1
Chapter 16 — Chemical Equilibria 3
Last Class
• Use of ICE tables to solve equilibrium
problems
• Calculate Q and K
• Relation between Kc and Kp
• Product-favored and reactant-favored
reactions
• Is the reaction at the equilibrium?
Heterogeneous equilibrium
Solids and liquids NEVER appear in equilibrium
expressions.
S(s) + O2(g) ' SO2(g)
K=
Writing and Manipulating K Expressions
[SO2 ]
[O2 ]
Writing and Manipulating K Expressions
K using concentration and pressure units
Concentration Units
We have been writing K in terms of mol/L.
These are designated by Kc
Kp = Kc (RT)∆n
where ∆n is the change in the number of moles of gas during the reaction
For
H2(g) + I2(g) ' 2 HI(g)
But with gases, P = (n/V)•RT = [conc]•RT
P is proportional to concentration, so we can write K in terms of P.
These are designated by Kp.
For H2(g) + I2(g) ' 2 HI(g)
Kc and Kp may or may not be the same.
For SO2(g) + 1/2 O2(g) ' SO3(g)
2
Chapter 16 — Chemical Equilibria 3
Product or Reactant
Favored Reactions
aA + bB
'
cC + dD
K >> 1
Products
Reactants
Q
K
K
Q
K
Q
K << 1
Reactants
Products
Reaction forms
products
Today’s Topics
Equilibrium
Reaction forms
reactants
Nitrogen Dioxide Equilibrium
N2O4(g) ' 2 NO2(g)
• Using
U i quadratic
d ti equations
ti
to
t solve
l equilibrium
ilib i
problems. 16.4
• Equilibrium Constant expressions. 16.5
'
3
Chapter 16 — Chemical Equilibria 3
N2O4(g) ' 2 NO2(g)
[NO2 ]2
Kc =
= 0.0059 at 298 K
[N2O4 ]
If initial concentration of N2O4 is 0.50 M, what are
the equilibrium concentrations?
Step 1. Set up an ICE table
[NO2]
[N2O4]
Initial
Change
Equilibrium
Step 2. Substitute into Kc expression and solve.
K c = 0.0059 =
[NO2 ]2
(2x) 2
=
[N2O 4 ]
(0.50 - x)
Rearrange:
This is a QUADRATIC
EQUATION
•
K c = 0.0059 =
[NO2 ]2
(2x) 2
=
[N2O 4 ]
(0.50 - x)
4x2 + 0.0059x - 0.0029 = 0
a = 4
b = 0.0059
x =
x =
-0.0059 ±
-b ±
c = -0.0029
b2 - 4ac
2a
(0.0059)2 - 4(4)(-0.0029)
2(4)
x =
-0.0059 ±
(0.0059)2 - 4(4)(-0.0029)
2(4)
4
Chapter 16 — Chemical Equilibria 3
N2O4(g) ' 2 NO2(g)
If initial concentration of N2O4 is 0.50 M, what are
the equilibrium concentrations?
p Get the answers!
Final Step.
[N2O4]
[NO2]
Initial
0.50
0
Change
-x
+2x
Equilibrium
0.5--x
0.5
2x
When do we need to use the
quadratic equation?
N2O4(g) ' 2 NO2(g)
• [A]0 – x ≈ [A]0
when 100 x K < [A]0
x=0
0.026
026
Writing and Manipulating K Expressions
Writing and Manipulating K Expressions
1. Adding equations for reactions
2. Changing stoichiometric coefficients
S(s) + O2(g) ' SO2(g)
S(s) + 3/2 O2(g) ' SO3(g)
SO2(g) + 1/2 O2(g) ' SO3(g)
2 S(s) + 3 O2(g) ' 2 SO3(g)
Net equation:
S( ) + 3/2 O2(g)
S(s)
( ) ' SO3(g)
( )
Knet = K1 • K 2
Adding two balanced equations multiplies their Ks
Doubling the balanced equation squares the K
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Chapter 16 — Chemical Equilibria 3
Writing and Manipulating K Expressions
Writing and Manipulating K Expressions
3. Changing direction
S(s) + O2(g) ' SO2(g)
1. Adding equations for reactions
SO2(g) ' S(s) + O2(g)
2. Changing stoichiometric coefficients
3. Changing direction
Reversing the balanced equation inverts the K