The statement that is incorrect about Ellingham diagram is
(a) provides idea about the reaction rate

Q: Why does the slope of the metal oxidation line change abruptly?

When a metal undergoes a phase change (like melting or boiling), its entropy ($S$) increases significantly. Since the slope of the Ellingham diagram is $-\Delta S^\circ$, this causes a sudden change in the slope.

Q: How do we know which element acts as a reducing agent?

In the Ellingham diagram, the element whose oxidation line lies lower (more negative $\Delta G^\circ$) at a given temperature can reduce the oxide of the element whose line lies above it.

Q: Why does the carbon oxidation line slope downwards?

The reaction $2C(s) + O_2(g) \rightarrow 2CO(g)$ produces 2 moles of gas from 1 mole of gas. This increases the entropy ($\Delta S^\circ > 0$), making the slope ($-\Delta S^\circ$) negative, so the line goes down.

Q: Can a reaction be thermodynamically feasible but practically useless?

Yes! A reaction might have a very negative $\Delta G^\circ$ (highly feasible), but its activation energy might be so high that it occurs at an imperceptibly slow rate at room temperature. This is why kinetics is just as important as thermodynamics.

Q: What exactly happens at the intersection point of two lines?

At the intersection point, the $\Delta G^\circ$ for both oxidation reactions is the same. This means the $\Delta G^\circ$ for the overall displacement (reduction) reaction is exactly zero, representing an equilibrium state.

Question

The statement that is incorrect about Ellingham diagram is

Options:

(A) provides idea about the reaction rate.

(B) provides idea about free energy change.

(C) provides idea about changes in the phases during the reaction.

(D) provides idea about reduction of metal oxide.

Sigma Percentile · Accepted Answer

### The Thermodynamic Map

To truly master metallurgy, one must understand the Ellingham diagram. It is essentially a thermodynamic map that guides chemists in choosing the right reducing agent for extracting metals from their ores.

Imagine a graph where we plot the standard Gibbs free energy change, $\Delta G^\circ$, on the y-axis, against temperature, $T$, on the x-axis. Because the y-axis is literally the free energy change, the diagram inherently **provides an idea about the free energy change** of the oxidation reaction. This immediately validates one of our options.

### Decoding the Lines

When we plot the oxidation of a metal (e.g., $2M + O_2 \rightarrow 2MO$), we typically get a straight line sloping upwards. The governing equation here is $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. The slope of this line is $-\Delta S^\circ$.

If you look closely at these lines, you will occasionally notice a sudden, sharp upward bend. This abrupt change in slope occurs when the metal undergoes a phase change, such as melting or boiling. During a phase change, the entropy ($\Delta S^\circ$) increases significantly, which alters the slope. Therefore, the diagram beautifully **provides an idea about changes in the phases during the reaction**.

Furthermore, if we introduce a reducing agent like carbon, its oxidation line will often cross the metal's oxidation line. The intersection point is crucial. Below this temperature, the metal oxide is more stable. However, above this intersection temperature, the reducing agent's line falls below the metal's line, meaning it can successfully reduce the metal oxide. Thus, the diagram absolutely **provides an idea about the reduction of metal oxides**.

### The Missing Piece: Time

Here is the catch. Scan the entire Ellingham diagram. Do you see a 'time' ($t$) axis anywhere? No!

The Ellingham diagram is a purely thermodynamic tool. Thermodynamics is like a strict bouncer at a club—it tells you if a reaction is *allowed* to happen (feasibility). However, it says absolutely nothing about *how fast* the reaction will occur. A reaction might be thermodynamically highly favorable (very negative $\Delta G^\circ$), but it could take a million years to complete if the activation energy is too high. The speed of a reaction is the exclusive domain of **chemical kinetics**.

Therefore, the statement that the Ellingham diagram **provides an idea about the reaction rate** is fundamentally incorrect.

Animated Solution for Chemistry - Metallurgy: The statement that is incorrect about Ellingham diagram is

Select Answer:

Visualized Solution

Ellingham Diagram Axes

Free Energy Change

Phase Changes

Reduction of Metal Oxide

The Missing Variable

Conclusion

The Sigma Insight: Principles of Metallurgy and Extraction

The Thermodynamic Map

Decoding the Lines

The Missing Piece

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