Question

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) 180°C atm

(b) 460°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

Answer 1

PbCO3(s) ------> PbO(s) + CO2(g) :G^o = ?

(a) At 180 oC = 180+273 = 453 K

G^o = H^o - TS^o

= (+88.3 kJ/mol) - (453 x 151.3x10^-3 kJ/(mol-K))

= +19.8 kJ/mol

= +19.8x10³ J/mol

We know that G^o = -RT lnK

Where

R = gas constant = 8.314 J/mol-K

K = Equilibrium constant = ?

Plug the values we get ln K = -G^o /(RT)

= -( 19.8x10³ ) / ( 8.314 x 453 )

= -5.25

K = e^-5.25

= 5.21x10^-3

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(b) At 460 oC = 180+273 = 733 K

G^o = H^o - TS^o

= (+88.3 kJ/mol) - (733 x 151.3x10^-3 kJ/(mol-K))

= -22.6 kJ/mol

= -22.6x10³ J/mol

We know that G^o = -RT lnK

Where

R = gas constant = 8.314 J/mol-K

K = Equilibrium constant = ?

Plug the values we get ln K = -G^o /(RT)

= -( -22.6x10³ ) / ( 8.314 x 733 )

= -3.71

K = e^-3.71

= 0.0245