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Topic: The Equilibrium Constant and Extent of Reaction  (Read 312 times)

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Offline Greek0000

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The Equilibrium Constant and Extent of Reaction
« on: August 11, 2019, 09:39:48 AM »
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)   Kc=1.8x10-10
AgBr(s) ::equil::Ag+(aq)+Br-(aq)   Kc=3.3x10-13

Keq=[Ag+][Cl-]/[AgCl] = x2/(1.00-x)=1.8x10-10
and
Keq=[Ag+][Br-]/[AgBr] = x2/(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.

Offline AWK

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Offline sunkal

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Re: The Equilibrium Constant and Extent of Reaction
« Reply #2 on: August 11, 2019, 05:20:04 PM »
Greek0000,

Your book seems to have big issues. In your other question I pointed out an issue. Here also, looking at your attachment, an issue becomes clear.

For the equilibrium AgCl (s) = Ag+ (aq) + Cl- (aq) Keq = Ksp = [Ag+]eq [Cl-]eq. [AgCl] will not appear in the denominator, since the molarity of a solid is a constant at a given temperature, similar to its density. Therefore, Keq = x2, not what the book shows after the ice table.

Incidentally, in the ice table they are not moles (like initial moles), but molarities (like initial molarity).

Offline Vidya

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Re: The Equilibrium Constant and Extent of Reaction
« Reply #3 on: August 22, 2019, 01:43:20 AM »
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)   Kc=1.8x10-10
AgBr(s) ::equil::Ag+(aq)+Br-(aq)   Kc=3.3x10-13

Keq=[Ag+][Cl-]/[AgCl] = x2/(1.00-x)=1.8x10-10
and
Keq=[Ag+][Br-]/[AgBr] = x2/(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 =x2
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.

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