February 22, 2017

Homework Help: Chemistry

Posted by Saira on Thursday, February 5, 2009 at 8:40pm.

Given the two reactions

1.PbCl2 <--->Pb^2+ + 2 Cl^-,
K1= 1.82×10^−10

2. AgCl <----> Ag^+ + Cl^-,
K2 = 1.15×10−4

what is the equilibrium constant Kfinal for the following reaction?

PbCl2+ 2 Ag^+ <---> 2AgCl+ Pb^2+

This is what i have soo far:

PbCl2<--->Pb2+ + 2Cl- K1= 1.82×10^−10
AgCl <---> Ag+ + Cl- K2= 1/1.15×10^−4

(Pb2+)(Cl)^2/(PbCl2)* (Ag)(Cl-)/(AgCl)

I dunt know what to do next because, i did :

= 1.82×10^−10 * 1/1.15×10^−4
and the answer is wrong,

It says When the stoichiometry of a reaction changes, the new equilibrium constant is raised to the same power. Thus, if a reaction is multiplied by 2, the equilibrium constant K is squared.

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