Question

A mixture of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at \(1000 \mathrm{K}\) contains the gases at the following concentrations: \(\left[\mathrm{SO}_{2}\right]=5.0 \times 10^{-3}\) \(\mathrm{mol} / \mathrm{L},\left[\mathrm{O}_{2}\right]=1.9 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) and \(\left[\mathrm{SO}_{3}\right]=6.9 \times 10^{-3}\) mol/L. Is the reaction at equilibrium? If not, which way will the reaction proceed to reach equilibrium? $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{3}(\mathrm{g}) \quad K=279 $$

Short answer

The reaction is not at equilibrium; it will proceed in the reverse direction.

Step-by-step solution

Write the Reaction Quotient Expression

For the reaction \( 2 \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{3}(\mathrm{g}) \), the reaction quotient \( Q \) is given by the formula: \[ Q = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2[\mathrm{O}_2]} \]

Substitute Initial Concentrations into Reaction Quotient

Substitute the given concentrations into the expression for \( Q \):\[ Q = \frac{(6.9 \times 10^{-3})^2}{(5.0 \times 10^{-3})^2 \times (1.9 \times 10^{-3})} \]

Calculate the Reaction Quotient, Q

Calculate \( Q \) using the substituted values:\[ Q = \frac{4.761 \times 10^{-5}}{4.75 \times 10^{-8} \times 1.9 \times 10^{-3}} = \frac{4.761 \times 10^{-5}}{9.025 \times 10^{-11}} \approx 527 \]

Compare Q to K to Determine Equilibrium Status

The equilibrium constant \( K \) is given as 279. Compare \( Q \) and \( K \):Since \( Q = 527 \) and \( K = 279 \), \( Q > K \).

Determine Direction of Reaction to Reach Equilibrium

Since \( Q > K \), the reaction is not at equilibrium. It will proceed in the reverse direction until equilibrium is reached.

Key concepts

Reaction Quotient

The reaction quotient, denoted as "Q," is a helpful tool in chemistry that tells us where a reaction currently stands in relation to equilibrium. Specifically, it compares the current concentrations of reactants and products in a chemical reaction. How does it do this? By using a specific mathematical formula that looks quite similar to the equilibrium constant (K).In the chemical reaction represented by \[2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftarrows 2 \text{SO}_3(\text{g})\]we calculate Q using the concentrations given:\[Q = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}\]This formula involves raising the concentration of each chemical species to the power of its coefficient in the balanced equation, utilizing multiplication for reactants and products. For the concentrations given in the problem, substituting these values into Q tells us whether we're at equilibrium or not. If Q equals the equilibrium constant K, the system is at equilibrium. If not, it indicates whether the reaction needs to shift left or right to reach equilibrium.

Equilibrium Constant

The equilibrium constant, represented by "K," is a numerical value that reflects the ratio of product to reactant concentrations at chemical equilibrium for a given temperature. It is unique for each chemical reaction and is essential for understanding the direction that a chemical reaction will prefer to proceed towards.For the equation \[2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftarrows 2 \text{SO}_3(\text{g})\]K is given as 279 at 1000 K. This value is derived similarly to the reaction quotient Q, except that it is specifically calculated when the system is at equilibrium:\[K = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]}\]Knowing K is critical because it informs us about the equilibrium position:- If K is much larger than 1, products predominate at equilibrium.- If K is much smaller than 1, reactants predominate.- If K is around 1, there is a balance between products and reactants. In this exercise, since Q (527) is greater than K (279), the reaction must shift in the reverse direction to reach equilibrium.

Le Chatelier's Principle

Le Chatelier's principle provides insight into how a system at equilibrium responds to changes in concentration, temperature, and pressure. It states that if an external change is applied to a system at equilibrium, the system adjusts itself to counteract that change and reestablish equilibrium.When we analyze the problem where Q = 527 and K = 279:- **Because Q > K**, according to Le Chatelier's principle, the system is not at equilibrium and must adjust by reducing the concentration of products and increasing the concentration of reactants.This adjustment means the reaction will proceed in the reverse direction:- Product formation of \(\text{SO}_3\) will decrease.- Conversely, the concentrations of \(\text{SO}_2\) and \(\text{O}_2\) will increase until Q matches K.Le Chatelier's principle predicts that the system compensates for the excess of products (\(\text{SO}_3\)) by favoring the reverse reaction — indicative of the system's natural tendency to restore equilibrium.