Wood alcohol i.e. methanol CH3OH(l) might be a cheap source of methane CH4(g) (which is natural gas) where 1 mol of liquid methanol decomposes into 1 mol of methane gas and 1/2 mol of oxygen O2 (g). Write out the reaction (as described) and using the data below determine if it is thermodynamically feasible at 1 atm and 25oC (i.e. spontaneous in the forward direction as written)? What drives this reaction ? Calculate DHorxn, DSorxn and the standard Gibbs free energy for the reaction and use these values in your answer.

Question

Wood alcohol i.e. methanol CH3OH(l) might be a cheap source of methane CH4(g) (which is natural gas) where 1 mol of liquid methanol decomposes into 1 mol of methane gas and 1/2 mol of oxygen O2 (g). Write out the reaction (as described) and using the data below determine if it is thermodynamically feasible at 1 atm and 25oC (i.e. spontaneous in the forward direction as written)? What drives this reaction ? Calculate DHorxn, DSorxn and the standard Gibbs free energy for the reaction and use these values in your answer.

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DG= DH - TDS
TK = T oC + 273.15
species DHoF (kJ/mol) DHoAC (kJ/mol) DSoAC (J/mol K) SoF (J/mol-K)
CH3OH -238.66 -2075.11 -651.2 126.8
CH4 (g) -74.81 -1662.09 -430.684 186.264
O2 0 -498.340 -116.972 205.138
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2. Determine (from data above) a value for the enthalpy of atomization of methanol. Compare this value to the enthalpies of atom combination for methane and oxygen in the context of the reaction. In terms of bond breaking and bond formation, how do these values explain the overall thermodynamic feasibility of the reaction? Hint: what role does the bond enthalpy of CH3OH play in determining DGo for the reaction.

Answer 1

To be honest about it your post is confusing at best. Your spacing of the table just doesn't work on an html board. You can do the first one this way.

dHorxn = (n*dHoproducts) - (n*dHoreactants)
dSorxn = (n*dSoproducts) - (n*(dSoreactants)
dGorxn = dHorxn - TdSorxn
If dGo rxn is + the reaction is not spontaneous. If dGo is - the reaction as written is spontaneous (at the specified temperature). I didn't run the numbers because it is difficult to determine what they are; however, IF WE ASSUME YOU GET A NEGATIVE NUMBER FOR dGrxn (only then), the driving force will be because of increase of S because methanol liquid goes to CH4 (as gas) and O2 (as gas) and those changes are always entropy increases.
I won't touch #2.