Question

Consider the reaction. $$ 2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) $$ An equilibrium mixture of this reaction at a certain temperature has \(\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.562 \mathrm{M},\left[\mathrm{H}_{2}\right]=2.74 \times 10^{-2} \mathrm{M}\), and \(\left[\mathrm{S}_{2}\right]=7.54 \times 10^{-3} \mathrm{M}\). What is the value of the equilibrium constant at this temperature?

Short answer

The value of the equilibrium constant, K, at this temperature is \(1.79 \times 10^{-5}\).

Step-by-step solution

Write the expression for the equilibrium constant K

For the reaction: \(2 \mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons 2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g)\), the equilibrium constant expression, K, is given by \[K = \frac{[\mathrm{H}_{2}]^2[\mathrm{S}_{2}]}{[\mathrm{H}_{2} \mathrm{S}]^2}\]

Insert the concentrations into the K expression

Using the given equilibrium concentrations, substitute them into the K expression: \[K = \frac{(2.74 \times 10^{-2} \, M)^2 \times (7.54 \times 10^{-3} \, M)}{(0.562 \, M)^{2}}\]

Calculate the equilibrium constant K

Perform the calculation with the given values: \[K = \frac{(2.74 \times 10^{-2})^2 \times (7.54 \times 10^{-3})}{(0.562)^{2}}\] \[K = \frac{(0.00075076) \times (7.54 \times 10^{-3})}{0.316} = \frac{5.6612 \times 10^{-6}}{0.316}\] \[K = 1.79 \times 10^{-5}\]

Key concepts

Chemical Equilibrium

Understanding chemical equilibrium is crucial for grasping many concepts in chemistry, particularly when dealing with reversible reactions. At a basic level, chemical equilibrium occurs in a closed system when the rates of the forward and reverse reactions are equal, leading to a constant concentration of reactants and products over time.

For the reaction \(2 \mathrm{H}_2 \mathrm{S}(g) \rightleftharpoons 2 \mathrm{H}_2(g) + \mathrm{S}_2(g)\), equilibrium would be achieved when the rate at which the reactant \(\mathrm{H}_2 \mathrm{S}\) breaks down to form \(\mathrm{H}_2\) and \(\mathrm{S}_2\) is equal to the rate at which \(\mathrm{H}_2\) and \(\mathrm{S}_2\) combine to reform \(\mathrm{H}_2 \mathrm{S}\).

This balance does not mean the reactants and products are in equal concentrations, but that their concentrations no longer change with time. Thus, the equilibrium state is dynamic, with both reactions still occurring but at a rate that does not alter the overall concentrations.

Reaction Quotient

The reaction quotient, denoted as \(Q\), is a measure closely related to the equilibrium constant but applies to any point in time during a reaction, not just at equilibrium. It's calculated using the same expression as the equilibrium constant, but with the instantaneous concentrations (or partial pressures) of the reactants and products at that particular moment.

In mathematical terms, for our given reaction \(2 \mathrm{H}_2 \mathrm{S}(g) \rightleftharpoons 2 \mathrm{H}_2(g) + \mathrm{S}_2(g)\), the expression is \(Q = \frac{[\mathrm{H}_2]^2[\mathrm{S}_2]}{[\mathrm{H}_2 \mathrm{S}]^2}\).

Comparing the reaction quotient \(Q\) to the equilibrium constant \(K\) gives insight into the system's direction of change. If \(Q < K\), the system will proceed in the forward direction (towards the products), and if \(Q > K\), the reaction will go in the reverse direction (towards the reactants) to reach equilibrium.

Equilibrium Concentrations

Equilibrium concentrations are the amounts of reactants and products present when a chemical reaction achieves equilibrium. In our example, these are provided in the exercise: \([\mathrm{H}_2 \mathrm{S}] = 0.562 \mathrm{M}, [\mathrm{H}_2] = 2.74 \times 10^{-2} \mathrm{M}, [\mathrm{S}_2] = 7.54 \times 10^{-3} \mathrm{M}\). These concentrations are plugged into the equilibrium constant expression to calculate \(K\), the numerical value that reflects the ratio of product concentrations to reactant concentrations at equilibrium for a particular temperature.

It's important to note that equilibrium concentrations are influenced by factors such as temperature, pressure, and volume. A change in these conditions can lead to a shift in equilibrium, altering the concentrations. This phenomenon is described by Le Châtelier's principle, which predicts how equilibrium will shift in response to a change in conditions.