Funny Car Dragster

Enthalpies of Formation Chemistry Problem?

Ok, I cannot figure this one out. Please tell me how to set it up and solve! Thanks!

Top fuel dragsters and funny cars burn nitromethane as fuel according to the following balanced combustion equation:
2 CH3NO2(l) + 3/2 O2(g) --> 2 CO2(g) + 3 H2O(g)

The standard enthalpy of combustion for nitromethane is -709.2 kJ/mol.

Calculate the standard enthalpy of formation for nitromethane. Answer in kJ/mol.

Thanks again

Jen,

According to the problem, to get this you will also need the enthalpies of formation of CO2 and H2O (enthalpy of formation for elements like O2 are zero).

I notice that the equation is not really balanced (where did the nitrogen go?); I will solve it according to how the problem is written, but it is not likely to match standard values you might find.

The heat of combustion is equal to the heats of formation of the products minus the heats of formation of the reactants:

Heat of combustion = 2*Hf(CO2) + 3*Hf(H2O) - 2*Hf(CH3NO2) = 2*(-709.2 kJ/mol). Note that you have two moles of nitromethane, so we had to double the value for heat of combustion.

2*Hf(CH3NO2) = 2*Hf(CO2) + 3*Hf(H2O) + 2*(709.2)
Hf(CH3NO2) = Hf(CO2) + 3/2*Hf(H2O) + 709.2

The heat of formation for O2(g) is zero
The heat of formation for CO2(g) is -393.5 kJ/mol
The heat of formation of H2O(g) is -241.8 kJ/mole (if your teacher or text uses different values, use them!):

Hf(CH3NO2) = Hf(CO2) + 3/2*Hf(H2O) + 709.2 = -393.5 kJ/mol - 1.5*241.8 kJ/mole + 709.2 kJ/mol = -47.0 kJ/mol.

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I looked online for a source that gave the balanced reaction for nitromethane combustion:

4CH3NO2 + 3O2 → 4CO2 + 6H2O + 2N2

This will not change the outcome of the calculation, as the heat of formation of nitrogen gas is also zero.

========

One last thing (I promise): if you generate liquid water (probably the state for which the -709.2 kJ/mol for nitromethane actually corresponds), the value changes to that which you see online:

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

Heat of combustion = 2*Hf[CO2(g)] + 3*Hf[H2O(l)] - 2*Hf[CH3NO2(l)] = 2*(-709.2 kJ/mol)

The heat of formation for N2(g) is zero
The heat of formation for O2(g) is zero
The heat of formation for CO2(g) is -393.5 kJ/mol
***The heat of formation of H2O(l) is -285.83 kJ/mole***

Hf(CH3NO2) = Hf[CO2(g)] + 3/2*Hf[H2O(l)] + 709.2 = -393.5 kJ/mol - 1.5*285.83 kJ/mole + 709.2 kJ/mol = -113.0 kJ/mol.

This is exactly what is reported (see second link)!

Hope that helped!

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Top fuel dragsters and funny cars burn nitromethane as fuel according to the following balanced combustion equation: The standard enthalpy of combustion for nitromethane is •