T1 and T2 are melting points of pure metal A and pure stoichiometric oxide AO_{2}, respectively and T1<T2. The stoichiometric metal oxidation reaction A(s) + O_{2}(g) = AO_{2 }(s) is in equilibrium at 1 atm pressure at temperature less than T1. If the temperature increases, which schematic represents the correct standard free energy change versus temperature plot?

Correct answer is option C. As per question, A(s) + O2(g) = AO2 (s) For Ellingham diagram's slope = -ΔS0 Three cases arises in this question, Case 1: If T<T1 i.e. only O2(g) is present slope= -(-ΔS0) = ΔS0 i.e change in entropy=(products-reactants) Case 2: If T1<T<T2 i.e O2(g) and A (liquidRead more

Correct answer is option C.

As per question, A(s) + O

_{2}(g) = AO_{2 }(s)For Ellingham diagram’s slope = -ΔS

^{0}Three cases arises in this question,

Case 1:If T<T_{1}i.e. only O_{2}(g) is presentslope= -(-ΔS

^{0}) = ΔS^{0 }i.e change in entropy=(products-reactants)Case 2:If T_{1}<T<T_{2 }i.e O_{2}(g) and A (liquid) are present.Change in entropy will be more negative as compared to case 1, so that slope will be more positive than case 1.

slope= -(-ΔS

^{0}) = ΔS^{0}Thus, slope 2> slope 1

Case 3:If T<T_{2 }i.e (O_{2}(g), A(l) and AO_{2}(l) are present.Change will be negative, so slope will be positive.

Slope 3< slope 1<slope 2