## 8.61

104969854
Posts: 36
Joined: Fri Sep 29, 2017 7:05 am

### 8.61

Has anyone figured out how to manipulate the equations to create the equation needed for the problem? So far, I have reversed and doubled the enthaply of the first reaction in order to cancel the NH4Br from equation (c) and the NH3 from equation (b). I'm stuck at what to do now, as we need to end up with H2+Br2==>2HBr by the end. If anyone could explain how they manipulated the equations that would be amazing!

Kaylin Krahn 1I
Posts: 57
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time

### Re: 8.61

So you know the first equation is multiplied by 2 to get 2HBr, the second equation is reversed, and the third is left alone.
If you write these out, it's easy to see which parts cancel. 2NH3, 2NH4Br, and N2 cancel leaving you with the desired H2+B2+2HBr. Remember to respectively change the delta H to get the total change in H.

Cooper1C
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: 8.61

Like balancing chemical reactions, I recommend manipulating the equations that have the least number of the same molecule first, and the should build on it.

Ashley Garcia 2L
Posts: 33
Joined: Fri Sep 29, 2017 7:05 am

### Re: 8.61

Reverse the first equation and multiply it by 2 to get HBr on the product side with a coefficient of 2.
Reverse the second equation to get the 2 NH3's to cancel out.
Leave the last equation as is.

You must also reverse the sign and multiply the first equations ΔH° by 2 and reverse the sign of the second equation's ΔH°.

See my attached work for the problem fully worked out. I like to keep track of my steps by writing "reverse" or "x2" to remember how I manipulated the equations.
Attachments IMG-0928.pdf

104969854
Posts: 36
Joined: Fri Sep 29, 2017 7:05 am

### Re: 8.61

Thank You!!

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