Question

Mechanism of a hypothetical reaction $X_2+Y_2\to 2XY$ is given below:
(i) $X_2\to X+X$ (fast)
(ii) $X+Y_2\rightleftharpoons XY+Y$ ( slow )
(iii) $X+Y\to XY$ (fast)
The overall order of the reaction will be

• A. 2
• B. 0
• C. 1.5
• D. 1

Answer 1

Answer: (C)

Take reaction (i) and (iii) to be equilibrium reaction where value of $k_f\gg k_b$
Therefore, $\frac{[X{]}^2}{\left[X_2\right]}=K_e^{\prime }$
$\therefore [X]=K_e^{\prime 0.5}\times {\left[X_2\right]}^{0.5}-(A)$
Since reaction (ii) is the slowest step, so the rate of this reaction will be reaction rate for the main reaction.
$\therefore r=K_e[X]\left[Y_2\right]-(B)$
Substituting equation (A) in equation (B)
We get $r=K_e\times K_e^{10.5}{\left[X_2\right]}^{0.5}\left[Y_2\right]$
Let $K_e\times K_e^{\prime }=K$
$\therefore r=K{\left[X_2\right]}^{0.5}\left[Y_2\right]$
Hence, order of reaction is $1+0.5=1.5$