From the reaction below, calculate the free energy of formation for C3H6(g) at 25°C, if ΔGfo(CH4(g)) is -50.790 kJ/mol.

3CH4(g) ↔ C3H6(g) + 3H2(g) ΔGo = 215.302 kJ/mol

i've tried doing final-initial, and products minus reactants but i'm not getting the right answer

3 answers

Instead of me flying in the dark why don't you post your work on the products-reactants you've tried and let me find the error in your work.
3(2.02) + 3(12.01) + 6.06 = 48.15 mol products

3(12.01)+ 3(4.04)=48.15 mol reactants
Prod-React=0
Where in the world are you getting your numbers. I don't see 2.02 or 12.01 in the problem. Wait, I suspect you are using molar mass H2 and molar mass C and molar mass. That isn't what the problem is all about.
dGrxn = (n*dGproducts) - (n*dGreactants)
They give you dGrxn = 215.302 so
215.302 = (1*dGC3H6 + 1*dGH2) - (3*dGCH4)
dGC3H6 is what we solve for.
dGH2 = 0 (dG for all elements is zero)
dGCH4 they give you as -50.790
215.302 = (1*dGC3H6 + 0) -(3*-50.790)
Solve for dGC3H6
Hope this helps.