The document discusses chemical equilibrium for the reaction of PCl3 and Cl2 to form PCl5. It provides definitions for the equilibrium constant Kc and reaction quotient Qc. When 0.39 moles of PCl5 are removed from the system, the reaction quotient Qc will be less than the equilibrium constant Kc. This means the reaction must shift in the forward direction to reestablish equilibrium, and the concentration of Cl2 will decrease as a result.

1. Consider the following system at equilibrium where -87.9 kJ/mol, and K 83.3, at 500 K 16.
Calcul 2re 17. Le Ch 2re Preparation PCl3 (g) + C2 (g) PCIs (g) nen o.39 moles of PCls (g) are
removed from the Question Question 18. Le Ch equilibrium system at constant temperature: The
value of Kc The value of Qc 2re 19. Le Ch 2re 20. Le Ch 2re The reaction must Orun in the
forward direction to restablish equilibrium. Orun in the reverse direction to restablish
equilibrium. Oremain the same. It is already at equilibrium. 21. Le Châtel 22. Le Ch 2re 23.
Equilib 2re The concentration of C2 will
Solution
Kc is equilibrium constant
Kc = [ PCl5] / [PCl3 ] *[Cl2] ( concentration at equilibrium )
Qc is reaction quotient .it is the same ratio as Kc at any point in the reaction at which equilibrium
is not attained.
Qc = [ PCl5] / [PCl3 ] *[Cl2] ( concentration at any point in the reaction when equilibrium is not
attained )
the equilibrium constant value will not change
Kc = 83.3
Kc = [ PCl5] / [PCl3 ] *[Cl2]
since the system is no longer at equilibrium at that point
Qc = [ PC 5] / [PCl3 ] *[Cl2]
= ( [PCl5 ] - 0.39 ) / [PCl3]*[Cl2])
=Â Â [ PCl5] / [PCl3 ] *[Cl2] - 0.39 /[ PCl3]*[Cl]
Qc= Kc - 0.39 /[ PCl3]*[Cl]

2. so the Qc is less than the Kc
since the Qc is less than Kc the reaction runs in forward direction and equilibrium attained
.
as the reaction goes in forward direction the conentration of Cl2 will decrease .