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5. Boris recently synthesized an explosive compound he named Badenoughium. The molecular formula for Bdenoughium is H5C3N3O9. The balance chemical equation for the ignition of Badenoughium is: 4 H5C3N3O9 (l) ? 12 CO2 (g) + 10 H2O (g) + O2 (g) + 6 N2 (g) a. When Badenoughium is ignited it has a ?H o rxn of - 5678 kJ. i. Determine the ?H o f for the compound Badenoughium. ii. Determine the ?H o rxn for when 1.25 grams of Badnenoughium is ignited. b. Boris ignites 1.25 grams of his compound in a 500.00 mL at a temperature of 100.0 oC. i. Determine the partial pressure for each of the gases produced when 1.25 grams of Badenoughium is ignited. ii. Determine the total pressure in the reaction vessel after 1.25 grams of Badenoughium has been igntited.
The balanced reaction
4 H5C3N3O9 (l) = 12 CO2 (g) + 10 H2O (g) + O2 (g) + 6 N2(g)
H rxn = - 5678 kJ
Part a
Enthalpy change = sum of Enthalpy of formation of products - sum of Enthalpy of formation of reactants
At standard state, enthalpy of formation of elements is zero
Hf(O2) = Hf(N2) = 0
H rxn = 10*Hf(H2O) + 12*Hf(CO2) - 4*Hf(H5C3N3O9)
-5678 = 10*(-241.8) + 12*(-393.5) - 4*Hf
-5678 = - 7140 - 4Hf
Hf = 365.5 kJ/mol
Part b
Moles of Badnenoughium = mass/molecular weight
= 1.25/227
= 0.0055 mol
4 moles of Badnenoughium ignited produces = - 5678 kJ
0.0055 mol produces = - 5678*0.0055/4 = 7.807 kJ
Part c
Partial pressure of Badnenoughium = nRT/V
= 0.0055 mol x 0.0821 L-atm/mol-K x (100+273)K / 0.500L
= 0.306 atm
Part d
At constant temperature and volume
Partial pressure of CO2 / partial pressure of Badnenoughium
= moles of CO2 / moles of Badnenoughium
Partial pressure of CO2 =( moles of CO2 / moles of Badnenoughium) x partial pressure of Badnenoughium
= [ (12*0.0055/4) / 0.0055] x 0.306
= 0.918 atm
Partial pressure of H2O = (10/4) x 0.306 = 0.765 atm
Partial pressure of O2 = (1/4) x 0.306 = 0.0765 atm
Partial pressure of N2= (6/4) x 0.306 = 0.459 atm
Total pressure after 1.25 grams of Badenoughium has been igntited
= PCO2 + PH2O + PO2 + PN2
= 0.918 + 0.765 + 0.0765 + 0.459
= 2.2185 atm