Consider the equilibrium between PCl 5 , PCl 3 and Cl 2 . The reaction is allowe
ID: 942140 • Letter: C
Question
Consider the equilibrium between PCl5, PCl3 and Cl2.
The reaction is allowed to reach equilibrium in a 19.4-L flask. At equilibrium, [PCl5] = 0.217 M, [PCl3] = 0.118 M and [Cl2] = 0.118 M.
(a) The equilibrium mixture is transferred to a 9.70-L flask. In which direction will the reaction proceed to reach equilibrium?
_________to the rightto the left
(b) Calculate the new equilibrium concentrations that result when the equilibrium mixture is transferred to a 9.70-L flask.
Explanation / Answer
given PCl5= 0.217M, PCl3= 0.118M and Cl2=0.118 M
concentrations when transferred to 9.7L Flask, PCl4 = 0.217*19.4/9.7=0.434M , PCl3= 0.118*19.4/9.7= 0.236= [Cl2]
new equilibrium constant= [0.236]2/ 0.434=0.1283 this value is higher than the Equilibrium value of 0.0646. This has to be reduced and this is possible only if more PCl3 is formed. So PCl5 concentration will have to increase and concentration of Cl2 and PCl3 will have to decrease
let x= Increase in molarit of PCl5
At equilibrium PCl5= 0.434+x , [PCl3] = [Cl2] =0.236-x
K= [0.236-x)2/ (0.434+x)= 6.46*10-2. This equation can be solved by trial and error using excel.
Assume some value of x and match LHS and RHS
x= 0.0578
New concentrations will be [PCl5]= 0.434+0.0578=0.4918 [PCL3]= [Cl2]= 0.236-0.0578=0.1782 M