Answer: Delta H = products - reactants
For 1 mole of CH3NO2 you need to adjust the coefficients in the chemical equation. Divde all by 2 since CH3NO2 has a coefficient of 2.
CH3NO2(l) + 1/2O2(g) ----> CO2(g) + 1/2N2(g) + 3/2H2O(g)
Delta H = the sum of the heat of formations of the products - the sum of the heat of formations for the reactants.
Remember to multiply by the coefficients in each case.
Delta H = -393.5 + 0 + (3/2)(-241.8) - [(112.0) + 0)]
Delta H = -756.2 - 112
Delta H = -868.2 kJ/mol
This is an exothermic reaction. Obviously because it is combustion and the delta H is a negative quantity.
For part C you have to determine the number of moles in 18.02 g H2O and then change the coefficients in the equation again.
18.02 g H2O x (1 mole H2O/18.02 g H2O) = 1 mole
Divide through the chemical equation by three since in the balanced equation of the reaction there are three moles of H2O.
2/3CH3NO2 (l) + 1/3O2(g) ---> 2/3CO2(g) + H2O(l)
Delta H = products - reactants
Delta H = 2/3(-393.5) + 0 + (-241.8) - [(2/3)(112) + 0]
Delta H = -262.33 - 241.8 - 74.67
Delta H = -95.2 kJ/mol
