Problem 31 - General Chemistry: Principles and Modern Applications

Question

In the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}), 0.455\) \(\mathrm{mol} \mathrm{SO}_{2}, 0.183 \mathrm{mol} \mathrm{O}_{2},\) and \(0.568 \mathrm{mol} \mathrm{SO}_{3}\) are introduced simultaneously into a 1.90 L vessel at \(1000 \mathrm{K}\). (a) If \(K_{c}=2.8 \times 10^{2},\) is this mixture at equilibrium? (b) If not, in which direction will a net change occur?

Step-by-Step Solution

Answer

a) The mixture is not in equilibrium. b) The reaction will occur in the direction of forming more SO3, that is to the right.

Step 1: Calculate Concentrations

Calculating concentrations, n(V), for SO2, O2, and SO3 to initiate the process. Concentration = Amount (in mol) / Volume (in L). Concentration of SO2 = 0.455 mol/1.9 L = 0.239 M. Concentration of O2 = 0.183 mol/1.9 L = 0.096 M. Concentration of SO3 = 0.568 mol/1.9 L = 0.299 M.

Step 2: Calculate Reaction Quotient

To determine whether a reaction is at equilibrium, a comparison between the Reaction Quotient, Qc, and the Equilibrium constant, Kc, has to be made. First, calculate the Reaction Quotient (Qc) using the formula Q = [Products] / [Reactants]. For the reaction, Q = [SO3]^2 /[SO2]^2 [O2] = (0.299)^2 / (0.239)^2*0.096 = 1.837.

Step 3: Compare Qc and Kc to Determine Equilibrium

Now we compare Qc and Kc. Here, Qc (1.837) is less than Kc (280), the reaction will therefore proceed to the right. This means the reaction isn't at equilibrium, hence more product (SO3) needs to be formed from the reactants (SO2 and O2).

Step 4: Determine the Reaction Direction

If a reaction is not at equilibrium, it will proceed in the forward or reverse direction in order to reach equilibrium. As our Qc < Kc, then our reaction will proceed to the right: SO2 and O2 will be used up, and more SO3 will be formed, until equilibrium is reached.

Key Concepts

Reaction Quotient

The Reaction Quotient, denoted as \(Q_c\), helps us understand the current state of a chemical reaction in relation to its equilibrium. It's particularly useful when you have initial concentrations of reactants and products and want to figure out if a reaction has reached equilibrium. By comparison with the equilibrium constant \(K_c\), we can ascertain the shift needed to achieve equilibrium. Here's how it works:

• For our reaction, \(Q_c\) is calculated using the formula \(Q_c = \frac{[\text{products}]}{[\text{reactants}]}\).
• If \(Q_c < K_c\), the reaction moves forward towards product formation.
• If \(Q_c > K_c\), the reaction shifts backward towards reactant formation.
• If \(Q_c = K_c\), then the reaction is at equilibrium and no shift is necessary.

In the given exercise, we found \(Q_c = 1.837\) which is less than \(K_c = 280\), indicating that the reaction will proceed towards more product formation.

Equilibrium Constant

The Equilibrium Constant, represented as \(K_c\), is a valuable figure in chemistry that signifies the ratio of product concentrations to reactant concentrations at equilibrium. It reflects where a balance is naturally achieved in a particular reaction at a given temperature. This constant is unique to each reaction:

• It depends only on the temperature, hence changing temperatures changes \(K_c\).
• A high \(K_c\) means that, at equilibrium, products are favored, while a low \(K_c\) suggests that reactants are more prevalent.

In the reaction highlighted in our exercise, \(K_c = 280\), which implies that at equilibrium, the formation of \(\text{SO}_3\) is highly favored over the reactants \(\text{SO}_2\) and \(\text{O}_2\). This information is crucial for determining the direction of the reaction.

Concentration Calculations

Before assessing whether a reaction is at equilibrium, you need to find the concentrations of all participants in the reaction. Concentration is essentially how much of a substance is present per unit volume. This step is basic but essential for calculating the Reaction Quotient. Here’s a quick reminder on how to compute these concentrations:

• Concentration \((\text{Mol/L or M})\) is calculated by dividing the amount in moles by the volume in liters.
• For example, the concentration of \(\text{SO}_2\) was found to be \( \frac{0.455 \, \text{mol}}{1.9 \, \text{L}} = 0.239 \, \text{M}\).

These concentrations allow you to then calculate the Reaction Quotient \(Q_c\) and check against \(K_c\) to assess the equilibrium state.

Direction of Reaction

The direction in which a reaction will shift when it is not at equilibrium is determined by comparing \(Q_c\) to \(K_c\). This is a pivotal aspect of chemical equilibrium, as it indicates whether more products or reactants will form as the system heads towards equilibrium. In practice, you apply it as follows:

• If \(Q_c < K_c\), as we have in our exercise \((1.837 < 280)\), the reaction will move to the right, meaning more \(\text{SO}_3\) will be formed from \(\text{SO}_2\) and \(\text{O}_2\).
• If \(Q_c > K_c\), it moves leftwards, converting products back to reactants.

This simple comparison helps predict changes in concentration over time, guiding a reaction to its equilibrium state efficiently.