Hess' cycle for C2H4(g) + HBr(g) → C2H5Br(g)

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TheonlyMrsHolmes
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C2H4(g) +52.3 ΔHf/kJ mol–1
HBr(g) –36.2 ΔHf/kJ mol–1

C2H5Br(g)–60.4 ΔHf/kJ mol–1


Use the data in the table above to calculate the standard enthalpy change for the following reaction.


C2H4(g) + HBr(g) → C2H5Br(g)


I already have the answer for this but I am very stuck on what the products would be at bottom of hess' cycle because there is a bromine attached to the hydrogen and not just hydrogen gas which forms straight H2O, can anyone help me please?
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TheonlyMrsHolmes
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(Original post by SeanFM)
It looks like it might be electrophilic addition, which might tell you something about the things that go on the bottom.
Oh okay, I was thinking it was just 2Co2 + 5H2O + Br but maybe not...
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Kevin De Bruyne
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(Original post by TheonlyMrsHolmes)
Oh okay, I was thinking it was just 2Co2 + 5H2O + Br but maybe not...
That seems a bit more sensible, but I'm not too familiar with Hess Cycles anymore - sorry, ignore me.
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Red_Inferno112
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It would be broken into constituent elements. For Br, for example, write 1/2Br2, that's generally acceptable, but only for diatomic molecules, i.e. you can do the same for H2, but don't write
1/2C! This is what I generally do.
It's a controversial concept in chemistry, also it probably doesn't matter if it's not balanced.
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Red_Inferno112
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Also, hydrogen gas can be modelled as free, remember the reverse of enthalpy of formation just gives you the elements, never any compounds, by definition! Hope this helped, if I'm not explaining it properly let me know.
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Rabadon
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Isn't it just 2C 2.5H2 and 0.5BR
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