Given (i) 2 Fe _2 O _3(s) \rightarrow 4 Fe ( s )+3 O _2(g); ∆_{r} G^{\circ}=+1487.0 kJ mol^{-1} (ii) 2 CO(g)+O_2(g) \rightarrow 2 CO_2(g) ; ∆_{r} G^{\circ}=-514.4 kJ mol^{-1} Free energy change, ∆_{ r } G ^{\circ} for the reaction 2 Fe_2 O_3(s)+6 CO(g) \rightarrow 4 Fe(s)+6 CO_2(g) will be :- [15 Apr 2018 Shift 2]

Correct Answer is : −56.2kJmol−1

JEE Main 15 Apr 2018 Shift 2 Solved Paper

Section: Chemistry

Question : 31 of 90

Marks: +1, -0

Given
(i) 2Fe2O3(s)⟶4Fe(s)+3O2(g);
∆rG∘=+1487.0kJmol−1
(ii) ‌2CO(g)+O2(g)⟶2CO2(g);
‌∆rG∘=−514.4kJmol−1
Free energy change, ∆rG∘
for the reaction 2Fe2O3(s)+6CO(g)⟶4Fe(s)+6CO2(g) will be :-

[15 Apr 2018 Shift 2]

−112.4kJmol−1
−56.2kJmol−1
168.2kJmol−1
208.0kJmol−1

Validate

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