Question

Using values from Appendix C of your textbook, calculate the value of Ho, So, and Go for each of the following reactions at standard temperature (298 K). Next calculate G if ALL of the GASSES in the reaction have partial pressures of 0.2 atm.

(a) NaOH(s) + CO2(g) NaHCO3(s) Ho = -127.44 kJ/mol

So = -176.02 J/mol-K

Go = -74.96 kJ/mol

If all gasses have partial pressures of 0.2 atm then G = ________ kJ/mol.

(b) 2 Fe(s) + 3 CO2(g) Fe2O3(s) + 3 CO(g)

Ho = ______kJ/mol

So = _______ J/mol-K

Go = ___________kJ/mol

If all gasses have partial pressures of 0.2 atm then G = ___________kJ/mol.

(c) 2 C2H6(g) + O2(g) 2 C2H4(g) + 2 H2O(g)

Ho = ___________ kJ/mol

So = ________ J/mol-K

Go = ___________ kJ/mol

If all gasses have partial pressures of 0.2 atm then G = _________ kJ/mol.

(d) 6 C(graphite) + 6 H2O(l) C6H12O6(s)

Ho = _______ kJ/mol

So =________ J/mol-K

Go =_________ kJ/mol

If all gasses have partial pressures of 0.2 atm then G = __________ kJ/mol.

Answer 1

G= GO+RTlnQ

Q= 1/partial pressure of CO2= 1/0.2= 5

G= -74.96 +0.0821*(298)*ln5= -35.38 Kj/mol

2. The reaction is 2Fe+3CO2----> Fe2O3+3CO

Enthalpy change = Sum of enthalpy of products- sum of enthalpy of reactants

for Fe(s), enthalpy change =0

Enthalpy change = 1* enthalpy of Fe2O3 + 3* enthalpy of CO- 3*enthalpy of CO2

=1*(-824.2)+ {3*-110.53)- {3*(-393.51)} =24.74 Kj/mol

entropy change =1* enthalpy of Fe2O3+3* enthalpy of CO- {3* enthalpy of CO2+2* entropy of Fe)

entropies : Fe2O3= 87.4 CO :197.4   CO2 = 213.6 and Fe= 27.3 ( all in J/mole.K)

entropy change = 1*87.4+3*197.4- (3*213.6+2*27.3)=-15.8J/K

GO= HO-TS0= 24.74+298*15.8/1000 =29.45 KJ

G = G+RTlnQ, Q = [CO]3/[CO2]3 =1

G=GO= 29.45 KJ

3. 2 C2H6(g) + O2(g) 2 C2H4(g) + 2 H2O(g)

enthalpy change = 2*52.3+2*(-241.8) -{ 2*-84.7}=-209.6 Kj

entropy change = 2*219.5+2*188.7- (2*229.5+205}= 152.4 J/K

GO= HO-TSO=-209.6-298*152.4/1000 =-255KJ

G= GO+RT ln Q

Q= [H2O]² [C2H4]² /[C2H6]2 [O2] = 0.4*0.4/0.2= 0.8 ( since partial pressure of H2O= 2*0.2 =0.4 atm)

G= -255+298*0.0821*ln(0.8)=-249.5 KJ/mol

d) Enthalpy change = enthalpy change of C6H12O6-6* enthalpy change of water

=-1273.1+6*285.3= 438.7 KJ

entropy change =   209.2-6*69.9= -210.2 J/K =-0.2102 Kj/K

GO= HO-TSO= 438.7 +298.15*0.2102 =501.3711KJ

since there are no gases

G= GO+RTlnQ= G0= 501.371 KJ/mol