Question

Write the equilibrium expression for each of the following reactions. a. \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) c. \(\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)\)

Short answer

The equilibrium expressions for the given reactions are as follows: a. \(K_c = \frac{[\mathrm{NO}]^4[\mathrm{H}_{2}\mathrm{O}]^6}{[\mathrm{NH}_{3}]^4[\mathrm{O}_{2}]^5}\) b. \(K_c = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{NO}]^2[\mathrm{O}_{2}]}\) c. \(K_c = \frac{[\mathrm{CH}_{2}\mathrm{O}][\mathrm{H}_{2}]}{[\mathrm{CH}_{3}\mathrm{OH}]}\)

Step-by-step solution

Reaction a: Identify products and reactants

In this reaction: Reactants: \(4 \mathrm{NH}_{3}(g)\) and \(5 \mathrm{O}_{2}(g)\) Products: \(4\mathrm{NO}(g)\) and \(6 \mathrm{H}_{2}\mathrm{O}(g)\)

Reaction a: Write the equilibrium expression

The general equation for the equilibrium expression: $$K_c = \frac{[P]^n}{[R]^m}$$ For reaction a, the expression is: $$K_c = \frac{[\mathrm{NO}]^4[\mathrm{H}_{2}\mathrm{O}]^6}{[\mathrm{NH}_{3}]^4[\mathrm{O}_{2}]^5}$$

Reaction b: Identify products and reactants

In this reaction: Reactants: \(2\mathrm{NO}(g)\) and \(\mathrm{O}_{2}(g)\) Products: \(2\mathrm{NO}_{2}(g)\)

Reaction b: Write the equilibrium expression

The general equation for the equilibrium expression: $$K_c = \frac{[P]^n}{[R]^m}$$ For reaction b, the expression is: $$K_c = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{NO}]^2[\mathrm{O}_{2}]}$$

Reaction c: Identify products and reactants

In this reaction: Reactants: \(\mathrm{CH}_{3}\mathrm{OH}(g)\) Products: \(\mathrm{CH}_{2}\mathrm{O}(g)\) and \(\mathrm{H}_{2}(g)\)

Reaction c: Write the equilibrium expression

The general equation for the equilibrium expression: $$K_c = \frac{[P]^n}{[R]^m}$$ For reaction c, the expression is: $$K_c = \frac{[\mathrm{CH}_{2}\mathrm{O}][\mathrm{H}_{2}]}{[\mathrm{CH}_{3}\mathrm{OH}]}$$

Key concepts

Equilibrium Expression

In chemical reactions, the equilibrium expression gives us a snapshot of the balance between reactants and products at equilibrium. This is vital in predicting how the concentrations of substances change when they reach this state of balance.
The equilibrium expression is mathematically represented using the formula:

• \[K_c = \frac{[P]^n}{[R]^m}\]

Here, \(K_c\) is the equilibrium constant.

• \([P]\) represents the concentration of the products.
• \([R]\) represents the concentration of the reactants.
• \(n\) and \(m\) are the stoichiometric coefficients from the balanced chemical equation.

A high \(K_c\) value points to a reaction that favors products, while a low \(K_c\) suggests more reactants are present at equilibrium. Understanding this helps in predicting the direction a reaction will shift if disturbed.

Reactants and Products

Reactants and products are integral to any chemical reaction. To write the equilibrium expression, we need to identify these participants first.

**Reactants:** These are substances you begin with in a chemical reaction. They are found on the left side of the equation. For example:

• In the reaction \(4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\), the reactants are \(\mathrm{NH}_{3}\) and \(\mathrm{O}_{2}\).
• In reaction \(2 \mathrm{NO}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\), the reactants are \(\mathrm{NO}\) and \(\mathrm{O}_{2}\).

**Products:** These are the substances formed as a result of the reaction, found on the right side.

• For \(4 \mathrm{NH}_{3}(g) + 5 \mathrm{O}_{2}(g) \rightleftharpoons 4 \mathrm{NO}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\), the products are \(\mathrm{NO}\) and \(\mathrm{H}_{2} \mathrm{O}\).
• In the reaction \(2 \mathrm{NO}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\), the product is \(\mathrm{NO}_{2}\).

Recognizing reactants and products allows us to appropriately set up the equilibrium expression and analyze the chemical dynamics of the reaction.

Balanced Chemical Equations

Balanced chemical equations are essential when evaluating equilibrium because they reflect the atoms' conservation in a reaction.

Each equation must have the same number of each type of atom on both sides of the reaction arrow. This ensures that mass is conserved, as stated by the law of conservation of mass. Here's why it's important:

• **Accuracy:** A balanced equation reflects the stoichiometry of a reaction, which is crucial for calculating the equilibrium expression.
• **Coefficients:** The coefficients from a balanced equation are used as the powers in the equilibrium expression. For example, in the expression \(\frac{[\mathrm{NO}]^4[\mathrm{H}_{2}\mathrm{O}]^6}{[\mathrm{NH}_{3}]^4[\mathrm{O}_{2}]^5}\), the numbers 4, 6, 4, and 5 come from the balanced equation.

Balancing equations ensures that all calculations based on them, including those for equilibrium, are correct and reliable.