The equilibrium constant K p for the reaction H 2(g) + CO 2(g) H 2 O (g) + CO (g

ID: 975837 • Letter: T

Question

The equilibrium constant Kp for the reaction

H2(g) + CO2(g) H2O(g) + CO(g)

is 4.40 at 2000K

a) Calculate Go for the reaction.

HINT Since Go is at standard state of 25oC (the temperature of this reaction was done at 2000K), you will need to do find the free energy change of formation from the values in an Appendix and use the appropriate version of sum(products)-sum(reactants) mentioned in the textbook.

b) Using the Kp, Go ,and temperature above, calculate G for the reaction when the partial pressures are PH2 = 0.25atm, PCO2 = 0.78atm, PH2O = 0.66atm, and PCO = 1.20 atm

HINT: you can use G = Go + RTln(Q) to figure out the free energy of a reaction under non-standard state conditions.

Explanation / Answer

You haven’t included the appendix; hence I have used the values obtained from internet. I found the following values for G°f as

Compound (state)

G°f (kJ/mol)

H2 (g)

0.00

CO2 (g)

-394.4

H2O (g)

-228.59

CO (g)

-137.23

The standard Gibb’s free energy change of the reaction is given as

G° = (nG°f)products - (nG°f)reactants where n is the number of moles.

For the given reaction, we have,

G° = {(-228.59) + (-137.23)} kJ/mol – {(0.00) + (-394.4)} kJ/mol = (-365.82) kJ/mol – (-394.4) kJ/mol = -(365.82 + 394.4) kJ/mol = 28.58 kJ/mol

The standard Gibb’s free energy for the reaction at 298 K (i.e, 25°C) is 28.58 kJ/mol (ans)

For non-standard conditions, the Gibb’s free energy change is given as

G = G° + RTlnQ

where T is the reaction temperature, R is the gas constant and Q is the reaction quotient. For the present reaction, Q is given as

Q = (pH2O)(pCO)/(pH2)(pCO2) = (0.66 atm)(1.20 atm)/(0.25 atm)(0.78 atm) = 4.061

Therefore,

G = G° + RTln(4.061) = 28.58 kJ/mol + (8.314 J/mol.K)(2000 K)ln(4.061)

= 28.58 kJ/mol + (16.628 kJ/mol)1.401 = 28.58 kJ/mol + 23.295 kJ/mol

= 51.875 kJ/mol 51.88 kJ/mol (ans)