Question

Assume that an \(\mathrm{A}\) molecule reacts with two \(\mathrm{B}\) molecules in a one-step process to give \(\mathrm{AB}_{2}\). (a) Write a rate law for this reaction. (b) If the initial rate of formation of \(\mathrm{AB}_{2}\) is \(2.0 \times 10^{-5}\) M/sec and the initial concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) are \(0.30 \mathrm{M}\), what is the value of the specific rate constant?

Short answer

The rate law for this reaction is: \(rate = k[A][B]^2\). Using the given data, the specific rate constant (k) is calculated to be approximately \(7.41 \times 10^{-4} \mathrm{M^{-2}s^{-1}}\).

Step-by-step solution

Write the balanced chemical equation

For the given reaction, one A molecule reacts with two B molecules to form an AB2 compound. The balanced chemical equation for this reaction is: \[A + 2B \rightarrow AB_2\]

Write the rate law expression

Considering this a one-step process, the rate law expression for this reaction, based on the stoichiometry is: \[rate = k[A]^m[B]^n\] Here, \(rate\) is the reaction rate, \(k\) is the specific rate constant, and \(m\) and \(n\) are the order of the reaction with respect to A and B respectively. For this exercise, since it's given that the A molecule reacts with two B molecules in a one-step process, the rate law expression will be: \[rate = k[A][B]^2\]

Calculate the specific rate constant

Given the initial rate of the formation of AB2 (\(2.0 \times 10^{-5} \mathrm{M}\) /s), and the initial concentrations of A and B (\(0.30 \mathrm{M}\)), we can calculate the value of the specific rate constant (\(k\)) using the rate law expression: \[2.0 \times 10^{-5} \mathrm{M/s} = k(0.30 \mathrm{M})(0.30 \mathrm{M})^2\] Now, let's isolate the specific rate constant, k, on one side of the equation: \[k = \frac{2.0 \times 10^{-5} \mathrm{M/s}}{(0.30 \mathrm{M})(0.30 \mathrm{M})^2}\] Then, plug in the values and solve for \(k\): \[k = \frac{2.0 \times 10^{-5} \mathrm{M/s}}{(0.30 \mathrm{M})(0.09 \mathrm{M^2})}\] \[k = \frac{2.0 \times 10^{-5} \mathrm{M/s}}{0.027 \mathrm{M^3}}\] \[k \approx 7.41 \times 10^{-4} \mathrm{M^{-2}s^{-1}}\] Thus, the value of the specific rate constant (k) is approximately \(7.41 \times 10^{-4} \mathrm{M^{-2}s^{-1}}\).

Key concepts

Reaction Rate

Understanding the speed at which chemical reactions occur is crucial in the study of chemical kinetics. This speed, known as the reaction rate, is defined as the change in concentration of a reactant or product over a specific period of time. We typically express it in units of molarity per second (M/s). For instance, when the concentration of a product increases by a specific amount per second, we're observing the reaction rate in action. It's important to note that reaction rates can be affected by various factors, including temperature, concentration of reactants, and the presence of a catalyst.

In practical terms, a higher reaction rate means the reaction proceeds faster. When you deal with chemical reactions, either in a lab or in industrial processes, controlling the reaction rate is often essential to ensure safety and efficiency. In the case of our exercise, the reaction rate of forming \(\mathrm{AB}_{2}\) was given as \(2.0 \times 10^{-5}\) M/s, which provided us a starting point to find the specific rate constant.

Rate Law

The rate law is a mathematical expression that relates the rate of a chemical reaction to the concentration of its reactants. It's determined experimentally and can't be deduced solely from a balanced chemical equation. In its general form, the rate law is expressed as \[rate = k[Reactant]^m[Reactant]^n...\], where \(k\) is the specific rate constant, and \(m\) and \(n\) are the orders of the reaction with respect to each reactant.

These exponents indicate how sensitive the rate is to changes in concentration; they are not necessarily related to the coefficients in the balanced equation. In our exercise, the derived rate law was \[rate = k[A][B]^2\], revealing that the reaction rate is directly proportional to the concentration of \(A\) and the square of the concentration of \(B\).

Specific Rate Constant

A pivotal component in the rate law equation is the specific rate constant, denoted by \(k\). This constant is unique for every reaction at a given temperature and it essentially incorporates factors like the properties of the reactants and the activation energy. Unlike the coefficients in a balanced equation or the exponents in a rate law expression, \(k\) provides a direct measure of the reaction's inherent tendency to proceed.

While \(k\) remains constant at a specific temperature, it can change if the temperature changes, implying it's also an indicator of how temperature affects the rate of the reaction. The calculation of \(k\) from our exercise involved rearranging the rate law and inserting the given concentrations and initial rate, yielding a value of approximately \(7.41 \times 10^{-4} \(\mathrm{M^{-2}s^{-1}}\)\).

Stoichiometry

At the heart of understanding chemical reactions is stoichiometry, the study of the quantitative relationships between the amounts of reactants and products involved in a reaction based on the balanced chemical equation. Stoichiometry allows us to predict how much product we will get from a given amount of reactant and vice versa.

These relationships are fundamental when you're scaling up a reaction from a lab bench to industrial production, as they ensure the correct proportions of substances are used. In our exercise, the stoichiometry indicated one molecule of \(A\) reacts with two molecules of \(B\) to form \(AB_2\), an essential piece of information for writing the correct rate law. Always remember, the coefficients in the balanced chemical equation are the starting point for understanding the stoichiometric ratios of a reaction.

Balanced Chemical Equation

The balanced chemical equation provides a symbolic representation of a chemical reaction where the number of atoms for each element, and the total charge, is the same on both the reactant and product sides. Balancing equations is a fundamental skill in Chemistry as it respects the law of conservation of mass and charge.

In terms of kinetics, while the balanced equation gives the overall picture, it doesn't tell the whole story about how the reaction proceeds. That's where kinetics comes in with the rate law and other concepts to flesh out the details of the reaction mechanism. Our exercise started with the balanced equation \[A + 2B \rightarrow AB_2\], which provided a foundation upon which the rate law was constructed and the specific rate constant was determined.