What rate law corresponds to the proposed mechanism for the forma...

Question

What rate law corresponds to the proposed mechanism for the formation of hydrogen iodide, which can be written in simplified form as: I2 Δk1k-1 2I (Fast), I + H2 Δk2k-2 H2I (Fast), H2I + I ¡k3 2HI (Slow)?

Accepted Answer

Identify the slow step in the mechanism, as it determines the rate law. In this case, the slow step is: $ \text{H}_2\text{I} + \text{I} \xrightarrow{k_3} 2\text{HI} $.. Write the rate law based on the slow step. The rate of the reaction is determined by the concentration of the reactants in the slow step: $ \text{Rate} = k_3 [\text{H}_2\text{I}][\text{I}] $.. Express the intermediate $ \text{H}_2\text{I} $ in terms of the initial reactants using the fast equilibrium steps. From the second fast step: $ \text{I} + \text{H}_2 \rightleftharpoons \text{H}_2\text{I} $, we have $ K_2 = \frac{[\text{H}_2\text{I}]}{[\text{I}][\text{H}_2]} $.. Solve for $ [\text{H}_2\text{I}] $ in terms of $ [\text{I}] $ and $ [\text{H}_2] $ using the equilibrium constant $ K_2 $: $ [\text{H}_2\text{I}] = K_2 [\text{I}][\text{H}_2] $.. Substitute $ [\text{H}_2\text{I}] $ from the previous step into the rate law: $ \text{Rate} = k_3 K_2 [\text{I}]^2 [\text{H}_2] $. This is the rate law for the overall reaction.