Question

Mechanism of a hypothetical reaction \(X_{2}+Y_{2} \rightarrow 2 X Y\), is given below:
(i) \(X_{2} \rightarrow X+X\) (fast)
(ii) \(X+Y_{2} \rightleftarrows X Y+Y\) (slow)
(iii) \(X+Y \rightarrow X Y\) (fast)
The overall order of the reaction will be
(a) 2
(b) 0
(c) \(1.5\)
(d) 1

Answer 1

Correct option (c) \(1.5\)
Explanation:
Correct the reactions given in Chemical Kinetics question as
$$
\begin{array}{lr}
X_{2} \rightleftharpoons X+X & \text { (fast) } \\
X+Y_{2} \rightarrow X Y+Y & \text { (slow) }
\end{array}
$$
Slow step is the rate determining step.
$$
\text { Rate }=k[X]\left[Y_{2}\right] \quad \ldots \text { (i) }
$$
Equilibrium constant for fast step,
$$ K=\frac{[X]^{2}}{\left[X_{2}\right]} $$ \([X]=\sqrt{K[} \overline{\left.X_{2}\right]}\) By substituting \([X]\) in equation (i), we get Rate \(=k \sqrt{K\left[X_{2}\right]}\left[Y_{2}\right]\) \(=k^{\prime}\left[X_{2}\right]^{1 / 2}\left[Y_{2}\right]\) \(\therefore \quad\) Order of reaction \(=\frac{1}{2}+1\) \(=\frac{3}{2}=1.5\) equation (i), we get