I need help with problem 4. I have included problems 7-4, 8-7, and problem three
ID: 699255 • Letter: I
Question
I need help with problem 4. I have included problems 7-4, 8-7, and problem three for reference.
Explanation / Answer
the reaction is CH4+CO2----->2CO+ 2H2
standard heats of formation ( KJ): CO= -110.5, CO2= -393.5 and CH4=-76, H2=0
enthalpy change ,deltaH= sum of standard heat of formation of products- sum of standard heat of formation of reactants =
2*(-110.5)+ 0- {-76+(-393.5)}=248.5Kj
standard entropy (J/mole.K) : CO=197.6, H2=130.6, CO2=213.6 and CH4=186.2
entropy change, detlaS = 2*197.6+2*130.6- (213.6+186.2) =256.6 J/K
standard gibbs free eneergy change= deltaH-T*deltaS= 248.5- 298*256.6/1000 Kj=172 KJ
since deltaG is +ve, the process is not spontaneous. since deltaH is +ve, the reaction is endothermic and increased temperature as per lechatlier principle favors the backward reaction to compensate the effect of temperature increases. So low temperarues are preferred.
there are more moles on products than reactants. Since decreased pressure favors the forward reaction.
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