A particular constant-pressure reaction is barely spontaneous at 390 K. The enthalpy change for the reaction is +23.7 kJ. Estimate ΔS for the reaction.

Verified step by step guidance

Step 1: Recall the Gibbs free energy equation for a reaction at constant pressure: \( \Delta G = \Delta H - T \Delta S \).

Step 2: Understand that a reaction is barely spontaneous when \( \Delta G = 0 \). Therefore, set \( \Delta G = 0 \) in the equation.

Step 3: Substitute the given values into the equation: \( 0 = 23.7 \text{ kJ} - 390 \text{ K} \times \Delta S \).

Step 4: Rearrange the equation to solve for \( \Delta S \): \( \Delta S = \frac{23.7 \text{ kJ}}{390 \text{ K}} \).

Step 5: Convert \( \Delta S \) from kJ/K to J/K by multiplying by 1000, since 1 kJ = 1000 J.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gibbs Free Energy

Gibbs Free Energy (G) is a thermodynamic potential that helps predict the spontaneity of a reaction at constant temperature and pressure. A reaction is spontaneous when the change in Gibbs Free Energy (ΔG) is negative. The relationship between ΔG, enthalpy change (ΔH), and entropy change (ΔS) is given by the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin.

Enthalpy Change

Enthalpy change (ΔH) is the heat content change of a system at constant pressure. A positive ΔH indicates that the reaction is endothermic, meaning it absorbs heat from the surroundings. In the context of the question, the given ΔH of +23.7 kJ suggests that the reaction requires energy input, which influences its spontaneity and the calculation of ΔS.

Entropy Change

Entropy change (ΔS) is a measure of the disorder or randomness in a system. In thermodynamics, a positive ΔS indicates an increase in disorder, which can favor spontaneity. To estimate ΔS for the reaction, we can rearrange the Gibbs Free Energy equation to ΔS = (ΔH - ΔG) / T, using the fact that the reaction is barely spontaneous (ΔG ≈ 0) at the given temperature.

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Related Practice

Textbook Question

Classify each of the following reactions as one of the four possible types summarized in Table 19.3: (i) spontanous at all temperatures; (ii) not spontaneous at any temperature; (iii) spontaneous at low T but not spontaneous at high T; (iv) spontaneous at high T but not spontaneous at low T.

(a) N₂(g) + 3 F₂(g) → 2 NF₃₍g) ΔH° = -249 kJ; ΔS° = -278 J/K

(b) N₂(g) + 3 Cl₂(g) → 2 NCl₃(g) ΔH° = 460 kJ; ΔS° = -275 J/K

Textbook Question

Classify each of the following reactions as one of the four possible types summarized in Table 19.3: (i) spontaneous at all temperatures; (ii) not spontaneous at any temperature; (iii) spontaneous at low T but not spontaneous at high T; (iv) spontaneous at high T but not spontaneous at low T.

Textbook Question

From the values given for ΔH° and ΔS°, calculate ΔG° for each of the following reactions at 298 K. If the reaction is not spontaneous under standard conditions at 298 K, at what temperature (if any) would the reaction become spontaneous?

a. 2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g) ΔH° = −844 kJ; ΔS° = −165 J/K

b. 2 POCl₃(g) → 2 PCl₃(g) + O₂(g) ΔH° = 572 kJ; ΔS° = 179 J/K

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Textbook Question

A certain constant-pressure reaction is barely nonspontaneous at 45 °C. The entropy change for the reaction is 72 J/K. Estimate ΔH.

Textbook Question

Reactions in which a substance decomposes by losing CO are called decarbonylation reactions. The decarbonylation of acetic acid proceeds according to: CH₃COOH(l) → CH₃OH(g) + CO(g) By using data from Appendix C, calculate the minimum temperature at which this process will be spontaneous under standard conditions. Assume that ΔH° and ΔS° do not vary with temperature.