Question

Consider the equilibrium system \(3 \mathrm{~A} \rightleftharpoons \mathrm{B}\). Sketch the change in concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) with time for these situations: (a) Initially only A is present; (b) initially only B is present; (c) initially both \(\mathrm{A}\) and \(\mathrm{B}\) are present (with \(\mathrm{A}\) in higher concentration). In each case, assume that the concentration of \(\mathrm{B}\) is higher than that of \(\mathrm{A}\) at equilibrium.

Short answer

The given examples show how the concentration of reactants and products change over time in an equilibrium system. In all cases, as the reaction progresses, the concentration of A will decrease as it gets converted into B, until equilibrium is reached where the concentration of B is more than A. The rate of this conversion varies depending on the initial concentrations of A and B.

Step-by-step solution

Analyzing the Initial Concentration of A and B

The given information is that initially, only A is present in the first case. This means that the concentration of B is nil, to begin with. As the reaction proceeds, A will begin to decrease as it gets converted into B.

Plotting the Change in Concentration Over Time for Case (a)

As the reaction continues, the concentration of A will decrease and the concentration of B will increase as A gets converted into B. This will continue until the equilibrium is reached, where the concentration of B will become higher than that of A.

Analyzing the Initial Concentration of A and B for Case (b)

In this case, it is given that solely B is present initially. So, basically, A is nil. As the reaction proceeds, in the reverse direction since B is present, the concentration of B decreases while that of A increases.

Plotting the Change in Concentration Over Time for Case (b)

As the reaction continues, the concentration of B will decrease and the concentration of A will increase. This will continue until we attain equilibrium, where the concentration of B is evidently higher than A.

Analyzing the Initial Concentration of A and B for Case (c)

Initially, both A and B are present but the concentration of A is higher. As the reaction proceeds, A will start getting converted into B.

Plotting the Change in Concentration Over Time for Case (c)

As the reaction continues, the concentration of A will decrease and the concentration of B will increase until we reach a point where B's concentration is greater than A's. After equilibrium, the rate of conversion of A to B will be equal to the rate of conversion of B to A, thus, their concentrations will remain constant.

Key concepts

Concentration Changes

In a chemical reaction, concentration refers to how much of a substance is present in a mixture. Concentration changes are vital in studying reactions because they show how molecules transform over time. When starting with a particular component, such as only \(\text{A}\) or only \(\text{B}\), their concentrations shift due to ongoing chemical processes.

For example:

• When only \(\text{A}\) is present initially, its concentration decreases as it transforms into \(\text{B}\).
• Conversely, when only \(\text{B}\) is present at the start, its concentration diminishes as \(\text{A}\) forms.

These changes persist until equilibrium is achieved. Understanding concentration alterations helps predict the amount of each substance at any given point during the reaction.

Reaction Progression

The progression of a reaction refers to how reactants are converted into products over time. Initially, reactions may favor the formation of products or reactants depending upon the starting conditions.

In the exercise, the reaction begins differently across three cases:

• Only \(\text{A}\) starts, leading to more products (\text{B}) forming as the reaction moves forward.
• When the reaction initiates with only \(\text{B}\), a reverse reaction takes place initially, creating more \(\text{A}\).
• With both \(\text{A}\) and \(\text{B}\) initially present, but \(\text{A}\) higher, the forward reaction slightly outweighs the backward reaction, and \(\text{B}\) will increase more at first.

As the reaction progresses, the system continuously shifts until reaching a state where no further large changes in concentrations are observed.

Equilibrium System

An equilibrium system in chemistry is where the reactants and products of a reaction are balanced in such a way that their concentrations remain constant over time. This does not mean that reactions stop happening; instead, the rate of the forward reaction equals the rate of the reverse reaction.

In our equilibrium system involving \(\text{3A} \rightleftharpoons \text{B}\), changes in concentrations occur respectively but stabilize at equilibrium:

• The two reactions happen simultaneously, leading eventually to a steady state.
• If started with only \(\text{A}\), \(\text{B}\) will increase until a point of balance is struck.
• Similarly, beginning with only \(\text{B}\), \(\text{A}\) forms until equilibrium is reached.
• With both visible initially, the concentrations find equilibrium naturally over time.

In all these cases, the final equilibrium concentration of \(\text{B}\) is higher compared to \(\text{A}\) as specified in the exercise.

Forward and Reverse Reactions

The forward reaction is when reactants convert into products, while the reverse reaction is when products revert back to reactants. Understanding both directions is crucial as they work together to ensure a chemical equilibrium.

Consider these points in the given scenarios:

• The \(\text{forward reaction}\) converts \(\text{3A}\) into \(\text{B}\), primarily occurring when starting mainly with \(\text{A}\).
• The \(\text{reverse reaction}\) is predominant if \(\text{B}\) is abundant initially, converting some of \(\text{B}\) back to \(\text{A}\).
• Here, as the system approaches equilibrium, both reactions run at equal rates—keeping concentration stable.

Forward and reverse reactions contribute equally at equilibrium in a closed system, assisting in maintaining constant concentrations over time.