Problem 23 - Chemistry The Molecular Science

Question

Consider this reaction at \(122^{\circ} \mathrm{C}\) : $$ 2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) $$ (a) Write an equilibrium constant expression for the reaction and call the constant \(K_{1}\). (b) Write an equilibrium constant expression for the decomposition of \(1 \mathrm{~mol} \mathrm{SO}_{3}\) to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) and call the constant \(K_{2}\) (c) Relate \(K_{1}\) and \(K_{2}\).

Step-by-Step Solution

Answer

(a) \(K_1 = \frac{[\mathrm{SO}_2]^2 [\mathrm{O}_2]}{[\mathrm{SO}_3]^2}\); (b) \(K_2 = \frac{[\mathrm{SO}_2][\mathrm{O}_2]^{1/2}}{[\mathrm{SO}_3]}\); (c) \(K_1 = (K_2)^2\).

Step 1: Writing the Equilibrium Constant Expression for the Given Reaction (K1)

The equilibrium constant expression for a reaction is written using the concentrations of the products raised to the power of their coefficients, divided by the concentrations of the reactants raised to the power of their coefficients. For the reaction \( 2 \mathrm{SO}_{3}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g}) + \mathrm{O}_{2}(\mathrm{~g}) \), the expression for \( K_1 \) is:\[K_1 = \frac{[\mathrm{SO}_2]^2 [\mathrm{O}_2]}{[\mathrm{SO}_3]^2}\]

Step 2: Writing the Equilibrium Constant Expression for the Decomposition of 1 mol SO3 (K2)

For the decomposition of \( 1 \mathrm{~mol} \mathrm{SO}_{3} \) to \( \mathrm{SO}_{2} \) and \( \mathrm{O}_{2} \), the reaction is: \( \mathrm{SO}_{3} \rightleftharpoons \mathrm{SO}_{2} + \frac{1}{2} \mathrm{O}_{2} \). The equilibrium constant expression \( K_2 \) for this reaction is:\[K_2 = \frac{[\mathrm{SO}_2][\mathrm{O}_2]^{1/2}}{[\mathrm{SO}_3]}\]

Step 3: Relating K1 and K2

The reaction described by \( K_1 \) involves twice the number of each species compared to the reaction described by \( K_2 \). Therefore, the equilibrium constants are related as follows:\[K_1 = (K_2)^2\]This is because the reaction for \( K_1 \) is effectively doubling the reaction for \( K_2 \).

Key Concepts

Chemical Equilibrium

Chemical equilibrium occurs when a chemical reaction reaches a state where the forward and reverse reactions occur at the same rate.
This results in the concentrations of the reactants and products remaining constant over time.

In any reversible reaction, such as the one we are considering with sulfur trioxide (\( \text{SO}_3 \)), sulfur dioxide (\( \text{SO}_2 \)), and oxygen (\( \text{O}_2 \)), equilibrium is a critical concept.
It implies that the molecules are continuing to react, yet no net change in concentration is observed.

To represent chemical equilibrium quantitatively, we use the equilibrium constant, denoted as \( K \).

• If \( K \) is large, the products are favored, indicating the reaction goes nearly to completion.
• If \( K \) is small, the reactants are favored, indicating little reaction occurs.

Understanding and calculating \( K \) helps us predict the direction and extent of chemical reactions.

Le Chatelier's Principle

Le Chatelier's Principle is vital for understanding how a system at equilibrium responds to external changes.
When a system in equilibrium is disturbed, it will adjust in such a way as to counteract that disturbance, thus re-establishing equilibrium.

For example, in the decomposition of \( \text{SO}_3 \) to \( \text{SO}_2 \) and \( \text{O}_2 \), if we increase the temperature, equilibrium will shift to favor the endothermic direction, which is forward for this reaction.
Similarly, if we remove \( \text{SO}_3 \), the system will shift to the left, favoring the formation of more \( \text{SO}_3 \).

This principle is crucial, as it explains how changes in concentration, temperature, and pressure can affect chemical equilibria.

• Adding more reactants or removing products generally shifts the equilibrium to the right.
• Increasing the pressure of a gaseous reaction will shift the equilibrium to the side with fewer moles of gas.

By applying Le Chatelier's Principle, we can manipulate conditions to obtain more of desired products.

Thermodynamics

Thermodynamics is the branch of physical science that deals with the relations between heat and other forms of energy.
It provides insight into whether a reaction is favorable and how energy changes during chemical processes.

In the context of chemical equilibrium, thermodynamics helps us understand the energy dynamics of reactions like \( \text{SO}_3 \rightleftharpoons \text{SO}_2 + \text{O}_2 \).
This reaction's favorability can depend heavily on whether it is exotheric (releases energy) or endothermic (absorbs energy).

Thermodynamic parameters, such as Gibbs free energy (\( \Delta G \)), are crucial in determining equilibrium.

• If \( \Delta G \) is negative, the reaction is spontaneous, favoring the products.
• If \( \Delta G \) is positive, the reactants are favored unless external energy is supplied.

At equilibrium, \( \Delta G = 0 \), which means there is no net change in energy.
Understanding these concepts helps predict reaction behavior under various conditions.