# calculate the equilibrium constany Kc (from the mass of a catalyst)

Submitted by izz on Mon, 03/26/2012 - 03:21

1. The following equilibrium was studied by analyzing the equilibrium mixture for the amount of HCl produced.

LaCl3(s)  +  H2O(g)                    LaOCl(s)  +  2HCl(g)

A vessel whose volume was 1.25 L was filled with 0.0125 mol of lanthanum(III) chloride, LaCl3, and 0.0250 mol H2O. After the mixture came to equilibrium in the closed vessel at 619ºC, the gaseous mixture was removed and dissolved in more water. Sufficient silver (I) ion was added to precipitate the chloride ion completely as silver chloride. If 3.59 g AgCl was obtained, what is the value of Kc at 619ºC ?

Assuming Ag= 108 and Cl=35.5, then molar mass of AgCl = 143.5g and 3.59g = 0.0268 moles AgCl. This means there are 0.0268moles of Cl present (as HCl) and this would comes from 0.0268/3 moles of LaCl3 = 0.0268 moles HCl in the gaseous mixture removed. Concentration of HCl = 0.0268/1.25 moles/L.

Initial concentration of LaCl3 = 0.0125/1.25L and initial concentration of water was 0.0250/1.25 moles per L. Initial concentration of HCl = 0

Change - for evey mole of LaCl3 that reacts, you get 2 mole of HCl. As you make 0.0268 moles of HCl the change in concentration of HCl = 0.0268/1.25 mole/L you will lose 0.0134 /1.25 mole/L LaCl3 and also 0.0134 /1.25 mole/L of water. so changes are

0.0134 /1.25 mole/L LaCl3 and 0.0134 /1.25 mole/L H2O and make 0.0268/1.25 mole/L HCl

Equilibrium concentrations LaCl3 = (0.0125/1.25L - .0134 /1.25 mole/L) , H2O = 0.0250/1.25 - 0.0134 /1.25 mole/L) and HCl= 0.0268/1.25 mole/L

Plug these into Kc expression

Check my logic