CH. 15 EQUILIBRIUMSystem of: Gases, Liquids, Solids (will look @ all gas sys to start)

Extent of a rxn, based on connect of reacts/pdts @ moment

K: equilibrium constantQ: rxn quotient

based on concent @ equilbvaries w/ rxn

m

n

rev

fwdc ]react[

]pdt[

k

kK m

n

c ]react[

]pdt[Q

Qc = Kc @ equilb

@ equilb: [react] & [pdt] not w/ time

N2O4 g, colorless <--------> 2 NO2 g, brown

ratefwd = raterev

k[react]meq = k[pdt]n

eq

k[N2O4]eq = k[NO2]2eq

kfwd[N2O4] = krev[NO2]2

K k

k

]ON[

]NO[

k

k

rev

fwd

42

22

rev

fwd

]ON[

]NO[K

42

22k: equilb const

ratio of [react]:[pdt] at oT

Magnitude of “K”

K <<<<<< sm (10-20) little pdt, mostly react @ equilbK >>>>>> lrg (10+18) little react, mostly pdt @ equilbsm < K < lrg (10-3-10+3) varying = amts of both reacts & pdts

write equilb const, K, for:

POCl3 <------> PCl3 + O2

Cu2S + O2 <-----> Cu2O + SO2

2 2

2 3 2 2

32

22

22

22

]O[]SCu[

]SO[]OCu[K

23

22

3

]POCl[

]O[]PCl[K

reverse rxn:

]O[]PCl[

]POCl[K

22

3

23

Overall RXN

1. N2O5 <------> NO2 + NO3

2. NO2 + NO3 <------> NO2 + NO + O2

3. NO3 + NO <-------> 2 NO2

]NO][NO[

]NO[

]NO][NO[

]O][NO][NO[

]ON[

]NO][NO[k*k*k

3

22

32

22

52

32321

N2O5 + NO3 <------> 3 NO2 + O2

k1 = 4.8*10-10

k2 = 1.1*10-5

k3 = 3.2*102

Qoverall = Q1*Q2*Q3 =

Koverall = k1*k2*k3 =

K = (4.8*10-10)*(1.1*10-5)*(3.2*102) = 1.69*10-12

NO2

N2O4

[ ]

time

Q = K Q = K

reverse Qfwd = 1/Qrev Kfwd = 1/Krev 2 SO2 + O2 <-----> 2 SO3

Kfwd = 261 Krev = 1/261 =

Coefficient Values

H2 + Cl2 <-----> 2 HCl Kc = 7.6*108

Find Kc: 0.5 H2 + 0.5 Cl2 <-----> HCl

40.58

5.02

5.02

21

22

2

10*2.8 10*7.6 ClH

HCl

ClH

HCl Kc

Find Kc: 4/3 HCl <-----> 2/3 H2 + 2/3 Cl2

6-

32

83

4

32

23

2

2

32

22

2 fwd

rev

10*1.4 10*7.6

1

HCl

ClH

ClHHCl

1

K

1 K

What did we see?

Specific K value has meaning only to that specific balanced equation

Diff. Phases: heterogeneous equilibriumSnO2 (s) + 2 H2 (g) <-----> Sn (s) + 2 H2O (g)

222

22

]H][SnO[

]OH][Sn[Qc

solids: same concen @ given temp; same # mols/L; no Vliquids: same applies therfore, we ignore pure solids & liquids

22

22

]H[

]OH[Qc

only concerned w/ those that will [ ] in rxn

What about liquids & solids?

PRESSURE, KP

nreact

mpdt

P P

PK ideal gas behavior; Pgas easier to measure than [gas]

PV = nRT ===> P = (n/v)RT ===> (P/RT) = (n/v) & (n/v) = M

P ∞ M since R * T constant

Kc – Kp related but not “=“

2 SO2 (g) + O2 (g) <-----> 2 SO3 (g) Qc = ?

]O[]SO[

]SO[Qc

22

2

23

Vn

V

nV

n

Qc22

3

O

2

2SO

2

2SO

[ ] : concen, M

)RT(

P

RT

PRT

P

22

3

O2

2SO

2

2SO

22

3

O2

SO

2SO

PP

RTP

Qc = Qp*RT @ equilibriumKc = Kp/RTKp = (Kc)(RT)

If no in n, then = 0No “RT” term, so Kp = Kc

Kp = (Kc)(RT)ngas

calculate Kp: PCl3 (g) + Cl2 (g) <-----> PCl5 (g) Kc = 1.67 @ 501 K

Kp = (Kc)(RT)= (1.67)(0.0821*501)-1

= 0.0406

COMPARE Q <----> K

Q < K more pdtsQ = K no , equilibriumQ > K more reactant

CH4 + Cl2 <-----> CH3Cl + HCl @ 1500 KPatm 0.13 0.035 0.24 0.47 Kp = 1.6*104

Find Qp, which direction?25

QQ

QQ Q

24

3

ClCH

HClClCHp

Qp < Kp more PDT

ngas = #mols gas pdt - # mols gas react 1 - 2 = -1

SOLVE EQUILIBRIUM

1. Given equil [ ] or P then find “K”2. Given K, initial [ ] or P then find [equilb]

What did we see?

Qc = QpRT @ equilibrium

Qfwd = Qn Qrev = (1/Q)n

Kfwd = Kn Krev = (1/K)n

find quantity from know equil [ ] & K

2 NO (g) <----> N2 (g) + O2 (g) Kc = 4.6*10-1

[.976] [.781] X222

]NO[

]O][N[Kc

0.561 ]N[

]NO[Kc]O[

2

2

2

TABLE SET UP -- find equil quantities & K

write balanced eqn, list given & unkn [ ] or P label conditions [initial, change, equlib]

0.500 mol ICl gas decomposes into two diatomic gases in a 5.00 L container.1) construct a reaction table 2) concen @ equilb, Kc = 0.110

2 ICl (g) <------> Cl2 (g) + I2 (g)

initial use/make equilibrium

Determine initial [ ]: ICl = 0.500 mol/5.00 L = 0.100 M since no rxn started; Cl2 & I2 = 0 mol

0.100 0 0 -2x +x +x0.100-2x +x +x

2

22

]ICl[

]I][Cl[Kc

2]x2100.0[

]x][x[ 110.0

2

2

]x2100.0[

]x[ 110.0

2 ICl <------> Cl2 + I2

used/made

equilibrium

0.332[0.100-2x] = 0.332x = 0.02

]x2100.0[

x 0.332

0.100-0.04 0.0+0.02 since Cl2 = I2

[0.06] [0.02] [0.02]

LE CHATELIER’S PRINCIPLE

When a sys at equilb is stressed, the sys will react in the directionto alleviate that stress & return to a new equilb position

Q1 = K1 -------------> Q = K -----------> Q1 = K1 again

* equilb shifts, Kc same

Effects[ ] add/remove either react/pdtP; T; add catalyst/inert gas

Shift RGT: add react; remove pdtShift LEFT: add pdt; remove reactShift???: P, V

Remember: sys adjusts thru [ ] changes, but Qc @ equilb same value as in orginal sys @ same given T

es [ ]

+CO2; shifts rxn to ----> ; [H2CO3/CO2] 1:99 es to 0.5:99.5 forms more H2CO3 shifts ratio back to 1:99

es TEMP

2 SO2 + O2 <----> 2 SO3 + heat incr T shifts dir that absorbs heat think heat as pdt, exo shifts to left

Rise T, adds heat, favors endo dir

drop T, removes heat, favors exo dirIncr Kc for +H

decr Kc for -H

es P

3 H2 (g) + N2 (g) <----> 2 NH3 (g)

equilb w/ only unequal # mols of gas 4 mols react 2 mols pdt effect incr/decr P???

Rise P, favors fewer gas mols dir

drop P, favors more gas mols dir

Add Catalyst

3 H2 (g) + N2 (g) + catalyst <----> 2 NH3 (g)

no effect on equilb, no Kc effects rate fwd & rate rev equilb reached quicker

Add Inert Gas

2 SO2 + O2 + inert gas <----> 2 SO3

inert gas not participate in equilb P’s don’t , no equilb