Question

Given the two reactions

1. PbCl2 <=> Pb²⁺ + 2Cl^- K1 = 1.77×10−10, and
2. AgCl <=> Ag⁺ + Cl^- K2 = 1.18×10−4,
what is the equilibrium constant for the following reaction?

$\rm PbCl_2+ 2 Ag^+ \rightleftharpoons 2 AgCl+ Pb^{2+}$

Answer 1

Things to note: When you reverse a reaction you take 1/K
When you add two reactions you multiply their K values
When you multiply a reaction by a constant you take K to the POWER of the constant you multiplied the reaction by:

(1) PbCl₂ <=> Pb²⁺ + 2Cl^- K1 = 1.77×10⁻¹⁰
(2) AgCl <=> Ag⁺ + Cl^- K2 = 1.18×10⁻⁴

We want to manipulate these reactions to get the final reaction so we know we need PbCl2 and 2Ag+ on the reactant side and 2AgCl and Pb2+ on the product side

We keep reaction (1) the same because it already has what we want where we want it but for the second reaction we see that AgCl is on the reactant side while we want it on the product side so first we reverse the equation (and take 1/K2) to get

(2) Ag⁺ + Cl^- <--> AgCl K2= 1/(1.18*10^-4)

Now we have it have to multiply the entire equation by 2 because we want 2AgCl and that means we have to take K2 squared:

(2) 2Ag⁺ + 2Cl^- <--> 2AgCl

Now adding the two equations together we get:

(1) PbCl₂ <=> Pb²⁺ + 2Cl^- K1 = 1.77×10−10

(2) 2Ag⁺ + 2Cl^- <--> 2AgCl

The 2Cl- on each side of the reactions cancel out so we end up with:

PbCl₂ + Ag⁺ <--> Pb²⁺ + 2AgCl

This is what we wanted. To determine K we multiply K1 and K2 together to get:

Feel free to let me know if you have any questions and please don't forget to rate!

Answer 2

k1= 1.77*10^-10             eq: PbCl2 <> Pb + 2Cl ((first reaction))

k2= 1.18*10^-4               eq: AgCl <> Ag + Cl   ((second reaction))

kfinal for the reaction: PbCl2 + 2Ag <> 2AgCl + Pb ((final reaction))

notice that there is 2 AgCl and 2 Ag, meaning that the second equation was multipied by 2. According to equilibrium constant rules, when you multiply an equation by n, raise k to the nth power. therefore, raise k2 to the second power. (k2)^2 =  1.392*10^-8. The reaction is also reversed since AgCl is on the right side in the final reaction, and on the left side in the second reaction. According to equilibrium constant rules, when a reaction is reversed, take the inverse of k. 1/(k2)^2= 7.182*10^7. 

now just multiply k1 by this new k2 value of 7.182*10^7.

k1*k2= 0.0127

kfinal = 0.0127

<3 hope this helps <3

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