Question

For the reaction between hydrogen and iodine, $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g) $$ the experimental rate expression is rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right] .\) Show that this expression is consistent with the mechanism $$ \begin{aligned} \mathrm{I}_{2}(g) & \rightleftharpoons 2 \mathrm{I}(g) & & \text { (fast) } \\\ \mathrm{H}_{2}(g)+\mathrm{I}(g)+\mathrm{I}(g) & \longrightarrow 2 \mathrm{HI}(g) & & \text { (slow) } \end{aligned} $$

Short answer

Answer: Yes, the given experimental rate expression is consistent with the proposed mechanism.

Step-by-step solution

Identify the rate-determining step

The rate-determining step is usually the slowest step in the mechanism. In this case, the second step is the slow step: $$ \mathrm{H}_{2}(g)+\mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) $$

Derive the rate law expression for the rate-determining step

The rate law expression can be written as the product of the reaction rate constant, k, and the concentrations of reactants raised to their stoichiometric coefficients. In this case, the rate law expression for the second step is: $$ \text{Rate} = k_2[\mathrm{H}_2][\mathrm{I}]^2 $$ where \(k_2\) is the rate constant for the second step.

Express the intermediate concentration in terms of reactants and products

In order to compare the derived rate law expression with the given experimental expression, we need to express the intermediate \(\mathrm{I}\) in terms of reactants and products. From the first step (the fast step), we can write the equilibrium expression as: $$ K_\text{eq} = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]} $$ Then solve for the intermediate concentration, \(\mathrm{I}\): $$ [\mathrm{I}] = \sqrt{K_\text{eq}[\mathrm{I}_2]} $$

Substitute intermediate concentration in the rate law expression derived in step 2

Now we will substitute the expression for the intermediate concentration, \([\mathrm{I}]\), found in the last step, into the rate law expression derived in step 2: $$ \text{Rate}=k_2[\mathrm{H}_2]\left(\sqrt{K_\text{eq}[\mathrm{I}_2]}\right)^2 $$ Simplify the expression: $$ \text{Rate} = k_2 [\mathrm{H}_2]K_\text{eq}[\mathrm{I}_2] $$

Comparing the given experimental rate expression and the derived rate expression

Now, we can compare the derived rate expression with the given experimental rate expression. Given experimental rate expression: $$ \text{Rate}=k[\mathrm{H}_2][\mathrm{I}_2] $$ Derived rate expression: $$ \text{Rate} = k_2 [\mathrm{H}_2]K_\text{eq}[\mathrm{I}_2] $$ We can observe that both expressions have the same form. The rate constants, \(k\) and \(k_2K_\text{eq}\), will be different, but the exponent associated with each reactant is the same in both expressions. This demonstrates that the given experimental rate expression is consistent with the proposed mechanism.

Key concepts

Rate-Determining Step

Understanding the rate-determining step is crucial when studying chemical kinetics. This step is the slowest phase in a reaction mechanism and it essentially sets the speed at which the entire reaction proceeds. Imagine it as the bottleneck in a multi-step sequence, slowing down the traffic of reactants turning into products.

To illustrate, consider the reaction of hydrogen with iodine to form hydrogen iodide. In the provided exercise, the mechanism consists of two steps - the rapid formation of iodine atoms from molecular iodine and a second, slower step where hydrogen gas reacts with iodine atoms to produce hydrogen iodide. The slow production of hydrogen iodide defines the rate at which the overall reaction can occur because the other reactions wait for this step to complete before they can proceed.

Why is this concept important? Well, it allows chemists to focus on the slowest part of a reaction to understand the overall rate. This can lead to improvements in industrial chemical processes by optimizing the slow steps, leading to more efficient production.

Rate Law Expression

The rate law expression articulates how the rate of a reaction depends on the concentration of the reactants. It essentially translates the stoichiometry and the reaction mechanism into a mathematical formula that can predict the rate of a reaction. In the provided exercise, we derived the rate law from the second step, which was the rate-determining step, with the rate given by \( k_2[\mathrm{H}_2][\mathrm{I}]^2 \) where \( k_2 \) is the rate constant and the reactants' concentrations were raised to powers matching their stoichiometric coefficients in this step.

The rate law is a powerful tool as it encapsulates the essence of the reaction's kinetics in a simple equation. This allows chemists and students alike to calculate reaction rates under different conditions or to infer mechanisms from observed rate laws. However, the true form of a rate law can only be determined by experiment, and sometimes the concentration of an intermediate species, such as iodine atoms (I) in this exercise, must be reckoned with by invoking the equilibrium conditions of preceding fast steps.

Reaction Mechanism

The reaction mechanism is the step-by-step sequence of elementary reactions by which overall chemical change occurs. It's the intricate dance of atoms and molecules on their way from reactants to products. As in our hydrogen-iodine example, the reaction proceeds via a rapid pre-equilibrium that produces iodine atoms followed by a slow reaction of iodine atoms with hydrogen molecules.

In-depth knowledge of the reaction mechanism is essential. It helps us to understand not only the 'how' of chemical transformations but also provides insights into 'why' certain products are favored, 'how' energy changes during reactions, and 'how' changing conditions may alter the reaction pathway. Mechanistic details allow chemists to design specific catalysts that accelerate desired pathways or to develop new synthetic routes to chemicals. They are, in many respects, the blueprint by which we understand and control chemical processes at the molecular level.