Question

Write the equilibrium expression for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\) b. \(2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{S}(g)\) c. \(\mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{HCN}(g)\)

Short answer

The equilibrium expressions for the given reactions are: a. \(K = \frac{[\mathrm{HBr}]^2 }{[\mathrm{H}_{2}][\mathrm{Br}_{2}]}\) b. \(K = \frac{[\mathrm{H}_{2}\mathrm{S}]^2}{[\mathrm{H}_2]^2[\mathrm{S}_2]}\) c. \(K = \frac{[\mathrm{HCN}]^2}{[\mathrm{H}_{2}][\mathrm{C}_{2}\mathrm{N}_{2}]}\)

Step-by-step solution

Write the equilibrium expression for reaction a.

In this reaction, the equilibrium constant K is represented as follows: K = \(\frac{[\mathrm{HBr}]^2 }{[\mathrm{H}_{2}][\mathrm{Br}_{2}]}\) b. Reaction: \(2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2}\mathrm{S}(g)\)

Write the equilibrium expression for reaction b.

In this reaction, the equilibrium constant K is represented as follows: K = \(\frac{[\mathrm{H}_{2}\mathrm{S}]^2}{[\mathrm{H}_2]^2[\mathrm{S}_2]}\) c. Reaction: \(\mathrm{H}_{2}(g)+\mathrm{C}_{2}\mathrm{N}_{2}(g) \rightleftharpoons 2\mathrm{HCN}(g)\)

Write the equilibrium expression for reaction c.

In this reaction, the equilibrium constant K is represented as follows: K = \(\frac{[\mathrm{HCN}]^2}{[\mathrm{H}_{2}][\mathrm{C}_{2}\mathrm{N}_{2}]}\)

Key concepts

Chemical Equilibrium

Chemical equilibrium describes a state in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction. This balance means that the concentrations of reactants and products remain constant over time, although they are not necessarily equal. Imagine a busy crosswalk where people are crossing at an equal pace from both sides; nobody gets 'stuck' in the middle, similar to molecules in a system at equilibrium.

It's crucial to understand that reaching equilibrium does not indicate that reactants and products are gone; they're just being converted at an equal rate so their concentrations don't change anymore. Visualizing a seesaw that is perfectly balanced can help; the seesaw doesn't stop moving, but it stays level despite the ongoing motion.

Equilibrium Constant

The equilibrium constant, denoted as K, is a number that expresses the ratio of the concentrations of products to reactants each raised to the power of their coefficients as they appear in the balanced chemical equation. It provides a quantitative measure of the position of equilibrium.

For example, looking at the reaction \( \mathrm{H}_{2}(g) + \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) \), the equilibrium constant expression is \( K = \frac{{[\mathrm{HBr}]^2 }}{{[\mathrm{H}_{2}][\mathrm{Br}_{2}]}} \). The squared term arises because there are two molecules of HBr produced per one set of reactants consumed. This expression indicates how the equilibrium constant is intimately tied to the stoichiometry of the reaction. The K value is mostly dependent on temperature; a higher K means the equilibrium favors the formation of products, while a lower K favors the reactants.

• The square brackets denote the molarity (concentration) of the species.
• Only species in the gaseous or aqueous phases are included.
• Pure solids or liquids do not appear in the expression as their concentration is constant.

Reaction Quotient

The reaction quotient, Q, is a measure designed to determine the direction in which a reaction will proceed to reach equilibrium. It is calculated using the same formula as the equilibrium constant, but with the initial concentrations rather than equilibrium concentrations.

For instance, if the initial concentrations of reactants and products in the reaction \( \mathrm{H}_{2}(g) + \mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) \) are substituted into the equilibrium expression, you get the reaction quotient Q. This value can then be compared to the equilibrium constant K to predict the behavior of the reaction:

• If \( Q < K \), the forward reaction is favored, and the concentration of products will increase.
• If \( Q > K \), the reverse reaction is favored, and the concentration of reactants will increase.
• If \( Q = K \), the system is already at equilibrium.

By analyzing Q relative to K, a chemist can predict how a reaction mixture will change over time, which is invaluable for controlling industrial chemical processes and understanding natural phenomena.