A reaction starts with 1.00 mol each of PCl3 and Cl2 in a 1.00-L flask. When equilibrium is established at 250 degree Celsius in the reaction PCl3(g)+ Cl2(g)-- PCl5(g), the amount of PCl5 present is 0.82 mol. What is Kc for this reaction?

Question

A reaction starts with 1.00 mol each of PCl3 and Cl2 in a 1.00-L flask. When equilibrium is established at 250 degree Celsius in the reaction PCl3(g)+ Cl2(g)--> PCl5(g), the amount of PCl5 present is 0.82 mol. What is Kc for this reaction?

DrBob222 · Answer

PCl3 + Cl2 ==> PCl5.

initial:
PCl3 = 1 mole/1 L = 1 M
Cl2 = 1 M
PCl5 = 0

change:
PCl5 = +x
Cl2 = -x
PCl3 = -x

equilibrium:
PCl3 = 1-x
Cl2 = 1-x
PCl5 = x = 0.82M
Redo PCl3 = 1-x = 1-0.82 = ??
Redo Cl2 = 1-x = 1 - 0.82 = ??

Substitute the equilibrium values into Kc and solve for Keq.

LoVEKOreA · Answer

keq is 1.53 and the Kc isproducts over reactants