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) 300°C ____atm

(b) 550°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

A) Delta H° in Joules.

Delta G° = Delta H° - T * Delta S°

Delta G° = (88.3*1000) - (300+273)(151.3)

Delta G° = 1605.1 J

R is gas constant with value 8.314. Kp is equilibrium constant.

Delta G° = - R * T * ln Kp

1605.1 = - 8.314 * (300+273) ln Kp

ln Kp = - 0.337

Kp = 0.714 ...Answer

B)

Delta G° = Delta H° - T * Delta S°

Delta G° = (88.3*1000) - (550+273)*(151.3) = -36219.9 J

Again,

Delta G° = - R * T * ln Kp

-36219.9 = - 8.314 * (550+273) * ln Kp

ln Kp = 5.2934

Kp = 199.03

If,

ln Kp = 5.3

Then Kp = 200.3 .... Answer

Let me know if any doubts.