The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{array}{l} \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ \mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{array} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?

To find the overall balanced equation for the given mechanism, we should add both reactions and then simplify the equation by canceling out any species that appear on both the reactant and product sides, as these are considered intermediates. Let's add both reactions: $$ \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ +\mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) $$ Now combine and cancel out the intermediate (HI): $$ \mathrm{H}_{2}(g)+2\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+2\mathrm{HCl}(g) $$ The balanced equation for the overall reaction is: $$ \mathrm{H}_{2}(g)+2\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+2\mathrm{HCl}(g) $$

Intermediates are species that are formed in one step of a mechanism and then consumed in a subsequent step. They are important for understanding a reaction mechanism, but do not appear in the overall balanced equation. In our mechanism, we can see that the species HI appears as a product in the first reaction and as a reactant in the second reaction. Thus, HI is an intermediate in this mechanism.

According to the exercise, the first step is slow and the second step is fast. In mechanisms where one reaction step is much slower compared to others, the overall rate of the reaction is determined by the slowest step, known as the rate-limiting step or rate-determining step. In our case, the slow (rate-determining) step is: $$ \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) $$ The rate law for this elementary reaction is simply determined by the order with respect to each reactant. As it is a bimolecular reaction, the rate law will be: $$ \text{rate} = k[\mathrm{H}_{2}][\mathrm{ICl}] $$ Since this step is determining the overall rate, its rate law is the same for the overall reaction: $$ \text{rate} = k_{overall}[\mathrm{H}_{2}][\mathrm{ICl}] $$ So, the observed rate law for the overall reaction is: $$ \text{rate} = k_{overall}[\mathrm{H}_{2}][\mathrm{ICl}] $$