April 29, 2024, 12:22:48 PM
Forum Rules: Read This Before Posting


Topic: enthalopy  (Read 2775 times)

0 Members and 1 Guest are viewing this topic.

Offline Bjc51192

  • Regular Member
  • ***
  • Posts: 63
  • Mole Snacks: +0/-0
enthalopy
« on: April 05, 2011, 07:59:40 PM »
Given the following data:
2O3 → 3O2 ΔH = -430 kJ
O2 → 2O ΔH = +500 kJ
NO + O3 → NO2 + O2 ΔH = -203 kJ


Calculate ΔH for the reaction
NO + O → NO2


What I did first was I saw that they wanted O as a reactant instead of a product so I switched equation 2 around.

2O----->O2

we need O by itself so I divide by 2

O-------1/2O

I understand that we have to keep equation 3 the same. So I know im making a mistake in equation one.

I am dividing by 2

O3-------------->3/2O2

but the equation doesnt seem to be cancling out.

What exacly am I missing?


Offline Schrödinger

  • Chemist
  • Sr. Member
  • *
  • Posts: 1162
  • Mole Snacks: +138/-98
  • Gender: Male
Re: enthalopy
« Reply #1 on: April 06, 2011, 12:29:29 AM »
Try this :
Reverse equations 1 and 2. Keep equation 3 the same. And do some intuitive algebra.
"Destiny is not a matter of chance; but a matter of choice. It is not a thing to be waited for; it is a thing to be achieved."
- William Jennings Bryan

Offline opti384

  • Full Member
  • ****
  • Posts: 434
  • Mole Snacks: +33/-25
  • Gender: Male
    • In the Search for the Laws of Nature
Re: enthalopy
« Reply #2 on: April 06, 2011, 12:57:07 AM »
If you fully understand the Hess's Law and do some practices, such kind of questions will be piece of cake.

Offline Bjc51192

  • Regular Member
  • ***
  • Posts: 63
  • Mole Snacks: +0/-0
Re: enthalopy
« Reply #3 on: April 06, 2011, 09:02:41 AM »
Yeap got it ty. My problem is finding questions to do =/

Sponsored Links