Question

If the cquilibrium constants of the following cquilibrium \(\mathrm{SO}_{3} \rightleftharpoons \mathrm{SO}_{2}+\mathrm{O}_{2}\) and \(\mathrm{SO}_{2} \mid \mathrm{O}_{2} \rightleftharpoons \mathrm{SO}_{3}\) are given by \(\mathrm{K}_{\mathrm{l}}\) and \(\mathrm{K}_{2}\), respectively, which relation is corrcct? (1) \(\mathrm{K}_{1}=\left(1 / \mathrm{K}_{2}\right)^{2}\) (2) \(\mathrm{K}_{2}=\left|1 / \mathrm{K}_{\mathrm{l}}\right|^{2}\) (3) \(K_{1}=1 / K_{2}\) (4) \(\mathrm{K}_{1}=\left(\mathrm{K}_{2}\right)^{2}\)

Short answer

Option (3): \(K_1 = \frac{1}{K_2}\)

Step-by-step solution

Write down the given equilibria

The given equilibria are:1. \(\text{SO}_3 \rightleftharpoons \text{SO}_2 + \text{O}_2\)2. \(\text{SO}_2 + \text{O}_2 \rightleftharpoons \text{SO}_3\)

Define the equilibrium constants

Let \(K_1\) be the equilibrium constant for \(\text{SO}_3 \rightleftharpoons \text{SO}_2 + \text{O}_2\).Let \(K_2\) be the equilibrium constant for \(\text{SO}_2 + \text{O}_2 \rightleftharpoons \text{SO}_3\).

Understand the relationship between the two equilibria

The equilibrium \(\text{SO}_3 \rightleftharpoons \text{SO}_2 + \text{O}_2\) is the reverse of the equilibrium \(\text{SO}_2 + \text{O}_2 \rightleftharpoons \text{SO}_3\).

Use the relation between forward and reverse reaction constants

For a reaction and its reverse, the equilibrium constants are reciprocals:\(K_{\text{reverse}} = \frac{1}{K_{\text{forward}}}\)

Apply this concept to the given equilibria

For the given equilibria, \(K_1\) for \(\text{SO}_3 \rightleftharpoons \text{SO}_2 + \text{O}_2\) and \(K_2\) for \(\text{SO}_2 + \text{O}_2 \rightleftharpoons \text{SO}_3\):\(K_2 = \frac{1}{K_1}\)

Identify the correct relation

From the options given, the correct relation is:Option (3) \(K_1 = \frac{1}{K_2}\)

Key concepts

Equilibrium Constants

An equilibrium constant, denoted as \(K\), describes the ratio of the concentrations of products to reactants for a reversible reaction at equilibrium. For a general reaction \(\text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD} \), the equilibrium constant expression is given by:
\([K = \frac{[\text{C}]^{\text{c}} [\text{D}]^{\text{d}}}{[\text{A}]^{\text{a}} [\text{B}]^{\text{b}}}]\).
Here, the square brackets \[...\] represent the concentrations of the substances at equilibrium.
Making sense of \(K\):

• A large \(K\) (greater than 1) implies that the reaction favors the formation of products.
• A small \(K\) (less than 1) means that reactants are favored.

It's crucial to understand that the equilibrium constant is specific to a particular reaction at a given temperature. So, even slight changes in temperature can affect the value of \(K\).

Reversible Reactions

A reversible reaction is one where the reactants can form products, which can then react together to give back the original reactants. This is depicted as a double-headed arrow \(\rightarrow\leftarrow\) in a chemical equation. For example, \( \text{SO}_{3} \rightleftharpoons \text{SO}_{2} + \text{O}_{2} \).
The term 'reversible' means that the reaction can proceed in both forward and backward directions until it reaches a state of balance known as equilibrium. At this point, the rates of the forward and reverse reactions are equal.
This concept is vital because:

• It explains why some reactions never go to completion but instead reach an equilibrium point where reactants and products coexist.
• It allows us to understand and predict how changes in conditions (like concentration or pressure) will impact the reaction.

Chemical Equilibrium

Chemical equilibrium represents a state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time. This does not mean the reactions stop; they continue to occur but at the same rate in both directions.
Key points about chemical equilibrium include:

• The system must be closed, meaning no reactants or products can enter or leave the system.
• The values of temperature, pressure, and concentration affect the equilibrium state.
• Le Chatelier's Principle helps predict how the equilibrium shifts when these conditions change.

In terms of equilibrium constants, for the reaction \(\text{SO}_{3} \rightleftharpoons \text{SO}_{2} + \text{O}_{2}\), \(K_{1}\) and \(K_{2}\) describe the constants for the forward and reverse reactions respectively. They are reciprocals: \(K_{\text{1}} = \frac{1}{K_{\text{2}}} \). This reciprocal relationship helps us understand how the system balances itself out in response to changes in both directions of the reversible process.