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!
I understand Hess's law and
I know I'm doing something silly wrong, thanks in advance!
1 Answer
Make sure you know that
Also remember to convert to
Write the formation reaction:
#4"C"(s) + 2"H"_2(g) -> "CH"_4(g)#
From this, for some state function
#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")#