Problem 27 - Chemistry A Molecular Approach

Question

For the reaction \(2 \mathrm{~A}(g)+\mathrm{B}(g) \longrightarrow 3 \mathrm{C}(g),\) a. determine the expression for the rate of the reaction in terms of the change in concentration of each of the reactants and products. b. when \(A\) is decreasing at a rate of \(0.100 \mathrm{M} / \mathrm{s},\) how fast is \(\mathrm{B}\) decreasing? How fast is C increasing?

Step-by-Step Solution

Answer

The rate of the reaction is expressed as Rate = -(1/2)(d[A]/dt) = -(d[B]/dt) = (1/3)(d[C]/dt). B is decreasing at 0.050 M/s and C is increasing at 0.150 M/s.

Step 1: Write the Rate Expression

The rate of a reaction is expressed in terms of the rate of change in concentration of either the reactants or the products over time. For the reaction given, this can be written as the rate of disappearance of A or B, or the rate of appearance of C. The rate can be written as: Rate = -(1/2)(d[A]/dt) = -(d[B]/dt) = (1/3)(d[C]/dt), where [A], [B], and [C] represent the molar concentrations of A, B, and C, respectively, and t represents time.

Step 2: Determine Rate of Decrease of B

Given that A is decreasing at a rate of 0.100 M/s, and the stoichiometry of A to B is 2:1, B must decrease at half that rate. Therefore, rate of decrease of B = (1/2) * (rate of decrease of A) = (1/2) * 0.100 M/s = 0.050 M/s.

Step 3: Determine Rate of Increase of C

The stoichiometry of A to C is 2:3. Since A is decreasing at 0.100 M/s, C will increase at a rate of (3/2) times the rate of decrease of A. Therefore, rate of increase of C = (3/2) * (rate of decrease of A) = (3/2) * 0.100 M/s = 0.150 M/s.

Key Concepts

Rate of Reaction

Understanding the rate of reaction is crucial for students studying chemical kinetics. It is the speed at which reactants are converted into products in a chemical reaction. Determining this rate involves measuring how the concentration of reactants or products changes over time. In a general sense, if the concentration decreases rapidly, the reaction rate is high; conversely, if the concentration changes little over a long period, the reaction rate is low.

To make this concept clearer, imagine observing a piece of paper burning. The rate at which the paper disappears is analogous to the rate of reaction – the faster it burns, the higher the rate. In the case of our exercise, where reactant A is decreasing at a rate of 0.100 M/s, this would signal a relatively fast reaction under many circumstances. However, quantifying this reaction rate requires more precise mathematical expressions, as detailed in the solution provided.

Reaction Rate Expression

The reaction rate expression illustrates the relationship between the rate of reaction and the concentrations of reactants and products. It is a mathematical way to express the speed of a chemical reaction, often involving derivatives to indicate the time rate of change for each substance involved in the reaction.

In our exercise, the rate of reaction is given by the equation:
\[\text{Rate} = -\frac{1}{2}\frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{1}{3}\frac{d[C]}{dt}\]
Each term represents the rate of change of concentration for a specific reactant or product over time. The negative signs in the equation indicate the consumption of reactants A and B. The coefficients, such as 1/2 for A and 1/3 for C, stem from the stoichiometry of the reaction and ensure that the rate is correctly represented for each component. This relationship is key to linking observations and measurements to the fundamental chemical processes taking place.

Stoichiometry

Stoichiometry is the mathematical relationship between the quantities of reactants and products in a chemical reaction. It is rooted in the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. As a result, stoichiometry allows us to predict the amounts of substances consumed and produced.

In the context of chemical kinetics, stoichiometry affects the rate of reaction expressions, as seen in our textbook exercise. The coefficients in the balanced chemical equation provide the ratio at which reactants are consumed and products are formed. Here, two molecules of A react with one molecule of B to produce three molecules of C. These coefficients are inverted and used to relate the change in concentration of A, B, and C to the overall reaction rate.

For example, since the stoichiometric ratio of A to B is 2 to 1, B is consumed at half the rate that A is consumed. This is why, if A decreases at 0.100 M/s, B will decrease at 0.050 M/s. Similarly, C is produced at a 3:2 ratio to the consumption of A, which is why C increases at 0.150 M/s. Stoichiometry not only guides the balancing of chemical equations but also provides the essential link connecting chemical quantities with the kinetics of reaction rates.