Problem 30 - Chemistry Structure and Properties

Question

For the reaction \(\mathrm{A}(g)+\frac{1}{2} \mathrm{B}(g) \longrightarrow 2 \mathrm{C}(g),\) \begin{equation} \begin{array}{l}{\text { a. determine the expression for the rate of the reaction in terms of }} \\ {\text { the change in concentration of cach of the reactants and products. }} \\ {\text { b. when } C \text { is increasing at a rate of } 0.0025 \mathrm{M} / \mathrm{s} \text { , how fast is } \mathrm{B}} \\ {\text { decreasing? How fast is A decreasing? }}\end{array} \end{equation}

Step-by-Step Solution

Answer

The rate of reaction in terms of the change in concentration of reactants A and B and product C is as follows: \(\text{Rate} = -2\frac{d[A]}{dt} = -4\frac{d[B]}{dt} = \frac{1}{2}\frac{d[C]}{dt}\). When C is increasing at a rate of 0.0025 M/s, B is decreasing at a rate of 0.0003125 M/s and A is decreasing at a rate of 0.000625 M/s.

Step 1 - Determine Rate Law

The rate of the reaction can be described by the rate law. Since the reaction is given by \(\mathrm{A}(g) + \frac{1}{2} \mathrm{B}(g) \rightarrow 2 \mathrm{C}(g)\), the rate of appearance of C is twice the rate of disappearance of A and four times the rate of disappearance of B due to the stoichiometry.

Step 2 - Establish the Rate of Reaction

Express the rate of the reaction in terms of the disappearance of A and B and the appearance of C. For reactants, the rate is negative as they are consumed: \[\text{Rate} = -\frac{1}{\text{Stoichiometric coefficient of } A}\frac{d[A]}{dt} = -2\frac{d[A]}{dt}\] \[\text{Rate} = -\frac{1}{\text{Stoichiometric coefficient of } B}\frac{d[B]}{dt} = -4\frac{d[B]}{dt}\] For the product C, the rate is positive as it is formed: \[\text{Rate} = \frac{1}{\text{Stoichiometric coefficient of } C}\frac{d[C]}{dt} = \frac{1}{2}\frac{d[C]}{dt}\]

Step 3 - Calculate the Rate of Decrease of B

Given that C is increasing at a rate of 0.0025 M/s, we can find the rate of decrease of B using its own relationship with the rate: \[\text{Rate} = -4\frac{d[B]}{dt} = \frac{1}{2}\frac{d[C]}{dt}\] Solve for \(\frac{d[B]}{dt}\): \[\frac{d[B]}{dt} = -\frac{1}{8} \frac{d[C]}{dt} = -\frac{1}{8}(0.0025)\] \[\frac{d[B]}{dt} = -0.0003125 \text{ M/s}\]

Step 4 - Calculate the Rate of Decrease of A

Use the rate law to find how fast A is decreasing: \[\text{Rate} = -2\frac{d[A]}{dt} = \frac{1}{2}\frac{d[C]}{dt}\] Solve for \(\frac{d[A]}{dt}\): \[\frac{d[A]}{dt} = -\frac{1}{4} \frac{d[C]}{dt} = -\frac{1}{4}(0.0025)\] \[\frac{d[A]}{dt} = -0.000625 \text{ M/s}\]

Key Concepts

Chemical Kinetics

Understanding chemical kinetics is crucial for analyzing how reactions occur and determining the factors that influence their rates. It is the branch of chemistry that deals with the speed or rate at which a chemical reaction occurs and the mechanisms by which reactions proceed.

The rate of a chemical reaction is usually expressed in terms of the change in concentration of a reactant or product over time. In our exercise, the speed of formation of product C was given, and from this information, we were asked to determine how quickly reactants A and B are being consumed. This exemplifies a fundamental concept in kinetics: the rate at which reactants are consumed and products are formed is interdependent, and can be figured out if the stoichiometry of the reaction is known.

With knowledge of kinetics, we can not just predict how long a reaction will take to complete under certain conditions but also design processes and reactors that optimize reactant conversion, energy use, and safety.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. It allows chemists to predict the amounts of substances consumed and produced in a given reaction, based on the balanced chemical equation.

In our exercise example, stoichiometry is used to understand the relationship between the rates of consumption of reactants A and B and the rate of production of product C. According to the stoichiometry of the reaction, one mole of A and half a mole of B combine to produce two moles of C.

Therefore, the rate at which C appears is directly related to the rate at which A and B disappear. This stoichiometric relationship helps us solve the problem by setting up proportionate expressions for the rates of decrease of A and B, highlighting the importance of understanding reaction stoichiometry to solve kinetics problems.

Rate Law

The rate law is an expression that relates the rate of a chemical reaction to the concentration of its reactants. It is determined experimentally and is unique for each reaction. A rate law has the general form: \(\text{Rate} = k[\text{Reactant}]^{n}\), where \(k\) is the rate constant, the concentration of the reactant is given by \[\text{Reactant}\], and \(n\) is the order of the reaction with respect to that reactant.

In the provided exercise, we used the rate law to relate the rates of disappearance of reactants A and B to the rate of appearance of product C. We assigned stoichiometric coefficients as powers in the rate expressions, showing a direct connection between the stoichiometry of the reaction and the form of the rate law. Understanding the rate law is essential for predicting how changes in concentration affect reaction rate and for calculating the specific rates at which reactants are used up and products are formed.