Question

Mechanism of a hypothetical reaction $X _{2}+ Y _{2} \rightarrow 2 XY$ is given below:
(i) $X_{2} \rightarrow X+X$ (fast)
(ii) $X + Y _{2} \rightleftharpoons XY + Y$ ( slow )
(iii) $X + Y \rightarrow XY$ (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, $\frac{[ X ]^{2}}{\left[ X _{2}\right]}= K _{ e }^{\prime}$
$\therefore[ X ]= K _{ e }^{\prime 0.5} \times\left[ X _{2}\right]^{0.5} \quad-( A )$
Since reaction (ii) is the slowest step, so the rate of this reaction will be reaction rate for the main reaction.
$\therefore r = K _{ e }[ X ]\left[ Y _{2}\right] \quad-( B )$
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 $1+0.5=1.5$