H_2(g) + CO_2(g) leftrighttharpoons H_2O(g) + CO(g) is 4.40 at 2000 K. (a) Calcu

Question

H_2(g) + CO_2(g) leftrighttharpoons H_2O(g) + CO(g) is 4.40 at 2000 K. (a) Calculate deltaGdegrees for the reaction (b) Calculate deltaG for the reaction when the partial pressures arc P_H_2 = 0.25 atm, P_co_2 = 0.78 atm. P_H_2O =0.66 atm. P_CO 1.20 atm 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 degree C using the information below. CuO(s) leftrighttharpoons Cu(s) +1/2O_2(g) Delta G degree=127.2 KJ/mol C(graphite) + 1/2 O_2(g) leftrighttharpoons Co(g) delta G degree=-137.3 KJ/mol Even though the coupled reaction is now spontaneous, it still does not produce Cu without heating the reaction to about 400 degreeC. Explain.

Explanation / Answer

7) Relation between G and Go is given by

   G = Go + RT In Kp

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

Kp = PH2O PCO / PH2 PCO2

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

= 4.06

K = 4.06

Go

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

Gfo [H2O(g)] = -228.4 kJ/mol

Gfo [CO(g)] = -137.2 kJ/mol

Gfo [CO2(g)] = -394.5 kJ/mol

Gfo [H2(g)] = 0 kJ/mol

Gorxn = Gfo(products) - Gfo(reactants)

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

= + 28.9 kJ/mol

Therefore, Go = + 28.9 kJ/mol

G

Go = + 28.9 kJ/mol = + 28900 kJ/mol

Temperature T = 2000 K

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

G =  Go + 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)

   Go = 127.2 kJ/mol - 137.3 kJ/mol

= -10.1 kJ/mol

  Go =  -10.1 kJ/mol

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.