Question

Consider this elementary step: $$ X+2 Y \longrightarrow X Y_{2} $$ (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 \(X\) and \(Y\) are \(0.26 M\) and \(0.88 M\), what is the rate constant of the reaction?

Short answer

The rate law for the reaction is \(r = k[X][Y]^2\). The rate constant for the reaction, calculated using the given initial rate and the initial concentrations, is approximately \(20.51 M^{-1}s^{-1}\).

Step-by-step solution

Formulation of the Rate Law

The rate law for a reaction relates the reaction rate to the concentrations of the reactants. For the given elementary reaction equation, \(X + 2Y \longrightarrow XY_{2}\), the rate law can be written in terms of the molar concentrations of the reactants in the solution. The rate, \(r\), is given as a product of rate constant, \(k\), and the concentrations of reactants raised to the power of their respective stoichiometric coefficients in the balanced chemical equation. Since the reaction is an elementary step, the coefficients for the reactants in the equation also represent the order of the reaction with respect to each reactant. Hence, the rate law for this reaction is: \(r = k[X][Y]^2\).

Calculation of the Rate Constant

Now, we know the rate of the reaction, the concentrations of both reactants, and the rate law. We can use this information to solve for the rate constant, \(k\). Given, the reaction rate \(r = 3.8 \times 10^{-3} M/s\), the concentration of \(X\) is \(0.26 M\), and the concentration of \(Y\) is \(0.88 M\). Plugging these values into the rate law: \(3.8 \times 10^{-3} M/s = k(0.26 M)(0.88 M)^2\), we can solve for \(k\).

Solution for the Rate Constant

Rearranging the equation to isolate \(k\), we get \[k = \frac{3.8 \times 10^{-3}}{(0.26)(0.88^2)} M^{-1}s^{-1}\]. Solving for \(k\), we get \(k \approx 20.51 M^{-1}s^{-1}\).

Key concepts

Rate Constant Calculation

The rate constant is a crucial element in chemical kinetics, as it quantifies the speed at which a reaction occurs. It's determined experimentally by observing how the rate of formation or consumption of a product or reactant changes with varying concentrations. For an elementary reaction like \( X + 2Y \longrightarrow XY_2 \), the rate constant calculation involves the use of the rate law to establish a relationship between the observed rate and the concentrations of reactants.

Given the initial rate and concentrations, as shown in the exercise, we can calculate the rate constant \( k \) by rearranging the rate law equation. This process involves plugging in the known values and solving for \( k \), which results in an expression that correlates the progress of the reaction with time under the specified conditions. In simple terms, the rate constant tells us how quickly \( XY_2 \) forms when \( X \) and \( Y \) are mixed together in a reaction vessel at the specified concentrations.

Reaction Kinetics

Reaction kinetics delves into the rates at which chemical reactions occur and the steps involved in the transformation from reactants to products. It encompasses concepts such as reaction orders, half-lives, and activation energies. The rate law described in this exercise illustrates a fundamental aspect of kinetics, connecting the reaction rate to the concentrations of reactants and the rate constant.

Understanding how various factors like temperature, catalysts, and the physical state of reactants impact the reaction rate is pivotal in reaction kinetics. It's pertinent to note that the rates we discuss are not constants but variables that can change under different conditions. Kinetics helps chemists to optimize reaction conditions for industrial processes, to understand reaction mechanisms, and to predict how a reaction formula will behave over time.

Elementary Reaction Steps

Elementary reaction steps refer to the simplest types of reactions where reactants convert to products in a single process without any intermediate steps. Each step in a complex reaction mechanism can be considered an elementary reaction. The importance of these steps lies in the fact that they directly dictate the stoichiometry and the reaction rate law.

The coefficients for the reactants in the elementary reaction's balanced equation, like in the exercise, provide us with the reaction orders. These orders correspond to the exponents used in the rate law expression. Unlike more complex reactions, the rate laws of elementary steps do not need to be determined empirically as they are intuited directly from the reaction stoichiometry. The simplicity of elementary reaction steps makes them the building blocks for understanding more intricate reaction mechanisms in reaction kinetics.