Question

Ellingham diagram represents:
 (A) Change of $\Delta G$ with temperature
(B) Change of $\Delta H$ with temperature
(C) Change of $\Delta G$ with pressure
(D) Change of $(\Delta G-T\Delta S)$ with temperature

Answer 1

Hint: To predict the spontaneity of a reaction based on enthalpy and entropy values by the Gibbs equation. Prediction of the spontaneity of reduction of various metal oxides by Ellingham diagram proposed by H.G.T Ellingham.

Complete step by step answer:
The Ellingham diagram is a curve related to the value of the Gibbs free energy with temperature.
Gibbs energy change is, $\Delta G=\Delta H-T\Delta S$ ---- (1)
Where $\Delta H$ are the change in enthalpy and $\Delta S$ the change in entropy.
From the above reaction, Gibbs free energy related to the equilibrium constant as,
$\Delta {{G}^{0}}=-RT\ln K$ --- (2)
Where K = equilibrium constant

The Ellingham diagram is a plot between ${{\Delta }_{f}}{{G}^{o}}$ and T for the formation of metal oxides and sulfides, which is a graph of showing the temperature dependence of the stability of the compound.
The general reaction of oxides,
$2xM(s)+{{O}_{2}}(g)\to 2{{M}_{x}}O(s)$

From the reaction, the decreasing amount of gaseous amount from left to right as the product formed is solid metal oxide on the right side. Hence, $\Delta S$ is negative due to the molecule's randomness decreasing from left to right.
So, form the equation (1), $\Delta S<0,$ then $\Delta G$ shifts towards increasing the temperature.
Hence, the Ellingham diagram represents a change of $\Delta G$ with temperature.
So, the correct answer is “Option A”.

Note: If the reaction is exothermic, enthalpy of the system is also negative which makes Gibbs free energy negative. Hence, we can conclude that the forward reaction is favorable with the positive value of the equilibrium constant.