1

Chemical Equilibrium: “Big K”

kinetics: rate constant “little k”

kinetics “little k” told us how fast a reaction proceeds and is used to indicate a possible mechanism.

Eq. tells us to what extent a RXN proceeds to completion.

react. prod. @ Eq: rate forward = rate reverse

H2CO3(l) H2O(l) + CO2(g)

2

Law of Mass Action: Values of Kc are constant for a RXN at a given temperature. Any equilibrium mixture of the above system at that temperature should give the SAME Kc value.

The Equilibrium Constant: K (Temperature Dependent)(mechanism independent)

aA + bB dD + eE

coeff.their to raised

coeff.their to raisedc

][reactants

[products] K

ba

edc

[B][A]

[E][D] K

3

We also have Kp which is sometimes used when dealing with gases with the “P” referring to the pressure of the gases.

aA(g) + bB(g) dD(g) + eE(g)b

Ba

A

eE

dD

)(P)(P

)(P)(P pK

Since PV = nRT (ideal gas law)

RTV

n P .. and ionconcentrat

Liters

moles

RT

P

V

n

so pressure is proportional to concentration.

ncP RTK K moles of GAS

temp in K0.0821L•atm mol•K

4

Problem: PCl3(g) + Cl2(g) PCl5(g)

In a 5.00 L vessel (@ 230oC) an equilibrium mixture is found to contain: 0.0185 mol PCl3, 0.0158 mol PCl5 and 0.0870 mol of Cl2.

Question: determine Kc and Kp

Solution:]][Cl[PCl

][PCl K

23

5c

L 5.00

mol 0.0870

L 5.00

mol 0.0185

L 5.00

mol 0.0158

Kc = 49.08

1.19 -503) 2149.08(0.08 (RT)K K 1ncP

Question: what does the value of K mean?

one mol less gason product side.

5

Using I.C.E. (Initial, Change, Equilibrium)

Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc?

CO(g) + 2H2(g) CH3OH(g)I.

C.

E.

0.1500 0.3000 0

-x -2x +x

0.1500 - x 0.3000 - 2x x

Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187

Therefore x = 0.0313

We can now solve for each of the other Eq. terms.

6

Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc?

CO(g) + 2H2(g) CH3OH(g)I.

C.

E.

0.1500 0.3000 0

-x -2x +x

0.1500 - x 0.3000 - 2x x

Since @Eq. there were 0.1187 mol CO present, 0.1500 - x = 0.1187

Therefore x = 0.0313We can now solve for each of the other Eq. terms.

H2: 0.3000 - 2x = 0.2374 moles

CH3OH: x = 0.0313 moles

Therefore E. 0.1187 0.2374 0.0313CO(g) + 2H2(g) CH3OH(g)

7

Problem: Manufacture of Wood Alcohol. A 1.500 L Vessel was filled with 0.1500 mol of CO and and 0.300 mol of H2. @ Eq. @500K, 0.1187 mol of CO were present. How many moles of each species were present @ Eq. and what is the value of Kc?

CO(g) + 2H2(g) CH3OH(g)I.

C.

E.

0.1500 0.3000 0

-x -2x +x

0.1500 - x 0.3000 - 2x x

Therefore E. 0.1187 0.2374 0.0313CO(g) + 2H2(g) CH3OH(g)

Now find Kc: 2

2

3c

][CO][H

OH][CH K

2

1.500

0.2374

1.500

0.1187

1.500

0.0313

Kc = 10.52

8

More I.C.E.

Problem: @ 77oC, 2.00 mol of NOBr (nitrosyl bromide) placed in a 1.000L flask dissociates to the extent of 9.4%. Find Kc.

2NOBr(g) 2NO(g) + Br2(g)I.

C.

E.

2.00 0 0

+x+2x-2x

2.00 - 2x 2x x

Now what?Since 9.4% dissociates, the Change in NOBr, -2x = 2.00(0.094)

and.... |x| = 0.0940

so substituting the x value into the “E. term” gives:

E. 1.812 0.188 0.0940

2NOBr(g) 2NO(g) + Br2(g)

9

More I.C.E.

Problem: @ 77oC, 2.00 mol of NOBr (nitrosyl bromide) placed in a 1.000L flask dissociates to the extent of 9.4%. Find Kc.

2NOBr(g) 2NO(g) + Br2(g)I.

C.

E.

2.00 0 0

+x+2x-2x

2.00 - 2x 2x x

E. 1.812 0.188 0.0940

2NOBr(g) 2NO(g) + Br2(g)

22

2

cOBr][N

][Br[NO] K

2

2

1

1.812

1

0.0940

1

0.188

= 1.01 x 10-3

What does the value of Kc tell us?

10

Treatment of Pure Solids and Liquids (as solvents) in K expressions.

S(s) + O2(g) SO2(g)

would expect Kc = [SO2] [S(s)][O2]

but since M is meaningless for solids, solids are dropped out.

and...][O

][SOK

2

2c

Experimentally Kc is found to be 4.2 x 1052 @ 25oC and independent of S.

11

AgCl(s) Ag+(aq) + Cl-(aq)

Ksp = [Ag+(aq)][Cl-(aq)] = 1.8 x 10-10 @ 25oC

This is an EQUILIBRIUM value independent of the amount of solid AgCl left sitting on the bottom of the container.

12

PURE LIQUIDS (SOLVENTS)

NH3(g) + H2O(l) NH4+(aq) + OH-(aq)

5-

3

4 10x 1.8][NH

]][OH[NHK

Note: by convention the water is ignored.

HCOOH(aq) + H2O(l) HCOO-(aq) + H3O+(aq)

43 10x 1.8[HCOOH]

]O][H[HCOOK

13

CaCO3(s) CaO(s) + CO2(g)

What is the Kc expression?

Kc = [CO2]

What is the Kp expression?

Kp = PCO2

14

Reversing Equations: Reactants become Products and Products become Reactants.

What is the relationship between the two K values?

1. 2H2(g) + O2(g) 2H2O(g)81

22

2

22

1 10x 3.4][O][H

O][HK

2. 2H2O(g) 2H2(g) + O2(g) 82

22

22

22 2.9x10

O][H

][O][HK

K2 = 1/K1 = K1-1 Relationship:

15

Knet for summing RXN’s:

If a RXN can be obtained from the sum of RXN’s,KRXN = K1K2

RXN 1: S(s) + O2 (g) SO2(g) 52

2

21 10x 4.2

][O

][SO K

RXN 2: SO2(g) + 1/2O2(g) SO3(g) 12

1/222

32 10x 2.6

]][O[SO

][SO K

Net RXN: S(s) + 3/2O2(g) SO3(g) 653/2

2

3net 10x 1.1

][O

][SOK

3/22

31/2

22

3

2

221

][O

][SO

]][O[SO

][SOx

][O

][SOKK

KRXN = K1K2

16

The RXN Quotient: Qc

Consider a system that may not yet be @ Equilibrium.

aA + bB dD + eEba

edc

[B][A]

[E][D] Q

If Qc = Kc ? @ Equilibrium

If Qc < Kc ? ratioReact.

Prod.is too small so RXN

If Qc > Kc ? ratioReact.

Prod. is too large so RXN

17

Example: PCl5(g) PCl3(g) + Cl2(g)@ 250oC Kc= 4.0 x 10-2

If: [Cl2] and [PCl3] = 0.30M and [PCl5] = 3.0M, is the system at Equilibrium? If not, which direction will it proceed?

Qc = [3.0]

0][0.30][0.3

][PCl

]][Cl[PCl

5

23 = 3.0 x 10-2

Qc < Kc (not @ Eq.)

Which way must the RXN go to achieve Equilbrium?

Remember ratio is prod./React

more products makes the number bigger

RXN goes

Find Qc and compare to Kc to decide.

18

Calculations Using Kc: (1st case....Perfect Square)

@ 699K H2(g) + I2(g) 2HI(g) Kc = 55.17

Experiment: 1.00 mol of each H2 and I2 in a 0.500 L flask. Find [ ] of products and reactants @ Equilibrium.

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

I.

C.

E.

[ ] [ ] [ ]

2.00 2.00 0

-x -x 2x

2.00 -x 2.00 - x 2x]][I[H

[HI]K

22

2

c

x]-x][2.00-[2.00

[2x]K

2

c

2

2

x]-[2.00

[2x]

“perfect square”

= 55.17

conc. in mol/L

19

Calculations Using Kc: (1st case....Perfect Square--continued)

@ 699K H2(g) + I2(g) 2HI(g) Kc = 55.17

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

E.[ ] [ ] [ ]2.00 -x 2.00 - x 2x]][I[H

[HI]K

22

2

c

x]-x][2.00-[2.00

[2x]K

2

c 2

2

x]-[2.00

[2x]

“perfect square”

= 55.17

x]-[2.00

[2x]55.17

7.428(2.00 - x) = 2x

1.58 = x

20

Calculations Using Kc: (1st case....Perfect Square)

@ 699K H2(g) + I2(g) 2HI(g) Kc = 55.17

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

E.[ ] [ ] [ ]2.00 -x 2.00 - x 2x]][I[H

[HI]K

22

2

c

7.428(2.00 - x) = 2x

1.58 = x

[H2] = 2.00 - 1.58 = 0.42M

[I2] = 0.42M

[HI] = 2(1.58) = 3.16M

21

Kc Problems with Quadratic Equation

2a

4acbbx

2 ax2 + bx +c = 0

If equilibrium expression is not a perfect square must usequadratic equation.

22

Problem: H2(g) + I2(g) 2HI(g) @ 458 oC Kc = 49.7

Experiment: 1.00 mol H2, 2.00 mol I2 in a 1.00 L flask.

Find: conc. of the equilibrium mixture.

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

I.

C.

E.

[ ] [ ] [ ]

1.00 2.00 0

-x -x 2x

1.00 - x 2.00 - x 2x]][I[H

[HI]K

22

2

c

K

x]-x][2.00-[1.00

[2x]49.7K

2

c

0.920x2 - 3.00x + 2.00 = 0

23

Problem: H2(g) + I2(g) 2HI(g) @ 458 oC Kc = 49.7

Experiment: 1.00 mol H2, 2.00 mol I2 in a 1.00 L flask.

Find: conc. of the equilibrium mixture.

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

I.

C.

E.

[ ] [ ] [ ]

1.00 2.00 0

-x -x 2x

1.00 - x 2.00 - x 2x

]][I[H

[HI]K

22

2

c

x]-x][2.00-[1.00

[2x]49.7K

2

c

0.920x2 - 3.00x + 2.00 = 0

2 solutions for x: 1.63 0.70

x = 2.33 or 0.93 will give positive solution for Eq. Conc.