November 28, 2021, 06:57:32 PM
Forum Rules: Read This Before Posting


Topic: Gibbs free energy for spontaneous reaction  (Read 7283 times)

0 Members and 1 Guest are viewing this topic.

thasan

  • Guest
Gibbs free energy for spontaneous reaction
« on: June 03, 2005, 03:02:45 AM »
Hi,


I am a bit confused about calculation of Gibbs free energy for the folowing equation:

CH4 = C + 2H2

I considered C as graphite, CH4 and H2 as gas.

According to my calculation, DH = 74, DS = 0.080844

So the reaction will be spontaneous at 915 k (642 C) according to DG = DH - T*DS (putting DG =0)

But I have found two papers where they indirectly mentioned that they found DH =90 for this equation which gives spontaneous reaction temperature at 1113 k (840 C).

Can anyone please explain??




Also, why do we consider C as graphite, not as gaseous? Is it because of temperature??

 
regards,
TH

Offline xiankai

  • Chemist
  • Full Member
  • *
  • Posts: 786
  • Mole Snacks: +77/-37
  • Gender: Male
Re:Gibbs free energy for spontaneous reaction
« Reply #1 on: June 03, 2005, 05:47:15 AM »
carbon is naturally found as a solid, not gas

i believe they consider C to be graphite because its the allotrope of carbon at rtp.
one learns best by teaching

thasan

  • Guest
Re:Gibbs free energy for spontaneous reaction
« Reply #2 on: June 03, 2005, 06:51:33 AM »
thanx xiankai.

it might be the case. but then in which situations we need to consider C as gaseous (with different thermodynamic properties than graphitic C)?

and do you have any idea about the spontaneous reaction temperature here??


Offline xiankai

  • Chemist
  • Full Member
  • *
  • Posts: 786
  • Mole Snacks: +77/-37
  • Gender: Male
Re:Gibbs free energy for spontaneous reaction
« Reply #3 on: June 03, 2005, 09:29:01 AM »
to my knowledge so far, graphite the lowest melting/boiling point among the allotropes of carbon.

sometimes, different compounds in different environments(presence of catalyst for example) will yield different enthalpy changes. it is hard to find a constant, and a difference of a few DH is usually common betweem different papers.

u may have to post the entire question for us to get a better idea of how u calculated DS and DH
one learns best by teaching

thasan

  • Guest
Re:Gibbs free energy for spontaneous reaction
« Reply #4 on: June 03, 2005, 11:04:02 AM »
ok, i have used the thermodynamic values from

1. http://www.codata.org/codata/databases/key1.html
2. http://dwb.unl.edu/Teacher/NSF/C09/C09Links/www.chem.ualberta.ca/courses/plambeck/p101/p00403.htm

other related information that i used are from

http://www.cartage.org.lb/en/themes/Sciences/Chemistry/Miscellenous/Helpfile/spontaneityentropy/GibbsFreeEnergy.htm

and other related sources like wikipedia (which are not really necessary).


how i calculated the values:

DH = DH(C) + 2DH(H2) - DH(CH4) = 0 + 2*0 - (-74) = 74 kJ/mol

DS = DS(C) + 2DS(H2) - DS(CH4) = 5.74 + 2*130.684 - 186.264 = 80.844 J/mol*K = 0.080844 kJ/mol*K

for spontaneous reaction,
DG = 0

so, DG = DH - T*DS = 0 = 74 -T*0.080844
hence, T = 74/0.080844 = 915.XXX K == 642.XXX C

i hope this helps.
if u require, i can mail you a single pdf file with all the information from the webpages above. if you need further information, please let me know here or mail me at tawfique.hasan@gmail.com
regards,
TH



GCT

  • Guest
Re:Gibbs free energy for spontaneous reaction
« Reply #5 on: June 03, 2005, 01:51:12 PM »
Hi,


I am a bit confused about calculation of Gibbs free energy for the folowing equation:

CH4 = C + 2H2

I considered C as graphite, CH4 and H2 as gas.

According to my calculation, DH = 74, DS = 0.080844

So the reaction will be spontaneous at 915 k (642 C) according to DG = DH - T*DS (putting DG =0)

But I have found two papers where they indirectly mentioned that they found DH =90 for this equation which gives spontaneous reaction temperature at 1113 k (840 C).

Can anyone please explain??




Also, why do we consider C as graphite, not as gaseous? Is it because of temperature??

 
regards,
TH

You need to know the specifics of the equation, such as it's exact form.  From there, find the standard enthalpy of formation and entropy for each compound, and calculate the net entropy.  It should be straightforward, be sure to pay attention to the molar values of the enthalpy and entropy.

thasan

  • Guest
Re:Gibbs free energy for spontaneous reaction
« Reply #6 on: June 03, 2005, 04:45:33 PM »
yes, please. i do understand what you meant.
but the equation i posted is a simple one, i want to find the temp at which CH4 decomposed to C and H2.
Did I make any mistake in my last post which shows my calculation? i dont know whether i am correct or not  ???
please help

Sponsored Links