Question

Mechanism of a hypothetical reaction $X_2+Y_2\to 2XY$ is given below:

(i) $X_2\rightleftharpoons X+X\left(fast\right)$

(ii) $X+Y_2\to XY+Y\left(slow\right)$
(iii) $X+Y\to XY\left(fast\right)$

The overall order of the reaction will be:

• A. $1.5$
• B. $1$
• C. $2$
• D. $0$

Answer 1

Answer: (A)

$1.5$

Take reaction (i) and (iii) to be equilibrium reaction where value of $k_f>>>k_b$

Therefore, $\frac{[X{]}^2}{\left[X_2\right]}=K_e^{\prime }$

$\therefore \left[X\right]=K_e^{\prime }{}^{0.5}\times [X_2{]}^{0.5}$ $-\left(A\right)$

Since reaction (ii) is the slowest step, so the rate of this reaction will be reaction rate for the main reaction.

$\therefore r=K_e\left[X\right]\left[Y_2\right]$ $-\left(B\right)$

Substituting equation $\left(A\right)$ in equation $\left(B\right)$

We get $r=K_e\times K_e^{\prime }{}^{0.5}\left[X_2{]}^{0.5}\right[Y_2]$

Let $K_e\times K_e^{\prime }=K$

$\therefore r=K\left[X_2{]}^{0.5}\right[Y_2]$

Hence, order of reaction is $1+0.5=1.5$