More equilibrium concepts. LeChâtelier’s Principle states that when a stress is applied to a...

more equilibrium concepts

LeChâtelier’s Principlestates that when a stress is applied to a system at equilibrium, the system will respond in a manner that attempts to bring the system back to equilibrium…stresses are:ConcentrationPressureTemperature

ConcentrationIf you add more reactant, then…

the rxn will shift to the right to use the excess reactant.

If you add more product, then…

the rxn will shift to the left to use the excess product.

If you remove reactant, then…

If you remove product, then…

Pressure3 means of affecting pressure of gaseous systemsConcentrationVolume of containerAddition of an inert gas

Pressure—concentration’s effectSame concept as concentration change

Pressure—volume’s effectIf you decrease the volume of the container, then the pressure will increase and the system will adjust by…

shifting to the side with the fewer number of moles in an effort to bringthe pressure back down to what it wasat equilibrium

Pressure—volume’s effectIf you increase the volume of the container, then the pressure will decrease and the system will adjust by…

shifting to the side with the greater number of moles in an effort to bringthe pressure back up to what it wasat equilibrium

Pressure—volume’s effectIf you increase or decrease the volume of the container for a reaction who has equal number of moles on reactant and product side, then…

the system cannot make the appropriatechanges and the rxn will not be able toreach equlibrium

Pressure—inert gas’s effectIf you add an inert gas to a system at equilibrium, then…

there will be no shift because the inert gas only increases the total pressure of the system (think Dalton’s Law) and has no effect on the partial pressures of the reactants or products

Temperatureeffect depends on whether the rxn is endothermic or exothermicendothermic—absorbs heat energy, thus the energy appears as a reactant or the change in enthalpy is positive

exothermic—releases heat energy, thus the energy appears as a product or the change in enthalpy is negative

Temperature—Endothermic RxnIf you add more heat, then…

the rxn will shift to the right to use the excess heat. (think of the heat as a reactant)

If you remove heat, then…

the rxn will shift to the left to replenish the missing heat.

Temperature—Exothermic RxnIf you add more heat, then…

the rxn will shift to the left to use the excess heat. (think of the heat as a product)

If you remove heat, then…

the rxn will shift to the right to replenish the missing heat.

1.Consider the reaction:N2(g) + 3H2(g) 2NH3(g) + 33.3kJExplain the direction of shift it willexperience when the following

stresses areapplied:

a.The system is heated.b.Its container is squeezed.c. Ammonia is added to the system.d.Neon is added to the system.

2.Consider the reaction:2SO3(g) + 33.3kJ 2SO2(g) + O2(g) Explain the direction of shift it willexperience when the following

stresses areapplied:

a.The system is cooled.b.Its container expands.c. Sulfur trioxide is added to the

system.d.Argon is added to the system.

applying equilibrium stuff to salts

Ksp

Unlike NaCl, not all salts dissolve completely in aqueous solutions.

The level to which a salt dissolves is expressed as its solubility product constant and is represented by Ksp (similar to K)

Remember that only gaseous and aqueous thingies can be represented in K expressions.

Ksp

So, in a salt dissociation rxn, only the products are aqueous.

The reactant (or salt) is solid.Consider the salt, silver hydroxide. Write its formula.

AgOH

Ksp

Now, write its dissociation into its ions…

AgOH (s) Ag1+(aq) + OH1-(aq)

•Make sure you pay attention to the phases.

•Now write the Ksp expression.

Ksp

Ksp= [Ag1+][OH1-]

•Note that the AgOH(s) is absent because it is a solid.

•Also note that the power of the Ag1+ and OH1- are both one because the dissociation only yielded one of each ion.

Ksp

•This expression allows you to work three types of problems:•determine the concentrations of ions that have dissolved

•determine the solubility of the salt (it will always equal x in your equilibrium chart)

•determine the Ksp of a salt

Ksp Practice Problem #1

Determine the concentration ofeach ion if the Ksp of calcium

fluoride is 4.0 x 10-11.

Ksp Practice Problem #1Step 1—write the dissociation ofsalt.

CaF2(s) Ca2+(aq) + 2F1-(aq)

Recognize that the mol:mol for theions is 1:2. This is very importantinfo that you will use in theproblem, so make certain that the formula for the salt is correct!

Ksp Practice Problem #1Step 2—make a dissociation chartlike you did for the other equilibriumrxns. Remember…only (g) & (aq)!!

i -- 0 0

[CaF2] [Ca2+] [F1-]

Δ -- +x +2x

eq -- x 2x

Ksp Practice Problem #1Step 3—write your Ksp expressionand plug in your stuff

Ksp= [Ca2+][F1-]2

Note that the mol:mol is reflectedin the power as well as in your eqline of your chart.

Ksp Practice Problem #1Step 3—continued

Ksp= [Ca2+][F1-]2

4.0 x 10-11 = [x][2x]2

4.0 x 10-11 = 4x3

1.0 x 10-11 = x3

2.15 x 10-4 = x

Ksp Practice Problem #1Step 4—use your value for x andthe eq line of your chart to figureout the concentrations of the ions.

[Ca2+]= 2.15 x 10-4 M

[F1-]= 2(2.15 x 10-4) = 4.30 x 10-4 M

Ksp Practice Problem #2

Determine the solubility ofcalcium fluoride from your workin Problem #1.

Ksp Practice Problem #2Step 1—You’ve already done all ofthe work for this problem. Sincethe mol:mol:mol for the salt to itsions is 1:1:2, the solubility of thesalt in this case is equal to theconcentration of the calcium ion.Just remember that the solubility ofthe salt will always equal x fromyour eq line of your chart

Ksp Practice Problem #2Step 1—Thus, the answer to thisproblem is 2.15 x 10-4 M for thesolubility of the calcium fluoride.

Ksp Practice Problem #3Determine the value of the solubilityproduct constant of bismuth sulfidewhich has a solubility of1.0 x 10-15 M.

Ksp Practice Problem #3Step 1—write the dissociation ofsalt and the Ksp expression.

Bi2S3(s) 2Bi3+(aq) + 3S2-(aq)

Ksp= [Bi3+]2[S2-]3

Ksp= [2x]2[3x]3

Ksp= 108x5

consolidated

Ksp Practice Problem #3Step 2—Since your know thesolubility (or x) of the salt, you candeduce the concentrations of theions from the mol:mol of the ions. Therefore, [Bi3+] will be two timesthe solubility, and [S2-] will bethree times the solubility.

Ksp Practice Problem #3Step 3—Plug your solubility intothe expression as x and solve forKsp.

Ksp=[2(1.0 x 10-15)]2[3(1.0 x 10-15)]3

Or since,Ksp= 108x5…

108(1.0 x 10-15)5

Ksp Practice Problem #3Thus, the Ksp should be

1.08 x 10-73 M