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
(AgCl)^2(Pb)/(PbCl2)(Ag)2=
(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.
The Values are correct i doubled checked:
But when do
1.82 x 10^-10/ (1/1.15×10^−4)^2
= 2.47*10^-18
But it says this answer is wrong
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.
Here is the error
So K for the reaction requested is K1/K2^2. Check me out on that.
It's K1*(1/K2)^2
1.82 x 10^-10*(1/1.15 x 10^-4)^2 = 0.014
No.
K1 = 1.82 x 10^-10
K2 = (1/1.15 x 10^-4)^2
Then Keq for the reaction is
K1*(1/K2)^2 =
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.
I would round it ti 0.014 = Keq.
That's right thank you
To find the equilibrium constant, Kfinal, for the given reaction, you need to use the equilibrium constants (K1 and K2) for the individual reactions and apply the principles of equilibrium constant manipulation.
The given reaction is:
PbCl2 + 2 Ag^+ ⇌ 2 AgCl + Pb^2+
We can break down this reaction into two separate reactions:
1. PbCl2 ⇌ Pb^2+ + 2 Cl^-
2. AgCl ⇌ Ag^+ + Cl^-
To find Kfinal, we need to relate the equilibrium constants of these two reactions.
Considering the stoichiometry of the overall reaction, we can observe that the Cl^- ions are common on both sides of the equation. Therefore, we can cancel them out.
Now we have:
PbCl2 ⇌ Pb^2+
AgCl ⇌ Ag^+
To derive the overall equilibrium constant, Kfinal, we multiply the individual equilibrium constants, K1 and K2.
Kfinal = K1 ∗ K2
Substituting the given values of K1 and K2, we get:
Kfinal = (1.82×10^−10) ∗ (1.15×10^−4)
Taking the product of these values, we have:
Kfinal ≈ 2.093×10^−14
So, the equilibrium constant Kfinal for the given reaction is approximately 2.093×10^−14.