Question

7.

H2(g) + CO2(g) ⇌ H2O(g) + CO(g) 

 is 4.40 at 2000 K. (a) Calculate ΔG° for the reaction (b) Calculate ΔG for the reaction when the partial pressures arc PH2 = 0.25 atm, Pco2 = 0.78 atm. PH2O =0.66 atm. PCO 1.20 atm 

8. Heating copper(II) oxide does not produce an appreciable amount of Cu. However if this reaction is coupled to the conversion of graphite to carbon monoxide, it becomes spontaneous. Write the coupled reaction and calculate its Delta G degree at 25℃ using the information below. 

Even though the coupled reaction is now spontaneous, it still does not produce Cu without heating the reaction to about 400℃. Explain.

Answer 1

7) Relation between ∆G and ∆G^o is given by

   ∆G = ∆G^o + RT In Kp

Kp:   Given reaction is H2(g) + CO2 (g) <----------> H2O (g) + CO (g)

Kp = P_H2O P_CO / P_H2 P_CO2

= (0.66 atm x 1.2 atm) / ( 0.25 atm x 0.78 atm)

= 4.06

K = 4.06

∆G^o

H2(g) + CO2 (g) <----------> H2O (g) + CO (g)

ΔG_f^o [H2O(g)] = -228.4 kJ/mol

ΔG_f^o [CO(g)] = -137.2 kJ/mol

ΔG_f^o [CO2(g)] = -394.5 kJ/mol

ΔG_f^o [H2(g)] = 0 kJ/mol

ΔG^orxn = ΔG_f^o(products) - ΔG_f^o(reactants)

= -228.4-137.2 - [-394.5-0]

= + 28.9 kJ/mol

Therefore, ΔG^o = + 28.9 kJ/mol

∆G

∆G^o = + 28.9 kJ/mol = + 28900 kJ/mol

Temperature T = 2000 K

R = universal gas constant = 8.314 J/K/mol

∆G =  ∆G^o + RT In Keq

= 28900 J/mol + (8.314 J/K/mol) (2000 K) In (4.06)

= + 52198 J/mol

∆G = + 52198 J/mol

Therefore,    ∆G = + 52198 J/mol

8) Coupled reaction is

CuO(s) + C (graphite ) + 1/2 O2 (g) <------------> Cu(s) + 1/2 O2 (g) + CO (g)

Cancel similar terms on both sides,

CuO(s) + C (graphite ) <------------> Cu(s) + CO (g)

   ∆G^o = 127.2 kJ/mol - 137.3 kJ/mol

= -10.1 kJ/mol

  ∆G^o =  -10.1 kJ/mol