Question

nitromethane CH3NO2 is a good fuel it is liquid at ordinary temperatures when the liquid is burned the reaction involved is 2CH3NO2 (l) + 3/2O2(g) --> 2 CO2 (g)+ N2(g) + 3 H2O(g)

the standard enthalpy at 25 c is -112Kj/mol other relevent values can be found in appendix D

a) calsulate the enthalpy change in the burning of 1 mol liquid nitromethane to form gasious products at 25c state explicitly if the reaction is endothermic or exothermic

b) would more or less heat be involved if gasious methane where burned under the same conditions ? indicate what additional information (if any) you would need to calculate the exact abount of heat and show how you would just use the information

Answer 1

a) the enthalpy change of given reaction

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

will give us the enthalpy change in the burning of 1 mole liquid.

The enthalpy of reaction = Enthalpy of formation of products -enthalpy of formation of reactants

Enthalpy of formation of elements in pure state = 0 = Enthalpy of formation of O2 and N2

Enthalpy of formation of

CO2 = -393.5 KJ / mole

H2O = -241.8 KJ / mole

CH3NO2 (l) = -112Kj/mol

Enthalpy of reaction = [Enthalpy of CO2 + 3/2 Enthalpy of H2O] - [ Enthalpy of CH3NO2]

Enthalpy of reaction = [-393.5 + 3/2 (-241.8)] - [-112] = [-393.5 - 362.7] -[-112] = -644.2 KJ /mole

b) CH4 + 2O2 --> CO2 + 2H2O

the enthalpy of reaction = [Enthalpy of formation of CO2 + 2 enthalpy of formation of H2O] - [Enthalpy of formation of CH4]

So we need enthalpy of formation of CH4 = −74.87 KJ / mole

Enthalpy of reaction = [-393.5 + 2(-241.8)] - [-74.87 ] = -802.23 KJ / mole

So for each mole of nitromethane and methane methane will give more energy.

However the calorific value of nitromethane will be less.