1)calculate the standard enthalpy change of formation of CH3OH(l) at 298K based on the following thermochemical data at 298K
1)standard enthalpy change of combustion of CH3OH(l)=-7266kJ/mol
2)standard enthalpy change of formation of CO2(g)=-393.5kJ/mole
3)standard enthalpy change of formation of H2O(l)=-285.8kJ/mol
2)standard enthalpy change of formation of C6H12 is -156.4kJ/mol.calculate the heat of combustion of one mole of C6H12.the heat of formation of CO2(g)and H2O(l)are -394kJ/mol and -285.8kJ/mol,repectively
energetic
2009-10-26 4:32 pm
回答 (1)
2009-10-28 11:33 pm
✔ 最佳答案
1.CO2(g) + 2H2O(l) → CH3OH(l) + (3/2)O2(g) .. ΔH = -(-7266) kJ/mol
C(graphite) + O2(g) → CO2(g) .. ΔH = -393.5 kJ/mol
2H2(g) + O2(g) → 2H2O(l) .. ΔH = 2(-285.8) kJ/mol
Add the above three thermochemical equations and eliminate CO2(g), 2H2O(l) and (3/2)O2(g) on the both sides.
C(graphite) + 2H2(g) + (1/2)O2(g) → CH3OH(l)
ΔH = -(-7266) + (-393.5) + 2(-285.8) = +6300.9 kJ/mol
2.
C6H12(l) → 6C (graphite) + 6H2(g) .. ΔH = -(-156.4) kJ/mol
6C (graphite) + 6O2(g) → 6CO2(g) .. ΔH = 6(-394) kJ/mol
6H2(g) + 3O2(g) → 6H2O(l) .. ΔH = 6(-285.8) kJ/mol
Add the above three therochemical equation and eliminate 6C (graphite) and 6H2(g) on the both sides.
C6H12(l) + 9O2(g) → 6CO2(g) + 6H2O(l)
ΔH = -(-156.4) + 6(-394) + 6(-285.8) = -3922.4 kJ/mol
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