[jennielynn_1980]

Hi all.

I have been given the following question and I am trying to figure out if I am missing something or there is a typo in the question because if there is not a typo, I have no idea how to solve this.  So here it is:

calculate the heat of combustion for C₂H₆ from the following information:

C₂H₄ + 3O₂ --> 2CO₂ + 2H₂O  heat of formation = -1409.5 kJ
C₂H₄ + H₂ --> C₂H₂  heat of formation = -136.7 kJ
H₂ + 1/2O₂ --> H₂O  heat of formation = -285.5 kJ

Thanks

[Will]

calculate the heat of combustion for C₂H₆ from the following information:

C₂H₄ + 3O₂ --> 2CO₂ + 2H₂O  heat of formation = -1409.5 kJ
C₂H₄ + H₂ --> C₂H₂  heat of formation = -136.7 kJ
H₂ + 1/2O₂ --> H₂O  heat of formation = -285.5 kJ

AFAIK the second equation is complete rubbish and most definately a typo/mistake. And I don't get what compound the heat of formation is referring to. I think the question means just the enthalpy change (with the last being heat of formation of H₂O).
If you made the assumption that the second equation is supposed to be C₂H₂ + H₂ --> C₂H₄ then you can estimate the answer.

[jennielynn_1980]

Okay thanks.  I am glad there is a type because I was getting extremely confused trying to wrap my head around this question 

[jennielynn_1980]

I am still stuck on this question.  If someone could give me a hint that would be great.  Here is what I have so far:

The combustion equation is:
2C₂H₆ + CO₂ --> 4CO₂ + 4H₂O


equation #1 C₂H₄ + 3O₂ --> 2CO₂ + 2H₂O  heat of formation = -1409.5 kJ
equation #2 C₂H₄ + H₂ --> C₂H₆  heat of formation = -136.7 kJ
equation #3 H₂ + 1/2O₂ --> H₂O  heat of formation = -285.5 kJ


So from these equations we can tell that the heat of formation for:
C₂H₆ is = 136.7 kJ (got this by reversing equation #2)
H₂O is = -285.5 kJ (from equation#3)
O₂ I don't need to know the heat of formation because it is assumed to be 0

So if I multiply #1 by 2 then:
2C₂H₄ + 6O₂ --> 4CO₂ + 4H₂O
Then mulitply #2 by -2 then:
2C₂H₆ --> 2C₂H₄ + 2H₂
Then multiply #3 by 2:
2H₂ + O₂ --> 2H₂O

Then add all three together to get:

2C₂H₄ + 6O₂ + 2C₂H₆ + 2H₂ + O₂ --> 4CO₂ + 4H₂O + 2C₂H₄ + 2H₂ + 2H₂O

Now that I have this, I am unsure of the next step.

Sorry for how long this is.

[Borek]

2C₂H₄ + 6O₂ + 2C₂H₆ + 2H₂ + O₂ --> 4CO₂ + 4H₂O + 2C₂H₄ + 2H₂ + 2H₂O

No idea if you did everything correctly, but obvious step now is to sort and cancel out.

[jennielynn_1980]

So I can cancel out to simplify the equation?  I wasn't sure I could do that.  I think I have the answer then.

If I mulitply the heat of formations for each equation by 2 and simplify the long equation like you said, then I should have my final equation and heat of combustion.