Question

Delta G f(CO)= -137.3 kJ/mol
Delta G f(CH3OH)= -166.3 kJ/mol
Delta H f(CO)= -110.5 kJ/mol
Delta H f(CH3OH)= -238.7 kJ/mol
S(CO)= 197.9 J/K mol
S(CH3OH)= 126.8 J/K mol

Calculate Delta G at 25 Degrees Celcius.

Answer 1

ΔG=ΔH-TΔS

IMPORTANT: ΔHf and ΔGf are ZERO for elements in their standard states (H_2(g)) but S^o is NOT zero

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

ΔH= n(ΔHf products)-n(ΔHf reactants) Where n is for coefficients

ΔH= [1(-238.7 kJ/mol)]-[1(-110.5 kJ/mol)+2(0)]=-128.2 kJ

ΔS= n(S^o products)-n(S⁰ reactants) Where n is for coefficients

ΔS= [1(126.8 J/mol*K)]-[1(197.9 J/mol*K)+2(130.7 J/mol*K)] = -332.5 J/K

NOTE: ΔH is in kJ and ΔS is in J, convert one of them

ΔS= -332.5 J/k * (1kJ/1000J) = -0.3325 kJ/K

K= 25+273=298K

ΔG= -128.2kJ-(298K)(-0.3325 kJ/K) = -29.115 kJ

Answer 2

The given reaction

CO + 2H2 = CH3OH

At standard state G°f(H2) = 0

Standard free energy change of reaction

G° = G°f(CH3OH) - 2*G°f(H2) - G°f(CO)

= - 166.3 - 2*0 - (-137.3)

= - 29 kJ/mol