Given below are two statements :
Statement I : The choice of reducing agents for metals extraction can be made by using Ellingham diagram, a plot of ΔG vs temperature.
Statement II : The value of ΔS increases from left to right in Ellingham diagram.
In the light of the above statements, choose the most appropriate answer from the options given below :

Solution

Chemistry ma’am · 4 steps

On an Ellingham line, ΔS is the slope, and a straight line's slope does not change from left to right.

1. Statement I. An Ellingham diagram plots Δ G° for forming oxides against temperature. A metal whose line lies lower can reduce the oxide of a metal whose line lies higher at that temperature. This is how a reducing agent is chosen, so Statement I is true.

   Δ G° = Δ H° - T Δ S°

2. Read the slope. Plot this against T and you get a straight line with slope -Δ S°. For a given reaction Δ S° stays nearly constant, so it does not grow as you move right. Only TΔ S° grows with T. The slope changes only at a phase change, such as melting or boiling of the metal.
3. Statement II. For 2xM + O₂ → 2MₓO, a gas is used up, so Δ S° is negative and the lines slope upward. What rises from left to right is Δ G°, which becomes less negative. Δ S° itself does not rise, so Statement II is false.
4. So. Statement I is true and Statement II is false, which is option (d).

The idea to keep. In any Δ G° vs T plot, the slope is -Δ S° and the intercept is Δ H°. A straight line means constant Δ S°. A kink means a phase change.

Answer from the NTA final answer key. Solution by PrepWiser.