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r316
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#1
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#1
Heat of reaction of:

C2H6 + 3.5 O2 --> 2 CO2 +3 H2O

I'm not sure how to deal with the 3.5 oxygen.

I've doubled everything to get rid of it and make the answer as -2374kJ/mol. But surely doubling everything had an effect on the calculation. How should I approach 1/2 moles?
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john !!
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#2
Report 17 years ago
#2
What's the problem with half moles?

C2H6 + 3.5 O2 --> 2 CO2 +3 H2O

I don't understand. To find the heat, you subtract the enthalpies of formation of the reactants from that of the products:

3*/\H*f H20 + 2 * /\H*f CO2 - /\H*f C2H6

Since the enthalpy of formation of any element is 0, the 3.5 doesn't even come up!
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h_cube3000
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#3
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yeah, there isnt no need to calculate with the 3.5O2 as it is in its standard state (value is 0)
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r316
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#4
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#4
I think that you calculate the energy put in to break the original bonds, then substract the energy released in making new bonds. Perhaps my gcse level is done in an easier way.
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h_cube3000
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#5
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#5
its Bonds formed - bonds broken.
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IntegralAnomaly
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#6
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(Original post by r316)
Heat of reaction of:

C2H6 + 3.5 O2 --> 2 CO2 +3 H2O

I'm not sure how to deal with the 3.5 oxygen.

I've doubled everything to get rid of it and make the answer as -2374kJ/mol. But surely doubling everything had an effect on the calculation. How should I approach 1/2 moles?
multiplying by two has no change because the molecular ratios will still be the same.
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Ben.S.
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(Original post by IntegralAnomaly)
multiplying by two has no change because the molecular ratios will still be the same.
You can either do it using Delta(r)H(standard) formation values (products - reactants), or by breaking and formng individual bonds (they are both the same). I think you just multiply by the stoichiometric coefficient when calculating the values for each species. It's very straightforward.

Ben
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r316
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#8
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(Original post by Ben.S.)
You can either do it using Delta(r)H(standard) formation values (products - reactants), or by breaking and formng individual bonds (they are both the same). I think you just multiply by the stoichiometric coefficient when calculating the values for each species. It's very straightforward.

Ben
Thank you.

I haven't yet met the stoichiometric coefficient, but I think I'm coping now. I'll just work with 1/2 moles quite happily; I merely had a moment of thinking that 1/2 moles in these calculations cannot be right.
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r316
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#9
Report Thread starter 17 years ago
#9
Thank you all for your help.
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