Question

For the following endothermic reaction at equilibrium: $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ which of the following changes will increase the value of \(K ?\) a. increasing the temperature b. decreasing the temperature c. removing \(\mathrm{SO}_{3}(g)\) (constant \(T )\) d. decreasing the volume (constant \(T )\) e. adding \(\operatorname{Ne}(g)(\text { constant } T)\) f. adding \(\mathrm{SO}_{2}(g)(\text { constant } T)\) g. adding a catalyst (constant T)

Short answer

In conclusion, the only change that will increase the value of \(K\) in this endothermic reaction is: a. increasing the temperature

Step-by-step solution

Understanding Le Chatelier's Principle

Le Chatelier's Principle states that if a change is applied to a system in equilibrium, the system will adjust itself to minimize the effect of the change and restore the equilibrium. It applies to temperature, concentration, volume, and pressure changes in a chemical equilibrium, but not to the addition of a catalyst.

Evaluate the effect of increasing temperature

For an endothermic reaction, increased temperature will shift the equilibrium to the side with more products (right side). This leads to an increase in the number of moles of \(\mathrm{SO}_2(g)\) and \(\mathrm{O}_2(g)\) and therefore results in an increase in the value of \(K\). So, option a. increasing the temperature is correct.

Evaluate the effect of decreasing temperature

For an endothermic reaction, a decrease in temperature will shift the equilibrium to the side with more reactants (left side). This will lead to a decrease in the value of \(K\). Option b. decreasing the temperature is incorrect.

Evaluate the effect of removing SO\(_3(g)\)

If we remove \(\mathrm{SO}_3(g)\), the reaction will shift towards the left side to offset the decrease in reactant concentration. However, this change will not affect the value of the equilibrium constant \(K\). The concentrations of the species will change, but the ratio at equilibrium will remain the same. Option c. removing SO\(_3(g)\) is incorrect.

Evaluate the effect of decreasing volume

Decreasing the volume causes an increase in pressure, and the equilibrium will shift towards the side with lesser moles of gas. In this reaction, both sides have equal number of moles of gas, so the change in volume will not affect the value of the equilibrium constant \(K\). Option d. decreasing the volume is incorrect.

Evaluate the effect of adding Ne(g)

Adding a noble gas (such as Ne) will not affect the equilibrium, as it does not participate in the reaction. Thus, the value of \(K\) remains unchanged. Option e. adding Ne(g) is incorrect.

Evaluate the effect of adding SO\(_2(g)\)

Adding \(\mathrm{SO}_2(g)\) (constant T) will shift the equilibrium to the left to offset the increase in product concentration. However, just like in step 4, this change does not affect the value of \(K\), only the concentration of species. Option f. adding SO\(_2(g)\) is incorrect.

Evaluate the effect of adding a catalyst

Adding a catalyst speeds up both the forward and reverse reactions, but it does not affect the equilibrium position or the value of \(K\). Option g. adding a catalyst is incorrect. In conclusion, the only change that will increase the value of \(K\) in this endothermic reaction is: a. increasing the temperature

Key concepts

Endothermic Reaction

An endothermic reaction is a type of chemical reaction that absorbs energy from its surroundings. This energy is typically in the form of heat. Because these reactions require heat to proceed, they tend to feel cold to the touch when they occur. For this type of reaction, the heat is considered one of the reactants.
In the given chemical equation, \(2 \mathrm{SO}_3(g) \rightleftharpoons 2 \mathrm{SO}_2(g)+\mathrm{O}_2(g)\), the reaction absorbs heat as it converts sulfur trioxide (\(\mathrm{SO}_3\)) to sulfur dioxide (\(\mathrm{SO}_2\)) and oxygen (\(\mathrm{O}_2\)).

• Endothermic reactions are characterized by a positive change in enthalpy (\(\Delta H > 0\)).
• These reactions are generally driven by an increase in temperature, effectively supplying the energy (heat) required to progress the reaction towards products.
• When temperature increases in such a system, the equilibrium shifts to produce more products, as dictated by Le Chatelier's Principle.

This shift towards the products increases the concentration of \(\mathrm{SO}_2\) and \(\mathrm{O}_2\), and results in an increased value of the equilibrium constant, \(K\).

Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a value that expresses the ratio of the concentration of the products to the concentration of the reactants of a reversible reaction at equilibrium. Each concentration is typically raised to the power of its coefficient in the balanced chemical equation.

For the given reaction, the equilibrium expression for \(K\) is:
\[K = \frac{[\mathrm{SO}_2]^2 [\mathrm{O}_2]}{[\mathrm{SO}_3]^2}\]
This ratio remains constant at a given temperature, but can change if the temperature of the system changes.

• The equilibrium constant gives an indication of the extent of a reaction; a large \(K\) value implies more products than reactants at equilibrium.
• Conversely, a small \(K\) value suggests that reactants predominate at equilibrium.

Since the value of \(K\) varies with temperature, analyzing its behavior helps predict shifts in equilibrium. For endothermic reactions specifically, increasing the temperature results in an increase in \(K\), meaning the equilibrium favors the products more at higher temperatures.

Effect of Temperature on Equilibrium

Temperature is a crucial factor affecting chemical equilibria. In the context of Le Chatelier's Principle, a change in temperature is treated as a stress applied to the system at equilibrium. The system will respond by adjusting its position to counteract the change.
For the endothermic reaction \(2 \mathrm{SO}_3(g) \rightleftharpoons 2 \mathrm{SO}_2(g)+\mathrm{O}_2(g)\), increasing temperature shifts the equilibrium position to the right, towards the products.

• Increasing temperature supplies the necessary energy, favoring the forward reaction, thus enhancing the formation of \(\mathrm{SO}_2\) and \(\mathrm{O}_2\).
• Consequently, the equilibrium constant \(K\) increases as more products are formed.

The reverse is true if the temperature decreases. The equilibrium will shift to the left, favoring reactants. In such cases, the equilibrium constant \(K\) decreases. It's crucial to note that while changes in concentration, pressure, and volume shift the position of equilibrium, only temperature changes can alter the value of \(K\) itself.