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Chapter 13: Problem 106

Hydrogen and iodine monochloride react as follows: $$ \mathrm{H}_{2}(g)+2 \mathrm{ICl}(g) \longrightarrow 2 \mathrm{HCl}(g)+\mathrm{I}_{2}(g) $$ The rate law for the reaction is rate \(=k\left[\mathrm{H}_{2}\right][\mathrm{ICl}]\) Suggest a possible mechanism for the reaction.

Short Answer

Expert verified
A proposed mechanism for the reaction is: Step 1 (Slow): H2 + ICl -> HCl + HI; Step 2 (Fast): HI + ICl -> HCl + I2. This mechanism explains the given rate law, as the rate is dependent on the concentration of both H2 and ICl and leads to the observed products.

Step by step solution

01

Identify the slow (rate-determining) step

The rate law indicates that both [H2] and [ICl] influence the reaction rate. Therefore, the slow step in this reaction must involve both H2 and ICl. The order of each reactant is 1, so each reactant should appear once in the rate-determining step.
02

Propose the mechanism

The slow, rate-determining step must involve the collision between one molecule each of H2 and ICl, as this is what the rate law suggests. We also need to arrive at the final products: 2HCl and I2. A potential mechanism could be as follows: Step 1 (Slow): H2 + ICl -> HCl + HI, Step 2 (Fast): HI + ICl -> HCl + I2. Note that HI is an intermediate species, it is produced in the first step and consumed in the second.
03

Verify the mechanism

A proposed mechanism is plausible if it leads to the observed rate law and final products. The rate law for the suggested mechanism indeed will be rate = k[H2][ICl], as the rate of reaction is determined by the slowest step, i.e., Step 1. The final products (2HCl and I2) are also produced through the proposed mechanism.

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