Coupled Reactions And Thermodynamic Feasibility
The reduction of iron(III) oxide by carbon monoxide proceeds via the overall reaction (Fe_2O_3(s) + 3CO(g) \rightarrow 2Fe(s) + 3CO_2(g)). Given (\Delta H^\circ = -28.5) kJ/mol and (\Delta S^\circ = +52.0) J/mol·K, at what minimum temperature (approximately) does this reaction become non-spontaneous at standard conditions?
Select the correct option:
Solution
The reaction is spontaneous at all temperatures
Spontaneity is governed by (\Delta G = \Delta H - T\Delta S). To find where the reaction ceases to be spontaneous, we set (\Delta G = 0): (T = \Delta H / \Delta S). However, this crossover analysis applies when (\Delta H) and (\Delta S) have opposite signs — in that case a crossover temperature exists. Here (\Delta H = -28.5) kJ/mol (negative, exothermic) and (\Delta S = +52.0) J/mol·K (positive, entropy increases). When both (\Delta H < 0) and (\Delta S > 0), the Gibbs free energy (\Delta G = \Delta H - T\Delta S) is negative for ALL values of (T > 0): the negative (\Delta H) already drives (\Delta G < 0), and the positive (T\Delta S) term makes (\Delta G) even more negative at higher temperatures. Therefore the reaction is spontaneous at all temperatures. Option B (above 548 K non-spontaneous) would only apply if both terms had the same sign. Option C (above 1235 K) uses the wrong formula application. Option D contradicts the clearly negative (\Delta G) at all temperatures. This is a discriminating JEE Advanced question that tests whether students correctly identify the regime where no crossover exists. Plausibility check: the reduction of iron oxide by CO is the basis of blast furnace ironmaking, which operates at hundreds to over a thousand Kelvin — confirming spontaneity across a wide temperature range.
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About This Question
- Subject
- chemistry
- Chapter
- chemical thermodynamics
- Topic
- coupled reactions and thermodynamic feasibility
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
The reaction is spontaneous at all temperatures
Spontaneity is governed by (\Delta G = \Delta H - T\Delta S). To find where the reaction ceases to be spontaneous, we set (\Delta G = 0): (T = \Delta H / \Delta S). However, this crossover analysis applies when (\Delta H) and (\Delta S) have opposite signs — in that case a crossover temperature exists. Here (\Delta H = -28.5) kJ/mol (negative, exothermic) and (\Delta S = +52.0) J/mol·K (positive, entropy increases). When both (\Delta H < 0) and (\Delta S > 0), the Gibbs free energy (\Delta G = \Delta H - T\Delta S) is negative for ALL values of (T > 0): the negative (\Delta H) already drives (\Delta G < 0), and the positive (T\Delta S) term makes (\Delta G) even more negative at higher temperatures. Therefore the reaction is spontaneous at all temperatures. Option B (above 548 K non-spontaneous) would only apply if both terms had the same sign. Option C (above 1235 K) uses the wrong formula application. Option D contradicts the clearly negative (\Delta G) at all temperatures. This is a discriminating JEE Advanced question that tests whether students correctly identify the regime where no crossover exists. Plausibility check: the reduction of iron oxide by CO is the basis of blast furnace ironmaking, which operates at hundreds to over a thousand Kelvin — confirming spontaneity across a wide temperature range.
This hard difficulty chemistry question is from the chapter chemical thermodynamics, covering the topic of coupled reactions and thermodynamic feasibility. It appeared in the 2025 exam.
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