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

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??

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

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