Question

If the equilibrium constant for the reaction \(\mathrm{A} \rightleftharpoons \mathrm{B}\) is \(4.0\), what is the value of the equilibrium constant for the reaction \(\mathrm{B} \rightleftharpoons \mathrm{A}\) ?

Short answer

The equilibrium constant for the reverse reaction B↔A is 0.25.

Step-by-step solution

Understand equilibrium constants

The equilibrium constant (K) is the ratio of the concentrations of the products of a reaction to the concentrations of the reactants at equilibrium. This is assuming that all the species are in the same phase. In this exercise, the reaction is A↔B with a given K value, and the question is what the K value would be for the reverse reaction.

Calculate the equilibrium constant for the reverse reaction

The equilibrium constant for the reverse reaction is the reciprocal of the equilibrium constant for the forward reaction. So to find the equilibrium constant for the reaction B↔A, divide 1 by the equilibrium constant for A↔B. The equilibrium constant for A↔B is 4, so the equilibrium constant for B↔A is \(1 / 4.0 = 0.25\).

Confirm the result

The equilibrium constant for B↔A is 0.25. This means that at equilibrium, the concentration of A is larger than that of B, as we would expect from the original equilibrium constant of 4 for A↔B. This confirms that the answer is correct.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a crucial concept in the field of chemistry, reflecting a state in which the rates of the forward and reverse reactions are equal, leading to no net change in the concentration of reactants and products over time.

It’s often compared to a perfectly balanced scale, where the addition or subtraction of weight on one side has a corresponding action to maintain balance. In the context of our exercise, for the reaction \(\mathrm{A} \rightleftharpoons \mathrm{B}\), equilibrium means that as much \(\mathrm{A}\) is converted to \(\mathrm{B}\) as \(\mathrm{B}\) is converted back to \(\mathrm{A}\), even though the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) may not be equal. This point of equilibrium is critical as it allows us to define an equilibrium constant that can predict the behavior of the system under different conditions.

Reaction Concentrations

Reaction concentrations are the amounts of reactants and products present in the system at a given time and are usually expressed in molarity \(\text{(M)}\) — moles of substance per liter of solution.

Understanding how to measure and calculate concentrations is essential for predicting the direction in which a reaction will proceed to reach equilibrium. In our exercise, knowing that the equilibrium constant \(K\) is 4.0 for the reaction \(\mathrm{A} \rightleftharpoons \mathrm{B}\), informs us about the relative concentrations of \(\mathrm{A}\) and \(\mathrm{B}\) at equilibrium. This ratio suggests \(\mathrm{B}\) is favored in the forward reaction. To better grasp this concept, imagine having four times more \(\mathrm{B}\) than \(\mathrm{A}\) when the system reaches stability, which concisely explains the dynamic of the chemical reaction.

Reciprocal Relationships in Chemistry

Reciprocal relationships in chemistry often refer to how changing one aspect of a system inversely affects another, a concept frequently applied in the context of equilibrium constants.

For example, if the equilibrium constant for one direction of a reaction is \(K\), then for the reverse reaction, it is \(1/K\). In the exercise we’re reviewing, the equilibrium constant for the reaction \(\mathrm{A} \rightleftharpoons \mathrm{B}\) is given as 4.0. When we consider the reverse reaction \(\mathrm{B} \rightleftharpoons \mathrm{A}\), this reciprocal relationship comes into play, giving us an equilibrium constant of 0.25, which signals that the forward reaction is more favorable than the reverse.