Consider the equilibrium between PCl 5 , PCl 3 and Cl 2 . The reaction is allowe
ID: 954743 • Letter: C
Question
Consider the equilibrium between PCl5, PCl3 and Cl2.
The reaction is allowed to reach equilibrium in a 5.10-L flask. At equilibrium, [PCl5] = 0.383 M, [PCl3] = 0.211 M and [Cl2] = 0.211 M.
(a) The equilibrium mixture is transferred to a 10.2-L flask. In which direction will the reaction proceed to reach equilibrium?
*to the right or to the left
(b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 10.2-L flask.
Explanation / Answer
For the reaction PCl5 <-------> PCl3 + Cl2, given Kc = 0.116 at 547K
The reaction in 5.1 L flask has
PCl5 <-------> PCl3 + Cl2
0.383 0.211 0.211 equilibrium concentrations
1.9533 1.0761 1.0761 equilibrium moles
When the volume is 10.2 L
1.9533/10.2 1.076/10.2 1.076/10.2 new concentration
The reaction quotient Qc = [PCl3][Cl2] /[PCl5] = ( 1.076/10.2 )( 1.076/10.2 ) /( 1.9533/10.2 )
= 0.05833
Since the Qc < Kc , the reaction proceeds right side .
b) At new equilibrium
PCl5 <-------> PCl3 + Cl2
1.9533 1.0761 1.0761 moles initial
1.9533-x 1.076+x 1.076+x equilibrium moles
1.9533-x/10.2 1.076+x/10.2 1.076+x/10.2 equilibrium concentrations
Kc = 0.116 = (1.076+x)2/ 10.2 x 1.9533-x
solving x= 0.3159
Thus equilibrium moles of PCl5 = 1.6374 PCl3 = 1.3919 and Cl2 = 1.3919 and equilibrium concentrations are [PCl5] = 0.1605M [PCl3] = 0.1364M[Cl2] = 0.1364M