IB Chemistry on Equilibrium Constant, Kc and Equilibrium Law.

http://lawrencekok.blogspot.com

Prepared by

Lawrence Kok

Tutorial on Equilibrium Law, Equilibrium constant

Kc and Reaction quotient Qc.

http://lawrencekok.blogspot.com/

Dynamic Equilibrium

Reversible (closed system)

Forward Rate, K1 Reverse Rate, K-1

Kc = ratio of molar conc of product (raised to power of their respective stoichiometry coefficient) to molar conc of reactant (raised to power of their respective stoichiometry coefficient)

Conc of product and reactantat equilibrium

At Equilibrium

Forward rate = Backward rateConc reactants and products remain

CONSTANT/UNCHANGE

Equilibrium Constant Kc

aA(aq) + bB(aq) cC(aq) + dD(aq)

coefficient

Solid/liq not included in Kc

Conc represented by [ ]

K1

K-1

ba

dc

cBA

DCK

1

1

K

KKc


express in

Conc vs time Rate vs time

A + B

C + D

Conc

Time

Click here notes on dynamic equilibrium

Excellent Notes

reversetconsrate

forwardtconsrate

K

K

..tan..

..tan..

1

1

http://www.pmstllc.com/Laney/Chem_30A_Documents_files/Lecture_21_notes.pdf



Large Kc

• Position equilibrium shift to right• More products > reactants

Magnitude of Kc

ba

dc

cBA

DCK

Extend of reaction

How far rxn shift to right or left?

Not how fast

ba

dc

cBA

DCK

Small Kc

• Position equilibrium shift to left• More reactants > products

cKcK

Position of equilibrium

2CO2(g) ↔ 2CO(g) + O2(g)

92103 cK

2H2(g) + O2(g) ↔ 2H2O(g)

81103cK

H2(g) + I2(g) ↔ 2HI(g)

2107.8 cK

1

Kc

• Position equilibrium lies slightly right• Reactants and products equal amount

Reaction completion

Product favouredReactant favoured Reactant/Product equal

cK

Temp dependent

Extend of rxn

Not how fast


ba

dc

cBA

DCK


Conc of product and reactant at equilibrium

Equilibrium expression HOMOGENEOUS gaseous rxn

4NH3(g) + 5O2(g) ↔ 4NO(g) + 6H2O(g) N2(g) + 3H2(g) ↔ 2NH3(g)

NH4CI(s) ↔ NH3(g) + HCI(g)

2SO2(g) + O2(g) ↔ 2SO3(g)

52

4

3

6

2

4

ONH

OHNOKc

32

1

2

2

3

HN

NHKc

11

3 HCINHKc

04

11

3

CINH

HCINHKc

12

2

2

2

3

OSO

SOKc

Equilibrium expression HETEROGENOUS rxn

CaCO3(s) ↔ CaO(g) + CO2(g)

03

1

2

1

CaCO

COCaOK c

12

1COCaOKc

CH3COOH(l) + C2H5OH(l) ↔ CH3COOC2H5(l) + H2O(l)

152

1

3

1

2

1

523

OHHCCOOHCH

OHHCOOCCHKc

Equilibrium expression HOMOGENEOUS liquid rxn

Cu2+(aq) + 4NH3(aq) ↔ [Cu(NH3)4]

2+

43

12

2

43 )(

NHCu

NHCuKc

Reactant/product same phase

Reactant/product diff phase

aA bB

2aA 2bB

bB aA

aA bB

aA bB

a

b

cA

BK

aA bB

Equilibrium Constant Kc Equilibrium Constant Kc

b

a

cB

AK

'

c

cK

K1'

inverse

X2 coefficient

2'

cc KK

coefficient

2

1

a

b

c

A

BK

21

21

'

ccc KKK 21'

a

b

cA

BK

a

b

cA

BK

a

b

cA

BK

2

2'

2

1

aA bB bB cC

a

b

ciA

BK

b

c

ciiB

CK

+ 2 reactions+ aA cC

a

c

a

b

b

c

cA

C

A

B

B

CK

'

ciciic KKK '

Effect on Kc

Inverse Kc

Square Kc

Square root cK

Multiply both Kc

2

1

ciiK ciK

N2(g) + O2(g) ↔ 2NO(g)

2NO(g) + O2(g) ↔ 2NO2(g)

19103.2 ciK

6103ciiK

2NO2(g) ↔ N2(g) + 2O2(g)

13

619

107

103103.2

c

c

ciicic

K

K

KKKN2(g) + 2O2(g) ↔ 2NO2(g)

13107 cK

12'

13

'

1042.1

107

11

c

c

c

K

KK

HF(ag) ↔H+(aq) + F -(aq)

H2C2O4(ag) ↔ 2H+(aq) + C2O4

2 -(aq)

4108.6 ciK

6108.3 ciiK

2HF(ag) + C2O42- ↔ 2F -(aq) + H2C2O4(aq)

2HF(ag) ↔ 2H+(aq) + 2F -(aq)

2H+(ag) + C2O4

2- ↔H2C2O4(aq)

7242'106.4108.6 cic KK

5

6

''106.2

108.3

11

cii

cK

K

12.0106.2106.4 57

'''

c

ccc

K

KKK

Kc for diff rxnAdding 2 rxns

+

Inverse rxn

Adding 2 rxns

2HF(ag) + C2O42- ↔ 2F -(aq) + H2C2O4(aq)

+

HF(ag) ↔H+(aq) + F -(aq)

4108.6 ciK

x2 coefficient

H2C2O4(ag) ↔ 2H+(aq) + C2O4

2 -

Inverse rxn

6108.3 ciiK

2HF(ag) ↔ 2H+(aq) + 2F -(aq)

2H+(ag) + C2O4

2- ↔H2C2O4(aq)

Add 2 rxn

7'106.4 cK

5''106.2 cK+

Effect on Kc Effect on Kc

Inverse rxn Inverts expression

Doubling rxn coefficient Squares expression

Tripling rxn coefficient Cubes expression

Halving rxn coefficient Square root expression

Adding 2 reactions Multiplies 2 expression

cK

1

2

cK

3

cK

cK

ii

c

i

c KK

Square Kc

Invert Kc

Multiply Kc

1

2

3

N2(g) + 2O2(g) ↔ 2NO2(g)

H2 + I2 ↔ 2HI

50cK

12

1

2

2

IH

HIKc

2HI ↔ H2 + I2

2

1

2

1

2'

HI

IHKc

02.050

11'

c

cK

K

2SO2 + O2 ↔ 2SO3

12

2

2

2

3

OSO

SOKc

200cK

SO2 + O2 ↔ SO3

1.14200'

cc KK

2

1

4SO2 + 2O2 ↔ 4SO3

40000

200

,

22'

c

cc

K

KK

N2(g) + 3H2(g) ↔ 2NH3(g)

32

1

2

2

3

HN

NHKc

Kc is 170 at 500K Determine if rxn is at equilibrium when conc are at:[N2] =1.50, [H2] = 1.00, [NH3] = 8.00

00.150.1

00.8

3

2

1

2

2

3

c

c

Q

HN

NHQ

• Rxn not at equilibrium

• Shift to right, favour product

• Qc must increase, till equal to Kc

IB Questions

Determine Kc for inversing rxn

inverse

Determine Kc for halving rxn

2

1

1

2

2

2

2

3

OSO

SOK c

halving

Determine Kc for doubling rxn

2SO2 + O2 ↔ 2SO3

doubling

12

2

2

2

3

OSO

SOKc

200cK

2

1

2

2

2

2

3

OSO

SOK c

21

34

170cK7.42cQ

cc KQ

Kc and Qc

H2(g) + I2(g) ↔ 2HI(g)

cK

Constant at fixed Temp

12

1

2

2

IH

HIKc

At equilibrium

Independent of initial conc

Initial conc of H2 , I2 and HI

00.4cQ

12

1

2

2

IH

HIQc

4.461012.01014.1

1052.21212

22

cK 4.46cK

Expt InitialConc H2

InitialConc I2

Initial Conc HI

1 0.0500 0.0500 0.100


Expt InitialConc H2

InitialConc I2

Initial Conc HI

1 2.40 x 10-2 1.38 x 10-2 0

Expt EquilibriumConc H2

EquilibriumConc I2

EquilibriumConc HI

1 1.14 x 10-2 0.12 x 10-2 2.52 x 10-2

At equilibrium conc

Not at equilibrium

H2(g) + I2(g) ↔ 2HI(g)

00.4050.0050.0

100.02

cQ

Predict the direction of rxn

Difference between

cQ

Conc of product/reactant

at equilibruim conc

Reaction quotient at particular time

Not at equilibrium conc

Varies NOT constant

Kc and Qc

H2(g) + I2(g) ↔ 2HI(g)

12

1

2

2

IH

HIKc

4.461012.01014.1

1052.21212

22

cK 4.46cK

At equilibrium conc

cc KQ cc KQ

cc KQ

Reaction at equilibrium

More product > reactant

Rxn shift left more reactant

→

cc KQ

cQ

Bring Qc down

More reactant > product

Rxn shift right → more product

Bring Qc up cQ

cc KQ

cQ

Expt InitialConc H2

InitialConc I2

Initial Conc HI

1 0.0500 0.0500 0.100


12

1

2

2

IH

HIQc

00.4050.0050.0

100.02

cQ

cQ

Expt InitialConc H2

InitialConc I2

Initial Conc HI

1 0.0250 0.0350 0.300


12

1

2

2

IH

HIQc

1030350.00250.0

300.02

cQ

Click here to view notes

http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch16/equilib.html



Kc from reaction stoichiometry

H2(g) + I2(g) ↔ 2HI(g)

4.46 sameK c

12

1

2

2

IH

HIKc

Kc = 46.4 ( 730K)At equilibrium 4 diff initial conc of H2 , I2 and HI

4.461012.01014.1

1052.21212

22

cKRxn 1

same

Qc = Kc - rxn at equilibrium, no side/shift occurQc < Kc– rxn shift right, favour product Qc > Kc – rxn shift left, favour reactant

Rxn 2, 3, 4diff initial conc

more products

H2(g) + I2(g) ↔ 2HI(g)

cQ

Rxn shift to right

Rxn shift to leftmore reactants

treac

productQc

tan

treac

productQc

tan

cQ

cc KQ cc KQ cc KQ

12

1

2

2

IH

HIKc

How dynamic equilibrium is shifted when H2 is added ?

• Add H2 , Qc decrease• Position equilibrium shift to right• Rate forward and reverse increase• New equilibrium conc achieved when

Rate forward Kf = Rate reverse Kr

• More product NH3 ,but Kc unchanged

N2(g) + 3H2(g) ↔ 2NH3(g) 07.4cK

Equilibrium disturbedH2 added. More reactant

At equilibrium Conc reactant/product no change

At new equilibrium Conc reactant/product no change

24.2cQ

Equilibrium Conc H2 = 0.82MEquilibrium Conc N2 = 0.20MEquilibrium Conc NH3 = 0.67M

32

1

2

2

3

HN

NHKc

31

2

82.020.0

67.0cK

New Conc H2 = 1.00MConc N2 = 0.20MConc NH3 = 0.67M

32

1

2

2

3

HN

NHQc

31

2

00.120.0

67.0cQ

07.4cK

New Equilibrium Conc H2 = 0.90MNew Equilibrium Conc N2 = 0.19MNew Equilibrium Conc NH3 = 0.75M

31

2

90.019.0

75.0cK

32

1

2

2

3

HN

NHKc

07.4cK

Shift to the right - Increase product - New Conc achieve- Qc = Kc again

cc KQ

How dynamic equilibrium shift when H2 added ?

• Add H2 , Qc decrease• Position equilibrium shift to right• Rate forward and reverse increase• New equilibrium conc achieved when


• More product NH3 ,but Kc unchanged


N2(g) + 3H2(g) ↔ 2NH3(g) 07.4cK

07.4 cc KQ

Equilibrium disturbedH2 added. More reactant

cc KQ

Equilibrium shift to right Rate forward Kf > Rate reverse Kr

cQ

At equilibrium Conc reactant/product no change

At new equilibrium Conc reactant/product no change

Qc increase until Qc = Kc

cQ


cc KQ cc KQ cc KQ

Equilibrium Law (Using ICE Method)

Reactants species present Product species present

2HI 1H2 1I2

Initial 0.8 0.0 0.0

Change -2x +x +x

Equilibrium 0.8 – 2x +x +x

2HI ↔ H2 + I2Mol ratio 2↔ 1:1

↔

CO + H2O ↔ CO2 + H2

CO H2O I2 H2

Initial 3 3 0 0

Change -x -x +x +x

Equilibrium 3 - x 3 - x +x +x

Mol ratio 1:1↔ 1:1

2NO + 1O2 ↔ 2CO2

2NO 1O2 2CO2

Initial 0.3 0.3 0.0

Change -x -1/2x +x

Equilibrium 0.3 - x 0.3 – 1/2x +x

Mol ratio 2:1 ↔ 1

↔

↔

2NO2 2NO 1O2

Initial 0 1 1

Change +2x -2x -x

Equilibrium +2x 1 – 2x 1 - x

2NO2 ↔ 2NO + 1O2Mol ratio 2↔ 2:1

↔

2HI 1H2 1I2

Initial 0 3 1

Change +4x -2x -2x

Equilibrium +4x 3 – 2x 1 - 2x

2HI ↔ 1H2 + 1I2Mol ratio 2↔ 1:1

↔

2NO 1O2 2CO2

Initial 0 0 1

Change +2x +1x -2x

Equilibrium +2x +1x 1 - 2x

2NO + 1O2 ↔ 2CO2Mol ratio 2:1 ↔ 2

↔

ICE method"I" - initial conc for each species rxn mixture."C" – change in conc for each species as system move toward eq."E" - equilibrium conc of each species in equilibrium.

Change

React and Produce

x amtreact

x amtproduce


H2 added to system

2HI 1H2 1I2

Equilibrium 0.8 0.6 0.8

2HI ↔ 1H2 + 1I2Mol ratio 2↔ 1:1

↔

2HI 1H2 1I2

Initial 0.8 1.0 0.8

Change +2x -x -x

Equilibrium 0.8 + 2x 1 - x 0.8 - x

2H

Rxn shift to left

2NO2 2NO 1O2

Equilibrium 3 1 2

2NO2 ↔ 2NO + 1O2Mol ratio 2↔ 2:1

↔ NO2NO2 2NO 1O2

Initial 3 1.5 2

Change +2x -2x -x

Equilibrium 3 + 2x 1.5 – 2x 2 - x

CO H2O I2 H2

Equilibrium 3 3 2 2

CO + H2O ↔ I2 + H2Mol ratio 1:1↔ 1:1

CO1CO 1H2O 1I2 1H2

Initial 4 3 2 2

Change -x -x +x +x

Equilibrium 4 - x 3 - x 2 + x 2 + x

↔↔

Rxn shift to left

Rxn shift to right

cc KQ

cc KQ

cc KQ

At equilibrium

cQ cc QK

↔

At equilibrium NO added to system

cQ cc QK

↔cc KQ

At equilibrium CO added to system

cQ cc QK


ba

dc

cBA

DCK


Conc of product and reactant at equilibrium

Equilibrium Law Expression - Law of Mass Action

Cal Kc. (Vol vessel - 1 dm3 )Eq Conc : [CO] = 8.78 x 10-3 M

[Br2] = 4.90 x 10-3M[COBr2] = 3.40 x 10-3M

CO + Br2 ↔ COBr2

CO + Br2 ↔ COBr2

Eq conc 8.78 x 10-3 4.90 x 10-3 3.40 x 10-3

Eq Conc = Moles/Vol

12

1

1

2

BrCO

COBrKc

1313

3

1090.41078.8

1040.3

cK 79cK

Ex 1

Cal eq conc CI2

Kc = 0.19. (Vol vessel - 1 dm3 ) Eq Conc : [PCI5] = 0.20M

[PCI3] = 0.01M

PCI5 ↔ PCI3 + CI2Ex 2

PCI5 ↔ PCI3 + CI2

Eq conc 0.20 0.01 x

13

1

2

1

3

PCI

CIPCIK c

1

11

20.0

01.019.0

x 8.3x

Ex 3 N2O4 ↔ 2NO2

Cal eq conc N2O4

Kc = 1.06 x 10-5. (Vol vessel - 1 dm3 ) Eq Conc : [NO2] = 1.85 x 10-3M

N2O4 ↔ 2NO2

Eq conc x 1.85 x 10-3

142

2

2

ON

NOKc

1

235 1085.1

1006.1x

323.0cK

Cal Kc

Cal Eq Conc

Cal Eq Conc

All homogeneous gaseous


1 mol ethanoic acid add to 1 mol ethanol, at equilibrium, 0.67 mol ester produced.Cal Kc. (Vol vessel - 1 dm3)

CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O

Initial mole 1.0 1.0 0 0

Change (1 - 0.67) (1 - 0.67) 0.67 0.67

Mole at eq 0.33 0.33 0.67 0.67

Eq Conc 0.33/1 0.33/1 0.67/1 0.67/1

In esterification, 2 mol ethanoic acid, 3 mol ethanol and 2 mol H2O mixed.Cal eq conc, Kc - 4.0 (Vol vessel – 1 dm3 )

CH3COOH + C2H5OH ↔ CH3COOC2H5 + H2O

Initial mole 2 3 0 2

Change (2 - x ) (3 - x) +x 2 + x

Mole at eq (2 – x ) (3 – x) +x 2 + x

Eq Conc (2 – x )/1 (3 -x)/1 x/1 (2 + x)/1

Conc CH3COOC2H5 = 1.33MConc H2O = 3.33MConc CH3COOH = 0.67MConc C2H5OH = 1.67M

Change = React and ProduceEx 4

Ex 5

Mole ratio 1:1 ↔ 1:1

152

1

3

1

2

1

523

OHHCCOOHCH

OHHCOOCCHKc

11

11

33.033.0

67.067.0cK 12.4cK

- x amt reacted + x amt produced

152

1

3

1

2

1

523

OHHCCOOHCH

OHHCOOCCHKc

11

11

32

24

xx

xx

33.1x

I

C

E

Eq Conc = Mole/Vol

-0.67 +0.67

Eq Conc = Mole/Vol


Change = Reacted and Produced

I

C

E

Click here notes equilibrium law

http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/ICEchart.htm

http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/ICEchart.htm

http://www.knockhardy.org.uk/sci_htm_files/08equil2.pdf

Initial amt X (0.02) and Y (0.01). At eq conc – 0.009 Z produced Cal Kc. (Vol vessel – 600 cm3 )

X + Y ↔ Z

Initial mole 0.02 0.01 0

Change (0.02 – 0.009) (0.01 – 0.009) 0.009

Mole at eq 0.011 0.001 0.009

Eq Conc 0.011/0.6 0.001/0.6 0.009/0.6

Mixture 1.9 mol H2 and 1.9 mol I2 was added to vessel . At eq conc, 3mol HI formCal Kc . (Vol vessel – 1 dm3 )

H2 + I2 ↔ 2HI

Initial mole 1.9 1.9 0

Change (1.9 – 1.5) (1.9 – 1.5) 3

Mole at eq 0.4 0.4 3

Eq Conc 0.4/1 0.4/1 3/1

0.05 mol SO2CI2 was used, at equilibrium 0.0345 mol CI2 formedCal Kc. (Vol vessel – 2 dm3 )

SO2CI2 ↔ SO2 + CI2

Initial mole 0.05 0 0

Change (0.05 – 0.0345) 0.0345 0.0345

Mole at eq 0.0155 0.0345 0.0345

Eq Conc 0.0155/2 0.0345/2 0.0345/2

Ex 6

Ex 8


X + Y ↔ Z

11

1

YX

ZKc

11

1

0016.00183.0

015.0cK 491cK

12

1

2

2

IH

HIKc

11

2

4.04.0

3cK 56cK


122

1

2

1

2

CISO

CISOKc

1

11

00775.0

0173.00173.0cK

0386.0cK

H2 + I2 ↔ 2HI Ex 7

I

C

E

I

C

E

I

C

E

Mole ratio 1:1 ↔ 1

Mole ratio 1:1 ↔ 2

Mole ratio 1 ↔ 1:1

Change = React and Produce

-0.009 +0.009


-1.5 +3


-0.0345 +0.0345

Initial amt, 3 mol CO and H2O were used.Cal eq conc H2, Kc - 4.0 (Vol vessel - 1 dm3 )

Initial amt PCI5 - 0.01 mol. Kc = 0.19. (Vol vessel - 0.5 dm3 ) Cal eq conc PCI5, PCI3 and CI2

[PCI5] = (0.01 – 0.0091)/0.5 = 0.0018M[PCI3] = 0.0091/0.5 = 0.018M[CI2] = 0.0091/0.5 = 0.018M

Ex 9


12

1

1

2

1

2

OHCO

HCOKc

CO + H2O ↔ CO2 + H2

11

11

334

xx

xx

Mx 2

PCI5 ↔ PCI3 + CI2

15

1

2

1

3

PCI

CIPCIK c 1

2

5.0

01.0

5.0

x

x

Kc 1

2

5.0

01.0

5.019.0

x

x

0091.0x

CO + H2O ↔ CO2 + H2


Change (3 - x) (3 - x) +x +x

Eq Conc (3 - x)/1 (3 - x)/1 x/1 x/1

I

C

E

Ex 10

I

C

E

PCI5 ↔ PCI3 + CI2


Change 0.01 - x +x +x

Eq Conc (0.01 – x)/0.5 x/0.5 x/0.5

1 mol HI was used initially, and 0.78 mol of HI remain at equilibrium.Cal Kc. (Vol - 1 dm3 )

Ex 11

2HI ↔ H2 + I2

2HI ↔ H2 + I2

Initial mole 1 0 0

Change (1 - 0. 22) 0.11 0.11

Eq Conc 0.78/1 0.11/1 0.11/1

I

C

E

2

1

2

1

2

HI

IHKc

2

11

78.0

11.011.0cK 02.0cK



- x amt reacted + x amt produced


Mole ratio 2↔ 1:1

-0.22 +0.11

Quadratic eqn

Quadratic eqn

H2 + I2 ↔ 2HI

Eq conc reactant/product given.Cal Kc using vol vessel - 1 dm3 /2 dm3

Ex 12


H2 + I2 ↔ 2HI

H2 + I2 ↔ 2HI Mole at eq 0.4 0.4 3

Eq Conc 0.4/1 0.4 /1 3/1

Using approximation and assumption

Eqn with equal number molecule both sideEq expression – Independent of volume

12

1

2

2

IH

HIKc

12

1

2

2

IH

HIKc

11

2

4.04.0

3cK

11

2

24.0

24.0

23

cK

56cK

H2 + I2 ↔ 2HI Mole at eq 0.4 0.4 3

Eq Conc 0.4/2 0.4 /2 3/2

56cK


Mole at eq 0.011 0.001 0.009

Eq Conc 0.011/1 0.001/1 0.009/1


122

1

2

1

2

CISO

CISOKc

1

11

011.0

009.0001.0cK

41018.8 cK

1

11

2011.0

2009.0

2001.0

cK

111

0055.0

00045.00005.0cK

51009.4 cK

1 dm3 2 dm3

1 dm3 2 dm3


Ex 13

Eqn with diff number molecule both sideEq expression – Dependent of volume

Eq conc reactant/product given.Cal Kc using vol vessel - 1 dm3 /2 dm3


Mole at eq 0.011 0.001 0.009

Eq Conc 0.011/2 0.001/2 0.009/2

Eq amt used instead of eq conc


Equilibrium Law (Using ICE Method) Using approximation and assumption

2H2O ↔ 2H2 + O2

2H2O ↔ 2H2 + O2


Change - x +x +x/2

Eq Conc 0.1 - x +x +x/2

22

1

2

2

2

OH

OHK c

2

12

5

1.0

2/104

x

xx

Solve cubic eqn difficult !

523 1041.02 xx

2

12

5

1.0

2/104

xx

523 1041.02 x

0093.0x

PCI5 ↔ PCI3 + CI2

15

1

2

1

3

PCI

CIPCIK c

Eq conc reactant/product given.Cal eq conc, Kc - 1. Vol vessel - 1 dm3

Solve quadratic eqn difficult !

Kc very small < 10-3

Not much change in reactant concApproximation is VALID

SMALLKc BIGKc

Eq conc reactant/product given.Cal eq conc, Kc - 4 x 10-5. Vol vessel - 1 dm3

- Little product form

- Initial Conc reactant

unchanged

Using

approximation

0.1 – x ≈ 0.1

[reactant]initial ≈[reactant]eq

Cannot use

approximation

1 – x ≠ 1

Kc very BIG > 10-3

Change in reactant concApproximation NOT VALID

[reactant]initial ≠ [reactant]eq

- Lots product form

- Initial Conc reactant

changed

PCI5 ↔ PCI3 + CI2

Initial mole 1 0 0

Change 1 - x +x +x

Eq Conc 1 – x x x

1

2

11

x

x

012 xx

62.0x

Equilibrium Law (Using ICE Method) Using approximation and assumption

PCI5 ↔ PCI3 + CI2

PCI5 ↔ PCI3 + CI2

Initial mole 1 0 0

Change 1 - x +x +x

Eq Conc 1 – x x x

15

1

2

1

3

PCI

CIPCIK c

1

2

11

x

x

62.0x

Eq conc reactant/product given.Cal eq conc, Kc - 1. Vol vessel - 1 dm3

Solve quadratic eqn 012 xxKc very small < 10-3

Not much change reactant concApproximation is VALID

SMALLKc BIGKc

Eq conc reactant/product given.Cal eq conc, Kc - 1 x 10-10. Vol vessel - 1 dm3

- Little product form

- Initial conc reactant

unchanged

Using

approximation

1 – x ≈ 1

Cannot use

approximation

1 – x ≠ 1

Kc very BIG > 10-3

Change in reactant conc, BIG

[reactant]initial ≠ [reactant]eq

[ 1 ]initial ≠ [ 0.38 ]eq

- Lots product form

- Initial conc reactant

changed

[PCI5] = (1 – 0.62) = 0.38M[PCI3] = 0.62M[CI2] = 0.62M

Change in x - big

PCI5 ↔ PCI3 + CI2

Initial mole 1 0 0

Change 1 - x +x +x

Eq Conc 1 – x x x

PCI5 ↔ PCI3 + CI2

15

1

2

1

3

PCI

CIPCIK c

1

2

10

1101

x

x

1

2

10

1101

x

5101 x

[PCI5] = (1 – 0.00001) = 0.99999M

[PCI3] = 0.00001M[CI2] = 0.00001M

[reactant]initial ≈ [reactant]eq

[ 1 ]initial ≈ [0.99999]eq

Eqn with equal number molecule both sideEq expression – Independent of volume

Initial amt, 1 mol CO and H2O were used.Cal eq conc all component, Kc - 4.0 (Vol vessel - 1 dm3 )

Ex 14


12

1

1

2

1

2

OHCO

HCOKc

CO + H2O ↔ CO2 + H2

11

11

114

xx

xx

667.0x

CO + H2O ↔ CO2 + H2


Change - x - x +x +x

Eq Amt (1 - x) (1 - x) x x

I

C

E

0.05 mol SO2CI2 added to 2 dm3 vessel. At equilibrium, 0.0345 mol CI2 form. Cal Kc

I

C

E

0386.0cK

2

2

14

x

x

[CO] = 1 – 0.667 = 0.333 [H2O] = 1 – 0.667 = 0.333[CO2] = 0.667[H2O] = 0.667

Using square root method




Change -0.0345 +0.0345 + 0.0345

Mole at eq (0.05 – 0.0345) 0.0345 0.0345

Eq Conc 0.0155/2 0.0345/2 0.0345/2

122

1

2

1

2

CISO

CISOKc

1

11

00775.0

0173.00173.0cK

Using approximation and assumption

Cannot use

approximation

[1 – x] ≠ 1

BIGKc

1 dm3

2 dm3

Eqn with diff number molecule both sideEq expression – Dependent of volume

Ex 15


Eq amt cannot be used


H2 + I2 ↔ 2HI

H2 + I2 ↔ 2HI

Initial mole 2 2 0

Change -x -x +2x

Mole at eq 2 - x 2 - x +2x

Eq Conc 2 – x 2 – x +2x

12

1

2

2

IH

HIKc

11

2

22

249

xx

x

x

x

2

249

Initial amt, 2 mol H2 and I2 were used.Cal eq conc all component, Kc - 49 (Vol vessel - 2 dm3 )

56.1x

I

C

EEqn with equal number molecule both sideEq expression – Independent of volume

[H2] = (2 – 1.56) = 0.44M[I2] = (2 – 1.56) = 0.44M[HI] = 2 x 1.56 = 3.12M

Click here simulation equilibrium Click here simulation

Equilibrium was investigated in two expt at diff tempResult shown table below. Cal Kc for expt 1 and 2.

2HI ↔ H2 + I22HI ↔ H2 + I2


Change -0.02 +0.01 +0.01

Mole at eq (0.06 – 0.02) +0.01 +0.01

Eq Conc 0.04 0.01 0.01

EXPT 1

2HI ↔ H2 + I2

Initial mole 0 0.04 0.04

Change +0.04 -0.02 - 0.02

Mole at eq +0.04 0.04 – 0.02 0.04 - 0.02

Eq Conc 0.04 0.02 0.02

2

1

2

1

2

HI

IHKc

2

11

04.0

01.001.0cK21025.6 cK

2

11

04.0

02.002.0cK 25.0cK

EXPT 2

Ex 16

http://chemcollective.org/activities/tutorials/equil/qtok_prog



http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/simDownload/



66% N2O4 reacted

Mol ratio N2O4 : NO2, = 1 : 2

66% N2O4 react – 66% x 1 = 0.66 mol N2O4 react

0.66 mol N2O4 react - 1.32 mol NO2 produced

Amt N2O4 left = 1 – 0.66 = 0.34 mol

Initial amt -2 mol SO2 and O2 were used.Cal eq amt, if 90% of SO3 formed .

Equilibrium Law)

2SO2 + 1O2 ↔ 2SO3

Initial mole 2 2 0

Change - 1.8 -0.9 +1.8

Eq Amt (2 – 1.8) (2 – 0.9) +1.8

Eq Amt 0.2 1.1 1.8

15100

2331

2%

xxx

xNitrogen

90% SO3 produce

Mol ratio SO2 : SO3, = 1 : 1

90% SO2 react – 90% SO3 produced

90% SO2 react - 90% x 2 = 1.8 mol SO3 produced

Amt SO2 = (2 – 1.8) = 0.2 mol left

Mol ratio SO2 : O2. = 2 : 1

1.8 mol SO2 react with 0.9 mol O2

Amt O2 = (2 – 0.9) = 1.1 mol left

Initial amt, 2 mol SO2 and O2 were used.Cal eq amt, if 20% of SO2 react.

20% SO3 react

Mol ratio SO2 : SO3, = 1 : 1

20% SO2 react – 20% SO3 produced

20% SO2 react - 20% x 2 = 0.4 molSO3 produced

Amt SO2 left = 2 – 0.4 = 1.6 mol left

Mol ratio SO2 : O2. = 2 : 1

0.4 mol SO2 react with 0.2 mol O2

Amt O2 left = 2 – 0.2 = 1.8 mol

Initial amt, 1 mol N2O4 used. 66% dissociated.Cal eq amt.

1N2O4 ↔ 2NO2

Initial mole 1 0

Change -0.66 +1.32

Eq Amt (1 – 0.66) +1.32

Eq Amt 0.34 1.32

N2 + H2 react ratio 1 : 3 to produce NH3. Equilibrium mixture contain 15% NH3. Cal eq amt for each species.

N2 + 3H2 ↔ 2NH3

Initial mole 1 3 0

Change -x -3x 2x

Eq Amt 1 – x 3 – 3x 2x

Eq Amt 0.74 2.22 0.52

Let x N2 reacted

X = 0.26

[N5] = 0.74, [H2] = 2.22, [NH3] = 0.52

2SO2 + 1O2 ↔ 2SO3

Initial mole 2 2 0

Change -0.4 -0.2 +0.4

Eq Amt (2 – 0.4) (2 – 0.2) +0.4

Eq Amt 1.6 1.8 0.4

Change

React and Produce

x amtreact

x amtproduce

Ex 17 Ex 18

Ex 19Ex 20

Click here to view simulationClick here simulation using paper clips Click here simulation on reversible rxn

Click here on reversible rxn

Simulation on Dynamic equilibrium

Click here equilibrium constant

Click here simulation equilibrium Click here simulation

http://www.ejmste.com/022006/d8.pdf



http://www3.fed.cuhk.edu.hk/chemistry/files/Simulation.pdf



https://www.youtube.com/watch?v=60Sylqei2DY


http://phet.colorado.edu/en/simulation/reversible-reactions




https://www.youtube.com/watch?v=cHAjhM3y3ds









Acknowledgements

Thanks to source of pictures and video used in this presentation

Thanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/

Prepared by Lawrence Kok

Check out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com

http://creativecommons.org/licenses/

http://lawrencekok.blogspot.com/

IB Chemistry on Equilibrium Constant, Kc and Equilibrium Law.

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Transcript of IB Chemistry on Equilibrium Constant, Kc and Equilibrium Law.