Question

14)

Top fuel dragsters and funny cars burn nitromethane as fuel according to the following balanced combustion equation:
2CH3NO2(l)+3/2O2(g)→2CO2(g)+3H2O(l)+N2(g)
The standard enthalpy of combustion for nitromethane is −1418kJ.

Calculate the standard enthalpy of formation (ΔH∘f) for nitromethane.

Answer 1

Given,

2 CH₃NO₂(l) + 3/2 O₂(g) --> 2 CO₂(g) + 3 H₂O(l) + N₂(g)

Standard enthalpy of combustion for nitromethane is -1418kJ.

Standard Enthalpy of formation (ΔH∘f) for Nitromethane = ?

Now,
Heat of combustion = Heat of formation of the products - Heat of formation of the reactants:

therefore,

Heat of combustion = { 2 * Hf[CO₂(g)] + 3 * Hf[H₂O(l)] + 1 * Hf[N₂(g)] } - { 2 * Hf[CH₃NO₂(l)] + 3/2 * Hf[O₂(g)] } = - 2* 1418 KJ

we need to multiply by 2 since 2 moles of nitromethane

but

The heat of formation for N₂(g) is zero
The heat of formation for O₂(g) is zero

and we know that,

The heat of formation for CO₂(g) is -393.5 kJ
The heat of formation of H₂O(l) is -285.83 kJ

the equation summaries to

{ 2*Hf[CO₂(g)] + 3*Hf[H₂O(l)] } - { 2*Hf[CH₃NO₂(l)] } = - 2*1418 kJ

substitute the values then

Hf(CH₃NO₂) = Hf[CO₂(g)] + 3/2*Hf[H₂O(l)] + 1418 = -393.5 kJ - 3/2 * 285.83 kJ + 1418 kJ = 595.755 kJ

therefore,

Standard Enthalpy of formation (ΔH∘f) for Nitromethane = 595.755 kJ