What is the Gibbs free energy of formation of methane at 298 K, using the given information?

I understand Hess's law and #DeltaG^o# laws, but every time I plug in values I'm not getting the answer of -50.7 kJ/mol. Here's what I'm given:

#CH_4(g)#: #DeltaH_f^o# = -74.8 kJ, #S^o# = 186.3 J/mol*K

#H_2(g)#: #S^o# = 130.7 J/mol* K

#CO_2(g)#: #DeltaH_f^o# = -393.5 kJ/mol, #S^o# = 213.7 J/mol*K, #DeltaG_f^o# = -394.4 kJ/mol

#H_2O(g)#: #DeltaH_f^o# = -241.8 kJ/mol, #S^o# = 188.8 J/mol*K, #DeltaG_f^o# = -228.6 kJ/mol

I know I'm doing something silly wrong, thanks in advance!

1 Answer
Truong-Son N.
Jan 26, 2018

Make sure you know that #S^@# at #25^@ "C"# and #"1 atm"# may NOT be zero for elements in their standard states, since the reference state for entropy is #"0 K"# and #"1 atm"#.

Also remember to convert to #"kJ/K"# before using it.

Write the formation reaction:

#4"C"(s) + 2"H"_2(g) -> "CH"_4(g)#

From this, for some state function #Y#,

#DeltaY_(rxn)^@ = sum_P n_P DeltaY_(f,P)^@ - sum_R n_R DeltaY_(f,R)^@#

if #Y = H,G# (enthalpy, Gibbs), or

#= sum_P n_P Y_(P)^@ - sum_R n_R Y_(R)^@#

if #Y = S#, entropy.

#n# is the mols of stuff, #R# is reactants and #P# is products.

So, the change in enthalpy is:

#DeltaH_(rxn)^@ = overbrace("1 mol" cdot -"74.8 kJ/mol")^(CH_4(g)) - [overbrace("1 mol" cdot "0 kJ/mol")^(C(gr)) + overbrace("2 mol" cdot "0 kJ/mol")^(H_2(g))]#

#= -"74.8 kJ"#

And the change in entropy is

#DeltaS_(rxn)^@ = overbrace("1 mol" cdot "186.3 J/mol"cdot"K")^(CH_4(g)) - [overbrace("1 mol" cdot "5.833 J/mol"cdot"K")^(C(gr)) + overbrace("2 mol" cdot "130.7 J/mol"cdot"K")^(H_2(g))]#

#= -"80.933 J/K" = -"0.0809 kJ/K"#

So, the change in Gibbs's free energy is

#color(blue)(DeltaG_(rxn)^@) = DeltaH_(rxn)^@ - TDeltaS_(rxn)^@#

#= -"74.8 kJ" - "298 K" cdot -"0.0809 kJ/K"#

#= -"50.68 kJ/mol" -> color(blue)(-"50.7 kJ/mol")#

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