For the formation of which compound in Ellingham diagram ΔG° becomes more and more negative with increase in temperature ?
- CO
- FeO
- ZnO
- Cu2O
The Correct Option is A
Approach Solution - 1
The question asks which compound in the Ellingham diagram has a negative change in Gibbs free energy (\( \Delta G \)) that becomes more and more negative with increasing temperature.
Ellingham Diagram Explanation:
The Ellingham diagram plots the change in Gibbs free energy (\( \Delta G \)) for various reactions as a function of temperature. The slope of the line represents how the favorability of the reaction changes with temperature. - If the slope is negative, the reaction becomes more favorable (more negative \( \Delta G \)) at higher temperatures. - If the slope is positive, the reaction becomes less favorable as the temperature increases.
Analysis of Options:
-(A) CO: Carbon monoxide is formed by reducing carbon dioxide. In the Ellingham diagram, the formation of CO has a **negative slope**, meaning its formation becomes more favorable as the temperature increases, and \( \Delta G \) becomes more negative at higher temperatures.
- (B) FeO: The formation of iron oxide (FeO) typically has a **positive slope**, meaning its formation becomes less favorable with increasing temperature.
- (C) ZnO: Zinc oxide formation also has a **positive slope**, so it becomes less favorable as the temperature increases.
- (D) Cu₂O: Copper(I) oxide formation similarly shows a **positive slope**, meaning it becomes less favorable at higher temperatures.
Conclusion:
The compound whose formation in the Ellingham diagram becomes more negative with increasing temperature is CO, as it has a negative slope.
Correct Answer: Option (A): CO
Approach Solution -2
In an Ellingham diagram, the change in the Gibbs free energy (ΔGo) is plotted against temperature. The slope of the graph is an indicator of the entropy change (ΔSo) for the reaction. A negative slope indicates that the reaction becomes more favorable at higher temperatures.
The reaction for the formation of CO from carbon involves a decrease in entropy (ΔSo is negative), which means the ΔGo becomes more negative with increasing temperature.
Therefore, in the Ellingham diagram, the line for the formation of CO shows a decreasing value of ΔGo as temperature increases, indicating that CO is increasingly favored at higher temperatures.
Hence, the correct answer is (A): CO.
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