For a spontaneous reaction, the free energy change must be negative. , is the enthalpy change during the reaction. T is the absolute temperature, and is the change in entropy during the reaction. Cons

Question

For a spontaneous reaction, the free energy change must be negative. $\Delta G =
\Delta H - T\Delta S$, $\Delta H$ is the enthalpy change during the reaction. T 
is the absolute temperature, and $\Delta S$ is the change in entropy during the 
reaction. Consider a reaction such as the formation of an oxide

$M + O_2 \rightarrow MO$

Dioxygen is used up in the course of this reaction. Gases have a more random 
structure (less ordered) than liquid or solids. Consequently gases have a higher
entropy than liquids and solids. In this reaction S (entropy or randomness) 
decreases, hence $\Delta S$ is negative. Thus, if the temperature is raised then
$T\Delta S$ becomes more negative. Since, $T\Delta S$ is substracted in the 
equation, then $\Delta G$ becomes less negative. Thus, the free energy change 
increases with the increase in temperature. 
The free energy changes that occur when one mole of common reactant (in this 
case dioxygen) is used may be plotted graphically against temperature for a 
number of reactions of metals to their oxides. The following plot is called an 
Ellingham diagram for metal oxide. Understanding of Ellingham diagram is 
extremely important for the efficient extraction of metals.
[SEE DIAGRAM]

Prashant Kotian · Accepted Answer

Correct answer: A. Step 1: Write down the net reaction for the reduction of ZnO by Carbon: ZnO(s) + C(s) -> Zn(s) + CO(g). Step 2: Calculate the standard Gibbs free energy change (ΔG°_reaction) for this net reaction: ΔG°_reaction = ΔG°_f(CO) - ΔG°_f(ZnO) = -460 kJ/mol - (-360 kJ/mol) = -100 kJ/mol. Step 3: Since the net Gibbs free energy change is negative (-100 kJ/mol), the reduction of ZnO by Carbon is thermodynamically spontaneous at 1000°C, meaning ZnO can be reduced to Zn by C (option b).

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