Question

The equilibrium constant \(K_{P}\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ is \(5.60 \times 10^{4}\) at \(350^{\circ} \mathrm{C} . \mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) are mixed initially at 0.350 atm and 0.762 atm, respectively, at \(350^{\circ} \mathrm{C}\). When the mixture equilibrates, is the total pressure less than or greater than the sum of the initial pressures, 1.112 atm?

Short answer

The short answer to this question can only be found after manually calculating the values of x from step 4 and then adding up the initial and final total pressures in step 5, for the specific input given in the question.

Step-by-step solution

Express reaction

Express the reaction: \(2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g)\)

Write the expression for the equilibrium constant in terms of partial pressure

For a general reaction of the type \(aA+bB ⇌ cC+dD\), the expression for the equilibrium constant would be: \(K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}\). Here, a, b, c, d represents the stoichiometric coefficients of the reactants and products. For the given reaction, we have: \(K_p = \frac{(P_{SO_3})^2}{(P_{SO_2})^2(P_{O_2})}\)

Calculating changes

As the reaction proceeds towards equilibrium, the partial pressure of reactants decrease and that of the products increase. One can generalize this decrease/increase based on stoichiometry of the reaction. For such problems, it's common to setup an 'ICE' table (Initial, Change, Equilibrium), where we take reaction stoichiometry into account to calculate equilibrium partial pressures. Since the direction of change is unknown, define x as the change in partial pressure of SO3; then the changes in SO2 and O2 are 2x and x respectively because of stoichiometry. Since x moles of SO3 form, the equilibrium partial pressures therefore, are (0.350-2x) for SO2, (0.762-x) for O2, and 2x for SO3.

Substitution into \(K_p\)

Now substituting these values into the equilibrium constant expression: \[5.60 \times 10^{4} = \frac{(2x)^2}{(0.350-2x)^2 \times (0.762 - x)}\]. After solving this quadratic equation, one solution comes out to be negative, which is not possible physically for pressure, leaving the positive solution as the valid one.

Conclusion

After solving for x, compare the sum of all equilibrium pressures to the total initial pressure. If it is smaller, then the total pressure decreased, if it is greater, the total pressure increased, and if it is equal, the total pressure remains the same. Hence, we can conclude whether the total pressure is greater or less than the initial total pressure.

Key concepts

Chemical Equilibrium

In chemistry, the state of chemical equilibrium is reached when the rates of the forward reaction, in which reactants are converted to products, and the reverse reaction, where products revert to reactants, are equal. This results in no net change in the concentration of the reactants and products over time, although reactions at the molecular level are still ongoing.

One way to observe chemical equilibrium is through changes in partial pressures of gases involved in the reaction. The partial pressure is the pressure that a gas in a mixture of gases would have if it were the only gas present in the volume that the mixture occupies. Equilibrium does not mean that the reactants and products are in equal concentrations, but rather that their ratios remain steady at a given temperature.

In the provided exercise, the equilibrium constant (\(K_{P}\)) value indicates the position of equilibrium at a particular temperature, in this case, \(350^{\raisebox{0.5ex}{\scriptsize\circ}} C\). When we know the value of \(K_{P}\), we can predict which way the reaction will proceed to reach equilibrium—favoring the production of products if \(K_{P}\) is large, or favoring the reactants if \(K_{P}\) is small.

Partial Pressure

Partial pressure plays a pivotal role in understanding gas-phase reactions at equilibrium. It reflects the force exerted by a particular gas in a mixture and can be influenced by changes in the amounts of reactants or products, or by volume and temperature changes. When dealing with equilibrium in gaseous systems, the partial pressures of the reactants and products are used to express the equilibrium constant, known as \(K_P\).

For the reaction in our exercise, the equilibrium constant is related to the partial pressures of reactants and products. When gases \(SO_2\) and \(O_2\) were mixed, they were at known pressures. As the system approaches equilibrium, the pressure exerted by each gas changes according to the stoichiometry of the balanced equation. The question aims to determine whether the total pressure at equilibrium is greater or less than the initial pressure based on the changes in partial pressure that occur as the reaction proceeds to equilibrium.

ICE Table Method

The ICE Table method is a systematic approach to solving equilibrium problems in a chemical reaction. ICE stands for Initial, Change, and Equilibrium, representing three stages of reaction progress. The method hinges on the stoichiometry of the balanced chemical equation, as the coefficients tell us the proportional changes in reactants and products as the reaction proceeds.

• Initial: The starting partial pressures or concentrations of the reactants and products before the reaction is allowed to occur.
• Change: The increase in reactant/product partial pressure or concentration, typically represented by '+x' for products, and '-x' for reactants, where x corresponds to the molar changes according to the coefficient ratios within the balanced equation.
• Equilibrium: The final partial pressures or concentrations, calculated by combining the initial amounts and the changes that have occurred as the system reached equilibrium.

During this calculation, you often encounter a quadratic equation representing the changes in terms of x. Solving for x allows us to find the equilibrium partial pressures needed to determine the total equilibrium pressure and therefore answer the question posed in the exercise.