Question

Given the two reactions PbCl2(aq)⇌Pb2+(aq)+2Cl−(aq), K3 = 1.84×10−10, and AgCl(aq)⇌Ag+(aq)+Cl−(aq), K4 = 1.14×10−4, what is the equilibrium constant Kfinal for the following reaction? PbCl2(aq)+2Ag+(aq)⇌2AgCl(aq)+Pb2+(aq)

Answer 1

Answer: The value of equilibrium constant for reaction is, $1.42\times 10^{-2}$

Explanation:

The given chemical equations are:

(1) $PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$ ;  $K_3=1.84\times 10^{-10}$

(2) $AgCl(aq)\rightleftharpoons Ag^{+}(aq)+Cl^-(aq)$ ;  $K_4=1.14\times 10^{-4}$

Now we have to calculate the equilibrium constant for chemical equation as:

$PbCl_2(aq)+2Ag^{+}(aq)\rightleftharpoons 2AgCl(aq)+Pb^{2+}(aq)$ ;  $K=?$

We are reversing reaction 2 and multiplying reaction 2 by 2 and then adding both reaction, we get the final reaction.

The equilibrium constant for the reverse reaction will be the reciprocal of that reaction.

If the equation is multiplied by a factor of '2', the equilibrium constant of that reaction will be the square of the equilibrium constant.

If we are adding equations then the equilibrium constants will be multiplied.

The value of equilibrium constant for reaction is:

$K=(\frac{1}{K_4})^2\times K_3$

Now put all the given values in this expression, we get:

$K=(\frac{1}{1.14\times 10^{-4}})^2\times (1.84\times 10^{-10})$

$K=1.42\times 10^{-2}$

Hence, the value of equilibrium constant for reaction is, $1.42\times 10^{-2}$