Question

For the system $$ \mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ at \(1000 \mathrm{~K}, K=0.45 .\) Sulfur trioxide, originally at \(1.00 \mathrm{~atm}\) pressure, partially dissociates to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) at \(1000 \mathrm{~K}\). What is its partial pressure at equilibrium?

Short answer

Answer: The partial pressure of sulfur trioxide (SO3) at equilibrium is approximately 0.641 atm.

Step-by-step solution

Write the equilibrium expression based on the chemical reaction

For the given reaction: $$ \mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ The equilibrium expression (in terms of partial pressures) can be written as: $$ K = \frac{P_{\mathrm{SO_2}} \cdot P_{\mathrm{O_2}}^{\frac{1}{2}}}{P_{\mathrm{SO_3}}} $$

Set up the initial conditions and unknowns

Initially, we have three different partial pressures: \(P_{\mathrm{SO_3}}\), \(P_{\mathrm{SO_2}}\), and \(P_{\mathrm{O_2}}\). We are given that the initial pressure of sulfur trioxide is 1 atm: $$ P_{\mathrm{SO_3}}^{\mathrm{initial}} = 1 \mathrm{~atm} $$ Let the change in the pressures during the dissociation be \(x\). Then, the partial pressures at equilibrium can be expressed as: $$ P_{\mathrm{SO_3}}^{\mathrm{eq}} = 1 - x $$ $$ P_{\mathrm{SO_2}}^{\mathrm{eq}} = x $$ $$ P_{\mathrm{O}_2}^{\mathrm{eq}} = \frac{1}{2} x $$

Substitute the unknowns into the equilibrium expression

Substitute the expressions for the equilibrium pressures into the equilibrium constant expression: $$ 0.45 = \frac{x \cdot (\frac{1}{2}x)^{\frac{1}{2}}}{1-x} $$

Solve for the change in pressure, x

To solve for x, we will first square both sides and then rearrange the equation. Squaring both sides leads to: $$ 0.45^2 = \frac{x^2 \cdot \frac{1}{4}x}{(1-x)^2} $$ Rearranging the equation to isolate x, we get: $$ x^3 = 0.2025(1-x)^2 $$ Solving this equation numerically (using trial and error method or a numerical solver) for x, we obtain: $$ x \approx 0.359 $$

Determine the equilibrium partial pressure of sulfur trioxide

Now that we have the value of \(x\), we can determine the partial pressure of sulfur trioxide at equilibrium: $$ P_{\mathrm{SO_3}}^{\mathrm{eq}} = 1 - x = 1 - 0.359 \approx 0.641 \mathrm{~atm} $$ Thus, the partial pressure of sulfur trioxide at equilibrium is approximately \(0.641 \mathrm{~atm}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a key concept in the study of chemical reactions and occurs when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations—or in the case of gases, the partial pressures—of reactants and products remain constant over time, even though both reactions continue to occur. It's important to note that equilibrium does not mean the reactants and products are present in equal amounts, but rather that their proportions do not change with further passage of time.

Consider a sealed container filled with sulfur trioxide gas at high temperature. It partially breaks down into sulfur dioxide and oxygen. Over time, the breakdown of sulfur trioxide and the formation of sulfur dioxide and oxygen occur at the same rate as their reverse reactions. Once this state is reached, we describe the system as being at equilibrium. In a dynamic equilibrium, both the forward and reverse reactions are ongoing processes, but because they occur at the same rate, the system's composition remains steady.

In applying chemical equilibrium to our textbook exercise, remember that we started with pure sulfur trioxide at a known pressure. As the system reached equilibrium, the partial pressures of the sulfur trioxide, sulfur dioxide, and oxygen changed, settling to constant values when the forward and reverse reactions balanced each other out.

Equilibrium Constant

The equilibrium constant, represented by the symbol 'K', quantifies the balance between reactants and products at chemical equilibrium for a given temperature. In the context of gases, this balance is expressed in terms of partial pressures. The equilibrium constant expression for a given chemical reaction is the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients to the product of the partial pressures of the reactants raised to their respective stoichiometric coefficients.

In the exercise we're examining, the equilibrium expression for the dissociation of sulfur trioxide into sulfur dioxide and oxygen is:
\[ K = \frac{P_{\mathrm{SO_2}} \cdot P_{\mathrm{O_2}}^{\frac{1}{2}}}{P_{\mathrm{SO_3}}} \]
Using this formula, we can then set up a relationship between the known equilibrium constant and the equilibrium partial pressures of the involved substances. Plugging in the initial conditions and changes in partial pressures due to the reaction, we can algebraically solve for the unknowns—ultimately, determining the equilibrium partial pressures of each gas. In the example we worked on, knowing the value of 'K' and the initial pressure of sulfur trioxide, we calculated the change in pressure to solve for the equilibrium partial pressures.

Le Chatelier's Principle

Le Chatelier's principle is a fundamental tool for predicting how a change in conditions can affect the position of equilibrium. According to this principle, if a dynamic equilibrium is disturbed by changing the conditions, such as pressure, temperature, or concentration, the system will adjust itself to counteract that change and establish a new equilibrium.

Considering our exercise, if the pressure were increased on the system after it has reached equilibrium, Le Chatelier's principle suggests that the equilibrium position would shift to reduce the pressure. Since increasing pressure favors the formation of fewer moles of gas, the reaction would shift to the left, producing more sulfur trioxide. On the other hand, if the temperature is increased, the equilibrium will shift to favor the endothermic direction, absorbing the added heat, which in this case would be the dissociation of sulfur trioxide.

Le Chatelier's principle helps explain why changing the conditions can lead to different equilibrium partial pressures. It also helps us make qualitative predictions about how the equilibrium will shift in response to changes, providing a deeper understanding of the dynamic behavior of chemical reactions under varying conditions.