Question

# The standard enthalpy of formation of FeO and Fe2O3 is −65kcalmol−1 and −197kcalmol−1 respectively. If a mixture containing FeO and FeO3 in 2:1 mole ratio on oxidation is changed into 1:2 mole ratio, thermal energy in kcal released per mole of mixture will be :

A
13.4
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B
26.8
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C
30.2
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D
none of these
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Solution

## The correct option is B 13.4 Fe+12O2⟶FeO;ΔH=−65kcalmol−1 . ..(i) 2Fe+32O2⟶Fe2O3;ΔH=−197kcalmol−1 ...(ii)By eq, [(ii)−2×(i)]Also, 2FeO+12O2⟶Fe2O3;ΔH=−67kcalmol−1 ...(iii)Let a mole of FeO and b mole of Fe2O3, are present such that a+b=1andab=2 a=23andb=13 2FeO+12O2⟶Fe2O3Initial a bAfter oxidation (a−2a′) b+a′Given, ab=2anda−2a′b+a′=12∴a−2a′a2+a′=12∴2a−4a′=a2+a′∴a′=3a10=3×210×3=15∴FeOused=2a′=2/5mol∵2moleFeOgivesheat=67kcal∴25moleFeOgivesheat=67×22×5=13.4kcal

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