Problem 29 - Chemistry Structure and Properties

Question

For the reaction \(2 \mathrm{A}(g)+\mathrm{B}(g) \longrightarrow 3 \mathrm{C}(g),\) \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 each of the reactants and products. }} \\ {\text { b. when A is decreasing at a rate of } 0.100 \mathrm{M} / \mathrm{s} \text { , how fast is } \mathrm{B} \text { decreas- }} \\ {\text { ing? How fast is Cincreasing? }}\end{array} \end{equation}

Step-by-Step Solution

Answer

For reaction A and B, rate = -0.100 M/s; For C, rate = +0.150 M/s.

Step 1: Determining the Rate Expression

The rate of the reaction can be expressed in terms of the decrease in concentration of reactants or the increase in concentration of products. Using the stoichiometry of the reaction, rate = - (1/2) * (d[A]/dt) = - (d[B]/dt) = (1/3) * (d[C]/dt).

Step 2: Calculating the Rate of Disappearance of B

If A is decreasing at a rate of 0.100 M/s, the rate of disappearance of B is the same because their stoichiometric coefficients are 1:1. So the rate of disappearance of B is also 0.100 M/s.

Step 3: Calculating the Rate of Increase of C

Since three moles of C are produced for every two moles of A consumed, we use the relationship rate = (1/3) * (d[C]/dt) = - (1/2) * (d[A]/dt). Plugging in the rate of disappearance of A gives (1/3) * (d[C]/dt) = - (1/2) * (-0.100 M/s), resulting in d[C]/dt = 0.150 M/s.

Key Concepts

Reaction Stoichiometry

Understanding the concept of reaction stoichiometry is pivotal for grasping how the quantities of reactants and products relate in a chemical reaction. Stoichiometry derives from the Greek words 'stoicheion' (element) and 'metron' (measure), hence it is the measurement of elements. In chemical reactions, stoichiometry provides the quantitative relationship between reactants and products. This is a critical foundation in chemistry since it allows us to predict the amounts of substances consumed and produced.

Following the stoichiometry of the given reaction, every 2 moles of A and 1 mole of B will react to form 3 moles of C. This ratio must be consistent throughout the analysis of the reaction, whether we are investigating the rate, yield, or any other aspect of the reaction dynamics. Understanding stoichiometry is not just a mere memorization of ratios; it involves a deep comprehension of how these ratios translate into actual experimental or environmental conditions.

Rate of Reaction Expression

Moving on to the rate of reaction expression, it essentially captures how rapidly a reaction proceeds. This is expressed in terms of the rate of change of concentration of the reactants or products over time. For the provided reaction, the reaction rate can be expressed in multiple ways due to the stoichiometry of the reactants and products involved.

The general form of the rate expression is \[ \text{rate} = -\frac{1}{\text{stoichiometric coefficient}} \times \frac{d[\text{Reactant}]}{dt} = \frac{1}{\text{stoichiometric coefficient}} \times \frac{d[\text{Product}]}{dt} \. \]The negative sign indicates the consumption of reactants, whereas the positive sign shows the formation of products. Therefore, the rate of reaction for the loss of A and B and the gain of C is directly connected to their respective stoichiometric coefficients. This is a crucial concept as it allows us to quantify the dynamics of a chemical process in a clear and standardized manner.

Concentration Change Rate

Finally, the concentration change rate embodies the speed at which reactants are being used up or products are being generated in a chemical reaction. In our example, the rate at which A is consumed directly influences the rate at which B is used and at which C is produced because of the stoichiometric relationships. Hence, we can determine the rate of change for each substance involved in the reaction.

By saying that A is decreasing at a rate of 0.100 M/s, this implies a certain rate of decrease for B and a rate of increase for C that must be consistent with the reaction's stoichiometry. In practical terms:

• For B, with a 1:1 stoichiometric ratio with A, B will also decrease at 0.100 M/s.
• For C, for every 2 moles of A decreased, 3 moles of C are produced, thus the rate of increase for C will be 0.150 M/s.

Students should not only be able to calculate these values but also understand them conceptually, thereby linking the stoichiometric ratios to the practical change in concentrations observed during a reaction.