In: Chemistry
Consider the following system at equilibrium where Delta H° =
108 kJ, and Kc =
1.29×10-2, at 600
K.
COCl2(g) <--->
CO(g)
+ Cl2(g)
If the VOLUME of the equilibrium system is
suddenly increased at constant
temperature:
The value of Kc | A. increases. | |
B. decreases. | ||
C. remains the same. |
The value of Qc | A. is greater than Kc. | |
B. is equal to Kc. | ||
C. is less than Kc. |
The reaction must: | A. run in the forward direction to reestablish equilibrium. | |
B. run in the reverse direction to reestablish equilibrium. | ||
C. remain the same. It is already at equilibrium. |
The number of moles of Cl2will: | A. increase. | |
B. decrease. | ||
C. remain the same. |
Solution:
As ideal gas equation, PV=nRT
So, v n where, V= volume of gas, n= no. of molee of gas
P=pressure of gas
No. of moles in reactant side = 1
No. of moles in product side = 2
(i) option (C), value of Kc will remains the same as according to Le Chateiler's principle whenever there is change in volume, equilibrium will try to resist by adjusting the pressure of gas and will maintain equilibrium constant Kc.
Equilibrium constant Kc only get affected by temperature.
(ii) option (C), value of Qc is less than Kc because with increase in volume, pressure of products side will decrease more as there are more moles in product side hence, there will be decrease in pressure. So, will decrease in Qc by using its formula,
as Qc partial pressure of products.
Qc= where, P(CO)= partial pressure of CO
P(Cl2) = partial pressure of Cl2
P(COCl2) = partial pressure of COCl2
(iii) option (A), run in the forward direction to reestablish equilibrium as increase in volume will decrease more in pressure in products sides as there are more moles in products, so reaction will move forward to reestablish products.
(iv) option (A), moles of Cl2 will increase as there are increase in volume will tend to increase in moles of gas of CO(g) and Cl2(g) as according to ideal gas equation volume is directly proportional to moles.