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Postby JChen_2I » Thu Jan 30, 2020 10:58 am

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 was wondering why you don't have to divide 92.22 by 3 in order to get the reaction enthalpy for one mole of H2 since there's only one mole in the synthesis of hydrogen bromide gas?

Jordan Young 2J
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Re: 4D19

Postby Jordan Young 2J » Thu Jan 30, 2020 11:04 am

In this problem you are using Hess's Law to find the reaction enthalpy. You have to rearrange/multiply the given equations to eventually end up with the equation H2 + Br2 --> 2HBr when you add all of the reactions together. To do this you will need to reverse the first equation and multiply it by a coefficient of 2. For the second equation, you need to reverse the equation. And for the third equation you just leave it as it is. By doing this, most of the reactants and products will cancel and you will end up with the reaction H2(g)+Br2(l)->2HBr(g) (The 3H2 will be subtracted from the 4H2 as they will be on opposite sides of the reaction and you will end up with just H2). After you end up with the correct equation you need to alter the corresponding delta H (if you multiply a reaction by a coefficient, you multiply the delta H by that coefficient and if you reverse the equation, you change the sign of the delta H). After you do this you can add up all of the delta Hs of the three equations and you will end up with the correct answer.

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