ple Atempts This test allows 3 attempts. This is attempt number 1 ce Completion
ID: 576331 • Letter: P
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ple Atempts This test allows 3 attempts. This is attempt number 1 ce Completion This test can be saved and resumed later uestion Completion Status Close Window s Moving to another question will save this response Question 2 of3 estion 2 3 points Save Answe The standard enthalpy of formation of propane gas, C3Helg), is-104 7klmol at 298 K. Given the entralpies of formation ol combustion of propane (in kirnol·&Mcfor; C3n8g) fr Cog) 393 5 klimol and He for o-265 ulimat clculate he enthupy Question 2 ol 3 Moving to another question will save this response SteamPurityConsidp TEN-201-3-21337pat 0 Fun with Cheeistry pptxExplanation / Answer
The standard enthalpies of formation of the compounds are given as below.
Compound
Standard Enthalpy of formation, Hf (kJ/mol)
Propane, C3H8
-104.7
Carbon dioxide, CO2
-393.5
Water, H2O (l)
-285.8
The balanced chemical equation for the reaction is
C3H8 (g) + 5 O2 (g) --------> 3 CO2 (g) + 4 H2O (l)
The enthalpy of combustion is given as
Hc = n.Hf (products) - n.Hf(reactants) where n = number of moles
====> Hc = [(3 mole CO2)*Hf(CO2, g) + (4 mole H2O)*Hf(H2O, l)] – [(1 mole C3H8)*Hf(C3H8, g)] (standard enthalpy of formation of O2 = 0)
====> Hc = [(3 mole)*(-393.5 kJ/mol) + (4 mole)*(-285.8 kJ/mol)] – [(1 mole)*(-104.7 kJ/mol)]
====> Hc = [(-1180.5) + (-1143.2)] kJ – (-104.7 kJ) = -2323.7 kJ + 104.7 kJ = -2219.0 kJ (ans).
Compound
Standard Enthalpy of formation, Hf (kJ/mol)
Propane, C3H8
-104.7
Carbon dioxide, CO2
-393.5
Water, H2O (l)
-285.8
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