1

EquilibriumEquilibrium pp

Online HW notes for the next 2 weeks:Online HW notes for the next 2 weeks:

Some problems use ‘bar” or “millibar” for Some problems use ‘bar” or “millibar” for pressure. They are just another unit.pressure. They are just another unit.

One problem might give you -[ ]s. It’s bad One problem might give you -[ ]s. It’s bad chem, but good math & the problem is chem, but good math & the problem is “doable.”“doable.”

Do not round when doing online HW Do not round when doing online HW (even if adding or subtracting).(even if adding or subtracting).

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13.1 Reactions are reversibleA + B C + D ( forward)A + B C + D ( forward)C + D A + B (reverse)C + D A + B (reverse) InitiallyInitially there is there is only A and Bonly A and B so only so only

the forward reaction is possiblethe forward reaction is possibleAs C and D build up, the reverse As C and D build up, the reverse

reaction speeds up while the forward reaction speeds up while the forward reaction slows down.reaction slows down.

Eventually the rates are Eventually the rates are equalequal

3

Rea

ctio

n R

ate

Time

Forward Reaction

Reverse reaction

Equilibrium

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What is equal at Equilibrium?RatesRates are equal are equalConcentrationsConcentrations are not! are not!Rates are determined by Rates are determined by

concentrations and activation energy.concentrations and activation energy.The concentrations do The concentrations do notnot change at change at

equilibrium.equilibrium.Or, if the reaction is verrrry slooooow.Or, if the reaction is verrrry slooooow.

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13.2 Law of Mass Action pp

For any For any solution or gaseoussolution or gaseous reaction reaction jjA + A + kkB B llC + C + mmDD

KKcc = [C] = [C]ll[D][D]mm PRODUCTS PRODUCTSpowerpower

[A] [A]jj[B][B]kk REACTANTS REACTANTSpowerpower

KKcc is called the equilibrium constant. is called the equilibrium constant.

KKcc often is written as just K often is written as just K

is how we indicate a reversible is how we indicate a reversible

reactionreaction

Does Does notnot apply to apply to solidssolids or or pure liquidspure liquids!!

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Playing with K pp

If we write the reaction in reverse.lC + mD jA + kB

Then the new equilibrium constant isK’ = [A]j[B]k = 1/K

[C]l[D]m

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Playing with K pp

If we multiply the equation by a constant

njA + nkB nlC + nmDThen the equilibrium constant is

K =

[C]nl[D]nm = ([C]l[D]m)n

= K

n

[A]nj[B]nk ([A] j[B]k)n

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Calculate K - all are gases, given NN22 + 3H + 3H22 2NH 2NH33 where K = 1.3 x 10 where K = 1.3 x 10-2-2

1/2N1/2N22 + 3/2H + 3/2H22 NHNH33 K = ? . . .K = ? . . .

K1/2 = 0.11 = 0.11 2NH2NH33 N N22 + 3H + 3H22 K = ? . . .K = ? . . .

K-1 = 77 = 77 NHNH33 1/2N 1/2N22 + 3/2H + 3/2H22 K = ? . . .K = ? . . .

(K-1)1/2 = (1/K)1/2 = 8.8 2N2N22 + 6H + 6H22 4NH 4NH33 K = ? . . .K = ? . . .

(K)2 = 1.7 x 10-4

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The units for KAre determined by the various Are determined by the various

powers and units of concentrations.powers and units of concentrations.They depend on the reaction.They depend on the reaction.K values usually written without unitsK values usually written without units

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K is CONSTANTAt any constant temperature.Temperature affects rate.The equilibrium concentrations don’t

have to be the same; only K.Equilibrium position is a set of

concentrations at equilibrium.Indicates reaction shift to left or right.There are an unlimited number.

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Equilibrium Constant

One for each Temperature

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Calculate Kc

N2 + 3H2 2NH3

Initial At Equilibrium[N2]0 =1.000 M [N2] = 0.921 M

[H2]0 =1.000 M [H2] = 0.763 M

[NH3]0 =0 M [NH3] = 0.157 M

Now do the calculation to get . . . K = 6.02 x 10-2

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Calculate KN2 + 3H2 2NH3

Initial At Equilibrium [N2]0 = 0 M [N2] = 0.399 M [H2]0

= 0 M [H2] = 1.197 M [NH3]0 =

1.000 M [NH3] = 0.203 M

Your answer . . .Kc is the same (6.02 x 10-2 ) no matter

what the amount of starting materials

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13.3 Equilibrium and PressureSome reactions are gaseousSome reactions are gaseousPV = nRTPV = nRTP = (n/V)RT where n/V = [ ] of gasP = (n/V)RT where n/V = [ ] of gasP = CRTP = CRT C is C is

a concentration in moles/Litera concentration in moles/LiterC = P/RTC = P/RT

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Equilibrium and Pressure2SO2(g) + O2(g) 2SO3(g)

Kp = (PSO3)2

(PSO2)2 (PO2)

Kc = [SO3]2

[SO2]2 [O2]

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Deriving a relationship between Kc & Kp

2SO2(g) + O2(g) 2SO3(g)

Since C = [ ] = P/RT

Kc = (PSO3/RT)2

(PSO2/RT)2(PO2/RT)

Kc = (PSO3)2 (1/RT)2

(PSO2)2(PO2) (1/RT)3

Kc = Kp (1/RT)2 = Kp RT

(1/RT)3

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General Equation - Deriving: jjA + A + kkB B llC + C + mmDD

KKpp= (P= (PCC))l l (P(PDD))mm= = (C(CCCxRTxRT))l l (C(CDDxRTxRT))m m

(P(PAA))j j (P(PBB))k k (C(CAAxRTxRT))jj(C(CBBxRTxRT))kk

KKpp== (C(CCC))l l (C(CDD))mmx(RT)x(RT)l+ml+m

(C(CAA))jj(C(CBB))kkx(RT)x(RT)j+kj+k

KKp p == K K (RT)(RT)((l+m)-(j+k) l+m)-(j+k) == KKcc(RT)(RT)nn

nn==(l+m)-(j+k)(l+m)-(j+k)==Change in moles of Change in moles of gasgas

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General Equation pp

KKp p == KKcc(RT)(RT)nn (R = .08206 l atm/K mol & T in Kelvin)(R = .08206 l atm/K mol & T in Kelvin)

Where Where n = change in moles of gas.n = change in moles of gas.Watch for mixed equilibriaWatch for mixed equilibria (solids & (solids &

liquids in reaction)liquids in reaction)Do Do notnot count their moles in count their moles in

calculating calculating n.n.Be sure to Be sure to convert C to Kconvert C to K for temp. for temp.

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Homogeneous EquilibriaSo far every example dealt with

reactants and products where all were in the same phase.

So, we can use K in terms of either concentration or pressure if gas phase.

Units depend on the reaction.Try # 29, 39 & 56, p 615 (5’ then WB).

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13.4 Heterogeneous EquilibriaIf the reaction involves pure solids or

pure liquids the concentration of the solid or the liquid does not change.

As long as they are not used up we can leave them out of the equilibrium expression.

For example:

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For ExampleH2(g) + I2(s) 2HI(g)

K = [HI]2

[H2][I2]

But the [ ] of I2 does not change since it’s a

solid (treat as = to 1). So . . .

K[I2]= [HI]2 = K’

[H2]

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E.g., Write the Kp Expression:2Fe(s) + 3/2 O2(g) Fe2O3(s)

Kp = 1 (PO2)3/2

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13.5 Applications of the Equilibrium Constant

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The Extent of a ReactionSize of K (i.e, extent of rxn) and time

to reach =m are not directly related.Time to reach =m determined by

kinetics; i.e., activation energy.Large K means strong shift to the

rightI.e., at =m have mostly products

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The Reaction QuotientTells you the direction the reaction

will go to reach equilibrium (lt vs. rt)Calculated the same as the

equilibrium constant, but for a system not at equilibrium (e.g., time = 0)

Q = [Products]coefficient

[Reactants] coefficient

Compare value to equilibrium constant

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What Q tells us pp

If Q < K Not enough products, so adjusts by Shift to right (to make more)

If Q > K Too many products, so adjusts by Shift to left (to reduce products)

If Q = K system at =m and no shift

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ExampleFor the reactionFor the reaction

22NOClNOCl(g) 2(g) 2NONO(g) + (g) + ClCl22(g) (g)

K = 1.55 x 10K = 1.55 x 10-5 -5 MM at 35ºC at 35ºC In an experiment 0.10 mol In an experiment 0.10 mol NOClNOCl, ,

0.0010 mol 0.0010 mol NONO and 0.00010 mol and 0.00010 mol ClCl22

are mixed in 2.0 L flask.are mixed in 2.0 L flask.Which direction will the reaction Which direction will the reaction

proceed to reach equilibrium? (Use proceed to reach equilibrium? (Use initial concentrations) Answer . . .initial concentrations) Answer . . .

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ExampleQ is not 1.0 x 10-8. Why?You were given moles in a 2 liter flaskSo [ ]s are halvedQ = 5.0 x 10-9

Since Q < K (which was 1.55 x 10-5)Reaction shifts to the right.

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13.6 Solving Equilibrium Problems pp

Given the starting concentrations and one equilibrium concentration.

Use stoichiometry to figure out other concentrations and K.

Learn to create a table of initial and final conditions.

Use “RICE”

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Consider the following at 600ºCConsider the following at 600ºC pp

22SOSO22(g)(g) + + OO22(g)(g) 2SO2SO33((g)g)

In a certain experiment 2.00 mol of In a certain experiment 2.00 mol of SOSO22, 1.50 mol of , 1.50 mol of OO22 and 3.00 mol of and 3.00 mol of SOSO33 were placed in a 1.00 L flask. At were placed in a 1.00 L flask. At equilibrium 3.50 mol equilibrium 3.50 mol SOSO33 were found were found to be present. Calculate: to be present. Calculate:

The equilibrium concentrations of OThe equilibrium concentrations of O22 and SOand SO22, K, Kcc and K and KPP (procedure follows (procedure follows . . .). . .)

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Strategy pp

Make RICE table to show initial, change and final molarity (convert moles to M by dividing by volume)

Use stoich & mole ratios as needed.Plug into =m expression to get Kc

Use Kp = Kc(RT)∆n

Remember: ∆n is change in gaseous moles only!!!

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Solving pp

2SO2(g) + O2(g 2SO3(g)

initial 2.00 M 1.5 M 3.00 M change

equil. 3.50 M

Kc = [SO3]2

[SO2]2[O2]

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Solving pp

2SO2(g) + O2(g 2SO3(g)

initial 2.00 M 1.5 M 3.00 M change 0.50 M 0.25 M 0.50 M final 3.50 M

Kc = [SO3]2

[SO2]2[O2]

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Solving pp

2SO2(g) + O2(g 2SO3(g)

initial 2.00 M 1.5 M 3.00 M change -0.50 M -0.25 M +0.50 M final 1.50 M 1.25 M 3.50 M

Kc = [SO3]2 = (3.50)2

[SO2]2[O2] (1.50)2(1.25)

Kc = 4.36

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Solving pp

600º C600º C 2SO2SO22(g)(g) + O + O22(g(g 2SO 2SO33(g)(g)

initial initial 2.00 2.00 MM 1.5 1.5 MM 3.00 3.00 MM change change -0.50 -0.50 MM -0.25 -0.25 MM +0.50 +0.50 MM

finalfinal 1.50 1.50 MM 1.25 1.25 MM 3.50 3.50 MM

KKcc = 4.36 = 4.36

KKpp = K = Kcc(RT)(RT)∆n∆n and ∆n = -1 and ∆n = -1

KKpp = 4.36(0.08206 x 873) = 4.36(0.08206 x 873)-1-1 = 6.09 x 10 = 6.09 x 10-2-2

atm (or 3.80 torr).atm (or 3.80 torr).

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2SO2SO22(g)(g) + O + O22(g)(g) 2SO2SO33(g)(g) @ 600ºC @ 600ºC

In a In a differentdifferent experiment experiment .500 mol SO.500 mol SO22

and and .350 mol SO.350 mol SO33 were placed in a were placed in a

1.000 L container. When the system 1.000 L container. When the system reaches equilibrium reaches equilibrium 0.045 mol of O0.045 mol of O22

are present.are present.Calculate the final concentrations of Calculate the final concentrations of

SOSO22 and SO and SO33 and K. Do on own -- and K. Do on own --

your answers are . . .your answers are . . .

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In a In a differentdifferent expt at 600 ºC .500 mol SO expt at 600 ºC .500 mol SO22 & .350 mol SO& .350 mol SO33 placed in a 1.000 L placed in a 1.000 L container. At =m 0.045 mol of Ocontainer. At =m 0.045 mol of O22 are present. are present.

2SO2SO22(g)(g) + O + O22(g)(g) 2SO 2SO33(g)(g)

Same temp.Same temp., so K, so Kcc is the same!! (4.36) is the same!! (4.36)Since no OSince no O2 2 initially, its change is +0.045 and initially, its change is +0.045 and

the reaction shifted to the the reaction shifted to the leftleft..Plug in to get SOPlug in to get SO22 and SO and SO33 =m [ ]s (derived =m [ ]s (derived

from the Ofrom the O22 change in [ ]) change in [ ])Your answer for their Your answer for their concentrationsconcentrations . . . . . . [SO[SO22] = .590 ] = .590 MM & [SO & [SO33] = .260 (mole ratios)] = .260 (mole ratios)

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What if not given equilibrium concentration? pp

Easy if you get a “perfect square.”3.000 mol each of hydrogen, fluorine

and hydrofluoric acid were added to a 1.5000 L flask. Calculate the equilibrium concentrations of all species.

Keq = 1.15 x 102. . .

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Example continued pp

3.00 mol each of hydrogen, fluorine and hydrofluoric acid were added to a 1.50 L flask. Calculate the equilibrium concentrations of all species.

Keq = 1.15 x 102.

Write balanced equation, determine Q Write balanced equation, determine Q and direction of shift, set up RICEand direction of shift, set up RICE

Steps on next slides.

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Example cont. where K = 1.15 x 102 pp

3.00 mol each of H2, F2, HF added to a 1.50 L flask. So, M = 2.00 for each.

H2(g) + F2(g) 2HF(g)

Q = (2.00)2 = 1.00 < Keq (2.00)(2.00)

So, shifts to right. So, Left side is losing stuff and right side is gaining.

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Example continued pp

H2(g) + F2(g) 2HF(g) initial 2.00 M 2.00 M 2.00 M change -x -x + 2x final 2.00-x 2.00-x 2.00+2x

K = 1.15 x 102 = (2.00 + 2x)2

(2.00 - x)(2.00 - x)K = 1.15 x 102 = (2.00 + 2x)2

(2.00 - x)2

Perfect square, so take square root of both sides.

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Example continued pp

√√K = 10.74 = (2.00 + 2x)K = 10.74 = (2.00 + 2x)

(2.00 - x) (2.00 - x)x = 1.528x = 1.528

HH22(g) +(g) + F F22(g) 2HF(g)(g) 2HF(g)

finalfinal 2.00-x 2.00-x 2.00-x 2.00-x 2.00+2x 2.00+2x

2.00 - x = [H2.00 - x = [H22]]eqeq & [F & [F22]]eqeq = = 0.472 0.472 MM

(2.00 + (2.00 + 22x) = [HF]x) = [HF]eqeq = = 5.056 5.056 MM

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What if you don’t get a “perfect square”? pp

We are then stuck with having to solve a quadratic expression unless we can make simplifying assumptions.

To avoid a quadratic, we will take different approaches depending on the size of Kc.

44

General Rules to avoid quadratics pp

When KWhen Kcc is large - find limiting reactant is large - find limiting reactant and set it as “x” in the “equilibrium” line of and set it as “x” in the “equilibrium” line of RICE. (Example follows).RICE. (Example follows).

When KWhen Kcc is small - easiest is to put “x” in is small - easiest is to put “x” in the “change” line or RICE (see example).the “change” line or RICE (see example).

If putting “x” in “change line then requires If putting “x” in “change line then requires a quadratic or you get a a quadratic or you get a negativenegative [ ], put [ ], put it in the “equilibrium line” (see example).it in the “equilibrium line” (see example).

When KWhen Kcc is “middle” then must do is “middle” then must do quadratic.quadratic.

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General Rules to avoid quadratics pp

Why don’t we just always put “x” in the Why don’t we just always put “x” in the “equilibrium” line since it always works?“equilibrium” line since it always works?

We could, but most textbooks don’t show We could, but most textbooks don’t show this method because the vast majority of this method because the vast majority of problems involve a small Kproblems involve a small Kcc where putting where putting it the “change” line works.it the “change” line works.

Also, when we do buffer equilibria we will Also, when we do buffer equilibria we will put “x” in the “change” line so we mostly put “x” in the “change” line so we mostly want to use that method unless we can’t want to use that method unless we can’t because of a quadratic or we get a -[ ].because of a quadratic or we get a -[ ].

46

Large Kc example pp

HH22(g) + I(g) + I22(g) 2HI(g)(g) 2HI(g) K = 7.1 x 10K = 7.1 x 1022 at 25ºC at 25ºC Calculate the equilibrium concentrations if a Calculate the equilibrium concentrations if a

5.00 L5.00 L container initially contains 15.7 container initially contains 15.7 gg of H of H22 and 294 and 294 gg I I22 . .

What is Q? . . . (can do without a calculator)What is Q? . . . (can do without a calculator) Q = 0Q = 0 because initially no product. because initially no product. [[HH22]]00 = (15.7g/2.02)/5.00 L = 1.56 = (15.7g/2.02)/5.00 L = 1.56 MM [I[I22]]00 = (294g/253.8)/5.00L = 0.232 = (294g/253.8)/5.00L = 0.232 MM [HI][HI]00 = 0 = 0 Problem: Need to know Problem: Need to know limiting reactantlimiting reactant..

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H2(g) + I2(g) 2HI(g) pp

Q = 0 < K so more product formed.Assumption: Since K is large, reaction will

go to completion (I.e., forward arrow only).But, it doesn’t quite go to completion.Use Stoich to find I2 is LR, so it will be

smallest at =m (since essentially used up. So, let it be x in “equilibrium” line.

Set up RICE table in concentrations (use moles with stoich,but M in RICE)

48

Choose X so it is small (so use I2 since it was limiting reactant.

For I2 the change in X must be X - 0.232 M

Final must = initial + change

H2(g) + I2(g) 2HI(g) pp initial 1.56 M 0.232 M 0 M change Equilibrium X

49

H2(g) + I2(g) 2HI(g)

Using stoichiometry we can find:Change in H2 = X-0.232 M

Change in HI = - twice change in H2

Change in HI = 0.464-2X

H2(g) I2(g) 2HI(g) pp initial 1.56 M 0.232 M 0 M change X-0.232final X

50

Now we can determine the final concentrations by adding.

H2(g) I2(g) 2HI(g) pp initial 1.56 M 0.232 M 0 M

change X-0.232 X-0.232 0.464-2Xfinal X

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Now plug these values into the equilibrium expression

K = (0.464-2X)2 = 7.1 x 102 (1.328+X)(X)

Now you can see why we made sure X was small (by using the LR).

H2(g) I2(g) 2HI(g) pp initial 1.56 M 0.232 M 0 M

change X-0.232 X-0.232 0.464-2X final 1.328+X X 0.464-2X

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Why We Chose x to be Small pp

K = (0.464-2X)2 = 7.1 x 102 (1.328+X)(X)

Since x is going to be small, we can ignore it in relation to 0.464 and 1.328

So we can rewrite the equation7.1 x 102 = (0.464)2

(1.328)(X)Makes the algebra easy

53

When we solve for X we get 2.3 x 10-4

So we can find the other concentrations

I2 = 2.3 x 10-4 M

H2 = 1.328 M

HI = 0.464 M Note: If had set “x” in change line we would Note: If had set “x” in change line we would

have got have got (-)(-) =m concentrations for H =m concentrations for H22 & I & I22..

HH22(g) (g) I I22(g) 2HI(g)(g) 2HI(g) pppp

initial 1.56 initial 1.56 MM 0.232 0.232 MM 0 0 MM change X-0.232 change X-0.232 X-0.232X-0.232

0.464-2X final0.464-2X final 1.328+X 1.328+X X X 0.464-2X0.464-2X

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Checking the assumption pp

Rule of thumb: If the value of X is less than 5% of all the other concentrations, assumption is valid.

If not we would have had to use the quadratic equation

More on this later.Our assumption was valid.

55

HH22(g) (g) I I22(g) 2HI(g) (g) 2HI(g)

initial 1.56 initial 1.56 MM 0.232 0.232 MM 0 0 MM change -xchange -x -x -x +2x +2x finalfinal 1.56-x 1.56-x 0.232-x0.232-x 0+2x 0+2x

Why we couldn’t put x in “change” line pppp

(2x)2 = 7.1 x 102 (flies on

Matt) (1.56-x)(.232-x)x = 8.01, and plugging into “final” we get: -6.45 -7.78 +16.02

(-) concentrations for H2 & I2 !!!

56

Practice (put “x” in final line)

For the reaction For the reaction ClCl22 + O + O22 2ClO(g) 2ClO(g)

K = 156K = 156 In an experiment 0.100 mol ClO, 1.00 In an experiment 0.100 mol ClO, 1.00

mol Omol O22 and 0.0100 mol and 0.0100 mol ClCl22 are mixed in are mixed in

a a 4.00 L4.00 L flask. flask. If the reaction is not at equilibrium, which If the reaction is not at equilibrium, which

way will it shift?way will it shift?Calculate the equilibrium concentrations.Calculate the equilibrium concentrations.Try first, then we’ll review (next slides)Try first, then we’ll review (next slides)

57

Practice ClCl22 + O + O22 2ClO(g)2ClO(g) K = 156K = 156

Remember to divide by 4 L to get molarity.Remember to divide by 4 L to get molarity. Q = 1 = products/reactants. Since less than K, Q = 1 = products/reactants. Since less than K,

reaction shifts to right.reaction shifts to right. Use RICE table of molarities (divide by 4 to get Use RICE table of molarities (divide by 4 to get

0.025M ClO0.025M ClO, , .25.25MM O O22 and and .0025.0025MM Cl Cl22 Since shift to right, make [=m] ClSince shift to right, make [=m] Cl22 = x (since it = x (since it

is the LR)is the LR)

58

Practice K = 156 Cl2 + O2 2ClO Init . 0025 .25 .025

Chg x-.0025 x-.0025 .0050-2x Final x x+.2475 .0300-2x

add above Simplify & solve for x. Your answer . . . X = 2.33 x 10-5 = [Cl2]

59

Problems with small K pp

K< .01

60

Example of small Kc pp

90% of the time you can x in “change” line with a small Kc,

For the reactionFor the reaction22NOClNOCl(g) 2(g) 2NONO(g) + (g) + ClCl22(g) (g)

K = 1.6 x 10K = 1.6 x 10-5 -5 MM at 35ºC at 35ºC In an experiment 1.00 mol In an experiment 1.00 mol NOClNOCl, is , is

placed in a 2.0 L flask.placed in a 2.0 L flask.Find the =m concentrationsFind the =m concentrations

61

Example of small Kc pp

1 mol NOCl in 2 L, K = 1.6 x 101 mol NOCl in 2 L, K = 1.6 x 10-5 -5

2NOCl2NOCl(g) 2(g) 2NONO(g) + (g) + ClCl22(g) (g) 0.50 0.50 MM 00 00 -2x -2x +2x +2x +x+x0.5 - 2x0.5 - 2x +2x +x +2x +x

KKcc = 1.6 x 10 = 1.6 x 10-5 -5 == (2x)(2x)22(x)(x) (0.50 - 2x) (0.50 - 2x)22

KKcc ≈ (2x) ≈ (2x)22(x)(x) ≈ 4x ≈ 4x33

(0.50) (0.50)22 (0.50) (0.50)22

x = 1.0 x 10x = 1.0 x 10-2-2

62

Example of small Kc pp

1 mol NOCl in 2 L, K = 1.6 x 101 mol NOCl in 2 L, K = 1.6 x 10-5 -5

2NOCl2NOCl(g) 2(g) 2NONO(g) + (g) + ClCl22(g) (g) 0.50 0.50 MM 00 00

-2x -2x +2x +2x+x+x 0.5 - 2x0.5 - 2x +2x +2x

+x +xSince x = 1.0 x 10Since x = 1.0 x 10-2-2, =m [ ]s are, =m [ ]s are 0.480.48

0.02 0.02 0.010.01

63

If can’t solve with “x” in change line pp

Set up table of initial, change, and final concentrations (RICE).

Choose X to be small (find LR) and put in “final” line of RICE.

For this case it will be a product because Kc is small.

The following problem must be done this way to avoid a quadratic.

64

For example pp

For the reaction For the reaction 2NOCl2NOCl 2NO2NO + + ClCl22

K= 1.6 x 10K= 1.6 x 10-5-5 If If 1.20 mol NOCl1.20 mol NOCl, , 0.45 mol of NO0.45 mol of NO and and 0.87 mol Cl0.87 mol Cl22

are mixed in a 1 L containerare mixed in a 1 L container What are the equilibrium concentrationsWhat are the equilibrium concentrations

Q = [NO]Q = [NO]22[Cl[Cl22] = (0.45)] = (0.45)22(0.87) = 0.15 M(0.87) = 0.15 M

[NOCl][NOCl]22 (1.20) (1.20)22

Since Q > K, shift to leftSince Q > K, shift to left

65

Choose X to be smallNO will be LRChoose NO to be X

2NOCl 2NO + Cl2 pp

Initial 1.20 0.45 0.87

Change

Final

66

Figure out change in NOChange = final - initialchange = X-0.45

2NOCl 2NO Cl2 pp

Initial 1.20 0.45 0.87

Change

Final X

67

Now figure out the other changesUse stoichiometryChange in Cl2 is 1/2 change in NO

Change in NOCl is - change in NO

2NOCl 2NO Cl2 pp

Initial 1.20 0.45 0.87

Change X-.45

Final X

68

Now we can determine final concentrations

Add

2NOCl 2NO Cl2 pp

Initial 1.20 0.45 0.87

Change 0.45-X X-.45 0.5X - .225

Final X

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Now we can write equilibrium constant K = (X)2(0.5X+0.645)

(1.65-X)2 Now we can test our assumption X is

small so ignore it in (+) and (-)

2NOCl 2NO Cl2 pp

Initial 1.20 0.45 0.87

Change 0.45-X X-.45 0.5X - .225

Final 1.65-X X 0.5X + 0.645

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K = (X)2(0.645) = 1.6 x 10-5

(1.65)2 X= 8.2 x 10-3 Figure out final concentrations

2NOCl 2NO Cl2 pp

Initial 1.20 0.45 0.87

Change 0.45-X X-.45 0.5X - .225

Final 1.65-X X 0.5X +0.645

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X= 8.2 x 10X= 8.2 x 10-3-3

[NOCl] = 1.6[NOCl] = 1.655

[Cl[Cl22] = 0.64] = 0.6499

Check assumptions (x < all other [ ]s)Check assumptions (x < all other [ ]s) .0082/.45 = 1.8 % OKAY!!!.0082/.45 = 1.8 % OKAY!!!

2NOCl2NOCl 2NO 2NO ++ ClCl22 pp

InitialInitial 1.20 1.20 0.45 0.45 0.870.87

ChangeChange 0.45-X0.45-X X-.45 X-.45 0.5X - .2250.5X - .225

FinalFinal 1.65-X X1.65-X X 0.5X +0.645 0.5X +0.645

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2NOCl2NOCl 2NO 2NO ++ ClCl22 InitialInitial 1.20 1.20 0.45 0.45 0.870.87

ChangeChange -2x -2x +2x +2x +x+x

FinalFinal 1.20-2X 0.45+2X1.20-2X 0.45+2X 0.87+x 0.87+x

Quadratic if put x in “change” pp

(0.45+2x (0.87+x)2 = 1.6 x 10-5 (1.20-2x)2

So, here we need “x” in the “final” line

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Practice Problem - put “x” in final

For the reactionFor the reaction2ClO2ClO(g)(g) ClCl22(g)(g) + + OO22(g)(g)

K = 6.4 x 10K = 6.4 x 10-3-3 In an experiment In an experiment 0.100 mol ClO(g)0.100 mol ClO(g), ,

1.00 mol O1.00 mol O22 and and 1.00 x 101.00 x 10-2-2 mol mol ClCl22 are mixed in a 4.00 L container.are mixed in a 4.00 L container.

What are the equilibrium [ ]s? What are the equilibrium [ ]s? (Remember to change to (Remember to change to molaritymolarity.).)

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2ClO(g) Cl2 (g) + O2 (g)

Q = 1, > K so shift to leftQ = 1, > K so shift to leftChlorine = LR so make it XChlorine = LR so make it X 2ClO(g) 2ClO(g) Cl Cl22 (g) + (g) + OO22 (g) (g)

InitInit . .025. .025 .0025.0025 .25.25 Chg 0050 - 2xChg 0050 - 2x x-.0025x-.0025 x-.0025 x-.0025 FinFin .030 - 2x .030 - 2x xx x+.2475x+.2475

X = 2.3 x 10X = 2.3 x 10-5-5

5% rule (.000023/.025)(100%) = 0.09%5% rule (.000023/.025)(100%) = 0.09%

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Mid-range K’s

.01<K<102

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No Simplification Scenario

Choose X to be small.Can’t simplify so we will have to solve

the quadratic (we hope)H2(g) + I2 (g) HI(g) K=38.6

What is the equilibrium concentrations if 1.800 mol H2, 1.600

mol I2 and 2.600 mol HI are mixed in

a 2.000 L container?

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Problems Involving PressureSolved exactly the same, with same Solved exactly the same, with same

rules for choosing “x” depending on Krules for choosing “x” depending on KPP

For For NN22OO44(g) 2NO(g) 2NO22(g)(g) KKpp = .131atm = .131atm

What are the equilibrium pressures if a What are the equilibrium pressures if a flask initially contains 1.000 atm Nflask initially contains 1.000 atm N22OO44??

Assume no quadratic, put “x” in “change” Assume no quadratic, put “x” in “change” column.column.Your Answers are . . .Your Answers are . . .

0.819 atm NO0.819 atm NO22 & .362 atm NO & .362 atm NO22

78

Super Hints pp

Always look Always look firstfirst for a perfect square!! for a perfect square!! Large K - put “x” in “final rowLarge K - put “x” in “final row Small K - 98% of time can put “x” in change Small K - 98% of time can put “x” in change

row, but . . . row, but . . . If with small K you get (-) [ ]s or must use If with small K you get (-) [ ]s or must use

quadratic then put “x” in final row.quadratic then put “x” in final row. Putting “x” in final row always works, but is Putting “x” in final row always works, but is

difficult to use when we get to difficult to use when we get to acid/base/buffer equilibrium. acid/base/buffer equilibrium.

So for small K try “x” in change row first. So for small K try “x” in change row first.

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13.7 Le Chatelier’s PrincipleIf a stress is applied to a system at

equilibrium, the position of the equilibrium will shift to reduce the stress.

3 Types of stressChange in amounts, pressure or

temperature

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Changing the amounts of reactants and/or products

Adding product makes Q > KAdding product makes Q > KRemoving reactant makes Q > KRemoving reactant makes Q > KAdding reactant makes Q < KAdding reactant makes Q < KRemoving product makes Q < K Removing product makes Q < K Determining the effect on Q will tell Determining the effect on Q will tell

you the direction of shiftyou the direction of shift

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Can Change Pressure Because partial pressures (and Because partial pressures (and

concentrations) change a new equilibrium concentrations) change a new equilibrium positionposition must be reached. must be reached.

By By reducingreducing volumevolume System will move in the direction that has the System will move in the direction that has the

leastleast molesmoles of gas. of gas. System tries to System tries to minimize the molesminimize the moles of gas. of gas. By By increasingincreasing volumevolume, system moves in , system moves in

direction that has the direction that has the mostmost molesmoles of gas. of gas.

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Can Change Total PressureBy adding an By adding an inertinert gas. gas.BUT partial pressures of BUT partial pressures of reactants reactants

and productsand products are are notnot changed changedNo effectNo effect on equilibrium position (test on equilibrium position (test

concept and problem questions)concept and problem questions)

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Change in TemperatureAffects the Affects the ratesrates of of bothboth the forward the forward

and reverse reactions.and reverse reactions.So, doesn’t just change the So, doesn’t just change the

equilibrium equilibrium positionposition, , but alsobut also changes changes the equilibrium the equilibrium constantconstant..

The The direction of the shift dependsdirection of the shift depends on on whether it is exo- or endothermicwhether it is exo- or endothermic

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Exothermic H < 0H < 0Releases heatReleases heatThink of heat as a Think of heat as a productproduct in in

exothermic rxnsexothermic rxnsRaising temperature pushes Raising temperature pushes toward toward

reactantsreactants..Shifts to Shifts to leftleft..

85

EndothermicH > 0Requires heatThink of heat as a reactantRaising temperature push toward

products.Shifts to right.

86

Review 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)

∆H = 2820 kJSome CO2(g is added and causes . . .

Shift to right (Q < K) so get more C6H12O6(s) .

Temperature is raised . . .Q < K Shift to right, more C6H12O6(s) forms.

Volume is decreasedEqual moles of gaseous reactants and

products, so no change.

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Review 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)

∆H = 2820 kJ Some O2(g) is removed . . .

Q < K shift to right, moreC6H12O6(s) forms.

Some C6H12O6(s) is removed . . .

No change (glucose is a solid so is not in the equilbrium expression). No more C6H12O6(s) .

A catalyst is added . . . No effect on equilibrium, only allows it be

attained more quickly. No more C6H12O6(s) .

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Review 6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)

∆H = 2820 kJSome H2O is removed . . .

No change, water is a liquid. No additional C6H12O6(s) forms.

End, Chapter 13.