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
C4H4 (g) + 2 H2 (g) --> C4H8 (g)
Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for C4H4 (-2341 kJ/mol), C4H8 (-2755 kJ/'mol), and H2 (-286 kJ/mol), calculate the change in heat for this reaction.
We will be happy to critique your work on this. See Hess' Law in your text.
C4H4 (g) --> 4C + 2H2 -2341 kJ/mol
2 H2 (g) --> H2 + H2 2(-286 kJ/mol)
4C + 4H2 --> C4H8 2755 kJ/mol
C4H4 + 2H2 --> C4H8 -158 kJ/mol
No. You have been given the enthalpies of combustion. Write and balance those equations first.
C4H4 + O2 >>> H20 (g) + CO2 balance this
H2+ O2 >>> H2O (g)
C4H8 + O2 >>> H2O (g) + CO2
balance all these. YOu know the heat of combustion in kj/mole of reactant.
From that, you solve for the Heats of formation for C4H4, H2O, and C4H8
Thence, you can calculate the heat of reaction for the reaction. It is an exercise in logic and algebra. Remember, that the heat of combusion is equal to the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants.
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This question looks answered it looks as if the final answer is 128kJ/mol is it?
Forgot to account for an additional -286 kj when calculating the heat of combustion for H2 (you have two moles of H2 in the combustion equation, not just one). The final answer should be -158 kJ/mol.
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