Question

Write the equilibrium expression for each of the following reactions. a. \(\mathrm{N}_{2}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NCl}_{3}(g)\) b. \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\) c. \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{H}_{4}(g)\)

Short answer

The equilibrium expressions for the given reactions are as follows: a. \(K_a = \frac{[\mathrm{NCl}_3]^2}{[\mathrm{N}_2]\times[\mathrm{Cl}_2]^3}\) b. \(K_b = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2]\times[\mathrm{I}_2]}\) c. \(K_c = \frac{[\mathrm{N}_2\mathrm{H}_4]}{[\mathrm{N}_2]\times[\mathrm{H}_2]^2}\)

Step-by-step solution

Equilibrium expression for reaction a

For the given reaction a: \(\mathrm{N}_{2}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NCl}_{3}(g)\), we can write the equilibrium expression as: \[K_a = \frac{[\mathrm{NCl}_3]^2}{[\mathrm{N}_2]\times[\mathrm{Cl}_2]^3}\]

Equilibrium expression for reaction b

For the given reaction b: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\), we can write the equilibrium expression as: \[K_b = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_2]\times[\mathrm{I}_2]}\]

Equilibrium expression for reaction c

For the given reaction c: \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}\mathrm{H}_{4}(g)\), we can write the equilibrium expression as: \[K_c = \frac{[\mathrm{N}_2\mathrm{H}_4]}{[\mathrm{N}_2]\times[\mathrm{H}_2]^2}\]

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction reaches a state where its forward and reverse reactions proceed at the same rate, meaning the concentrations of reactants and products remain constant over time. This stable condition does not imply that the reactions stop; rather, the processes continue but without any net change in concentration. In a closed system, when a chemical equilibrium is achieved, no more reactants can be converted into products or vice versa without affecting this balance. This leads us to the concept of the equilibrium constant, which is derived from the concentrations of the products and reactants at this equilibrium state. A sound understanding of equilibrium is crucial because it helps chemists predict how a system will react to different changes, such as temperature, pressure, and concentration, which are pivotal in industrial chemical processes.

Law of Mass Action

The law of mass action is fundamental for writing equilibrium expressions. It states that for any reversible reaction at equilibrium, the rate of the forward reaction equals the rate of the reverse reaction. This leads to a relationship between the concentrations of reactants and products known as the equilibrium expression.Key points to understand include:

• The equilibrium expression takes the form \[K = \frac{[C]^c [D]^d}{[A]^a [B]^b}\] where \([A]\), \([B]\), \([C]\), and \([D]\) are concentrations of reactants \(A\), \(B\) and products \(C\), \(D\),
• \(a, b, c, \) and \(d\) are the coefficients from the balanced equation.
• \(K\) is the equilibrium constant at a given temperature, which tells us about the relative amounts of products and reactants at equilibrium.

The law of mass action simplifies understanding chemical equilibria by providing a straightforward framework to calculate the equilibrium constant, revealing the state of balance in a chemical process.

Reaction Stoichiometry

Reaction stoichiometry involves using the balanced chemical equation to predict how much of each substance is involved in a reaction. It's an essential component when formulating equilibrium expressions as it directly influences the powers to which concentrations are raised.For any given chemical equation, the stoichiometric coefficients (the numbers in front of molecules in the balanced equation) indicate the proportional amount of each substance involved. For instance, in the reaction:\[\mathrm{N}_2(g) + 3\, \mathrm{Cl}_2(g) \rightleftharpoons 2\, \mathrm{NCl}_3(g)\]the stoichiometry tells us that one molecule of \(\mathrm{N}_2\) will react with three molecules of \(\mathrm{Cl}_2\) to produce two molecules of \(\mathrm{NCl}_3\).Understanding the stoichiometry allows us to construct correct equilibrium expressions by using the coefficients as exponents. For the above reaction, the equilibrium expression is \[K = \frac{[\mathrm{NCl}_3]^2}{[\mathrm{N}_2][\mathrm{Cl}_2]^3}\], showing the relationship according to the balanced equation. This ensures that calculations adhere to the inherent ratios dictated by the reaction, a critical factor in precise chemical analysis and prediction.