Problem 74 - Chemistry: The Central Science

Question

Indicate whether \(\Delta G\) increases, decreases, or does not change when the partial pressure of \(\mathrm{H}_{2}\) is increased in each of the following reactions: (a) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (b) \(2 \mathrm{HBr}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) (c) \(2 \mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)\)

Step-by-Step Solution

Answer

For the given reactions, when the partial pressure of H₂(g) is increased: (a) ∆G increases in the reaction N₂(g) + 3H₂(g) → 2NH₃(g). (b) ∆G decreases in the reaction 2HBr(g) → H₂(g) + Br₂(g). (c) ∆G increases in the reaction 2H₂(g) + C₂H₂(g) → C₂H₆(g).

Step 1: Recall the relationship between ∆G, Q, and K

The Gibbs free energy change (∆G) helps us understand whether a reaction is spontaneous or not. When ∆G = 0, the system is at equilibrium, and when ∆G < 0, the reaction is spontaneous in the forward direction. Recall the relationship between ∆G, reaction quotient (Q), and equilibrium constant (K) as: \[ ∆G = RT \ln (\frac{Q}{K})\] where R is the gas constant, T is the temperature, Q is the reaction quotient, and K is the equilibrium constant.

Step 2: Use Le Chatelier's Principle

According to Le Chatelier's principle, if a change is imposed on a system at equilibrium, the system will adjust itself to counteract that change and reestablish equilibrium. When the partial pressure of H₂ is increased in the given reactions, the system will try to consume more H₂ gas to reestablish equilibrium. Let's analyze the reactions to predict how ∆G will change. (a) N₂(g) + 3H₂(g) → 2NH₃(g)

Step 3: Analyze the forward direction of Reaction (a)

An increase in the partial pressure of H₂(g) will cause the system to produce more NH₃(g) to consume the excess H₂(g). This will increase the number of moles of the products and decrease the number of moles of the reactants, which leads to an increase in the reaction quotient (Q). According to ∆G = RTln(Q/K), as Q increases, ∆G will increase.

Step 4: ∆G for Reaction (a)

For Reaction (a), an increase in the partial pressure of H₂(g) causes ∆G to increase. (b) 2HBr(g) → H₂(g) + Br₂(g)

Step 5: Analyze the forward direction of Reaction (b)

An increase in the partial pressure of H₂(g) will cause the system to consume more H₂(g) by shifting the equilibrium towards the reactants, which means it will convert the products back into the reactants. This will decrease the number of moles of the products and increase the number of moles of the reactants, which leads to a decrease in the reaction quotient (Q). According to ∆G = RTln(Q/K), as Q decreases, ∆G will decrease.

Step 6: ∆G for Reaction (b)

For Reaction (b), an increase in the partial pressure of H₂(g) causes ∆G to decrease. (c) 2H₂(g) + C₂H₂(g) → C₂H₆(g)

Step 7: Analyze the forward direction of Reaction (c)

An increase in the partial pressure of H₂(g) will cause the system to produce more C₂H₆(g) to consume the excess H₂(g). This will increase the number of moles of the products and decrease the number of moles of the reactants, which leads to an increase in the reaction quotient (Q). According to ∆G = RTln(Q/K), as Q increases, ∆G will increase.

Step 8: ∆G for Reaction (c)

For Reaction (c), an increase in the partial pressure of H₂(g) causes ∆G to increase.