Posted by Jayd on Sunday, May 20, 2007 at 6:52pm.
The enthalpies of combustion are determined by these equations:
C4H4(g) + 5 O2(g) ---> 4 CO2(g) + 2 H2O(g) : ÄH = -2341 kJ/mol
2 H2(g) + O2(g) ---> 2 H2O(g) : ÄH = -286 kJ/mol
C4H8(g) + 6 O2(g) ---> 4 CO2(g) + 4 H2O(g) : ÄH = -2755 kJ/mol
If we reverse the last equation, we get:
4 CO2(g) + 4 H2O(g) ---> C4H8(g) + 6 O2(g) : ÄH = 2755 kJ/mol - notice the change in sign
Adding the first 2 equation together, we get:
C4H4(g) + 5 O2(g) + 2 H2(g) + O2(g) ---> 4 CO2(g) + 2 H2O(g) + 2 H2O(g) : ÄH = -2341 kJ/mol + -286 kJ/mol
Combining like items, we get:
C4H4(g) + 2 H2(g) + 6 O2(g) ---> 4 CO2(g) + 4 H2O(g) : ÄH = -2627 kJ/mol
Adding the fourth equation to the last one, we get:
C4H4(g) + 2 H2(g) + 6 O2(g) + 4 CO2(g) + 4 H2O(g) ---> 4 CO2(g) + 4 H2O(g) + C4H8(g) + 6 O2(g) : ÄH = -2627 kJ/mol + 2755 kJ/mol
Removing like items from both sides and summing the energy, we get:
C4H4(g) + 2 H2(g) ---> C4H8(g): ÄH = 128 kJ/mol
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