Question

At constant temperature in a 1-L vessel when the reaction \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) is at equilibrium, the \(\mathrm{SO}_{2}\) concentration is \(0.6 \mathrm{M}\), initial concentration of \(\mathrm{SO}_{3}\) is \(1 \mathrm{M}\). The equilibrium constant is (1) \(2.7\) (2) \(1.36\) (3) \(0.34\) (4) \(0.675\)

Short answer

The equilibrium constant is 0.675.

Step-by-step solution

- Write the equilibrium expression

The reaction given is \[2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\]. The equilibrium expression for this reaction is \[K_{c} = \frac{[\mathrm{SO}_{2}]^2 \cdot [\mathrm{O}_{2}]}{[\mathrm{SO}_{3}]^2}\].

- Determine the change in concentrations

Let the change in concentration of \(\mathrm{SO}_{3}\) be \(2x\). Since the stoichiometry is 2:2:1, at equilibrium, the concentration of \(\mathrm{SO}_2\) formed is twice the change. Therefore, the concentration of \(\mathrm{SO}_{2}\) will be \(0.6 \mathrm{M}\) which means the change \(2x = 0.6\). This implies \(x = 0.3\).

- Find equilibrium concentrations

The concentration of \(\mathrm{SO}_{3}\) at equilibrium is the initial concentration minus the change, which equals to \(1 - 0.6 = 0.4\, \mathrm{M}\). For \(\mathrm{O}_{2}\), the concentration at equilibrium will be \(x\), thus \([\mathrm{O}_2] = 0.3 \mathrm{M}\) since \(x = 0.3\).

- Substitute into equilibrium expression

Now, substitute the equilibrium concentrations into the equilibrium expression: \[K_{c} = \frac{[\mathrm{SO}_{2}]^2[\mathrm{O}_{2}]}{[\mathrm{SO}_{3}]^2} = \frac{(0.6)^2 \cdot 0.3}{(0.4)^2} = \frac{0.36 \cdot 0.3}{0.16} \].

- Calculate the equilibrium constant

Finally, calculate the value: \[K_{c} = \frac{0.108}{0.16} = 0.675\].

Key concepts

equilibrium constant

The equilibrium constant, denoted as \(K_c\), is a number that expresses the ratio of the concentrations of products to reactants at equilibrium. Each concentration is raised to the power of its respective coefficient from the balanced chemical equation.
For the reaction: \[2 \text{SO}_3(\text{g}) \rightleftharpoons 2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g})\]The equilibrium expression is: \[K_c = \frac{[\text{SO}_2]^2 \times [\text{O}_2]}{[\text{SO}_3]^2}\]Here, the brackets represent the concentration of each gas. The exponents correspond to the coefficients in the balanced equation.
Values of \(K_c\) provide insights into the position of equilibrium.

• When \(K_c >> 1\), the reaction favors products.
• When \(K_c << 1\), the reaction favors reactants.
• When \(K_c \text{ is around } 1\), both reactants and products are present in comparable amounts.

Understanding \(K_c\) helps predict the extent of a reaction and is crucial in chemical equilibrium calculations.

reaction stoichiometry

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction based on the balanced equation. The coefficients indicate the molar ratio in which substances react and form.
In the given reaction: \[2 \text{SO}_3(\text{g}) \rightleftharpoons 2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g})\]The stoichiometry tells us that:

• 2 moles of \( \text{SO}_3\) decompose to form 2 moles of \( \text{SO}_2\) and 1 mole of \( \text{O}_2\).

This means for every 2 moles of \( \text{SO}_3\) that react, we produce 2 moles of \( \text{SO}_2\) and 1 mole of \( \text{O}_2\).
Knowing this stoichiometry is essential for calculating changes in concentration. For example, if the concentration of \(\text{SO}_2\) increases by 0.6 M, then based on stoichiometry, the concentration of \(\text{SO}_3\) decreases by 0.6 M/2 = 0.3 M, and the concentration of \(\text{O}_2\) increases by 0.3 M.

concentration changes in equilibrium

Concentration changes at equilibrium help us understand how the amounts of reactants and products adjust during the reaction. The change in concentration for each substance is directly linked to the stoichiometry of the reaction.
For our specific reaction: \[2 \text{SO}_3(\text{g}) \rightleftharpoons 2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g})\]We start with an initial concentration of 1 M for \(\text{SO}_3\). At equilibrium, the concentration of \(\text{SO}_2\) is given as 0.6 M. This change is expressed as \[2x = 0.6\],where \(x\) is the change in concentration.
Solving for \(x\), we get \[x = 0.3 \text{ M}\].This means that the concentration of \(\text{SO}_3\) decreases by 0.6 M (or 2x), leaving us with \[1 \text{ M} - 0.6 \text{ M} = 0.4 \text{ M}\] at equilibrium. The concentration of \(\text{O}_2\) formed, due to the stoichiometric ratio 2:1, is \[x = 0.3 \text{ M}\].Substituting these values into the equilibrium expression, we calculate the equilibrium constant to understand how these concentrations relate to one another.
Recognizing these concentration changes allows for accurate equilibrium constant calculations and offers insights into the dynamics of the chemical reaction at equilibrium.