Question

Elemental sulfur is formed by the reaction of zinc sulfide with oxygen: $$2 \mathrm{ZnS}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)+2 \mathrm{~S}(s)$$ (a) If \(\Delta H^{\circ}=-289.0 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta S^{\circ}=-169.4 \mathrm{~J} / \mathrm{K}\), what is \(\Delta S_{\text {total }}\) for this reaction? Is the reaction spontaneous under standard-state conditions at \(25^{\circ} \mathrm{C} ?\) (b) At what temperature, if any, will the reaction become nonspontaneous?

Short answer

The reaction is spontaneous at 25°C with \( \Delta S_{\text{total}} = 799.9 \, \text{J/K} \) and becomes non-spontaneous above 1706.6 K.

Step-by-step solution

Define the Given Variables

We have the given values:- \( \Delta H^{\circ} = -289.0 \, \text{kJ/mol} \)- \( \Delta S^{\circ} = -169.4 \, \text{J/K} \)Note: We need to convert \( \Delta H^{\circ} \) to the same units as \( \Delta S^{\circ} \), which are J, by multiplying by 1000: \( \Delta H^{\circ} = -289.0 \, \text{kJ/mol} \times 1000 = -289000 \, \text{J/mol} \).

Calculate ΔG°

To find the spontaneity under standard state conditions, use the Gibbs free energy equation:\[\Delta G^{\circ} = \Delta H^{\circ} - T \cdot \Delta S^{\circ}\]Substitute the given values at \( 25^{\circ} \text{C} = 298.15 \text{K} \):\[\Delta G^{\circ} = -289000 \, \text{J/mol} - (298.15 \, \text{K} \times (-169.4 \, \text{J/K}))\]\[\Delta G^{\circ} = -289000 + 50507.7 = -238492.3 \, \text{J/mol}\]

Determine Spontaneity at 25°C

A reaction is spontaneous if \( \Delta G^{\circ} < 0 \). Since \( \Delta G^{\circ} = -238492.3 \text{ J/mol} \), the reaction is spontaneous at 25°C.

Calculate ΔS_total

The total entropy change \( \Delta S_{\text{total}} \) is calculated using the relation:\[ \Delta S_{\text{total}} = \frac{-\Delta G^{\circ}}{T} \] Substitute the values:\[ \Delta S_{\text{total}} = \frac{-(-238492.3 \, \text{J/mol})}{298.15 \, \text{K}} = 799.9 \, \text{J/K} \]

Find Temperature for Non-Spontaneity

A reaction becomes non-spontaneous when \( \Delta G^{\circ} > 0 \). Set \( \Delta G^{\circ} = 0 \) to find the crossover temperature:\[ 0 = \Delta H^{\circ} - T \cdot \Delta S^{\circ} \]Rearrange to find \( T \):\[ T = \frac{\Delta H^{\circ}}{\Delta S^{\circ}} = \frac{-289000 \, \text{J/mol}}{-169.4 \, \text{J/K}} \approx 1706.6 \, \text{K} \]

Conclusion on Spontaneity with Temperature

At temperatures above 1706.6 K, the reaction will become non-spontaneous, as \( \Delta G^{\circ} \) will be greater than zero.

Key concepts

Gibbs Free Energy

Gibbs free energy (Gibbs energy) is a valuable concept in thermodynamics that helps predict the spontaneity of a reaction. Imagine it as a measure of the maximum useful work obtainable from a thermodynamic process at constant temperature and pressure. Spontaneous reactions tend to occur without external intervention. Whether a reaction is spontaneous is determined using the Gibbs free energy change (\( \Delta G \)).

The formula to calculate Gibbs free energy is:

• \( \Delta G = \Delta H - T \Delta S \)

Here, \( \Delta H \) represents the enthalpy change, \( T \) is the temperature in Kelvin, and \( \Delta S \) is the entropy change.

A key takeaway is:

• If \( \Delta G \) is negative, the reaction is spontaneous.
• If \( \Delta G \) is zero, the system is at equilibrium.
• If \( \Delta G \) is positive, the reaction is non-spontaneous.

In the exercise, we calculated \( \Delta G \) as \(-238492.3 \), which indicates the process is spontaneous at standard conditions.

Entropy Change

Entropy, symbolized as \( S \), describes the amount of disorder or randomness in a system. The change in entropy (\( \Delta S \)) influences the spontaneity of a reaction. It reflects how energy spreads or disperses during a chemical process.

In the given reaction, the entropy change \( \Delta S^{\circ} \) is \(-169.4 \, \text{J/K} \), suggesting a decrease in system disorder.

When considering entropy:

• A positive \( \Delta S \) usually indicates an increase in disorder, which often supports spontaneous reactions.
• A negative \( \Delta S \) suggests a decrease in disorder, potentially requiring external energy to proceed spontaneously.

Despite the decrease in entropy, the reaction can still be spontaneous if favored by other factors, such as a high negative enthalpy (\( \Delta H \)). Thus, entropy alone doesn't determine spontaneity; it must be considered alongside other factors like enthalpy and temperature.

Enthalpy

Enthalpy, represented as \( H \), can be thought of as the total heat content of a system. The enthalpy change (\( \Delta H \)) indicates whether a reaction releases or absorbs heat under constant pressure.

In the exercise, \( \Delta H^{\circ} \) is \(-289.0 \, \text{kJ/mol} \), meaning the reaction is exothermic and releases heat. An exothermic process typically helps a reaction become spontaneous because the excess energy doesn't require external input.

Key insights about enthalpy:

• Negative \( \Delta H \): The reaction is exothermic and usually favorable for spontaneity.
• Positive \( \Delta H \): The reaction is endothermic and less likely spontaneous without accommodating conditions like temperature increase.

Enthalpy isn't the sole factor in determining spontaneity. It works alongside entropy and temperature to influence the system's overall Gibbs free energy.