Chapter 8, Problem 61

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Fayt Sarreal 1G
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Joined: Fri Jul 22, 2016 3:00 am

Chapter 8, Problem 61

Postby Fayt Sarreal 1G » Thu Jan 19, 2017 7:57 pm

Calculate the reaction enthalpy for the synthesis of hydrogen bromide gas, H2(g)+Br2(l) -> 2HBr.
The problem gives you 3 equations and their enthalpies but how do you use the three equations to obtain the reaction enthalpy for H2(g)+Br2(l) -> 2HBr. I tried to combine the equations like I did in problem 57 but things were not quite cancelling out. I was left with HBr+2N2+7H2->HBr.
Any insight?

Diwana Lucero 3K
Posts: 26
Joined: Fri Jul 15, 2016 3:00 am

Re: Chapter 8, Problem 61

Postby Diwana Lucero 3K » Thu Jan 19, 2017 8:08 pm

What I did was double the first equation ( NH3+HBr→NH4Br) and reverse it, so you get the ΔH= 2(+188.32kJ).Then I added that first equation to the reverse of the second equation (N2+ 3H2→ 2NH3 to 2NH3→ N2+3H2) which gets you ΔH=+92.22 kJ. The third equation just gets added to the first two equations without any changes, so you're left with H2+Br2→2HBr, with ΔH=-72.80 kJ.

Fayt Sarreal 1G
Posts: 25
Joined: Fri Jul 22, 2016 3:00 am

Re: Chapter 8, Problem 61

Postby Fayt Sarreal 1G » Thu Jan 19, 2017 8:18 pm

How do you know when you have to reverse the equation?

Allison Suzuki 2B
Posts: 26
Joined: Wed Sep 21, 2016 2:57 pm

Re: Chapter 8, Problem 61

Postby Allison Suzuki 2B » Fri Jan 20, 2017 4:45 pm

You only have to reverse the equation so the molecules will match the final equation. For example, since you were left with HBr on the reactants side of the equation, you would reverse the equation so that HBr will be on the product side.

Suzanne Reyes-Ingwersen 3A
Posts: 11
Joined: Wed Sep 21, 2016 2:56 pm

Re: Chapter 8, Problem 61

Postby Suzanne Reyes-Ingwersen 3A » Fri Jan 20, 2017 6:02 pm

Why can't you just take the -92.22 from the second equation and divide it by three in order to get H2, which would match the final equation?

Aashi_Patel_3B
Posts: 20
Joined: Wed Sep 21, 2016 2:57 pm

Re: Chapter 8, Problem 61

Postby Aashi_Patel_3B » Sun Jan 22, 2017 11:46 am

I don't think the KJ value is equally divided by the atoms or molecules, so you can't just go through and divide. You must manipulate the reactants and products and reverse positive and negative signs depending on if you moved products and reactants on opposite sides.


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