The Equilibrium Constant and Extent of Reaction

Solid AgCl and AgBr were each placed in 1.0 L of water in separate beakers. Are the following reactions product- or reactant-favored? When equilibrium is achieved, in which beaker will the concentration of silver ion be larger?

AgCl(s) ::equil::Ag^{+}(aq)+Cl^{-}(aq) K_{c}=1.8x10^{-10}

AgBr(s) ::equil::Ag^{+}(aq)+Br^{-}(aq) K_{c}=3.3x10^{-13}

K_{eq}=[Ag^{+}][Cl^{-}]/[AgCl] = x^{2}/(1.00-x)=1.8x10^{-10}

and

K_{eq}=[Ag^{+}][Br^{-}]/[AgBr] = x^{2}/(1.00-x)=3.3x10^{-13}

there must be something wrong with my math because I cannot get the number of moles correct at equilibrium. I found that using whole numbers doesn't work with a significant figure of 10 digits. I tried using a quadratic formula but I only got a whole value. The answer in the book says

concentration of sivler in the AgCl beaker (1.3x10^{-5} M) is greater than in the AgBr beaker (5.7x10^{-7} M). Both are reactant-favored.

This question is definitely related to chemical equilibrium constant concept.

if you remember in chemical equilibrium pure liquids and solids are not the part of equilibrium constant calculations because their concentrations are considered to be unity.

Always remember equilibrium constant is never in terms of moles or grams and is always in terms of concentration that is moles per liter.

Let's pay attention to this question in this question AgCl and AgBr are solids so they should not be the part of your equilibrium constant and their concentrations are considered as unity

This changes the expression over here and this new expression is called as Ksp

Ksp will follow the same rules as Keq

(Solubility product-sp) Now this solubility product is the product of the concentration of ions in the solution.

Ice table which you have made is correct accept it should be in terms of molarity and we need to ignore solid reactant

Ksp expression for AgCl and AgBr will be

Ksp =x

^{2 }You need to calculate x in both cases and compare your answers.

There is no need of doing full calculations.

As you can see the solubility product of AgCl is more than AgBr so from here try to figure out answer of your question.