rate of formation of $\mathrm{D}$ in ${\mathrm{Ms}}^{-1}$
1	0.1	0.1	$6.0\times {10}^{-3}$
2	0.3	0.2	$7.2\times {10}^{-2}$
3	0.3	0.4	$2.88\times {10}^{-1}$
4	0.4	0.1	$2.40\times {10}^{-2}$

On the basis of the above data which one is correct?

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

Answer 1

The correct option is D

$\mathrm{r}=\mathrm{k}\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^2$

Explanation:

Step 1: Calculating the order with respect to A

For the reaction $2\mathrm{A}+\mathrm{B}\to \mathrm{C}+\mathrm{D}$

The rate equation can be written as $\mathrm{r}=\mathrm{k}{\left[\mathrm{A}\right]}^{\mathrm{x}}{\left[\mathrm{B}\right]}^{\mathrm{y}}$ ….(1)

Consider the experiment 1 and 4 where the concentration of $\left[\mathrm{B}\right]$ is constant,

The rate equation for experiment 1 is $6.0\times {10}^{-3}=\mathrm{K}{\left(0.1\right)}^{\mathrm{x}}{\left(0.1\right)}^{\mathrm{y}}$ ….(2)

The rate equation for experiment 2 is $2.4\times {10}^{-2}=\mathrm{K}{\left(0.4\right)}^{\mathrm{x}}{\left(0.1\right)}^{\mathrm{y}}$ ….(3)

Divide equation (2) by (3)

$$\begin{gathered} \frac{6.0\times {10}^{-3}}{2.4\times {10}^{-2}}=\frac{\mathrm{K}{\left(0.1\right)}^{\mathrm{x}}{\left(0.1\right)}^{\mathrm{y}}}{\mathrm{K}{\left(0.4\right)}^{\mathrm{x}}{\left(0.1\right)}^{\mathrm{y}}} \\ \frac{0.6}{2.4}={\left(\frac{1}{4}\right)}^x \\ \frac{1}{4}={\left(\frac{1}{4}\right)}^x \end{gathered}$$

That is, $x=1$

Step 2: Calculating the order with respect to B

Consider the experiments 2 and 3 where the concentration of $\left[\mathrm{A}\right]$ is constant,

The rate equation for experiment 2 is $7.2\times {10}^{-2}=\mathrm{K}{\left(0.3\right)}^{\mathrm{x}}{\left(0.2\right)}^{\mathrm{y}}$ ….(4)

The rate equation for experiment 3 is $2.88\times {10}^{-1}=\mathrm{K}{\left(0.3\right)}^{\mathrm{x}}{\left(0.4\right)}^{\mathrm{y}}$ ….(5)

Divide equation (4) by (5)

$$\begin{gathered} \frac{7.2\times {10}^{-2}}{2.88\times {10}^{-1}}=\frac{\mathrm{K}{\left(0.3\right)}^{\mathrm{x}}{\left(0.2\right)}^{\mathrm{y}}}{\mathrm{K}{\left(0.3\right)}^{\mathrm{x}}{\left(0.4\right)}^{\mathrm{y}}} \\ \frac{0.72}{2.88}={\left(\frac{2}{4}\right)}^y \\ \frac{1}{4}={\left(\frac{1}{2}\right)}^y \end{gathered}$$

That is $y=2$

Step 3: Rate equation

Substituting the value of $\mathrm{x}$ and $\mathrm{y}$ in the equation (1)

That is,$\mathrm{r}=\mathrm{k}\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^2$

Hence, option (D) is correct.