Question

At \(1000 \mathrm{K}, K_{p}=1.85\) for the reaction \(\mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{sO}_{3}(g)\) (a) What is the value of \(K_{p}\) for the reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons\) \(\mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) ?\) (c) What is the value of \(K_{c}\) for the reaction in part (b)?

Short answer

(a) For the reverse reaction, the Kp value is the reciprocal of the given Kp value: \(K_{p(reverse)} ≈ 0.54\). (b) For the reaction with coefficients multiplied, the Kp value is the original Kp value raised to the power of 2: \(K_{p(new)} ≈ 3.42\). (c) The Kc value for the reaction in part (b) is approximately \(2.80 × 10^4\).

Step-by-step solution

STEP 1: Determine Kp for the reverse reaction

Recall that for a reverse reaction, the Kp value of the reverse reaction is the reciprocal of the original Kp value. Therefore, to find the Kp value for the reaction: \(\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\), We just need to calculate the reciprocal of the given Kp value for the original reaction, 1.85: \(K_{p(reverse)} = \frac{1}{K_{p(original)}}\) \(K_{p(reverse)} = \frac{1}{1.85} ≈ 0.54\) So, the Kp value of the reverse reaction is 0.54.

STEP 2: Determine Kp for the reaction with coefficients multiplied

To find the Kp value for the reaction: \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\), We should notice that the reaction is the original reaction multiplied by 2. We need to recall that when reaction coefficients are multiplied by a factor n, the new Kp value is the original Kp value raised to the power of n: \(K_{p(new)} = {K_{p(original)}}^n\) In this case, n=2, and the original Kp=1.85: \(K_{p(new)} = {1.85}^2\) \(K_{p(new)} ≈ 3.42\) Therefore, the Kp value for the new reaction is 3.42.

STEP 3: Determine Kc for the reaction in part (b)

To find the Kc value for the reaction: \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\), we will use the relationship between Kp and Kc values: \(K_p = K_c(RT)^{Δn}\) where R is the gas constant (0.0821 L atm/(mol K)), T is the temperature (1000 K), and Δn is the change in the total number of moles of gas (the sum of moles of products minus the sum of moles of reactants). For this reaction, Δn = (2 - 2 - 1) = -1. We already found Kp in STEP 2, so we can solve for Kc: \(K_c = \frac{K_p}{(RT)^{Δn}}\) \(K_c = \frac{3.42}{(0.0821 \times 1000)^{-1}}\) \(K_c ≈ 2.80 × 10^4\) Thus, the Kc for the reaction in part (b) is approximately 2.80 × 10^4.

Key concepts

Equilibrium Constant (Kp)

Understanding the equilibrium constant at a given temperature is crucial for grasping the balance between products and reactants in a chemical reaction. For gas-phase reactions, this is often expressed as Kp, which relates to the partial pressures of the gases involved.

In the given exercise, the equilibrium constant (Kp) for the synthesis of sulfur trioxide at 1000 K is provided as 1.85. This value quantifies the ratio of the partial pressure of the products to that of the reactants at equilibrium. When a reaction is reversed, as in part (a), the new Kp is simply the inverse of the original Kp. This is due to the equilibrium favoring the opposite side upon reversal, and the mathematical representation of this phenomenon is taking the reciprocal. Therefore, for the reverse reaction, we calculated the Kp to be about 0.54.

A common misconception is that doubling the coefficients of a reaction will also double the Kp value. However, as shown in the step by step solution for part (b), when you double the reaction coefficients, you square the original Kp.

Le Chatelier's Principle

Now let's dip into Le Chatelier's Principle, it's a powerful tool that predicts how a system at equilibrium reacts to disturbances. This principle states that if an external change is applied to a system at equilibrium, the system adjusts to minimize that change and restore a new balance.

Changes can include variations in concentration, pressure, volume, or temperature. For instance, if you increase the pressure of a gas reaction at equilibrium by reducing the volume, the system will shift to favor the side with fewer moles of gas. This concept could assist students in predicting the direction of a reaction shift when the Kp value changes due to alteration in conditions.

Reaction Quotient (Kc)

The reaction quotient (Kc) is akin to the equilibrium constant, but instead of being applied at equilibrium, it's used for any point in the reaction. While Kp is used for gas-phase reactions and involves partial pressures, Kc is used for reactions in solution and relates to the concentrations of reactants and products.

In part (c), we convert the calculated Kp to Kc by using the equation Kp = Kc(RT)^{Δn}. Here, R is the ideal gas constant and T is the temperature in kelvins. The change in the number of moles of gas, Δn, is important since it reflects how changes in pressure or volume can affect reaction direction and extent. By knowing Kp, temperature, and Δn, we found that Kc for the doubled reaction is approximately 2.80 × 10^4. This highlights how equilibria are interconnected and how understanding one can help deduce another.