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)

Short 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-by-step solution

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.

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.

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.

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.

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.

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.

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.