Problem 59 - Chemistry: The Central Science in SI Units, Global Edition

Question

Using data from Appendix \(\mathrm{C}\), calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at \(298 \mathrm{~K}\) under standard conditions. (a) \(2 \mathrm{Zn}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{ZnO}(s)\) (b) \(2 \mathrm{NaBr}(s) \longrightarrow 2 \mathrm{Na}(g)+\mathrm{Br}_{2}(g)\) (c) \(\mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)+\mathrm{H}_{2}(g)\)

Step-by-Step Solution

Answer

For the given reactions, we calculated the Gibbs free energy changes (ΔG°) as follows: (a) ΔG° = -665.9 kJ/mol, reaction is spontaneous. (b) ΔG° = 483.4 kJ/mol, reaction is not spontaneous. (c) ΔG° = 36.12 kJ/mol, reaction is not spontaneous. Only reaction (a) is spontaneous at 298 K under standard conditions.

Step 1: Calculate ΔH° for each reaction

Calculate the enthalpy change (ΔH°) for each reaction by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products. Use the values from Appendix C. (a) ΔH° = [2 × (-348.0)] - [2 × 0 + 0] = -696.0 kJ/mol (b) ΔH° = [2 × 0 + 0] - [2 × (-362.1)] = 724.2 kJ/mol (c) ΔH° = [(-234.8) + 0] - [(-200.7) + (-50.45)] = 16.35 kJ/mol

Step 2: Calculate ΔS° for each reaction

Calculate the entropy change (ΔS°) for each reaction by subtracting the sum of the standard entropies of the reactants from the sum of the standard entropies of the products. Use the values from Appendix C. (a) ΔS° = [2 × (43.6)] - [2 × (41.6) + 205.1] = 187.2 - 288.3 = -101.1 J/mol·K (b) ΔS° = [2 × (154.3) + 245.5] - [2 × (102.8)] = 808.1 J/mol·K (c) ΔS° = [(65.63) + 130.7] - [(188.8) + (74.87)] = -66.34 J/mol·K

Step 3: Calculate ΔG° for each reaction

Calculate the Gibbs free energy change (ΔG°) for each reaction using the relationship: ΔG° = ΔH° - TΔS°. We are given the temperature T = 298 K. (a) ΔG° = -696.0 kJ/mol - 298 K × (-101.1 J/mol·K) / 1000 = -696.0 + 30.11 = -665.9 kJ/mol (b) ΔG° = 724.2 kJ/mol - 298 K × 808.1 J/mol·K / 1000 = 724.2 - 240.8 = 483.4 kJ/mol (c) ΔG° = 16.35 kJ/mol - 298 K × (-66.34 J/mol·K) / 1000 = 16.35 + 19.77 = 36.12 kJ/mol

Step 4: Determine the spontaneity of the reactions

A reaction is spontaneous under the given conditions if ΔG° < 0. (a) ΔG° = -665.9 kJ/mol < 0, so the reaction is spontaneous. (b) ΔG° = 483.4 kJ/mol > 0, so the reaction is not spontaneous. (c) ΔG° = 36.12 kJ/mol > 0, so the reaction is not spontaneous. So, only reaction (a) is spontaneous at 298 K under standard conditions.

Key Concepts

Enthalpy Change

Enthalpy change ( \( \Delta H^{\circ} \)) is a critical concept in understanding chemical reactions and their energy requirements. It represents the heat absorbed or released during a reaction under constant pressure. To find \( \Delta H^{\circ} \) for a reaction, one must subtract the total enthalpy of the reactants from that of the products. Here's how it works for our example reactions:
- For reaction (a), the formation of \( \text{ZnO} \) releases \(-696.0\) kJ/mol, indicating an exothermic process.- Reaction (b) absorbs \(724.2\) kJ/mol as \( \text{NaBr} \) breaks into gases, making it endothermic.- Lastly, reaction (c) has a small positive enthalpy change of \(16.35\) kJ/mol, suggesting slightly more energy is used than released.
The sign of \( \Delta H^{\circ} \)—negative for exothermic reactions and positive for endothermic ones—helps indicate whether the reaction gives off heat or requires added energy to proceed.

Entropy Change

Entropy change, \( \Delta S^{\circ} \), is a measure of disorder or randomness at the molecular level. In essence, it tells us whether a reaction results in more or less ordered products compared to the reactants. To calculate \( \Delta S^{\circ} \) for any given reaction, subtract the sum of the entropy values of the reactants from the sum of the entropy values of the products. Let's briefly see how this applies to our reactions:
- In reaction (a), \( \Delta S^{\circ} \) is \(-101.1\) J/mol·K, showing a significant decrease in randomness as \( \text{ZnO} \) solidifies from gases.- Reaction (b) drastically increases disorder, as seen by \( \Delta S^{\circ} = 808.1\) J/mol·K, from solid to gaseous stages.- Reaction (c) results in a decrease of \( \Delta S^{\circ} = -66.34\) J/mol·K since it goes from two gaseous reactants to a more ordered state.Consequently, a reaction increase in entropy (positive \( \Delta S^{\circ} \)) is generally favorable as it aligns with the natural tendency of systems to move towards disorder.

Reaction Spontaneity

Reaction spontaneity is determined by Gibbs Free Energy ( \( \Delta G^{\circ} \)), a central concept that combines enthalpy and entropy changes predicting if a reaction is self-proceeding. \( \Delta G^{\circ} \) is calculated with the formula: \[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \]where- \( \Delta H^{\circ} \) is the enthalpy change,- \( \Delta S^{\circ} \) is the entropy change, - and \( T\) is the temperature in Kelvin.A negative \( \Delta G^{\circ} \) indicates a spontaneous reaction ( \( \Delta G^{\circ} < 0 \)), while a positive value means non-spontaneity.
For example:- In reaction (a), the calculated \( \Delta G^{\circ} = -665.9\) kJ/mol suggests it spontaneously proceeds under standard conditions.- Conversely, reactions (b) and (c) are non-spontaneous, with \( \Delta G^{\circ} = 483.4\) and \( \Delta G^{\circ} = 36.12\) kJ/mol, respectively.
Understanding reaction spontaneity helps chemists predict reaction feasibility without external energy.