Problem 69 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The equilibrium constant \(K_{p}\) of the reaction: $$ 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) $$ is 900 atm. at \(800 \mathrm{~K}\). A mixture containing \(\mathrm{SO}_{3}\) and \(\mathrm{O}_{2}\) having initial partial pressure of 1 and 2 atm. respectively is heated at constant volume to equilibrate. Calculate the partial pressure of each gas at \(800 \mathrm{~K}\).

Step-by-Step Solution

Answer

At equilibrium: \( P_{\mathrm{SO}_3} = 0.28\) atm, \( P_{\mathrm{SO}_2} = 0.86\) atm, and \( P_{\mathrm{O}_2} = 1.14\) atm.

Step 1: Write the Expression for the Equilibrium Constant

The equilibrium constant for the reaction in terms of partial pressures is given by:\[K_p = \frac{P_{\mathrm{SO}_3}^2}{P_{\mathrm{SO}_2}^2 \cdot P_{\mathrm{O}_2}}\]Substitute the given value of \(K_p\), which is 900 atm, into the expression.

Step 2: Define Initial and Change in Partial Pressures

Initially, the partial pressures are \(P_{\mathrm{SO}_3} = 1\) atm and \(P_{\mathrm{O}_2} = 2\) atm, with \(P_{\mathrm{SO}_2} = 0\) atm. Let \(x\) be the change in partial pressure for \(\mathrm{SO}_3\). Then, at equilibrium:- \(P_{\mathrm{SO}_3} = 1 - 2x\) - \(P_{\mathrm{SO}_2} = 2x\)- \(P_{\mathrm{O}_2} = 2 - x\) (Consider stoichiometry for changes in pressures.)

Step 3: Substitute Variables into Equilibrium Expression

Plug the expressions for the equilibrium partial pressures back into the equilibrium constant equation:\[900 = \frac{(1 - 2x)^2}{(2x)^2 \cdot (2 - x)}\]This equation can be solved to find the value of \(x\).

Step 4: Solve the Equation

Multiply through to clear the fraction and simplify:\[900 \cdot (2x)^2 \cdot (2-x) = (1 - 2x)^2\]Simplifying further gives a polynomial in terms of \(x\). Solve this polynomial equation to find the equilibrium value of \(x\).

Step 5: Calculate Equilibrium Partial Pressures

Substitute the value of \(x\) back into the expressions for partial pressures:- \(P_{\mathrm{SO}_3} = 1 - 2x\) - \(P_{\mathrm{SO}_2} = 2x\)- \(P_{\mathrm{O}_2} = 2 - x\) Find the specific numerical values of all pressures.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted by \(K_p\) when dealing with gases, is a crucial concept in understanding chemical equilibrium. It tells us how far a reaction will go to reach equilibrium, based on the conditions provided. For the reaction \(2 \text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2 \text{SO}_3(\text{g})\), the equilibrium constant \(K_p\) at 800K is given as 900 atm.
This value implies that at equilibrium, the ratio of the partial pressures of the products to that of the reactants raised to their stoichiometric coefficients is equal to 900. Therefore, a high value of \(K_p\) such as 900 suggests the reaction favors the production of \(\text{SO}_3\), i.e., the products are more dominant at equilibrium.

• This concept helps predict the direction in which a reaction mixture needs to shift to achieve equilibrium, depending on the initial conditions of the system.
• It is calculated using the formula: \[K_p = \frac{P_{\mathrm{SO}_3}^2}{P_{\mathrm{SO}_2}^2 \cdot P_{\mathrm{O}_2}}\]

Partial Pressure

Partial pressure is the pressure exerted by individual gas components in a gaseous mixture. In equilibrium calculations, it is essential to determine the contribution of each gas to the total pressure. In our reaction, we are concerned with the partial pressures of \(\text{SO}_3\), \(\text{SO}_2\), and \(\text{O}_2\). At the onset, these partial pressures are specified: 1 atm for \(\text{SO}_3\) and 2 atm for \(\text{O}_2\).

• Partial pressures change as the reaction progresses towards equilibrium due to the conversion of reactants to products and vice versa.
• Understanding how to manipulate these changes allows us to predict what the equilibrium pressures will be, and consequently, confirms the achievement of equilibrium according to the equilibrium constant \(K_p\).

This process involves defining changes in partial pressures (denoted as \(x\)) and plugging these changes back into the initial pressures to understand their equilibrium state.

Stoichiometry

Stoichiometry is the study of the quantitative relationship between reactants and products in a chemical reaction. It is crucial in determining how the partial pressures of different gases change as the system reaches equilibrium.
In the reaction \(2 \text{SO}_2 + \text{O}_2 \rightleftharpoons 2 \text{SO}_3\), stoichiometry dictates the relationship between the molecules involved. The reaction shows that:

• Two moles of \(\text{SO}_3\) are formed for every two moles of \(\text{SO}_2\) reacted and one mole of \(\text{O}_2\) consumed.
• This provides a clear path to calculate equilibrium partial pressures, once the change in pressure \(x\) is determined.

Changes are expressed as multiples of \(x\) for the molecular species involved:

• \(\text{SO}_3\) decreases by \(2x\),
• \(\text{SO}_2\) increases by \(2x\),
• \(\text{O}_2\) decreases by \(x\).

These changes plug back into the equilibrium expression for \(K_p\) to solve for \(x\) and calculate equilibrium partial pressures.

Reaction Quotient

The reaction quotient, represented as \(Q_p\) for reactions involving gases, helps compare the current state of a system to its equilibrium position. It has the same mathematical form as the equilibrium constant but uses the initial conditions:
\[Q_p = \frac{P_{\mathrm{SO}_3}^2}{P_{\mathrm{SO}_2}^2 \cdot P_{\mathrm{O}_2}}\]
Calculating \(Q_p\) given initial partial pressures allows us to predict in which direction the reaction will move to achieve equilibrium. Should \(Q_p\) differ from \(K_p\):

• If \(Q_p < K_p\), the reaction will shift to the right, producing more products.
• If \(Q_p > K_p\), the reaction will shift to the left, producing more reactants.

In this exercise, understanding the relationship between \(Q_p\) and \(K_p\) is key to determining how initial partial pressures need to adjust, ultimately finding the equilibrium pressures of the gases in question. This balance is critical in chemical reactions, maintaining a state where both rates of forward and reverse reactions are equal.