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°.
Ch.19 - Chemical Thermodynamics
Chapter 19, Problem 64
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
Verified step by step guidance1
Step 1: Convert the ΔH° and ΔS° values to the same units. Since ΔH° is given in kJ and ΔS° is given in J/K, convert ΔH° to J by multiplying by 1000. So, ΔH° = -844 kJ * 1000 = -844000 J.
Step 2: Use the Gibbs free energy equation ΔG° = ΔH° - TΔS° to calculate ΔG° at 298 K. Remember to use the Kelvin temperature scale in this equation.
Step 3: If ΔG° is positive, the reaction is not spontaneous under standard conditions at 298 K. If ΔG° is negative, the reaction is spontaneous under standard conditions at 298 K.
Step 4: If the reaction is not spontaneous at 298 K (i.e., if ΔG° is positive), you can find the temperature at which the reaction becomes spontaneous by setting ΔG° to zero and solving for T. This gives the equation 0 = ΔH° - TΔS°.
Step 5: Solve the equation from step 4 for T to find the temperature at which the reaction becomes spontaneous. Remember to convert your answer back to degrees Celsius if necessary by subtracting 273.15 from your answer in Kelvin.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Gibbs Free Energy (ΔG)
Gibbs Free Energy (ΔG) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. It 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 that a reaction is spontaneous under the given conditions.
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Gibbs Free Energy of Reactions
Enthalpy (ΔH)
Enthalpy (ΔH) is a measure of the total heat content of a system. It reflects the energy required to break and form bonds during a chemical reaction. A negative ΔH value, as in the given reaction, indicates that the reaction is exothermic, meaning it releases heat to the surroundings, which can favor spontaneity.
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Entropy (ΔS)
Entropy (ΔS) is a measure of the disorder or randomness in a system. In thermodynamics, an increase in entropy is generally associated with the spontaneity of a process. A negative ΔS value, as provided in the question, suggests that the reaction leads to a decrease in disorder, which can hinder spontaneity unless compensated by a sufficiently negative ΔH at higher temperatures.
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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.
(c) N2F4(g) ⟶ 2 NF2(g) ΔH° = 85 kJ; ΔS° = 198 J/K
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.