nitromethane CH3NO2 is a good fuel it is liquid at ordinary temperatures when th
ID: 987430 • Letter: N
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
Explanation / Answer
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.
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