Equilibrium Constant (Keq)

Keq:# expressing ratio of [products] & [reactants] at equilibrium

constant for every reversible rxn at equilibrium

(@ constant T & P)

aA + bB cC + dD Keq = [C]c[D]d

[A]a[B]b

** pure liquids & solids are not included in the rate

expression because their effective concentrations

do not change

Magnitude of Keq

Keq = 1

Keq > 1

Keq < 1

means products = reactants

Forward rxn is favoured, products > reactants

reverse rxn is favoured, products < reactants

Keq = [Products]

[Reactants]

Given: 2NO2 N2O4 + E

To calculate Keq, the concentrations must be given at

equilibrium!

What is the Keq if the [N2O4] is 0.00140 M & [NO2] is 0.0172M

at equilibrium?

Keq = [N2O4]

[NO2]2

= [0.0014]

[0.0172]2

Keq = 4.73

Given:

2NO2 N2O4 + E

Calculate Keq, if the [N2O4 ] is 0.00455 M & [NO2] is

0.031M at equilibrium.

Keq = [N2O4]

[NO2]2

= [0.00455]

[0.031]2 Keq = 4.73

Same Keq!!!

Important!

Keq is the same for a given reaction that is:

at equilibrium

at the same temperature

no matter what the initial concentrations were.

Keq changes with temperature!

H2 (g) + I2 (g) ⇄ 2HI (g)

Find the Keq if the [H2] is 0.46M, [I2] is 0.39M & [HI] is 3.0M

at equilibrium .

Keq = [HI]2__

[H2][I2]

= __[3.0]2__

[0.46][0.39] Keq = 50

Ex.1

PCl5 (g) PCl3 (g) + Cl2 (g)

Find the Keq, if the initial [PCl5]is 0.70M & the equilibrium [Cl2] is

0.15M.

Keq expression for equilibrium concentrations only!!!

Use an ICE table

Ex.2

[PCl5] [PCl3] [Cl2]

IInitial

CChange

EEquilibrium

0.70

0.55

0 0

+ 0.15+ 0.15- 0.15

0.15 0.15

Keq = [PCl3][Cl2]__

[PCl5]

= [0.15][0.15]__

[0.55]= 0.041

Ex.2PCl5 (g) ⇄ PCl3 (g) + Cl2 (g)

Based on

ratio from

balanced

rxn!

P318 #1 to #7

NH3 (g) H2 (g) + N2 (g)

Initially a 5.0L flask contains 0.2M NH3 & 0.08M N2.

At equilibrium [NH3] is 0.156M. Find Keq

NH3 H2 N2

I

C

E

0.2 0.080

0.156

- 0.044 + 0.066 + 0.022MOLE

RATIO

0.066 0.102

Ex.3 2 3 1

Keq = [H2]3[N2]__

[NH3]2

= [0.066]3[0.102]__

[0.156]2

Keq = 0.0012

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

Find the [HI] at equilibrium, if [H2] is 0.50M &

[I2] is 0.50M at equilibrium (Keq = 50).

Keq = [HI]2

[H2] [I2]

50 = [x]2__

[0.50][0.50]

[HI] = 3.5 M at equi

Ex.4

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

What are the equilibrium concentrations of each

substance if a flask initially contains only 0.5M H2 & 0.5M I2?

Keq = 50.

H2 I2 HI

I

C

E

0.5 00.5

0.5 - x

- x - x + 2x

0.5 - x 2x

Ex.5

Keq = [HI]2

[H2] [I2]

50 = [2x]2__

[0.5-x][0.5-x]

At Equilibrium:

[H2] = 2x = 0.5 - 0.39 = 0.11M

[I2] = 0.5 – x = 0.5 - 0.39 = 0.11M

[HI] = 2x = 2 x 0.39 = 0.78M50 = [2x] 2_

[0.5-x] 2

7.07(0.5 – x) = 2x

3.54 - 7.07x = 2x

x = 0.39

7.07 = [2x]_

[0.5-x]

√ / /

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

What are the equilibrium concentrations of

each substance if a 0.5L flask initially contains

2 moles H2 and 2 moles of I2? Keq = 50

H2 I2 HI

I

C

E

4 04

4 - x

- x - x + 2x

4 - x 2x

Ex.6

Keq = [HI]2

[H2] [I2]

50 = [2x]2__

[4-x][4-x]

At Equi:

[H2] = 0.88M

[I2] = 0.88M

[HI] = 6.24M

7.07 = [2x]_

[4-x]

7.07(4 – x) = 2x

28. 28 - 7.07x = 2x

x = 3.12

If not same,

multiply out then

use quadratic formula

At 1100K the Keq = 25.

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

2 moles of H2 and 3 moles of I2 are placed in a 1 L

container. Find the concentration of each substance at

equilibrium?

Ex.7

H2 I2(g) HI

I

C

E

2 M 3 M 0 M

2x M

+2x-x

2-x M 3-x M

-x

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

Continuing the Problem

]][[

][

22

2

IH

HIKeq So..

)3)(2(

)2(25

2

xx

x

2)2()3)(2(25 xxx 22 4)56(25 xxx

22 425125150 xxx

015012521 2 xx Carried over to next slide

Using the Quadratic formula

015012521 2 xx

a

acbbx

2

42 Quadratic formula, where:

a=21, b=-125, c=150

)21(2

)150)(21(4)125()125( 2 x

29.4x 67.1xTwo solutions to formula

But only one of the solutions will give a real answer!

Choosing the Best Answers

29.4x 67.1xTwo solutions to formula

Try finding the correct concentrations using the solutions

xH E 2][ 2

xI E 3][ 2

xHI E 2][

29.229.42][ 2 H 33.067.12][ 2 H

33.167.13][ 2 I

34.3)67.1(2][ HI

At equilibrium:

[H2] = 0.33 mol/L,

[I2] = 1.33 mol/L and

[HI] = 3.34 mol/L

Concentration

cannot be

negative!

Solve using 4.29 Solve using 1.67

Simplifying I.C.E. tables(the 5% rule)

When doing an ICE table you may have to subtract a very small

value from a relatively large value…

for example 2.0 mol/L – 1.0x10-4 mol/L.

In this case, don’t bother doing the subtraction, since by the time

you change it to show significant digits, the result will be the

same:

2.0 mol/L – 0.0001 mol/L = 1.9999 mol/L ≈ 2.0 mol/L.

If the number you subtract is less than 5% of the original number,

you can usually skip the subtraction.