13. Consider the reaction shown below.
PbCO3(s) PbO(s) + CO2(g)
Calculate the equilibrium pressure of CO2 in the system at the following temperatures.
(a) 160°C
____ atm
(b) 490°C
____ atm
Note: To find the value of the equilibrium constant at each temperature you must first find the value of G0 at each temperature by using the equation
G0 = H0 - TS0
For this reaction the values are H0 = +88.3 kJ/mol and S0= 151.3 J/mol*K
(a)
T = 160 °C = (273 + 160)K = 433 K
G0 = H0 - TS0
= 88.3 kJ/mol - [ 433 K (151.3 J/mol*K) ]
= 88300 J/mol - 65512.9 J/mol
= 22787.1 J/mol
= 22.79 kJ/mol
Now
G0 = - R T ln Kp
22787.1 J/mol = - (8.314 J/Kmol) ( 433 K) ln Kp
22787.1 J/mol = (- 3600 J/mol) x ln Kp
ln Kp = (22787.1 J/mol) / (- 3600 J/mol)
ln Kp = - 6.3
Kp = e- 6.3 = 1.84
PbCO3(s) PbO(s) + CO2(g)
Kp = PCO2 Since, PbCO3(s) and PbO(s) are solids.
PCO2 = 1.84 atm
(b)
T = 490 °C = (273 + 490)K = 763 K
G0 = H0 - TS0
= 88.3 kJ/mol - [ 763 K (151.3 J/mol*K) ]
= 88300 J/mol - 115442 J/mol
= - 27142 J/mol
Now
G0 = - R T ln Kp
- 27142 J/mol = - (8.314 J/Kmol) ( 763 K) ln Kp
27142 J/mol = (6344 J/mol) x ln Kp
ln Kp = (27142 J/mol) / (6344 J/mol)
ln Kp = 4.3
Kp = e4.3 = 73.7
PbCO3(s) PbO(s) + CO2(g)
Kp = PCO2 Since, PbCO3(s) and PbO(s) are solids.
PCO2 = 73.7 atm
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