Problem 77 - Chemistry: Principles and Reactions

Question

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

Step-by-Step Solution

Answer

In this exercise, we were given a reaction between hydrogen and iodine, and an experimental rate expression. We were asked to verify whether the rate expression is consistent with a given two-step reaction mechanism. To do this, we determined the rate law for each step in the mechanism, applied the steady-state approximation method to find the overall rate law, and compared it with the given rate expression. We found that the rate expression is consistent with the mechanism, assuming that the rate constant has the observed relationship.

Step 1: Write the rate law for each step of the mechanism

For each step in the given mechanism, we can write the rate law based on the reactants involved in each step. For the first step (fast equilibrium): $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ The forward rate: \(rate_1 = k_1[\mathrm{I}_{2}]\) The reverse rate: \(rate_{-1} = k_{-1}[\mathrm{I}]^2\) For the second step (slow): $$\mathrm{H}_{2}(g)+\mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g)$$ The rate: \(rate_2 = k_2[\mathrm{H}_{2}][\mathrm{I}]^2\)

Step 2: Apply the steady-state approximation to find the overall rate law

The steady-state approximation states that the rate of the change in concentration of any intermediate species (in this case, I atoms) is zero. Therefore, for I atoms, we have: $$\frac{d[\mathrm{I}]}{dt} = rate_1 - rate_{-1} - rate_2 = 0$$ Substitute the rate laws from Step 1: $$k_1[\mathrm{I}_{2}] - k_{-1}[\mathrm{I}]^2 - k_2[\mathrm{H}_{2}][\mathrm{I}]^2 = 0$$ Now solve for the concentration of I atoms: $$[\mathrm{I}]^2 = \frac{k_1[\mathrm{I}_{2}]}{k_{-1} + k_2[\mathrm{H}_{2}]}$$

Step 3: Express the rate of the overall reaction in terms of the reactants

The overall rate of the reaction is the rate of the slow step, which is given by \(rate_2 = k_2[\mathrm{H}_{2}][\mathrm{I}]^2\). Substitute the expression for \([\mathrm{I}]^2\) obtained in Step 2: $$rate = k_2[\mathrm{H}_{2}]\frac{k_1[\mathrm{I}_{2}]}{k_{-1} + k_2[\mathrm{H}_{2}]}$$

Step 4: Compare the derived rate law with the given rate expression

Now we compare this derived rate law with the given experimental rate expression, which is: $$rate = k[\mathrm{H}_{2}][\mathrm{I}_{2}]$$ To make these rate expressions consistent, the rate constant \(k\) must be equal to: $$k = k_2\frac{k_1}{k_{-1} + k_2[\mathrm{H}_{2}]}$$ Thus, we have shown that the rate expression given in the exercise is consistent with the provided mechanism, assuming that the rate constant \(k\) has the relationship shown above.

Key Concepts

Rate Law

A rate law tells us how the speed of a chemical reaction depends on the concentrations of the reactants. For a simple reaction:

• \( A + B \rightarrow C \)

The rate law could be written as:

• \( ext{Rate} = k [A]^m[B]^n \)

Here, \([A]\) and \([B]\) stand for the concentrations of the reactants, while \(k\) represents the rate constant, and \(m\) and \(n\) are the reaction orders that indicate how the rate is affected by the reactant concentrations.
Rate laws must be experimentally determined, as they cannot be directly inferred from the balanced equation of the reaction.
In our exercise, the rate law is given as \( ext{Rate} = k[ ext{H}_2][ ext{I}_2] \). This shows a first-order dependence on both hydrogen and iodine, indicating a bimolecular reaction mechanism in the slow step.

Steady-State Approximation

The steady-state approximation is a useful method in reaction kinetics for simplifying the analysis of complex reactions. It assumes that the concentration of intermediate species remains constant over the course of a reaction.
This means the rate of formation of any intermediate is equal to the rate of its consumption.
In mathematical terms, for an intermediate \( X \), this assumption can be written as:\[ \frac{d[X]}{dt} = 0 \] In the problem we work through, the intermediate is atomic iodine, \( \text{I} \). Applying the steady-state approximation to \([\text{I}]\), the equation becomes:

• \( k_1[\text{I}_2] - k_{-1}[\text{I}]^2 - k_2[\text{H}_2][\text{I}]^2 = 0 \)

Through this step, we can find an expression for the concentration of \([\text{I}]^2\), which helps simplify the overall rate law to match the observed experimental expression.

Rate Constant

The rate constant \( k \) is a critical component of every rate law. It links the concentration terms to the rate of reaction and is a measure of how quickly a reaction proceeds.
The value of \( k \) is determined experimentally and:

• Depends on the temperature and the specific reaction.
• Has units that vary depending on the overall order of the reaction.

In the context of our exercise, \( k \) is the composite rate constant derived from the mechanism's slow step. By combining various constants for the individual steps, we achieve a form consistent with the experimental rate law. The equation \( k = k_2\frac{k_1}{k_{-1} + k_2[\text{H}_2]} \) articulates this relationship for the exercise.

Equilibrium

Equilibrium in a chemical reaction is reached when the forward and reverse reaction rates are equal, resulting in no net change in the concentrations of reactants and products.
For reactions that go back and forth, like the first step in our given mechanism:

• \( \text{I}_2(g) \rightleftharpoons 2 \text{I}(g) \)

This stage is fast and contributes to the reaction reaching equilibrium swiftly.
The ratio of the forward and reverse rate constants \( k_1 \) and \( k_{-1} \) determines the position of equilibrium for this step.
Although we primarily focus on the slow step to determine the rate law, understanding the equilibrium in the fast steps offers insight into the concentrations of intermediates like \( \text{I} \), which play a crucial role in the steady-state approximation.

Chemical Kinetics

Chemical kinetics involves studying how fast chemical reactions occur and the steps or paths taken in transforming reactants to products.
It includes the exploration of:

• Rate laws
• Mechanisms
• Activation energies
• Factors influencing reaction rates

Kinetics help us understand and predict the timing and sequence of events in a chemical reaction.
In the exercise, kinetics is applied to dissect a multi-step reaction into individual reactions, each contributing to the observed behavior of the system. By analyzing the mechanism given, through kinetics, the consistent rate law with the experimental data is derived, allowing us to ascertain what happens at the molecular level during the reaction of hydrogen and iodine.