The reaction of hydrogen and iodine monochloride is given as: $H_2(g) + 2ICl(g) \rightarrow 2HCl(g) + I_2(g)$.

Question

The reaction of hydrogen and iodine monochloride is given as: $H_2(g) + 2ICl(g) \rightarrow 2HCl(g) + I_2(g)$. This reaction is of first order with respect to $H_2(g)$ and $ICl(g)$, for which of the following proposed mechanisms:
Mechanism A: $H_2(g) + 2ICl(g) \rightarrow 2HCl(g) + I_2(g)$
Mechanism B: $H_2(g) + ICl(g) \rightarrow HCl(g) + HI(g)$; slow
$HI(g) + ICl(g) \rightarrow HCl(g) + I_2(g)$; fast

Accepted Answer

The order of a reaction and the rate law are determined experimentally and cannot be predicted simply from the balanced chemical equation unless it is an elementary reaction. For complex reactions, the overall rate is controlled by the slowest step, called the rate-determining step.

Given: The reaction is first order w.r.t $H_2$ and first order w.r.t $ICl$. This implies the Rate Law is $Rate = k[H_2][ICl]$.

**Mechanism A:** This suggests a single-step (elementary) trimolecular collision involving one $H_2$ and two $ICl$ molecules. The rate law would be $Rate = k[H_2][ICl]^2$, which is third order. This contradicts the given data. Furthermore, trimolecular collisions are very rare .

**Mechanism B:** The reaction occurs in two steps. The first step is slow and involves one $H_2$ molecule colliding with one $ICl$ molecule. Since the slow step is rate-determining, the predicted rate law is $Rate = k[H_2][ICl]$. This matches the experimentally observed first-order dependence on both reactants. The second step is fast and does not affect the rate law.

Therefore, Mechanism B is the correct mechanism .

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