ChemistryGiven that 2 C s + 2 Fe 2 O 3 s 4 Fe s + 3 CO 2 g ; ∆ H 0 = – 93657 kcal at 25 0 C 3 C s + 3 O g 3 CO 2 g ∆ H 0 = – 94050 kcal at 25 0 C The value of ∆ H f 0 Fe 2 O 3 is

Given that

2Cs +2Fe2O3s   4Fes + 3CO2g; H0=-93657 kcal at 250C 3Cs +3Og   3CO2g H0=-94050 kcal at 250C The value of Hf0 Fe2O3 is

  1. A
    16.750 kcal
  2. B
    -16.750 kcal
  3. C
    -196.5 kcal
  4. D
    -393 kcal

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    Solution:

    2Fes +3/2O3g   Fe2O3s; H=? Given, 3Cs +Fe2O3s   4Fes+3CO2g; H=-93657 kcal  And  3Cs +3O2g   3CO2g; H=-94050 kcal Equation ii-Equation i, we get 4Fes +3O2  2Fe2O3s; H=-393 kcal    2Fe +3/2O2   Fe2O3s; H=-196.5 kcal at 250C

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