The document provides notes on homework problems involving chemical equilibrium. It states that problems may use different pressure units and that negatives are possible. It also advises not rounding when doing online homework. Key concepts covered include reversible reactions reaching equilibrium when forward and reverse reaction rates are equal. The equilibrium constant K is defined using concentrations or pressures and describes the position of equilibrium. K is constant at a given temperature but the concentrations or pressures at equilibrium can vary.

1. 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” forSome 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 badOne problem might give you -[ ]s. It’s bad
chem, but good math & the problem ischem, but good math & the problem is
“doable.”“doable.”
Do not round when doing online HWDo not round when doing online HW
(even if adding or subtracting).(even if adding or subtracting).

2. 2
13.1 Reactions are reversible
A + B C + D ( forward)A + B C + D ( forward)
C + D A + B (reverse)C + D A + B (reverse)
InitiallyInitially there isthere is only A and Bonly A and B so onlyso only
the forward reaction is possiblethe forward reaction is possible
As C and D build up, the reverseAs C and D build up, the reverse
reaction speeds up while the forwardreaction speeds up while the forward
reaction slows down.reaction slows down.
Eventually the rates areEventually the rates are equalequal

3. 3
ReactionRate
Time
Forward Reaction
Reverse reaction
Equilibrium

4. 4
What is equal at Equilibrium?
RatesRates are equalare equal
ConcentrationsConcentrations are not!are not!
Rates are determined byRates are determined by
concentrations and activation energy.concentrations and activation energy.
The concentrations doThe concentrations do notnot change atchange at
equilibrium.equilibrium.
Or, if the reaction is verrrry slooooow.Or, if the reaction is verrrry slooooow.

5. 5
13.2 Law of Mass Action pp
For anyFor any solution or gaseoussolution or gaseous reactionreaction
jjA +A + kkBB llC +C + mmDD
KKcc = [C]= [C]ll
[D][D]mm
PRODUCTSPRODUCTSpowerpower
[A][A]jj
[B][B]kk
REACTANTSREACTANTSpowerpower
KKcc is called the equilibrium constant.is called the equilibrium constant.
KKcc often is written as just Koften is written as just K
is how we indicate a reversibleis how we indicate a reversible
reactionreaction
DoesDoes notnot apply toapply to solidssolids oror pure liquidspure liquids!!

6. 6
Playing with K pp
If we write the reaction in reverse.
lC + mD jA + kB
Then the new equilibrium constant is
K’ = [A]j
[B]k
= 1/K
[C]l
[D]m

7. 7
Playing with K pp
If we multiply the equation by a
constant
njA + nkB nlC + nmD
Then 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

8. 8
Calculate K - all are gases, given
NN22 + 3H+ 3H22 2NH2NH33 where K = 1.3 x 10where K = 1.3 x 10-2-2
1/2N1/2N22 + 3/2H+ 3/2H22 NHNH33 K = ? . . .K = ? . . .
K1/2
= 0.11= 0.11
2NH2NH33 NN22 + 3H+ 3H22 K = ? . . .K = ? . . .
K-1
= 77= 77
NHNH33 1/2N1/2N22 + 3/2H+ 3/2H22 K = ? . . .K = ? . . .
(K-1
)1/2
= (1/K)1/2
= 8.8
2N2N22 + 6H+ 6H22 4NH4NH33 K = ? . . .K = ? . . .
(K)2
= 1.7 x 10-4

9. 9
The units for K
Are determined by the variousAre 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

10. 10
K is CONSTANT
At 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.

11. 11
Equilibrium Constant
One for each Temperature

12. 12
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

13. 13
Calculate K
N2 + 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

14. 14
13.3 Equilibrium and Pressure
Some reactions are gaseousSome reactions are gaseous
PV = nRTPV = nRT
P = (n/V)RT where n/V = [ ] of gasP = (n/V)RT where n/V = [ ] of gas
P = CRTP = CRT
C is a concentration in moles/LiterC is a concentration in moles/Liter
C = P/RTC = P/RT

15. 15
Equilibrium and Pressure
2SO2(g) + O2(g) 2SO3(g)
Kp = (PSO3)2
(PSO2)2
(PO2)
Kc = [SO3]2
2

16. 16
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

17. 17
General Equation - Deriving:
jjA +A + kkBB llC +C + mmDD
KKpp= (P= (PCC))ll
(P(PDD))mm
== (C(CCCxRTxRT))ll
(C(CDDxRTxRT))mm
(P(PAA))jj
(P(PBB))kk
(C(CAAxRTxRT))jj
(C(CBBxRTxRT))kk
KKpp== (C(CCC))ll
(C(CDD))mm
x(RT)x(RT)l+ml+m
(C(CAA))jj
(C(CBB))kk
x(RT)x(RT)j+kj+k
KKpp == KK (RT)(RT)((l+m)-(j+k)l+m)-(j+k)
==KKcc(RT)(RT)∆∆nn

18. 18
General Equation pp
KKpp == KKcc(RT)(RT)∆∆nn
(R = .08206 l atm/K mol & T in Kelvin)(R = .08206 l atm/K mol & T in Kelvin)
WhereWhere ∆∆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)
DoDo notnot count their moles incount their moles in
calculatingcalculating ∆∆n.n.
Be sure toBe sure to convert C to Kconvert C to K for temp.for temp.

19. 19
Homogeneous Equilibria
So 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).

20. 20
13.4 Heterogeneous Equilibria
If 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:

21. 21
For Example
H2(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’
[H ]

22. 22
E.g., Write the Kp Expression:
2Fe(s) + 3/2 O2(g) Fe2O3(s)
Kp = 1
(PO2)3/2

23. 23
13.5 Applications of the
Equilibrium Constant

24. 24
The Extent of a Reaction
Size 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
right
I.e., at =m have mostly products

25. 25
The Reaction Quotient
Tells 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

26. 26
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

27. 27
Example
For 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 35ºCat 35ºC
In an experiment 0.10 molIn an experiment 0.10 mol NOClNOCl,,
0.0010 mol0.0010 mol NONO and 0.00010 moland 0.00010 mol ClCl22
are mixed in 2.0 L flask.are mixed in 2.0 L flask.
Which direction will the reactionWhich direction will the reaction
proceed to reach equilibrium? (Useproceed to reach equilibrium? (Use
initial concentrations) Answer . . .initial concentrations) Answer . . .

28. 28
Example
Q is not 1.0 x 10-8
. Why?
You were given moles in a 2 liter flask
So [ ]s are halved
Q = 5.0 x 10-9
Since Q < K (which was 1.55 x 10-5
)
Reaction shifts to the right.

29. 29
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”

30. 30
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 ofIn a certain experiment 2.00 mol of
SOSO22, 1.50 mol of, 1.50 mol of OO22 and 3.00 mol ofand 3.00 mol of
SOSO33 were placed in a 1.00 L flask. Atwere placed in a 1.00 L flask. At
equilibrium 3.50 molequilibrium 3.50 mol SOSO33 were foundwere found
to be present. Calculate:to be present. Calculate:
The equilibrium concentrations of OThe equilibrium concentrations of O22
and SOand SO22, K, Kcc and Kand KPP (procedure follows .(procedure follows .
. .). .)

31. 31
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!!!

32. 32
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]

33. 33
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]

34. 34
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

35. 35
Solving pp
600º C600º C 2SO2SO22(g)(g) + O+ O22(g(g 2SO2SO33(g)(g)
initialinitial 2.002.00 MM 1.51.5 MM 3.003.00 MM
changechange -0.50-0.50 MM -0.25-0.25 MM +0.50+0.50 MM
finalfinal 1.501.50 MM 1.251.25 MM 3.503.50 MM
KKcc = 4.36= 4.36
KKpp = K= Kcc(RT)(RT)∆n∆n
and ∆n = -1and ∆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).

36. 36
2SO2SO22(g)(g) + O+ O22(g)(g) 2SO2SO33(g)(g) @ 600ºC@ 600ºC
In aIn a differentdifferent experimentexperiment .500 mol SO.500 mol SO22
andand .350 mol SO.350 mol SO33 were placed in awere placed in a
1.000 L container. When the system1.000 L container. When the system
reaches equilibriumreaches equilibrium 0.045 mol of O0.045 mol of O22
are present.are present.
Calculate the final concentrations ofCalculate the final concentrations of
SOSO22 and SOand SO33 and K. Do on own --and K. Do on own --
your answers are . . .your answers are . . .

37. 37
In aIn a differentdifferent expt at 600 ºC .500 mol SOexpt at 600 ºC .500 mol SO22 & .& .
350 mol SO350 mol SO33 placed in a 1.000 L container. Atplaced in a 1.000 L container. At
=m 0.045 mol of O=m 0.045 mol of O22 are present.are present.
2SO2SO22(g)(g) + O+ O22(g)(g) 2SO2SO33(g)(g)
Same temp.Same temp., so K, so Kcc is the same!! (4.36)is the same!! (4.36)
Since no OSince no O22 initially, its change is +0.045 andinitially, its change is +0.045 and
the reaction shifted to thethe reaction shifted to the leftleft..
Plug in to get SOPlug in to get SO22 and SOand SO33 =m [ ]s (derived=m [ ]s (derived
from the Ofrom the O22 change in [ ])change in [ ])
Your answer for theirYour answer for their concentrationsconcentrations . . .. . .
[SO[SO22] = .590] = .590 MM & [SO& [SO33] = .260 (mole ratios)] = .260 (mole ratios)

38. 38
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
. . .

39. 39
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 QWrite balanced equation, determine Q
and direction of shift, set up RICEand direction of shift, set up RICE
Steps on next slides.

40. 40
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.

41. 41
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.

42. 42
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) + FF22(g) 2HF(g)(g) 2HF(g)
finalfinal 2.00-x2.00-x 2.00-x2.00-x 2.00+2x2.00+2x
2.00 - x = [H2.00 - x = [H22]]eqeq & [F& [F22]]eqeq == 0.4720.472 MM
(2.00 +(2.00 + 22x) = [HF]x) = [HF]eqeq == 5.0565.056 MM

43. 43
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. 44
General Rules to avoid quadratics pp
When KWhen Kcc is large - find limiting reactantis large - find limiting reactant
and set it as “x” in the “equilibrium” line ofand 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” inis 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 requiresIf putting “x” in “change line then requires
a quadratic or you get aa 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 dois “middle” then must do
quadratic.quadratic.

45. 45
General Rules to avoid quadratics pp
Why don’t we just always put “x” in theWhy 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 showWe could, but most textbooks don’t show
this method because the vast majority ofthis method because the vast majority of
problems involve a small Kproblems involve a small Kcc where puttingwhere putting
it the “change” line works.it the “change” line works.
Also, when we do buffer equilibria we willAlso, when we do buffer equilibria we will
put “x” in the “change” line so we mostlyput “x” in the “change” line so we mostly
want to use that method unless we can’twant 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. 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 25ºCat 25ºC
Calculate the equilibrium concentrations ifCalculate the equilibrium concentrations if
aa 5.00 L5.00 L container initially contains 15.7container initially contains 15.7 gg
of Hof H22 and 294and 294 gg II22 ..
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

47. 47
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. 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. 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-0.232
final X

50. 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-
2X
final X

51. 51
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

52. 52
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 equation
7.1 x 102
= (0.464)2
(1.328)(X)
Makes the algebra easy

53. 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 wouldNote: If had set “x” in change line we would
have gothave got (-)(-) =m concentrations for H=m concentrations for H22 & I& I22..
HH22(g)(g) II22(g) 2HI(g)(g) 2HI(g) pppp
initial 1.56initial 1.56 MM 0.2320.232 MM 00 MM
change X-0.232change X-0.232 X-0.232X-0.232
0.464-2X final0.464-2X final 1.328+X1.328+X XX
0.464-2X0.464-2X

54. 54
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. 55
HH22(g)(g) II22(g) 2HI(g)(g) 2HI(g)
initial 1.56initial 1.56 MM 0.2320.232 MM 00 MM
change -xchange -x -x-x +2x+2x
finalfinal 1.56-x1.56-x 0.232-x0.232-x 0+2x0+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. 56
Practice (put “x” in final line)
For the reactionFor the reaction ClCl22 + O+ O22 2ClO(g)2ClO(g)
K = 156K = 156
In an experiment 0.100 mol ClO, 1.00In an experiment 0.100 mol ClO, 1.00
mol Omol O22 and 0.0100 moland 0.0100 mol ClCl22 are mixed in aare mixed in a
4.00 L4.00 L flask.flask.
If the reaction is not at equilibrium, whichIf 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. 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 getUse RICE table of molarities (divide by 4 to get
0.025M ClO0.025M ClO,, .25.25MM OO22 andand .0025.0025MM ClCl22
Since shift to right, make [=m] ClSince shift to right, make [=m] Cl22 = x (since it is= x (since it is
the LR)the LR)

58. 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. 59
Problems with small K pp
K< .01

60. 60
Example of small Kc pp
90% of the time you can x in “change”
line with a small Kc,
For the reactionFor the reaction
22NOClNOCl(g) 2(g) 2NONO(g) +(g) + ClCl22(g)(g)
K = 1.6 x 10K = 1.6 x 10-5-5
MM at 35ºCat 35ºC
In an experiment 1.00 molIn 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. 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)
0.500.50 MM 00 00
-2x-2x +2x+2x +x+x
0.5 - 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. 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)
0.500.50 MM 00 00
-2x-2x +2x+2x +x+x
0.5 - 2x0.5 - 2x +2x +x+2x +x
Since x = 1.0 x 10Since x = 1.0 x 10-2-2
, =m [ ]s are, =m [ ]s are 0.480.48
0.020.02 0.010.01

63. 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. 64
For example pp
For the reactionFor the reaction
2NOCl2NOCl 2NO2NO ++ ClCl22
K= 1.6 x 10K= 1.6 x 10-5-5
IfIf 1.20 mol NOCl1.20 mol NOCl,, 0.45 mol of NO0.45 mol of NO andand 0.87 mol0.87 mol
ClCl22 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

65. 65
Choose X to be small
NO will be LR
Choose NO to be X
2NOCl 2NO + Cl2 pp
Initial 1.20 0.45 0.87
Change
Final

66. 66
Figure out change in NO
Change = final - initial
change = X-0.45
2NOCl 2NO Cl2 pp
Initial 1.20 0.45 0.87
Change
Final X

67. 67
Now figure out the other changes
Use stoichiometry
Change 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. 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

69. 69
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

70. 70
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

71. 71
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 2NO2NO ++ ClCl22 pp
InitialInitial 1.201.20 0.450.45 0.870.87
ChangeChange 0.45-X0.45-X X-.45X-.45 0.5X - .2250.5X - .225
FinalFinal 1.65-X X1.65-X X 0.5X +0.6450.5X +0.645

72. 72
2NOCl2NOCl 2NO2NO ++ ClCl22
InitialInitial 1.201.20 0.450.45 0.870.87
ChangeChange -2x-2x +2x+2x +x+x
FinalFinal 1.20-2X 0.45+2X1.20-2X 0.45+2X 0.87+x0.87+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

73. 73
Practice Problem - put “x” in final
For the reactionFor the reaction
2ClO2ClO(g)(g) ClCl22(g)(g) ++ OO22(g)(g)
K = 6.4 x 10K = 6.4 x 10-3-3
In an experimentIn an experiment 0.100 mol ClO(g)0.100 mol ClO(g),,
1.00 mol O1.00 mol O22 andand 1.00 x 101.00 x 10-2-2
molmol ClCl22 areare
mixed in a 4.00 L container.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.).)

74. 74
2ClO(g) Cl2 (g) + O2 (g)
Q = 1, > K so shift to leftQ = 1, > K so shift to left
Chlorine = LR so make it XChlorine = LR so make it X
2ClO(g)2ClO(g) ClCl22 (g) +(g) + OO22 (g)(g)
InitInit. .025. .025 .0025.0025 .25.25
Chg 0050 - 2xChg 0050 - 2x x-.0025x-.0025 x-.0025x-.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%

75. 75
Mid-range K’s
.01<K<102

76. 76
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?

77. 77
Problems Involving Pressure
Solved exactly the same, with sameSolved exactly the same, with same
rules for choosing “x” depending on Krules for choosing “x” depending on KPP
ForFor NN22OO44(g) 2NO(g) 2NO22(g)(g) KKpp = .131atm= .131atm
What are the equilibrium pressures if aWhat 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. 78
Super Hints pp
Always lookAlways 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 changeSmall K - 98% of time can put “x” in change
row, but . . .row, but . . .
If with small K you get (-) [ ]s or must useIf 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 isPutting “x” in final row always works, but is
difficult to use when we get todifficult 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.

79. 79
13.7 Le Chatelier’s Principle
If a stress is applied to a system at
equilibrium, the position of the
equilibrium will shift to reduce the
stress.
3 Types of stress
Change in amounts, pressure or
temperature

80. 80
Changing the amounts of reactants and/or
products
Adding product makes Q > KAdding product makes Q > K
Removing reactant makes Q > KRemoving reactant makes Q > K
Adding reactant makes Q < KAdding reactant makes Q < K
Removing product makes Q < KRemoving product makes Q < K
Determining the effect on Q will tellDetermining the effect on Q will tell
you the direction of shiftyou the direction of shift

81. 81
Can Change Pressure
Because partial pressures (andBecause partial pressures (and
concentrations) change a new equilibriumconcentrations) change a new equilibrium
positionposition must be reached.must be reached.
ByBy reducingreducing volumevolume
System will move in the direction that has theSystem will move in the direction that has the
leastleast molesmoles of gas.of gas.
System tries toSystem tries to minimize the molesminimize the moles of gas.of gas.
ByBy increasingincreasing volumevolume, system moves in, system moves in
direction that has thedirection that has the mostmost molesmoles of gas.of gas.

82. 82
Can Change Total Pressure
By adding anBy adding an inertinert gas.gas.
BUT partial pressures ofBUT partial pressures of reactantsreactants
and productsand products areare notnot changedchanged
No effectNo effect on equilibrium position (teston equilibrium position (test
concept and problem questions)concept and problem questions)

83. 83
Change in Temperature
Affects theAffects the ratesrates ofof bothboth the forwardthe forward
and reverse reactions.and reverse reactions.
So, doesn’t just change theSo, doesn’t just change the
equilibriumequilibrium positionposition,, but alsobut also changeschanges
the equilibriumthe equilibrium constantconstant..
TheThe direction of the shift dependsdirection of the shift depends onon
whether it is exo- or endothermicwhether it is exo- or endothermic

84. 84
Exothermic
H < 0H < 0
Releases heatReleases heat
Think of heat as aThink of heat as a productproduct inin
exothermic rxnsexothermic rxns
Raising temperature pushesRaising temperature pushes towardtoward
reactantsreactants..
Shifts toShifts to leftleft..

85. 85
Endothermic
H > 0
Requires heat
Think of heat as a reactant
Raising temperature push toward
products.
Shifts to right.

86. 86
Review
6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)
∆H = 2820 kJ
Some 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 decreased
Equal moles of gaseous reactants and
products, so no change.

87. 87
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) .

88. 88
Review
6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)
∆H = 2820 kJ
Some H2O is removed . . .
No change, water is a liquid. No additional
C6H12O6(s) forms.
End, Chapter 13.