## Homework 8.55

Sophia Kim 1C
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### Homework 8.55

The given reaction enthalpies are
2Ba (s) + O2 (g) ---> 2BaO (s) deltaH= -1107 kj
3Al (2) + O2 (g) ---> Al2O3 (s) deltaH = -1676 kj

and we have to calculate the reaction enthalpy for
3 BaO + 2Al ---> Al2O3 + 3Ba

In the solution manual the first reaction is reversed and multiplied by 3/2 so it becomes
3BaO --> 3Ba + 3/2 O2 which makes sense to me but then the second reaction is left exactly the same (deltaH = -1676 kj) but the equation changed to
3Al + 3/2 O2 --> Al2O3 so the O2 turned to 3/2O2 but the deltaH stayed the same. Are you supposed to just multiply 3/2 to cancel out the O2 but after the O2's cancel out you multiply the reaction by 3/2 to get rid of the changes made?

Johann Park 2B
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### Re: Homework 8.55

To balance the second equation, you must add the coefficient 3/2 to O2. This does not change the enthalpy of the reaction because the enthalpy of an element in its natural elemental state, such as O2 gas, is zero.

Anna Okabe
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

### Re: Homework 8.55

In case you haven't noticed, 8.55 is not included in the assigned problems of chapter 8

Return to “Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)”

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