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

Verified step by step guidance

Identify the Gibbs free energy equation: ΔG° = ΔH° - TΔS°. This equation helps determine the spontaneity of a reaction.

For reaction (a), N2(g) + 3 F2(g) → 2 NF3(g), note that ΔH° = -249 kJ and ΔS° = -278 J/K. Convert ΔS° to kJ by dividing by 1000, resulting in ΔS° = -0.278 kJ/K.

Substitute the values into the Gibbs free energy equation for reaction (a): ΔG° = -249 kJ - T(-0.278 kJ/K). Analyze the sign of ΔG° at different temperatures to determine spontaneity.

For reaction (b), N2(g) + 3 Cl2(g) → 2 NCl3(g), note that ΔH° = 460 kJ and ΔS° = -275 J/K. Convert ΔS° to kJ by dividing by 1000, resulting in ΔS° = -0.275 kJ/K.

Substitute the values into the Gibbs free energy equation for reaction (b): ΔG° = 460 kJ - T(-0.275 kJ/K). Analyze the sign of ΔG° at different temperatures to determine spontaneity.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

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. The change in Gibbs Free Energy (ΔG) is calculated using the equation ΔG = ΔH - TΔS, where ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy. A negative ΔG indicates a spontaneous reaction, while a positive ΔG suggests non-spontaneity.

Enthalpy and Entropy

Enthalpy (ΔH) is a measure of the total heat content of a system, while entropy (ΔS) quantifies the degree of disorder or randomness in a system. In chemical reactions, ΔH can be either positive (endothermic) or negative (exothermic), and ΔS can also be positive or negative depending on the change in disorder. The interplay between these two factors determines the spontaneity of a reaction across different temperature ranges.

Temperature Dependence of Spontaneity

The spontaneity of a reaction can depend on temperature due to the relationship between enthalpy and entropy. Reactions that are exothermic (negative ΔH) and have positive entropy (positive ΔS) are spontaneous at all temperatures. Conversely, reactions that are endothermic with negative entropy are non-spontaneous at all temperatures. The temperature can shift the balance for reactions that are spontaneous at low or high temperatures, depending on the signs of ΔH and ΔS.

Recommended video:

Guided course

01:24

Kw Temperature Dependence

Related Practice

Textbook Question

Using data from Appendix C, calculate the change in Gibbs free energy for each of the following reactions. In each case, indicate whether the reaction is spontaneous at 298 K under standard conditions.

(a) 2 Ag(s) + Cl2(g) → 2 AgCl(s)

(b) P₄O₁₀(s) + 16 H₂(g) → 4 PH₃(g) + 10 H₂O(g)

(d) 2 H₂O₂(l) → 2 H₂O(l) + O₂(g)

Textbook Question

Sulfur dioxide reacts with strontium oxide as follows: SO₂(g) + SrO(g) → SrSO₃(s) (a) Without using thermochemical data, predict whether ΔG° for this reaction is more negative or less negative than ΔH°.

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

views