ΔHof for nitromethane

Question

Nitromethane, CH₃NO₂, can be used as a fuel. When the liquid is burned, the (unbalanced) reaction is mainly

CH₃NO₂ (l) + O₂ (g) à CO₂ (g) + N₂ (g) + H₂O (g)

a.) The standard enthalpy change of reaction (ΔH^o_rxn) for the balanced reaction (with lowest whole- number coefficients) is -1288.5 kJ. Calculate the ΔH^o_f for nitromethane.

Answer 1

Apparently you didn't want to look up the values for the standard enthalpy of formation. How come? Just curious. Or was my original explanation not easy to understand? Just curious. Here is the solution with more detail.

4CH₃NO₂(l) + 7O₂(g) ==> 4CO₂(g) + 2N₂(g) + 6H₂O(g) ... balanced equation ∆Hº_rxn = -1288.5 kJ

∆Hº_rxn = -1288.5 kJ = ∆Hºf products - ∆Hºf reactants

∆Hº formation values obtained from the internet:

CO₂(g) = -393.5 kJ/mol

H₂O(g) = -241.8 kJ/mol

N₂ and O₂ are zero

∆Hº_rxn = -1288.5 kJ = ∆Hºf products - ∆Hºf reactants

-1288.5 kJ = [(4x-393.50 + (6x-241.8)] - [(4x∆H_CH3NO2)]

-1288.5 = -1574 + (-1450.8) - 4∆H_CH3NO2

-1288.5 = -3024.8 - 4∆H_CH3NO2

∆H_CH3NO2 = -431.1 kJ/mol