Question

Apply the law of mass action to the following equilibria: (i) formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) (ii) formation of \(\mathrm{NO}_{2}\) from nitric oxide and oxygen

Short answer

Question: Write the expressions for the equilibrium constants, \(K_c\), for the following reactions, using the law of mass action: (i) Formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) (ii) Formation of \(\mathrm{NO}_{2}\) from nitric oxide and oxygen Answer: (i) \(K_c = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_{2}]^2[\mathrm{O}_{2}]}\) (ii) \(K_c = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{NO}]^2[\mathrm{O}_{2}]}\)

Step-by-step solution

(i) Formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\)

The balanced chemical equation for the formation of \(\mathrm{SO}_{3}\) from \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) is: $$2\mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2\mathrm{SO}_{3}(g)$$ Now we can write the expression for the equilibrium constant, \(K_c\), using the law of mass action as: $$K_c = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_{2}]^2[\mathrm{O}_{2}]}$$ Where [A] denotes the concentration of species A at equilibrium.

(ii) Formation of \(\mathrm{NO}_{2}\) from nitric oxide and oxygen

The balanced chemical equation for the formation of \(\mathrm{NO}_{2}\) from nitric oxide and oxygen is: $$2\mathrm{NO}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2\mathrm{NO}_{2}(g)$$ Now we can write the expression for the equilibrium constant, \(K_c\), using the law of mass action as: $$K_c = \frac{[\mathrm{NO}_{2}]^2}{[\mathrm{NO}]^2[\mathrm{O}_{2}]}$$ Where [A] denotes the concentration of species A at equilibrium.

Key concepts

Equilibrium Constant

In a chemical reaction that reaches a state of equilibrium, the equilibrium constant, represented as \(K_c\), provides critical insight into the ratio of the concentration of products to reactants, each raised to the power of their stoichiometric coefficients. It's a snapshot of the system at equilibrium. For example, in the reaction forming \(\mathrm{SO}_3\) from \(\mathrm{SO}_2\) and \(\mathrm{O}_2\), the equilibrium constant \(K_c\) is calculated as:\[K_c = \frac{[\mathrm{SO}_{3}]^2}{[\mathrm{SO}_{2}]^2 [\mathrm{O}_{2}]}\]This formula helps us understand the proportions of reactants and products once the system has reached equilibrium. If \(K_c\) is large, it indicates a greater concentration of products compared to reactants, showing the reaction "lies to the right." Conversely, a small \(K_c\) suggests a reaction "lies to the left," with a prominence of reactants. Understanding \(K_c\) is essential for predicting the direction and extent of chemical reactions.Typically:

• A large \(K_c\) implies a strong tendency to form products.
• A small \(K_c\) suggests the reaction favors reactants.
• \(K_c\) is dimensionless and temperature-dependent.

Using \(K_c\) values, chemists can predict how a change in conditions (like concentration or temperature) can shift the equilibrium.

Chemical Equilibria

Chemical equilibria occur when a reaction goes in both directions at the same rate, leading to no net change in the amounts of reactants and products. This condition is called dynamic equilibrium. At this point, the forward and reverse reactions occur simultaneously and balance each other out.For forming \(\mathrm{SO}_3\) and \(\mathrm{NO}_2\) via their respective reactions:- \(2\mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{SO}_3(g)\)- \(2\mathrm{NO}(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}_2(g)\)The fact that these reactions are in equilibrium implies:

• The concentrations of all species (reactants and products) remain constant over time.
• The reaction system is balanced, not necessarily at equal concentrations, but in a steady state.
• Changes in pressure, temperature, or concentration can shift this equilibrium, described by Le Chatelier's Principle.

Understanding chemical equilibria is crucial for controlling and optimizing reaction conditions in industrial and laboratory settings.

Balanced Chemical Equation

A balanced chemical equation accurately represents a chemical reaction. It depicts the equality of atoms of each element in the reactants and the products, preserving the law of conservation of mass. This means that the atoms go through a transformation but don't just vanish or appear out of thin air. For a balanced equation, the number of each type of atom on both sides of the equation must be equal.Let's consider the reaction forming \(\mathrm{SO}_3\):\[2\mathrm{SO}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{SO}_3(g)\]Here, we see that there are four sulfur atoms and six oxygen atoms on both sides of the equation. This balance is achieved by adjusting the coefficients, which are the numbers in front of the formulas.

• Ensures matter is neither created nor destroyed.
• Coefficients are adjusted to maintain this balance.
• Helps in calculating the equilibrium constant \(K_c\) accurately.

Balancing is a foundational skill in chemistry, necessary for predicting how substances react and figuring out the role each participant plays in a reaction.