5. Boris recently synthesized an explosive compound he named Badenoughium. The m

Question

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.

Explanation / Answer

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

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