Question

Mechanism of a hypothetical reaction ${\mathrm{X}}_2+{\mathrm{Y}}_2\to 2\mathrm{XY}$ is given below.$\left(\mathrm{i}\right){\mathrm{X}}_2\rightleftharpoons \mathrm{X}+\mathrm{X}\left(\mathrm{fast}\right)\left(\mathrm{ii}\right)\mathrm{X}+{\mathrm{Y}}_2\to \mathrm{XY}+\mathrm{Y}\left(\mathrm{slow}\right)\left(\mathrm{iii}\right)\mathrm{X}+\mathrm{Y}\to \mathrm{XY}\left(\mathrm{fast}\right)$What will be the overall order of the reaction?

Answer 1

Order of Reaction:

• The reaction rate or rate of reaction is known as the speed at which a chemical reaction usually takes place.
• Also in other words, it can be also defined as proportional to the increase in the concentration of a product per unit time and to the decrease in the concentration of a reactant per unit time.

Step 1: Given data:

The hypothetical reaction is given as ${\mathrm{X}}_2+{\mathrm{Y}}_2\to 2\mathrm{XY}$

The mechanism for the above reaction is given as:

$$\begin{gathered} \left(\mathrm{i}\right){\mathrm{X}}_2\rightleftharpoons \mathrm{X}+\mathrm{X}\left(\mathrm{fast}\right) \\ \left(\mathrm{ii}\right)\mathrm{X}+{\mathrm{Y}}_2\to \mathrm{XY}+\mathrm{Y}\left(\mathrm{slow}\right) \\ \left(\mathrm{iii}\right)\mathrm{X}+\mathrm{Y}\to \mathrm{XY}\left(\mathrm{fast}\right) \end{gathered}$$

Step 2: Formula for rate of the reaction:

The rate of reaction is given by, $\mathrm{r}$ =$\mathrm{K}[{\mathrm{x}}_2{]}^{\mathrm{a}}[{\mathrm{y}}_2{]}^{\mathrm{b}}$ (Equation 1)

Where a and b are the order of the reaction in terms of X and Y respectively.
Step 3: Finding the value of a and b:
We know that the rate of reaction is always given in the slow step.

Hence, using the slow step( step 2) that is given as ; $\mathrm{X}+{\mathrm{Y}}_2\to \mathrm{XY}+\mathrm{Y}$

The rate of reaction of this equation can be given as:
$\mathrm{r}$ =$\mathrm{K}\left[\mathrm{x}\right]{\left[\mathrm{y}\right]}^2$ (Equation 2)

Hence, from the above equation, the value of$\mathrm{a}=1\mathrm{and}\mathrm{b}=2$
Step 4 : Finding the rate of reaction in terms of$x_2$:
We can write the ${\mathrm{K}}_{\mathrm{eq}}$ (equilibrium) for the step 1 , ${\mathrm{X}}_2\rightleftharpoons \mathrm{X}+\mathrm{X}$ given as:
$K_{eq}$ =$${\displaystyle \frac{[x][x]}{[x_2]}}$$=${\displaystyle \frac{{[x]}^2}{[x_2]}}$
=> $[x]$ =$(K_{eq}[x_2]{)}^{1/2}$
Now, this value in equation 2.
$r$ =$$K(K_{eq}[x_2]{)}^{1/2}[y_2{]}^1)$$
=$$K.{K_{eq}}^{1/2}[x_2{]}^{1/2}[y_2{]}^1$$

(Let $K.{K_{eq}}^{1/2}$ =$K$)
so, $r$ =$K[x_2{]}^{1/2}[y_2{]}^1$
Here, we got the rate of reaction in the format of equation 1.

Step 5: Calculating the order of the reaction:

so using the value of $\mathrm{a},\mathrm{b}$ ,We can calculate the order of the reaction as:
The overall order of the reaction $=\mathrm{a}+\mathrm{b}=\frac{1}{2}+1=1.5$.

Hence, the order of the reaction is calculated as$1.5.$