Equilibrium Calculations Lesson 5

How can we describe an

equilibrium

system

mathematically

?

reactants ⇌

products products

Keq =

=

3.0

reactants The

Keq

is the

equilibrium constant

- a number that does not change.

Providing the

temperature

is kept

constant

.

Equilibrium Calculations

An

equilibrium system

, at any given temperature, can be described by an

equilibrium expression

and equilibrium

constant

.

a

A +

b

B ⇌

c

C +

d

D

Keq = Products Reactants Keq Equilibrium Constant- a number

= [C]

c

[D]

d

[A]

a

[B]

b Expression mathematical equation (aq) and (g) are included in the expression!

(l) (pure liquids, meaning it is the only one in the equation) and (s) are not because they have constant concentration!

Only changes to

(aq)

and

(g)

reactants or products

cause

the equilibrium to

shift.

(s)

and

(l)

do

not!

Solids (s) and pure liquids (l) have constant concentrations.

Solids & Liquids have fixed densities, cannot be compressed

,

so their molar concentrations are constant

(s) And (l) concentrations are already included in Keq value so we don’t include them in the equation.

CaCO 3(s) + 2H + (aq) + 2Cl (aq)

⇌

Ca 2+ (aq) + Cl 2 (g) + CO 2(g) + H 2 O (l) no shift right left left right left no shift

1.

At equilibrium at 25 o C, [SO 3 ] = 0.200 M. [H 2 O] = 0.480 M, and [H 2 SO 4 ] = 24 M. Calculate the Keq.

No ICE SO 3(g) + H 2 O (g)

⇌

H 2 SO 4(l) Keq = = 1 [SO 3 ] [H 2 O] 1 (0.200)(0.480) don’t count (l)! Use 1 = 1 0.4

The Keq has no units but concentration units that go in the expression must be M !

2.

0.500 mole PCl 5 , 0.40 mole H 2 O, 0.200 mole HCl, and 0.400 mole POCl 3 a 2.0 L container at equilibrium at 125 o C. Calculate the Keq

.

are found in No ICE PCl 5(s) + H 2 O (g)

⇌

2HCl (g) + POCl 3(g) [HCl] =

0.200 moles 2.0 L

[POCl 3 ] =

0.400 moles 2.0 L

[H 2 O]

= 0.40 moles 2.0 L

=

= =

0.10 M 0.20 M 0.20 M Keq Keq

= =

[HCl] [ [H 0.10

] 2 2 2 [POCl O] [ 0.20

] 3 ] [ 0.20

] Keq

=

0.010

3.

[SO [SO 3 2

If 0.600 mole of SO 3 and 0.0200 mole of SO 2 equilibrium at 25 o C. Calculate the [O 2 ].

are found in a 2.00 L container at 2SO 2(g) + O 2(g)

⇌

2SO 3(g) Keq = 798

] = 0.600 mole/2.00 L = ] = 0.0200 mole/2.00 L =

0.300 M 0.0100 M Keq

798 1 (0.3) [O 2 ] 2 = [SO 3 ] 2 [SO 2 ] 2 [O 2 ] = (0.300) 2 (0.0100) 2 [O 2 ] = 798(0.01) 2 [O 2 ] = (0.3) 2 798(0.01) 2 =

1.14 M

Size of Keq & Effect of Temperature on Keq

Big Keq Keq =

products

reactants

= 10 Keq

Little Keq

products

Keq = 0.1

reactants

Keq = Note that the keq cannot be a negative number!

Keq = 1 Keq =

products reactants

Effect of Temp on Keq

Keq

Only a temperature change

can affect the value of Keq.

Changes in concentrations, pressure or surface area have

NO effect

on Keq. – These changes correspond to increase in number of reacting molecules per liter.

– Increased once and then equilibrium is re-established. – So ratios of products to reactants do not change.

PBr 3(g) + Br 2(g)

⇋

PBr 5(g) @ 100 o C Keq = 0.17 + energy @ 200 o C Keq = 0.091

How can you tell if Keq gets bigger or smaller?

Temperature

increased

Shifted

left, Keq decreased

Keq= [products] ------------ [reactants]

More questions…

If the value of K eq increases when the temperature decreases , is the reaction exothermic or endothermic?

More questions…

What will happen to the value of Keq in the following reaction if we added more [B]?

A + B ⇌ C + 100 KJ.

Hebden Practice

Page 60: Exercises 31-35 Page 62: Exercises 36-45