Question

For the formation of which compound in Ellingham diagram ΔG° becomes more and more negative with increase in temperature ?

• A. CO
• B. FeO
• C. ZnO
• D. Cu₂O

Answer 1

The Correct Option is A

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

Answer 2

In an Ellingham diagram, the change in the Gibbs free energy (ΔG^o) is plotted against temperature. The slope of the graph is an indicator of the entropy change (ΔS^o) 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 (ΔS^o is negative), which means the ΔG^o becomes more negative with increasing temperature. 

Therefore, in the Ellingham diagram, the line for the formation of CO shows a decreasing value of ΔG^o as temperature increases, indicating that CO is increasingly favored at higher temperatures. 

Hence, the correct answer is (A): CO.