Question

1) Calculate the enthalpy of reaction (ΔHrxn) for the following reaction:

Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)

Given: 4 Fe(s) + 3 O2(g) → 2 Fe2O3(s) ΔH = –1648 kJ 2 CO2(g) → 2 CO(g) + O2(g) ΔH = +565.4 kJ

2) Use standard enthalpies of formation to calculate the standard enthalpy of reaction for the following reaction:

2 H2S(g) + 3 O2(g) → 2 H2O(l) + 2 SO2(g) ΔH∘rxn

Answer 1

1)

4Fe_(s) + 3O_2(g)    2Fe₂O_3(s) -1648 KJ -------------------------------1

2CO_2(g) 2CO_(g) + O_2(g) + 565.4 KJ ---------------------------------2

Reverse the reaction 1 and divide by 2 and reverse reaction 2 and multiply by 1.5 then reaction become

Fe₂O_3(s)    2Fe_(s)   + 1.5O_2(g) +824 KJ ------------------------3

3CO_(g)   + 1.5 O_2(g)    3CO_2(g) - 848.1 KJ ------------------------4

in the reaction 3 and 4 opposite side 1.5O_2(g) get cancelled and after addition of both reaction we get reaction

Fe₂O_3(s) +   3CO_(g)    2Fe_(s) +  3CO_2(g) -24.5 KJ

H_rxn  = -24.5 KJ

2)

_fH⁰ H₂S_(g) = -20.63 KJ/mol

_fH⁰ O_2(g) = 0 KJ/mol

_fH⁰ SO_2(g) = -296.84 KJ/mol

_fH⁰ H₂O_(l) = -285. KJ/mol

H⁰ = _fH⁰ (Product) - _fH⁰ (Reactant)

H⁰ = [(2_fH⁰ SO₂) + (2_fH⁰ H₂O)] - [(2_fH⁰ H₂S_(g)) + (3_fH⁰ O_2(g))]

= [(2 (-296.84)) + (2 (-285.8))] - [2(-20.63) + (5 (0)]

= [-1165.28] - [-41.26]

H⁰ = -1124.02 KJ/mol

H⁰ for reaction = -1124.02 KJ/mol