C. Y. Yeung (CHW, 2009)

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Chemical Equilibrium Chemical Equilibrium and and Equilibrium LawEquilibrium Law

Chemical Equilibrium Chemical Equilibrium andand Equilibrium LawEquilibrium Law

It is the study of It is the study of Reversible RxnsReversible Rxns!!

A + B C + DA + B C + DThe rxn makes products which themselves The rxn makes products which themselves react to give back the reactants.react to give back the reactants.

The rxn never stops.The rxn never stops.

How does the How does the Concentrations Concentrations changchange with time in a reversible rxn?e with time in a reversible rxn?

p.02

concconc..

timetime

eqm reached: no further changeeqm reached: no further change

[product][product]

[reactant][reactant]

t’t’

At time = t’, an equilibrium is established. At time = t’, an equilibrium is established. No further change in the concentrations.No further change in the concentrations.

At eqm, [reactant] and [product] At eqm, [reactant] and [product] may not be the samemay not be the same..

[reactant] [reactant] 0 0

i.e. Not all reactants i.e. Not all reactants converted to converted to products.products.

How does theHow does the Rate Rate change with time in a change with time in a reversible rxn?reversible rxn?

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raterate

timetime

forward rateforward rate

backward ratebackward rate

t’t’

FWR and BWR FWR and BWR do not stopdo not stop, but have the same rate., but have the same rate.

At time = t’, an equilibrium is established. At time = t’, an equilibrium is established. Rate of FWR = Rate of BWR Rate of FWR = Rate of BWR 0 0

rate rate 0 0

i.e. The reaction is Dynamic!i.e. The reaction is Dynamic!

A chemical eqm occurs when we have a reversible A chemical eqm occurs when we have a reversible

reaction in a reaction in a closed systemclosed system and the and the cconditions do not changeonditions do not change ..

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““Chemical Eqm” Chemical Eqm” can exist only if:can exist only if:

When a condition changes, the eqm will When a condition changes, the eqm will shiftshift

to a new to a new “eqm position”“eqm position”..

A Set of [Reactant] anA Set of [Reactant] and [Product] at eqm.d [Product] at eqm.

(temperature / concentration / pressure)(temperature / concentration / pressure)

Dynamic EquilibriumDynamic Equilibrium: : Observe the Observe the “Eqm Shift” “Eqm Shift” frofrom the Colour Change in rxn.m the Colour Change in rxn.

e.g. Bre.g. Br22 + H + H22O HO H++ + Br + Br-- + HOBr + HOBr

Adding conc. Adding conc. OHOH-- turns the colour lighter (paler). turns the colour lighter (paler).

Adding conc. Adding conc. HH++ turns the colour darker (deeper). turns the colour darker (deeper).

[If no eqm is involved, no colour change would be observed.][If no eqm is involved, no colour change would be observed.]

[If no eqm is involved, no colour change would be observed.][If no eqm is involved, no colour change would be observed.]

(OH(OH-- decreases [H decreases [H++], eqm shifts forward to increase [H], eqm shifts forward to increase [H++], a], and decreases [Brnd decreases [Br22] at the same time).] at the same time).

([H([H++] increases, eqm shifts backward to decrease [H] increases, eqm shifts backward to decrease [H++], an], and increases [Brd increases [Br22] at the same time).] at the same time).

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(ref. p. 92 for another example)(ref. p. 92 for another example)

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Calculation about Equilibrium: Calculation about Equilibrium:

“Eqm Law” …. “Eqm Law” …. (1)(1)

When a condition changes, the eqm would When a condition changes, the eqm would shift to a new eqm positions.shift to a new eqm positions.

e.g. Ne.g. N22 + 3H + 3H22 2NH 2NH33

[H2]/mol dm-

3

[N2]/mol dm-3 [NH3]/mol dm-3

0.50 1.00 0.0871.35 1.15 0.4122.44 1.85 1.27

Consider the following 3 eqm positions at temp = T:Consider the following 3 eqm positions at temp = T:

[NH3]2/[N2][H2]3

0.0600.0600.060

eqm constant (Keqm constant (Kcc))

e.g.1 Ne.g.1 N22(g) + 3H(g) + 3H22(g) 2NH(g) 2NH33(g)(g)

Calculation about Equilibrium: Calculation about Equilibrium:

“Eqm Law” …. “Eqm Law” …. (2)(2)

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KKcc = =[NH[NH33(g)](g)]22

[N[N22(g)][H(g)][H22(g)](g)]33

KKcc = =[FeSCN[FeSCN2-2-(a(a

q)]q)][Fe[Fe3+3+(aq)][SCN(aq)][SCN--(aq)](aq)]

e.g.2 Fee.g.2 Fe3+3+(aq) + SCN(aq) + SCN--(aq) FeSCN(aq) FeSCN2-2-(aq)(aq)

KKcc = [CO = [CO22(g)](g)]e.g.3 CaCOe.g.3 CaCO33(s) CaO(s) + CO(s) CaO(s) + CO22(g)(g)

p. 95 Check Point 16-4p. 95 Check Point 16-4

Implication of Implication of “K“Kcc”” ? ?

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e.g.1 CHe.g.1 CH33COOH + HCOOH + H22O CHO CH33COOCOO-- + H + H33OO++

KKcc = 1.00 = 1.001010-5-5

Position lies to the LEFTPosition lies to the LEFT

i.e.i.e. reactants dominate. (reactants dominate. ( CH CH33COOH is a weak acid)COOH is a weak acid)

e.g.2 HCl + He.g.2 HCl + H22O HO H33OO++ + Cl + Cl--

KKcc = 5.56 = 5.56101088

Position lies to the RIGHTPosition lies to the RIGHT

i.e.i.e. products dominate. (products dominate. ( HCl is a strong acid) HCl is a strong acid)

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p. 98 Check Point 16-5Ap. 98 Check Point 16-5A

FeFe2+2+(aq)(aq) + Ag + Ag++

(aq)(aq) Fe Fe3+3+(aq)(aq) + Ag + Ag(s)(s)

0.05M0.05M 0.05M0.05M 0M0Mat startat start

at eqmat eqm 0.0122M0.0122M

(6.10/1000)(0.05)(6.10/1000)(0.05)

25/100025/1000

0.0378M0.0378M0.0122M0.0122M

KKcc = =(0.0378)(0.0378)

(0.0122)(0.0122)(0.0122)(0.0122) = 254 M= 254 M-1-1

Note that KNote that Kcc is quite large, the FWR is more favorable. is quite large, the FWR is more favorable.

p.10

p. 100 Check Point 16-5Bp. 100 Check Point 16-5B

PClPCl55(g) PCl(g) PCl33(g) + Cl(g) + Cl22(g)(g)

0.0090.009 0.250.25 00at start (at start (mol dmmol dm--

33))

at eqm (at eqm (mol dmmol dm--

33))0.0020.0020.0070.007 0.2520.252

KKcc = =(0.252)(0.002)(0.252)(0.002)

(0.007)(0.007) = 0.072 mol dm= 0.072 mol dm-3-3

Note that KNote that Kcc is quite small, the BWR is more favorable. is quite small, the BWR is more favorable.

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Calculation about Eqm involving GaCalculation about Eqm involving Ga

ses: ses: “K“Kpp” ” …. …. (partial pressures)(partial pressures)

e.g. Ne.g. N22OO44(g) 2NO(g) 2NO22(g)(g)

At 350K, the eqm pressure is 7At 350K, the eqm pressure is 7101044 Pa in a closed co Pa in a closed container. The mixture is pale brown in colour, and molntainer. The mixture is pale brown in colour, and mole fraction of Ne fraction of N22OO44(g) = 0.46 at eqm. Find K(g) = 0.46 at eqm. Find Kpp..

= 0.46(7= 0.46(7101044) = 3.22 ) = 3.22 101044 Pa PaPPNN22OO44

= 7= 7101044 – 3.22 – 3.22 101044 = 3.78 = 3.78 101044 Pa PaPPNONO22

(3.22 (3.22 101044))KKpp = =

(3.78 (3.78 101044))22

= 4.44= 4.44101044 Pa Pa

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p. 102 Check Point 16-6p. 102 Check Point 16-6

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

at start at start (mol)(mol)

at eqm at eqm (mol)(mol)

11 0011

0.20.2 0.20.2 1.61.6

total no. of mol = 0.2+0.2+1.6 = 2.0 total no. of mol = 0.2+0.2+1.6 = 2.0 molmolPPHH22

= P= PTT

0.20.22.02.0

= P= P II22= 0.10 P= 0.10 PTT

PPHIHI = P= PTT

1.61.62.02.0

= 0.80 P= 0.80 PTT

KKpp = = = 64= 64(0.80)(0.80)22

(0.10)(0.10)(0.10)(0.10)

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p. 230 Q.6(a)(i),(ii) (1996 --- Kp. 230 Q.6(a)(i),(ii) (1996 --- Kpp))

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

at start at start (mol)(mol)

at eqm at eqm (mol)(mol)

33 0011

1.51.5 1.51.50.250.25

(a)(a) Expression for KExpression for Kpp:: KKpp = = (P )(P )22

SOSO33

(P )(P )OO22(P )(P )22

SOSO22

(b(b))PPSOSO22

= 373 = 373 1.51.53.253.25

= P= PSOSO33= 172 kPa= 172 kPa

total no. of mol = 1.5 + 0.25 + 1.5 = 3.25 moltotal no. of mol = 1.5 + 0.25 + 1.5 = 3.25 mol

PPOO22= 373 = 373 0.250.25

3.253.25= 28.7 kPa= 28.7 kPa KKpp = =

(172)(172)22

(28.7)(28.7)(172)(172)22

= 0.035 kPa= 0.035 kPa-1-1

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p. 231 Q.10 (1998 --- Kp. 231 Q.10 (1998 --- Kcc))

NN22(g) + O(g) + O22(g) 2NO(g)(g) 2NO(g)

at start at start (mol)(mol)

at eqm at eqm (mol)(mol)

22 0011

2-x2-x 2x2x1-x1-x

KKcc = 1.2 = 1.21010-2-2

x = 0.077x = 0.077

KKcc = 1.2 = 1.21010-2-2 = =

(2x/2)(2x/2)22

[(2-x)/2][(1-x)/2][(2-x)/2][(1-x)/2]

4x4x22

(2-x)(1-x)(2-x)(1-x)= =

[N[N22]]eqmeqm = (2 – 0.077)/2 = 0.96 mol dm = (2 – 0.077)/2 = 0.96 mol dm-3-3

AssignmentAssignment

Study the examples in p. 98-104,Study the examples in p. 98-104,

p.104 p.104 Check Point 16-7Check Point 16-7

p.127 Q.1-5, 7 p.127 Q.1-5, 7

p. 231 Q.9(a) [due date: 9/3(Wed)] p. 231 Q.9(a) [due date: 9/3(Wed)]

Quiz on Chemical Kinetics (Ch. 13-15)Quiz on Chemical Kinetics (Ch. 13-15)

[9/3(Mon)] [9/3(Mon)]

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Next ….Next ….

Significances of Eqm Constant and Significances of Eqm Constant and Le Chatelier’s Principle (p. 110 - 123)Le Chatelier’s Principle (p. 110 - 123)

[Tuesday][Tuesday]

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