hw Q8.61

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Lizzie Zhang 2C
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

hw Q8.61

Postby Lizzie Zhang 2C » Thu Jan 18, 2018 11:05 pm

I have trouble solving question 8.61
Calculate the reaction enthalpy for the synthesis of hydrogen bromide gas, H2(g)+Br2(l)->2HBr(g),
NH3(g)+HBr(g)->NH4Br(s) dH=-188.32kJ
N2(g)+3H2(g)->2NH3(g) dH=-92.22kJ
N2(g)+4H2(g)+Br2(l) dH=-541.66kJ

I understand you have to arrange them in some way that the overall reaction shows as H2(g)+Br2(l)->2HBr(g), but my problem is just that I don't know how to rearrange the equations. Can someone explain to me step by step how to rearrange those equations?

Thanks a lot!

miznaakbar
Posts: 55
Joined: Thu Jul 13, 2017 3:00 am
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Re: hw Q8.61

Postby miznaakbar » Thu Jan 18, 2018 11:58 pm

From what I understand, rearranging is essentially trial and error, where you are trying to cancel out compounds that you do not want in your final equation through flipping equations and multiplying coefficients. First you would flip the 1st and 2nd rxns and write them in reverse so that the H2 and N2 would cancel out from the 3rd rxn. Since the second rxn given has the coefficient 2 in front of NH3, you would multiply the coefficients in the first rxn by 2 so that those NH3's would cancel out (since we do not want NH3 in our final rxn). With these changes, you will get H2(g)+Br2(l)->2HBr(g) as your overall rxn abd -72.80 as your rxn enthalpy.


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