Question

In cquilibrium \(2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\), the partial pressure of \(\mathrm{SO}_{2}, \mathrm{O}_{2}\) and \(\mathrm{SO}_{3}\) are \(0.662,0.101\) and \(0.331\) atm, respectively. What would be the partial pressurc of oxygen so that the equilibrium concentration of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) arc equal? (1) \(0.4 \mathrm{~atm}\) (2) \(1.0 \mathrm{~atm}\) (3) \(0.8 \mathrm{~atm}\) (4) \(0.25\) atm

Short answer

1 atm

Step-by-step solution

- Write the Equilibrium Expression

The equilibrium constant expression for the reaction \[2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2 \text{SO}_3(\text{g})\] is given by\[K_p = \frac{{P_{\text{SO}_3}^2}}{{P_{\text{SO}_2}^2 P_{\text{O}_2}}}\] where \(P\) denotes partial pressure.

- Calculate the Equilibrium Constant

Using the given partial pressures of \(P_{\text{SO}_2} = 0.662 \text{ atm}\), \(P_{\text{O}_2} = 0.101 \text{ atm}\), and \(P_{\text{SO}_3} = 0.331 \text{ atm}\), substitute these values into the equilibrium expression:\[K_p = \frac{{(0.331)^2}}{{(0.662)^2(0.101)}}\] Calculating this gives \[K_p \approx 0.25\]

- Set up New Equilibrium Condition

We need to find the partial pressure of O2 that makes the partial pressures of SO2 and SO3 equal. Let's assume this common partial pressure is \(P\).Then, our new equilibrium conditions are: \[P_{\text{SO}_2} = P \text{ atm}\]\[P_{\text{SO}_3} = P \text{ atm}\]Substitute these into the equilibrium expression\[K_p = \frac{{P^2}}{{P^2(P_{\text{O}_2}')}}\]where \(P_{\text{O}_2}'\) is the new partial pressure of O2.

- Solve for the New Partial Pressure of O2

Rearrange the equilibrium expression to solve for \(P_{\text{O}_2}'\):\[0.25 = \frac{{P^2}}{{P^2(P_{\text{O}_2}') }}\]Since \[K_p = 0.25\], we get:\[1 = 0.25(P_{\text{O}_2}')\]\[P_{\text{O}_2}' = 4 \times 0.25\]\[P_{\text{O}_2}' = 1\] atm.

Key concepts

equilibrium constant

Chemical equilibrium involves the balance of reactants and products in a reversible reaction. The equilibrium constant (K_p) is a measure of how far the reaction proceeds before reaching this balance. For our reaction \(2 \mathrm{SO}_{2}(\text{g}) + \mathrm{O}_{2}(\text{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\text{g})\), it’s represented by the ratio of the partial pressures of the products to the reactants, with each partial pressure raised to the power of its stoichiometric coefficient.
The equation is:
\[ K_p = \frac{P_{\text{SO}_3}^2}{P_{\text{SO}_2}^2 P_{\text{O}_2}} \] This equation shows a predictive relationship between the pressures of gases involved in the reaction. Knowing the equilibrium constant can help us understand how changes in conditions affect the reaction's position at equilibrium.

partial pressure

In a gaseous reaction, the partial pressure refers to the pressure that each gas exerts in a mixture. It’s an essential concept because the behavior of gases in reactions is often described in terms of their partial pressures.
For the given reaction, we calculate the partial pressures of individual gases using the equilibrium expression:
\[ K_p = \frac{P_{\text{SO}_3}^2}{P_{\text{SO}_2}^2 P_{\text{O}_2}} \]
Given that \( P_{\text{SO}_2} = 0.662 \) atm, \( P_{\text{O}_2} = 0.101 \) atm, and \( P_{\text{SO}_3} = 0.331 \) atm, the equilibrium constant is derived as:

\[ K_p = \frac{(0.331)^2}{(0.662)^2 (0.101)} \approx 0.25 \]
The partial pressure concept simplifies dealing with gas mixtures, especially in equilibrium context, by providing a manageable way to apply equilibrium rules.

Le Chatelier’s principle

Le Chatelier’s principle helps us predict how a change in conditions affects equilibrium. It states that if a dynamic equilibrium is disturbed, the system adjusts to restore equilibrium. For example, changing the partial pressure of a gas in our reaction:
\(2 \mathrm{SO}_{2}(\text{~g}) + \mathrm{O}_{2}(\text{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\text{~g})\), if we increase \( P_{\text{O}_2} \), the system will shift to the right to favor the production of more SO3 to reduce the disturbance. Similarly, a decrease in \( P_{\text{O}_2} \) will shift equilibrium to the left, making more SO2 and O2.
This principle shows us that equilibrium is dynamic and can be influenced by changes in pressure, temperature, and concentrations.

gaseous reactions

Understanding gaseous reactions involves analyzing how gases interact and reach equilibrium. Gaseous reactions are critical in both natural processes and industrial applications. In our example, we deal with the reaction:
\(2 \mathrm{SO}_{2}(\text{~g}) + \mathrm{O}_{2}(\text{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\text{~g})\),
where the species involved are all gases. The behavior of these gases is determined by their partial pressures and the equilibrium constant.
To solve for the new equilibrium partial pressure of \(O_2\) where \(P_{\text{SO}_2}\) and \(P_{\text{SO}_3}\) are equal, let’s assume this common partial pressure is \(P\). Then
\( K_p = \frac{P^2}{P^2(P_{\text{O}_2}')} \)
Given that \( K_p = 0.25 \), we can solve for the new partial pressure of \(O_2\). Overall, gaseous reactions require a solid understanding of how partial pressures and equilibrium constants interrelate to determine the balance of gases in the system.