Posted by Saira on Thursday, February 5, 2009 at 8:40pm.
Given the two reactions
1.PbCl2 <--->Pb^2+ + 2 Cl^-,
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
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.
- Chemistry - DrBob222, Thursday, February 5, 2009 at 8:58pm
Right. First, however, check your numbers for the k values. I think they are reversed because Ksp for AgCl is about 10^-10 and Ksp for PbCl2 is about 10^-4. But if the Ks were put in the problem on purpose, then
K1 = 1.82 x 10^-10
and K2 for the reverse direction is 1/1.15 x 10^-4 and you square that [that is (1/K2)^2].
Since you added the equations to get the final equation, now you multiply the new Ks. So K for the reaction requested is K1/K2^2. Check me out on that.
- Chemistry - Saira, Thursday, February 5, 2009 at 9:05pm
The Values are correct i doubled checked:
But when do
1.82 x 10^-10/ (1/1.15×10^−4)^2
But it says this answer is wrong
- Chemistry - DrBob222, Thursday, February 5, 2009 at 9:25pm
K1 = 1.82 x 10^-10
K2 = (1/1.15 x 10^-4)^2
Then Keq for the reaction is
1.82 x 10^-10*(1/1.15 x 10^-4)^2 =
1.82 x 10^-10*(7.56 x 10^7) =
0.01376. Check it for math. Check it for significant figures.
- Chemistry - DrBob222, Thursday, February 5, 2009 at 9:27pm
I would round it ti 0.014 = Keq.
- Chemistry - DrBob222, Thursday, February 5, 2009 at 9:35pm
Here is the error
So K for the reaction requested is K1/K2^2. Check me out on that.
1.82 x 10^-10*(1/1.15 x 10^-4)^2 = 0.014
- Chemistry - Saira, Thursday, February 5, 2009 at 9:45pm
That's right thank you
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