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.
aA
+
bB
⇌
cC
+
dD
Products
Keq
=
Reactants
[C]c[D]d
Keq
=
[A]a[B]b
Equilibrium Constant- a number
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.
CaCO3(s) + 2H+(aq) + 2Cl-(aq) ⇌ Ca2+(aq) + Cl2 (g) + CO2(g) + H2O(l)
no shift
right
left
left
right
left
no shift
1.
At equilibrium at 25oC, [SO3] = 0.200 M. [H2O] = 0.480 M, and [H2SO4] = 24 M.
Calculate the Keq.
No ICE
SO3(g)
⇌
+ H2O(g)
1
Keq
=
H2SO4(l)
don’t count (l)!
Use 1
[SO3] [H2O]
=
1
(0.200)(0.480)
=
10.4
The Keq has no units but concentration units that go in the expression must be M!
2.
0.500 mole PCl5, 0.40 mole H2O, 0.200 mole HCl, and 0.400 mole POCl3 are found in
a 2.0 L container at equilibrium at 125 oC. Calculate the Keq.
No ICE
PCl5(s) + H2O(g) ⇌ 2HCl(g) + POCl3(g)
= 0.200 moles
2.0 L
[HCl]
[POCl3] =
0.400 moles
= 0.10 M
= 0.20 M
Keq =
2.0 L
[H2O]
=
0.40 moles
2.0 L
[HCl]2[POCl3]
[H2O]
= 0.20 M
Keq =
[0.10]2[0.20]
[0.20]
Keq =
0.010
3.
If 0.600 mole of SO3 and 0.0200 mole of SO2 are found in a 2.00 L container at
equilibrium at 25 oC. Calculate the [O2].
2SO2(g) + O2(g) ⇌ 2SO3(g)
Keq = 798
Keq
[SO3] = 0.600 mole/2.00 L =
0.300 M
[SO2] = 0.0200 mole/2.00 L =
0.0100 M
=
[SO 3]2
[SO2]2[O2]
798
1
=
(0.3)2
=
[O2]
=
(0.300)2
(0.0100)2[O2]
798(0.01)2[O2]
(0.3)2
798(0.01)2
=
1.14 M
Size of Keq & Effect of
Temperature on Keq
Big Keq
products
Keq
=
reactants
Keq
=
10
Little Keq
products
Keq
=
reactants
Keq
=
0.1
Note that the keq cannot be a negative number!
Keq
=
1
products
Keq
=
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.
PBr3(g) + Br2(g) ⇋ PBr5(g)
@ 100 oC
+
Keq = 0.17
energy
@ 200 oC
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 Keq 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