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From an Ellingham diagram at 1500 K, $\Delta G^\circ$ for $2Fe+O_2\rightarrow 2FeO$ is $-350$ kJ and for $2C+O_2\rightarrow 2CO$ is $-480$ kJ. The free-energy change for reducing one mole of FeO by carbon is
A$-130$ kJ
B$-65$ kJ
C$-415$ kJ
D$+65$ kJ
Answer & Solution
Correct answer: B. $-65$ kJ
1. Reverse the iron line: $2FeO \rightarrow 2Fe + O_2$, $\Delta G = +350$ kJ.
2. Add the carbon line: $2FeO + 2C \rightarrow 2Fe + 2CO$, $\Delta G = +350 + (-480) = -130$ kJ for **2 mol** FeO.
3. Per mole of FeO: $-130/2 = -65$ kJ, which is negative, so the reduction is feasible.
_Source: Samacheer Kalvi Class 12 Chemistry Vol.1, Unit 1 'Metallurgy', §1.4.2 (worked coupling)_
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