Question

The cquilibrium constant for cquilibria \(\mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_{3}(\mathrm{~g})\) and \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\) are \(\mathrm{K}_{1}\) and \(\mathrm{K}_{2}\) respectively. Then (1) \(\mathrm{K}_{2}=\mathrm{K}_{1}\) (2) \(\mathrm{K}_{2}=\mathrm{K}_{1}^{2}\) (3) \(\mathrm{K}_{2}=\frac{1}{\mathrm{~K}_{1}}\) (4) \(\mathrm{K}_{2}=\frac{1}{\mathrm{~K}_{1}^{2}}\)

Short answer

\text{K}_{2} = \frac{1}{K}_{1}^{2}

Step-by-step solution

Write the equilibrium expression for the first reaction

For the equilibrium reaction: \text{SO}_{2}(\text{g}) + \frac{1}{2} \text{O}_{2}(\text{g}) \rightleftharpoons \text{SO}_{3}(\text{g})The equilibrium constant \text{K}_{1} is given by:\[ K_{1} = \frac{[\text{SO}_{3}]}{[\text{SO}_{2}][\text{O}_{2}]^{1/2}} \]

Write the equilibrium expression for the second reaction

For the equilibrium reaction: 2\text{SO}_{3}(\text{g}) \rightleftharpoons 2\text{SO}_{2}(\text{g}) + \text{O}_{2}(\text{g})The equilibrium constant \text{K}_{2} is given by:\[ K_{2} = \frac{[\text{SO}_{2}]^{2}[\text{O}_{2}]}{[\text{SO}_{3}]^{2}} \]

Relate the two equilibrium constants

Since the second reaction is simply the reverse of the first reaction with each species coefficient doubled: 2(\text{SO}_{2}(\text{g}) + \frac{1}{2} \text{O}_{2}(\text{g}) \rightleftharpoons \text{SO}_{3}(\text{g}))Thus, \[ K_{2} = \left(\frac{1}{K_{1}\right)^{2} \]This simplifies to \[ K_{2} = \frac{1}{K_{1}^{2}} \]

Key concepts

chemical equilibrium

In chemistry, an equilibrium state occurs when the rate of the forward reaction equals the rate of the reverse reaction. This balance means that the concentrations of reactants and products remain constant over time.

Imagine a scenario where no further changes happen because the processes of formation and decomposition occur at the same speed. This is what we mean by chemical equilibrium.

At equilibrium, it doesn't matter if you start with reactants or products. The system will adjust itself to reach the balance point. This state is described by the equilibrium constant, a value that helps chemists understand the position of equilibrium and how the system behaves under different conditions.

equilibrium expressions

Equilibrium expressions quantify the relationships between the concentrations of reactants and products at equilibrium. These expressions allow us to calculate the equilibrium constant, denoted as K.

For a generic reaction: \text{aA} + \text{bB} \rightleftharpoons \text{cC} + \text{dD}, the equilibrium expression is given by:

\[ K = \frac{[\text{C}]^{c}[\text{D}]^{d}}{[\text{A}]^{a}[\text{B}]^{b}} \]

This formula uses the concentrations of the products in the numerator and the concentrations of the reactants in the denominator.

Here’s how it works for the given exercise:

For \text{SO}_{2}(\text{g}) + \frac{1}{2} \text{O}_{2}(\text{g}) \rightleftharpoons \text{SO}_{3}(\text{g}), the equilibrium expression is:

\[ K_{1} = \frac{[\text{SO}_{3}]}{[\text{SO}_{2}][\text{O}_{2}]^{1/2}} \]

For 2\text{SO}_{3}(\text{g}) \rightleftharpoons 2\text{SO}_{2}(\text{g}) + \text{O}_{2}(\text{g}), the equilibrium expression is:

\[ K_{2} = \frac{[\text{SO}_{2}]^{2}[\text{O}_{2}]}{[\text{SO}_{3}]^{2}} \]

By understanding these expressions, you can solve for the equilibrium constants and relate them to each other for different reactions.

reaction coefficients

Reaction coefficients are the numbers in front of the chemical species in a balanced equation. They indicate the mole ratios of reactants and products involved in the reaction.

For instance, in the balanced reaction: 2\text{SO}_{2}(\text{g}) + \text{O}_{2}(\text{g}) \rightleftharpoons 2\text{SO}_{3}(\text{g}), the coefficients are 2 for \text{SO}_{2}, 1 for \text{O}_{2}, and 2 for \text{SO}_{3}.

Doubling all the coefficients in a balanced reaction affects the equilibrium expression. If the coefficients change, the concentrations in the equilibrium expression will have their exponents adjusted accordingly.

In the exercise, the second reaction is a scaled version of the first, where all species' coefficients are doubled. This means:

• Each species' concentration is squared in the equilibrium expression for K_{2}
• The expression is inverted because the reaction is reversed

As a result, for doubled coefficients, K_{2} relates to K_{1} as follows:
\[ K_{2} = \frac{1}{K_{1}^{2}} \]

Understanding reaction coefficients ensures you can correctly relate equilibrium expressions for different reactions and determine their constants accurately.