Problem 198 - A Complete Resource Book in Chemistry for JEE Main

Question

The equilibrium constant for the reaction \(\mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{~g})+1 / 2 \mathrm{O}_{2}(\mathrm{~g})\) is \(\mathrm{K}=4.9 \times 10^{-2}\). the value of \(\mathrm{K}\) for the reaction \(2 \mathrm{SO}_{2}\) \((\mathrm{g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{~g})\) will be \([2006]\) (a) 416 (b) \(2.40 \times 10^{-3}\) (c) \(9.8 \times 10^{-2}\) (d) \(4.9 \times 10^{-2}\)

Step-by-Step Solution

Answer

The equilibrium constant for the reaction is 416. (option a)

Step 1: Identify Original Reaction and Its Equilibrium Constant

The given reaction is \( \text{SO}_3(g) \rightleftharpoons \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \) with an equilibrium constant \( K = 4.9 \times 10^{-2} \).

Step 2: Identify Reverse Reaction

The reaction we need to find the equilibrium constant for is the reverse: \( 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \).

Step 3: Apply Equilibrium Constant Inversion

When a reaction is reversed, the equilibrium constant \( K' \) for the reverse reaction is the reciprocal of the original. Therefore, if \( K = 4.9 \times 10^{-2} \) for the forward reaction, then \( K' = \frac{1}{K} = \frac{1}{4.9 \times 10^{-2}} \).

Step 4: Calculate the Reciprocal of the Equilibrium Constant

Calculate \( K' \) by performing the division: \[ K' = \frac{1}{4.9 \times 10^{-2}} = \frac{1}{0.049} \approx 20.41. \]

Step 5: Adjust for Stoichiometry

The forward reaction given initially involves \( \text{SO}_3 \) forming \( 2\text{SO}_2 + \text{O}_2 \), which is the half-reaction. The reverse requires doubling mollies, thus equating to excluding the initial power performer of \[ K'' =(K')^2 =20.41^2\approx 416 \] ensuring to account for corrective measure based on stage alteration.

Key Concepts

Equilibrium Constant Calculations

The equilibrium constant, represented by the symbol \( K \), is a numerical value that indicates the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. It provides insight into the position of a reaction's equilibrium. If \( K \) is greater than 1, the reaction tends to have more products at equilibrium. Conversely, if \( K \) is less than 1, the reactants are more predominant.

In the problem, we have a given reaction: \( \text{SO}_3(g) \rightleftharpoons \text{SO}_2(g) + \frac{1}{2} \text{O}_2(g) \) with \( K = 4.9 \times 10^{-2} \). To find the equilibrium constant for the reverse reaction \( 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \), we use the reciprocal of the provided constant because reversing the reaction flips the products and reactants.

• The inverse of \( K \) is calculated as \( \frac{1}{K} = \frac{1}{4.9 \times 10^{-2}} \).
• This gives \( K' \approx 20.41 \) for the reversed reaction.

The final step involves adjusting for the stoichiometric coefficients when doubling the reaction coefficients.

Reversible Reactions

Reversible reactions are chemical reactions where the reactants can be converted to products and vice versa. This is symbolized by the special double arrow \( \rightleftharpoons \), representing both the forward and backward reactions.

In the given exercise, the notation \( \rightleftharpoons \) indicates the bidirectional nature of the reaction between \( \text{SO}_3 \) and \( \text{SO}_2 + \frac{1}{2} \text{O}_2 \). Understanding reversible reactions is crucial, as it shows that chemical reactions can reach a state of balance called chemical equilibrium, where the rate of the forward reaction equals the rate of the backward reaction.

• At equilibrium, there is no net change in the concentration of reactants and products.
• Both forms of the reaction can be described using an equilibrium constant specific to the direction of interest.

Ultimately, the characterization of reversible reactions highlights that chemical processes are dynamic, with ongoing formation and consumption of reactants and products.

Stoichiometry in Chemical Equilibrium

Stoichiometry in chemical equilibrium involves using the balanced chemical equation to understand the proportions of reactants and products involved during a chemical reaction.

Take for instance the conversion for the reaction \( 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \). To find the new equilibrium constant, after obtaining the reciprocal, we must raise \( K' \) to the power of the change in stoichiometry due to doubling of the reaction participants.

• The change from \( \text{SO}_3 \rightleftharpoons \text{SO}_2 + \frac{1}{2} \text{O}_2 \) to \( 2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 \) involves multiplying each stoichiometric coefficient by two.
• This requires squaring the inverse equilibrium constant: \( K'' = (K')^2 \).
• The squared value gives \( K'' \approx 416 \), reflecting how changes in stoichiometry impact equilibrium constants.

This step is essential to ensure that the equilibrium calculation respects the balance dictated by the reaction coefficients.