Question

Write an equilibrium expression for each chemical equation. (a) \(2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)\) (b) \(2 \mathrm{BrNO}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\) (c) \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\) (d) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\)

Short answer

\(K_{(a)} = \frac{[\mathrm{N}_{2}\mathrm{O}_{4}]}{[\mathrm{NO}_{2}]^2}\), \(K_{(b)} = \frac{[\mathrm{NO}]^2[\mathrm{Br}_{2}]}{[\mathrm{BrNO}]^2}\), \(K_{(c)} = \frac{[\mathrm{H}_{2}][\mathrm{CO}_{2}]}{[\mathrm{H}_{2}\mathrm{O}][\mathrm{CO}]}\), \(K_{(d)} = \frac{[\mathrm{CS}_{2}][\mathrm{H}_{2}]^4}{[\mathrm{CH}_{4}][\mathrm{H}_{2}\mathrm{S}]^2}\)

Step-by-step solution

Understanding Equilibrium Expressions

An equilibrium expression for a chemical reaction is written by taking the product of the concentrations of the products, each raised to the power of its coefficient in the balanced equation, divided by the product of the concentrations of the reactants, each also raised to the power of its coefficient. This is known as the law of mass action. For gases, concentrations are often replaced with partial pressures.

Write the Equilibrium Expression for Equation (a)

For the reaction \(2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)\), the equilibrium expression is: \[K = \frac{{[\mathrm{N}_{2}\mathrm{O}_{4}]}}{{[\mathrm{NO}_{2}]^2}}\] where \(K\) is the equilibrium constant.

Write the Equilibrium Expression for Equation (b)

For the reaction \(2 \mathrm{BrNO}(g) \rightleftharpoons 2\mathrm{NO}(g)+\mathrm{Br}_{2}(g)\), the equilibrium expression is: \[K = \frac{{[\mathrm{NO}]^2[\mathrm{Br}_{2}]}}{{[\mathrm{BrNO}]^2}}\]

Write the Equilibrium Expression for Equation (c)

For the reaction \(\mathrm{H}_{2}\mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g)\), the equilibrium expression is: \[K = \frac{{[\mathrm{H}_{2}][\mathrm{CO}_{2}]}}{{[\mathrm{H}_{2}\mathrm{O}][\mathrm{CO}]}}\]

Write the Equilibrium Expression for Equation (d)

For the reaction \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\), the equilibrium expression is: \[K = \frac{{[\mathrm{CS}_{2}][\mathrm{H}_{2}]^4}}{{[\mathrm{CH}_{4}][\mathrm{H}_{2}\mathrm{S}]^2}}\]

Key concepts

Chemical Equilibrium

Imagine a dance where partners are constantly swapping, but the number of dancing couples remains the same. This is similar to what occurs in a chemical equilibrium. At equilibrium, the rate of the forward reaction, where reactants turn into products, is equal to the rate of the reverse reaction, where products revert into reactants. This creates a dynamic balance where the concentration of reactants and products remains constant over time, although they are still reacting.

It's crucial to grasp that equilibrium does not mean that the reactants and products are present in equal amounts but rather that their ratios are steady. Factors such as temperature, pressure, and concentration can disrupt this harmony, leading to a shift in the equilibrium as described by Le Chatelier's Principle.

Law of Mass Action

The law of mass action is a rule that enables us to relate the concentrations of reactants and products to the progress of a reaction at a given temperature. Imagine it as a mathematical measure of that dance we mentioned before. It states that, for a balanced chemical equation at equilibrium, the product of the concentrations of the reactants and products, each raised to the power of their respective coefficients in the reaction, gives us a constant value when the system is at equilibrium.

For instance, in a simple reaction where one molecule A reacts with one molecule B to form C and D, the law can be expressed as: \( K = \frac{{[C][D]}}{{[A][B]}} \), where \(K\) is the equilibrium constant. This helps us predict the direction and extent of a reaction under equilibrium conditions.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a profound expression derived from the law of mass action. It is the ratio of the concentration of the products to the reactants at equilibrium, with each raised to the power of their coefficients from the balanced equation. Essentially, it's like a score that tells you how likely a reaction is to produce products or reactants at a given moment.

The magnitude of the equilibrium constant gives us insight into the position of equilibrium: a large \(K\) indicates a greater concentration of products, suggesting a reaction that favors the product side. Conversely, a small \(K\) signifies a larger concentration of reactants. It's essential to note that this constant only changes with temperature and is not influenced by changes in concentration or pressure.

Partial Pressures

When dealing with gases, we often use partial pressures instead of concentrations in equilibrium expressions. The partial pressure is the pressure that a gas would exert if it alone occupied the entire volume of the mixture. It's like assigning a specific 'zone' in a room filled with a gas mixture, where each gas has its own hypothetical zone pressure.

In the context of equilibrium, the equilibrium constant for gases can be given in terms of partial pressures, commonly represented as \(K_p\). For a reaction involving gases, the relation is similar to concentrations: \( K_p = \frac{{\text{product of partial pressures of products}}}{{\text{product of partial pressures of reactants}}} \), with the corresponding powers reflecting the stoichiometry from the balanced equation. Understanding how partial pressures factor into chemical equilibrium is key for predicting the behavior of gases in reactions.