The heat of formation of ethane (C2H6) at SRS conditions is -84.7 MJ/kmole. The heat of formation of water vapor (H2O) at SRS conditions is -241.8 MJ/kmole. The heat of formation of Carbon Dioxide (CO2) at SRS conditions is -393.5 MJ/kmole. What is the heat of reaction for stoichiometric combustion with air at SRS conditions?
-1599? -1429? -721? -551? +551?
Just curious if anyone can at least get me an equation or help me figure out how to solve
Roger the Mole2015-05-09T14:49:50Z
These are the reactions you have been given: 2 C(s) + 3 H2(g) → C2H6(g), ΔH = -84.7 kJ/mol H2(g) + 1/2 O2(g) → H2O(g), ΔH = -241.8 kJ/mol C(s) + O2(g) → CO2(g), ΔH = -393.5 kJ/mol
This is the reaction you're aiming at: C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(g)
You need C2H6(g) on the left, so write the first given equation backwards: C2H6(g) → 2 C(s) + 3 H2(g), ΔH = +84.7 kJ/mol You need 2 CO2(g) on the right, so multiply the third given equation by 2: 2 C(s) + 2 O2(g) → 2 CO2(g), ΔH = -787.0 kJ/mol You need 3 H2O(g) on the right, so multiply the second given equation by 3: 3 H2(g) + 3/2 O2(g) → 3 H2O(g), ΔH = -725.4 kJ/mol Add up the last three equations: C2H6(g) + 2 C(s) + 2 O2(g) + 3 H2(g) + 3/2 O2(g) → 2 C(s) + 3 H2(g) + 2 CO2(g) + 3 H2O(g), ΔH = +84.7 kJ/mol -787.0 kJ/mol -725.4 kJ/mol Cancel like amounts on opposite sides of the arrow, add like amounts on the same side of the arrow, and do the arithmetic for ΔH: C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(g), ΔH = -1427.7 kJ/mol or -1427.7 MJ/kmol