Question

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)2CH3NO2(l)+3/2O2(g)→2CO2(g)+3H2O(l)+N2(g)
The standard enthalpy of combustion for nitromethane is −709.2kJmol−1−709.2kJmol−1.

Calculate the standard enthalpy of formation ΔH∘fΔHf∘  for nitromethane.

ΔH∘fΔHf∘ ( kJmol−1kJmol−1 )  
CO2(g)CO2(g) -393.5
H2O(l)H2O(l) -285.8

Express your answer using one decimal place.

Answer 1

Given, enthalpy of reaction, = enthalpy of combustion = −709.2 kJ mol⁻¹

Now,

where,

= stoichiometric coefficients of products

= stoichiometric coefficients of reactants

Given,

of CO₂(g) = -393.5 kJ mol^-1

of H₂O(l) = -285.8 kJ mol^-1

of O₂(g) and N₂(g) are 0 as both are elements in their standard state.

Now, given,

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

Dividing both sides by 2 to get,

CH₃NO₂(l)+3/4O₂(g)CO₂(g)+3/2H₂O(l)+1/2N₂(g)

Thus,

or, −709.2 kJ mol⁻¹ = [(1 x -393.5 kJ mol^-1) + (3/2 x -285.8 kJ mol^-1) + (1/2 x 0)] - [(1 x of CH₃NO₂(l)) + (3/4 x 0)]

or, of CH₃NO₂(l) = -113.0 kJ mol^-1

Therefore, of CH₃NO₂(l) = -113.0 kJ mol^-1