CO2 DELTA_H = -393.5 ; DELTA_G = -394.4 ; S= 213.6 PbO DELTA_H = -217.3 ; DELTA_

Question

CO2 DELTA_H = -393.5 ; DELTA_G = -394.4 ; S= 213.6
PbO DELTA_H = -217.3 ; DELTA_G = -187.9 ;S= 68.7
PbCO3 DELTA_H = -699.1; DELTA_G=-625.5; S = 131.0

Consider the following reaction : Part A Using data in Appendix C in the textbook, calculate the equilibrium pressure of CO2 in the system at 180 Degree C. Express your answer using two significant figures. Part B Using data in Appendix C in the textbook, calculate the equilibrium pressure of CO2 in the system at 490 Degree C. Express your answer using two significant figures.

Explanation / Answer

The appropriate equation to begin with is:

Delta G = Delta Go + R T ln Q

where Q has the form of the equilibrium constant.

Now, at equilibrium, DeltaG = 0, so

Delta Go = - RT ln Keq

For this reaction, Kp = [PbO][CO2] / [PbCO3] = x2/x = x (where x is partial pressure of co2)

Now,

Delta Go = G [PbO]+G[CO2]-G[PbCO3] = -394.4-187.9+625.5 = 43.2

a) T = 180+273=453 K

Delta Go = - RT ln Kp = -8.314 J/molK*453K* ln[x]

=> 43.2 = -3766.2 ln[x] => x = 0.988

b) T = 490+273 = 763

Delta Go = - RT ln Kp = -8.314 J/molK*763K* ln[x]

=> ln[x] = -0.0068 => x = 0.993

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