Question

At 900 degrees celsius, Kc = 92.6 M^-1 for the euilibrium reaction of calcium oxide and carbon dioxide to make calcium carbonate

(a) write the balance equation for the reaction:

(b) For this exothermic reaction, indicate which way each of the following changes will shift the position of equilibrium:

(i) increase the pressure (three different possible ways):

(1) Decrease the volume (which increases pressure P):

(2) Add an inert gas:

(3) Add some more carbon dioxide gas:

(ii) Decreasing the temperature

(iii) Removing some of the calcium carbonate

(c) If a mixture of all three compounds in part (a) is placed in 1.0L vessel at 900 degrees celsius, for each of the following mixtures, will the amount of calciu carbonate increase, decrease, or remain the same?

(i) 0.0550 mol calcium carbonate, 0.645 mol calcium oxide, and 0.149 mol carbon dioxide?

(ii) 0.0304 mol calcium carbonate, 0.0454 mol calcium oxide, & 0.0108 mol carbon dioxide?

(iii) 0.250 mol calcium carbonate, 0.467 mol calcium oxide, & 0.00874 mol of carbon dioxide?

Answer 1

Solution :-

At 900 degrees celsius, Kc = 92.6 M^-1 for the euilibrium reaction of calcium oxide and carbon dioxide to make calcium carbonate

(a) write the balance equation for the reaction:

Solution :- Balanced reaction equation is as follos

CaO(s) + CO2(g) ------ > CaCO3(s)

(b) For this exothermic reaction, indicate which way each of the following changes will shift the position of equilibrium:

(i) increase the pressure (three different possible ways):

(1) Decrease the volume (which increases pressure P):

Decreasing volume will increase the concentration of the CO2 gas therefore equilibrium will shift to right side. Because increase in the concentration of the reactant shifts the equilibrium to product side.

(2) Add an inert gas:

Adding inert gas will help to increase the pressure of the CO2 therefore it will shift equilibrium to right side.

(3) Add some more carbon dioxide gas:

Addition of CO2 means adding more reactant therefore equilibrium will shift to right side.

(ii) Decreasing the temperature

Since reaction is exothermic therefore decreasing the temperature will shift the equilibrium to product side (right side)

(iii) Removing some of the calcium carbonate

CaCO3 is the solid product and the solid do not affect the equilibrium of the reaction therefore it will not have any effect on equilibrium.

(c) If a mixture of all three compounds in part (a) is placed in 1.0L vessel at 900 degrees celsius, for each of the following mixtures, will the amount of calciu carbonate increase, decrease, or remain the same?

Solution

For the reaction equilibrium constant equation is Kc= [CO2]

(i) 0.0550 mol calcium carbonate, 0.645 mol calcium oxide, and 0.149 mol carbon dioxide?

Since the concentration of the CO2 is less than Kc therefore it will shift to right side

(ii) 0.0304 mol calcium carbonate, 0.0454 mol calcium oxide, & 0.0108 mol carbon dioxide?

Since the concentration of the CO2 is less than Kc therefore it will shift to right side

(iii) 0.250 mol calcium carbonate, 0.467 mol calcium oxide, & 0.00874 mol of carbon dioxide?

Since the concentration of the CO2 is less than Kc therefore it will shift to right side