Problem 75 - Chemistry: The Central Science in SI Units, Global Edition

Question

Indicate whether \(\Delta G\) increases, decreases, or stays the same for each of the following reactions as the partial pressure of \(\mathrm{O}_{2}\) is increased: (a) \(\mathrm{HgO}(s) \longrightarrow \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c)

Step-by-Step Solution

Answer

In summary, for an increase in the partial pressure of O₂: - In reaction (a), ∆G increases. - In reaction (b), ∆G decreases. - In reaction (c), ∆G decreases.

Step 1: A) Analyze Reaction (a)

The reaction in this problem is: \[ \mathrm{HgO}(s) \longrightarrow \mathrm{Hg}(l) + \mathrm{O}_{2}(g) \] As the partial pressure of O₂ is increased, the only species in the reaction affected is O₂ as it is the only species in gaseous form. The reaction quotient Q for this reaction is: \[ Q = \frac{P_{\mathrm{O}_2}}{1} \] As the partial pressure of O₂ is increased, Q also increases.

Step 2: B) Determine the direction of Reaction (a)

For a reaction to be spontaneous, the Gibbs free energy change must be negative: \[ \Delta G < 0 \] If Q is less than the equilibrium constant K (Q < K), the forward reaction is spontaneous (favoring products), and ∆G will be negative. As ∆G becomes less negative, it means it's increasing. Since Q is increasing, we infer that, under the given conditions, ∆G will increase.

Step 3: C) Analyze Reaction (b)

The reaction in this problem is: \[ 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g) \] The reaction quotient Q for this reaction is: \[ Q = \frac{P_{\mathrm{SO_3}}^2}{P_{\mathrm{SO_2}}^2 \times P_{\mathrm{O_2}}} \] As the partial pressure of O₂ is increased, the denominator of Q increases.

Step 4: D) Determine the direction of Reaction (b)

If the denominator of Q increases, Q decreases. If Q < K, the forward reaction is spontaneous (favoring products), and ∆G will be negative. Since Q is decreasing, we infer that, under the given conditions, ∆G will decrease.

Step 5: E) Analyze Reaction (c)

The reaction in this problem is: \[ 2 \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \] The reaction quotient Q for this reaction is: \[ Q = \frac{1}{P_{\mathrm{H_2}}^2 \times P_{\mathrm{O_2}}} \] As the partial pressure of O₂ is increased, the denominator of Q increases.

Step 6: F) Determine the direction of Reaction (c)

If the denominator of Q increases, Q decreases. If Q < K, the forward reaction is spontaneous (favoring products), and ∆G will be negative. Since Q is decreasing, we infer that, under the given conditions, ∆G will decrease.

Step 7: G) Conclusion:

In summary, for an increase in the partial pressure of O₂: - In reaction (a), ∆G increases. - In reaction (b), ∆G decreases. - In reaction (c), ∆G decreases.

Key Concepts

Reaction Quotient

The reaction quotient, denoted as \( Q \), is a way to determine how a reaction is progressing at a given moment before it reaches equilibrium. It's a measure that compares the ratio of the concentrations of products to reactants at any point in a reaction.
For any general reaction \( aA + bB \rightarrow cC + dD \), the reaction quotient is given as:

• \( Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)

With primary focus on gases, we typically measure \( Q \) using partial pressures instead of concentrations, substituting \( P \) for pressure.When the pressure of a gas like \( O_2 \) is changed, \( Q \) will adjust because it directly involves the pressures of reactants and products. In the example of reaction \( a \), as \( P_{O_2} \) increases, so does \( Q \), since \( O_2 \) is a product in the reaction. In other reactions like \( b \) and \( c \), increasing \( P_{O_2} \) affects \( Q \) differently because \( O_2 \) is a reactant. A change in \( Q \) reflects an adjustment the system needs to attain equilibrium.

Equilibrium Constant

The equilibrium constant, \( K \), is a special value of the reaction quotient when the reaction is at equilibrium. It describes the ratio of the concentrations or partial pressures of products and reactants at equilibrium.Each reaction at a specific temperature has its own \( K \), which remains constant.

• For \( aA + bB \rightarrow cC + dD \), \( K \) is defined as: \( K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \).

The relationship between \( Q \) and \( K \) shows the direction in which the reaction needs to proceed to attain equilibrium:

• If \( Q < K \), the reaction will proceed forward (to the right), favoring the formation of products.
• If \( Q > K \), the reaction will proceed backward (to the left), favoring the formation of reactants.
• If \( Q = K \), the reaction is at equilibrium.

Changes in \( Q \) prompted by alterations in partial pressures lead to adjustments to return the system to its \( K \) value, indicating changes in \( \Delta G \), the Gibbs Free Energy.".

Spontaneity of Reactions

Spontaneity in chemical reactions is determined by the Gibbs Free Energy change \( (\Delta G) \). This value helps predict whether a reaction will occur under specific conditions without external input.- If \( \Delta G < 0 \), the reaction is spontaneous in the forward direction.- If \( \Delta G > 0 \), the reaction is non-spontaneous in the forward direction but spontaneous in the reverse.- If \( \Delta G = 0 \), the reaction is at equilibrium.The relationship between \( \Delta G \), \( Q \), and \( K \) is given by the equation:

• \( \Delta G = \Delta G^0 + RT\ln(Q) \)

Where \( \Delta G^0 \) is the standard Gibbs Free Energy change, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.In reactions (b) and (c) from the exercise, increasing the partial pressure of \( O_2 \) affects \( Q \) directly. Depending on how \( Q \) compares to \( K \), it indicates whether the Gibbs free energy change \( \Delta G \) becomes more negative or positive, hence determining if the reaction spontaneity shifts or not.