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T1 and T2 are melting points of pure metal A and pure stoichiometric oxide AO2, respectively and T1<T2. The stoichiometric metal oxidation reaction A(s) + O2(g) = AO2 (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?
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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) + 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 (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= -(-ΔS0) = ΔS0
Thus, slope 2> slope 1
Case 3: If T<T2 i.e (O2(g), A(l) and AO2(l) are present.
Change will be negative, so slope will be positive.
Slope 3< slope 1<slope 2
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