Question

Consider the following elementary step: $$ \mathrm{X}+2 \mathrm{Y} \longrightarrow \mathrm{XY}_{2} $$ (a) Write a rate law for this reaction. (b) If the initial rate of formation of \(\mathrm{XY}_{2}\) is \(3.8 \times 10^{-3} \mathrm{M} / \mathrm{s}\) and the initial concentrations of \(\mathrm{X}\) and \(\mathrm{Y}\) are \(0.26 \mathrm{M}\) and \(0.88 \mathrm{M}\), respectively, what is the rate constant of the reaction?

Short answer

(a) Rate law: \( \text{Rate} = k [\mathrm{X}][\mathrm{Y}]^2 \). (b) Rate constant: \( k \approx 1.887 \times 10^{-2} \mathrm{M}^{-2} \mathrm{s}^{-1} \).

Step-by-step solution

Identify the Rate Law Structure

The rate law for a reaction is based on the stoichiometry of the elementary step given. The reaction \[ \mathrm{X} + 2 \mathrm{Y} \longrightarrow \mathrm{XY}_{2} \]suggests that the rate law is directly proportional to the concentration of \( \mathrm{X} \) and \( \mathrm{Y} \). Therefore, the rate law can be expressed as:\[ \text{Rate} = k [\mathrm{X}]^a [\mathrm{Y}]^b \]For this elementary step, the stoichiometric coefficients can be used as the exponents \( a = 1 \) and \( b = 2 \). Thus, the rate law becomes:\[ \text{Rate} = k [\mathrm{X}][\mathrm{Y}]^2 \]

Use the Initial Rate and Concentrations to Find the Rate Constant

We are given the initial rate of formation of \( \mathrm{XY}_{2} \), which is \( 3.8 \times 10^{-3} \mathrm{M} / \mathrm{s} \), and the initial concentrations: \([\mathrm{X}] = 0.26 \mathrm{M}\) and \([\mathrm{Y}] = 0.88 \mathrm{M}\). Substitute these into the rate law:\[ 3.8 \times 10^{-3} = k (0.26)(0.88)^2 \]This equation relates the rate constant \( k \) to known quantities.

Calculate the Rate Constant

Solve the equation from Step 2 for the rate constant \( k \):\[ k = \frac{3.8 \times 10^{-3}}{(0.26)(0.88)^2} \]First, compute \( (0.88)^2 \), which is approximately \( 0.7744 \). Substitute this back into the equation:\[ k = \frac{3.8 \times 10^{-3}}{0.26 \times 0.7744} \]Calculate the denominator:\[ 0.26 \times 0.7744 \approx 0.201344 \]Now calculate \( k \):\[ k \approx \frac{3.8 \times 10^{-3}}{0.201344} \approx 1.887 \times 10^{-2} \mathrm{M}^{-2} \mathrm{s}^{-1} \]

Key concepts

Understanding Elementary Steps in a Reaction

An elementary step is a single stage in a more extensive chemical reaction mechanism. In chemistry, elementary steps are the individual processes by which a reaction proceeds. They are the simplest possible reactions and often directly reflect the rearrangement of atoms or molecules.
Understanding these steps is vital because the rate law of an elementary reaction can be directly derived from its molecular composition or stoichiometry.
For the reaction given,

• Reactant X combines with two moles of reactant Y to form a product XY₂.
• This single step describes a direct interaction and transformation of reactants into the product.

Elementary steps do not generally involve intermediates, even though the complete reaction can involve many such steps. Each step has its own rate law, which gives insight into the dynamics of the reaction.
The simplicity of elementary steps makes them a crucial tool for understanding complex reaction mechanisms.

The Role of the Rate Constant in Reactions

The rate constant, denoted as \( k \), is a crucial component of the rate law. It is an indicator of the speed at which a reaction proceeds in a given system under specific conditions.
The rate constant is unique for each reaction and can be influenced by factors such as temperature and the presence of catalysts.
In the rate law equation:

• \( ext{Rate} = k [ ext{X}][ ext{Y}]^2 \)
• \( k \) provides the proportional relationship between the concentrations of the reactants and the reaction rate.

A higher rate constant corresponds to a faster reaction. In our case, by substituting the provided concentrations and rate, we computed \( k \).
Once calculated, the rate constant gives insight into the kinetic properties of the reaction, allowing predictions about how different concentrations will affect the reaction speed.

Stoichiometry in Rate Laws of Elementary Steps

Stoichiometry involves the quantitative relationships of reactants and products in chemical reactions. In an elementary step, stoichiometry reflects the direct interaction ratios of the molecules involved.
For the reaction

• \( ext{X} + 2 ext{Y} \rightarrow ext{XY}_2 \)
• The coefficients (a = 1 for X and b = 2 for Y) are used directly as exponents in the rate law.

Therefore, the rate law derived from the stoichiometry is \( ext{Rate} = k [ ext{X}][ ext{Y}]^2 \).
This means the rate of reaction is directly dependent on the concentration of X and the square of the concentration of Y.

• Doubling X would double the rate.
• Doubling Y would increase the rate fourfold.

Thus, stoichiometry provides a clear guide for predicting how changes in concentration affect reaction speed in elementary processes.

Exploring Reaction Rate Dynamics

Reaction rate is the speed at which a chemical reaction proceeds, and it’s a key measure of how quickly reactants are transformed into products.
It is often expressed in terms of concentration change over time, for example, Molarity per second (M/s).
In the case of our example,

• The given reaction rate for the formation of \( ext{XY}_2 \) is \( 3.8 \times 10^{-3} \, ext{M/s} \).
• This value indicates the rate at which the concentration of \( ext{XY}_2 \) increases in the reaction mixture.

The rate of a reaction can be affected by various factors:

• Concentration of reactants: Higher concentrations typically result in higher reaction rates.
• Temperature: Increasing temperature generally increases the reaction rate.
• Catalysts: These substances can provide an alternative pathway with a lower activation energy.

Comprehending these dynamics is essential for controlling reactions in both laboratory and industrial settings.