# Chemistry problem

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Thread starter 17 years ago
#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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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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17 years ago
#3
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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Thread starter 17 years ago
#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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17 years ago
#5
its Bonds formed - bonds broken.
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17 years ago
#6
(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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17 years ago
#7
(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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Thread starter 17 years ago
#8
(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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Thread starter 17 years ago
#9
Thank you all for your help.
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