November 26, 2024, 05:18:12 PM
Forum Rules: Read This Before Posting


Topic: Enthalpy of Formation!!!  (Read 8989 times)

0 Members and 1 Guest are viewing this topic.

Offline lilns8252

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
Enthalpy of Formation!!!
« on: March 20, 2009, 04:33:02 PM »
I have a couple homework questions due tonight and i've tried everything but i dont seem to understand!!! ???
the question is:

Using enthalpies of formation, calculate the quantity of heat produced when 4.65  of butane is completely combusted in air under standard conditions. The answer should be in kJ

Offline typhoon2028

  • Full Member
  • ****
  • Posts: 251
  • Mole Snacks: +18/-12
Re: Enthalpy of Formation!!!
« Reply #1 on: March 20, 2009, 04:46:41 PM »
4.65 what?  Units?  (grams, kilograms, pounds, moles)

After you know that, calculate the heat of formation of butane.  This should be in kJ/mole.

Offline ARGOS++

  • Sr. Member
  • *****
  • Posts: 1489
  • Mole Snacks: +199/-56
  • Gender: Male
Re: Enthalpy of Formation!!!
« Reply #2 on: March 20, 2009, 04:47:33 PM »

Dear nimritsohal;

This page should be of help to you:
      http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation
And please use units!: What is 4.65 meaning grams, or kg, or moles?

Good Luck!
                    ARGOS++

Offline lilns8252

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
Re: Enthalpy of Formation!!!
« Reply #3 on: March 20, 2009, 04:58:47 PM »
sorry i forgotto add the units.. it was 4.65 grams

Offline lilns8252

  • Regular Member
  • ***
  • Posts: 10
  • Mole Snacks: +0/-0
Re: Enthalpy of Formation!!!
« Reply #4 on: March 20, 2009, 05:03:29 PM »
I solved for heat of formation of butane and got -5271kJ/mol... what do i do after that?? or is that the answer? because the units they want the answer in is just kJ.. wat happends to the moles then?

Offline ARGOS++

  • Sr. Member
  • *****
  • Posts: 1489
  • Mole Snacks: +199/-56
  • Gender: Male
Re: Enthalpy of Formation!!!
« Reply #5 on: March 20, 2009, 05:53:45 PM »

Dear nimritsohal;

If I’m correct, then Wiki reports another value than you:
            http://en.wikipedia.org/wiki/Butane_(data_page)

What's also required is a balanced reaction equation!

Good Luck!
                    ARGOS++

Sponsored Links