Problem 98 - Chemistry An Atoms First Approach

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 Ne(g) (constant \(T\) ) f. adding \(\mathrm{SO}_{2}(g)\) (constant \(T\) ) g. adding a catalyst (constant \(T\) )

Step-by-Step Solution

Answer

In conclusion, only option a (increasing the temperature) will increase the value of K for the given endothermic reaction at equilibrium.

Step 1: Analyze the effect of temperature changes

According to Le Chatelier's Principle, if an endothermic reaction is at equilibrium and the temperature is increased, the reaction will shift towards the products (right) to counteract the increase in temperature. Therefore, the value of K will increase. On the other hand, if the temperature is decreased, the reaction will shift towards the reactants (left), and thus the value of K will decrease. Hence, increasing the temperature (option a) will increase the value of K, while decreasing the temperature (option b) will not.

Step 2: Analyze the effect of concentration changes

According to Le Chatelier's principle, when the concentration of a reactant or product changes, the reaction will shift towards the side that opposes the change. Removing SO3 (option c) will cause the reaction to shift towards the reactants, but this change in the equilibrium position won't affect the value of K. Adding SO2 (option f) will cause the reaction to shift towards the products, but again, this change in the equilibrium position won't affect the value of K. Both options c and f will alter the equilibrium position, but they will not change the value of K.

Step 3: Analyze the effect of volume and pressure changes

Decreasing the volume (option d) at constant temperature will result in an increased pressure. According to Le Chatelier's Principle, the system will shift towards the side with fewer moles of gas to counteract the pressure increase. Since the number of moles of gas on both sides of the equilibrium is the same (i.e., 2 moles on both sides), the system will not shift in either direction, and the value of K will remain constant.

Step 4: Analyze the effect of adding an inert gas and catalyst

Adding an inert gas (option e) at constant temperature and volume won't affect the partial pressures or concentrations of the reactants and products, so the equilibrium position will remain unchanged, and the value of K will not be affected. Adding a catalyst (option g), on the other hand, will increase the rate of both the forward and reverse reactions equally, but it will not affect the relative amounts of reactants and products at equilibrium, and thus the value of K will remain constant. In conclusion, only option a (increasing the temperature) will increase the value of K for the given endothermic reaction at equilibrium.

Key Concepts

Endothermic Reaction

In an endothermic reaction, heat is absorbed from the surroundings as the reaction proceeds. This means that energy is needed to convert the reactants into products, making the surroundings cooler. For reactions at equilibrium, like the one given, which is \(2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g)\), heat can be considered a reactant. Thus, adding more heat pushes the equilibrium towards the product side.

• Heat is absorbed, making the reaction endothermic.
• Increasing temperature drives the reaction forward.
• Endothermic reactions typically have larger reactions with temperature changes.

Understanding this will be crucial for predicting how equilibrium shifts with temperature changes.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a value that describes the ratio of product concentrations to reactant concentrations at equilibrium for a given reaction. It is specific to a particular reaction at a specific temperature. For our reaction equation, \(K\) can be described by the formula:\[K = \frac{[\text{SO}_2]^2 [\text{O}_2]}{[\text{SO}_3]^2}\]At constant conditions of temperature, the value of \(K\) remains unchanged with modifications in concentration or pressure. Changes that are done to concentrations or pressures do not directly affect the value of \(K\); however, they may prompt a shift in equilibrium positions momentarily before returning to \(K\).

Effect of Temperature on Equilibrium

Temperature significantly impacts chemical equilibria. According to Le Chatelier's Principle, increasing the temperature of an endothermic reaction will cause the equilibrium to shift to the right, favoring product formation. This shift will result in an increase in the equilibrium constant \(K\). For our endothermic reaction, increasing the temperature effectively adds more 'reactant' in the form of heat, pushing the equilibrium:

• Towards the products (right).
• Increasing the value of \(K\).

Decreasing temperature has the opposite effect, shifting equilibrium towards reactants and reducing \(K\). Temperature changes are unique in that they can alter \(K\), unlike changes in pressure, volume, or concentration.

Effect of Concentration Changes

Shifting the concentration of reactants or products impacts the equilibrium position but not the equilibrium constant \(K\). If \(\text{SO}_3\) is removed, for instance, the equilibrium would shift left to produce more \(\text{SO}_3\) to counteract the change. Similarly, adding \(\text{SO}_2\) shifts the equilibrium right to reduce its concentration.

• The system counteracts concentration changes by shifting equilibrium.
• No change in \(K\) occurs with concentration changes.

In essence, while concentration changes can momentarily affect the position of equilibrium, the intrinsic balance described by \(K\) remains constant unless the temperature itself is altered.

Effect of Volume and Pressure Changes

For gaseous reactions like \(2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g)\), changes in volume and pressure can influence equilibrium positions according to the number of moles of gas on either side. Decreasing volume would increase pressure, which makes the system favor the side with fewer moles of gas. However, since our reaction has 3 moles of gas on both sides:

• There is no side with fewer moles of gas.
• Consequently, the equilibrium and \(K\) remain unchanged when volume or pressure is changed.

Thus, pressure and volume changes in this situation exert minimal to no effect on the equilibrium constant \(K\). This highlights the importance of examining the stoichiometry of reactions when discussing equilibrium.