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3. thermodynamics

Equilibria: Part III
Reviewing the rules
Things to remember
1. When reacting species (both the reactants
and the products) are in their most
condensed form (solids), the units are
– mol/L (molarity)
and the constant is expressed as
– Kc
Things to remember
2. When species are in a gaseous phase
state, the concentrations can be
expressed in:
– mol/L (Kc)
OR
– atm (Kp)
Things to remember
3. The equilibrium constant is a dimensionless
quantity.
– MEANING: the equilibrium constant (Kp or Kc)
has no units.
Things to remember
4. Because the constant only applies to the
reaction occurring at a specific temperature
that does not change throughout the whole
reaction, we MUST state the temperature
and the balanced equation to which the
equilibrium constant is referring.
EXAMPLE: "CO(g) + 2H2(g) ↔CH3OH(g)
has a equilibrium constant of 10.5 at
220°C"
Things to remember
5. If you know the equilibrium constant going
in one direction (say from left to right), the
equilibrium constant in the other direction
(say right to left) is the inverse of that value.
ClNO2(g) + NO(g) → NO2(g) + ClNO(g)
Kc = 1.3 x 10^4 at 25° C
NO2(g) + ClNO(g) → ClNO2(g) + NO(g)
Kc = 1/(1.3 x 10^4 at 25° C)
or...
7.7 x 10^-5 at 25° C
Things to remember
6. This is how you convert an Equilibrium
constant from Kc to Kp:
Kp = Kc (0.0821 T)^Δn
T is the temperature during the
reaction in Kelvin. Remember that
Kelvin has exactly the same degree
of difference between integers as
celsius, but it does have a different
starting point. Kelvin's zero is
273° less than celsius. For
example, 2° C is equal to 275° K.
27° C is equal to 300° K.
Δn is the sum of all stoichiometric
coefficients belonging to products
minus the sum of all stoichiometric
coefficients belonging to reactants
Things to remember
7. Heoterogeneous equilibria refers to when
a condensed solid or liquid interacts with a
gas. To calculate the equilibrium constant,
solve only for the gaseous state during
such a multi-phase state reaction.
NEW RULE!!!
8. If a reaction can be expressed as the sum of
two or more reactions, the equilibrium constant
for the overall reaction is given by the product
of the equilibrium of the individual reactions.
Kc = (K'c) (K"c)
Let me elaborate...
1st Step Reaction
n
2nd Step Reaction
n
n
n
[E] [F ]
[C ] [ D]
K
"

K 'c 
c
n
n
n
n
[C ] [ D]
[ A] [ B]
Overall
Reaction
A B  C  D
K'c
CD EF
K"c
__________________
A B  E  F
Kc
To find the overall equilibrium of a
multi-step reaction
n
n
n
n
[C ] [ D] [ E ] [ F ]
K 'c K "c 
x

K
c
n
n
n
n
[ A] [ B] [C ] [ D]
2 step reaction
1st step reaction
N2(g) + O2(g) ↔ 2 NO(g) Kc1 = 2.3 x 10^-19
2nd step reaction
2 NO(g) + O2(g) ↔ 2 NO2(g) Kc2 = 3 x 10^6
Write the equilibrium equation for this multi-step reaction:
Kc = Kc1 x Kc2 = (2.3 x 10^-19)(3 x 10^6)
Kc = 7 x 10^-13
B
P
1/[O2(g)]
D
WHY?
P
Putting it all together
• At the start of a reaction, there are 0.449 mol N2,
0.0221 mol of H2, and 0.000942 mol of NH3, in a
4.0-L reaction vessel at 375°C. If the equilibrium
constant (Kc) for the reaction
N2(g) + 3 H2 (g)→ 2 NH3 (g)
Is 2.2 at this temperature, decide whether the
system is at equilibrium. If not, predict which
way the net reaction will proceed.
Step 1
• find the molarity of each species in the reaction:
– molarity is mol/L
– divide all mole values by 4 to find mol/L (because the
total solution makes up 4 L).
0.449mol
N 2  
 0.112 M
4L
0.0221mol
3
[H 2 ] 
 5.525 x 10 M
4L
0.000942mol
4
[ NH 3 ] 
 2.355 x 10 M
4L
Step 2
• Set up these initial
values as a quotient
expression (in the
same manner as you
would in finding the
equilibrium constant)
2
[ NH 3 ]
Qc 
[ N 2 ][ H 2 ]3
(2.355 x10  4 ) 2
Qc 
(0.112)(5.525 x10 3 ) 3
Qc  2.9375
Step 3
• Compare Qc to Kc (Kc = 2.2)
– if Qc > Kc, then...
• the initial concentrations of all the species are not in
equilibrium. To reach equilibrium, the products must
convert to reactants. The reactions will travel from right
to left.
– if Kc > Qc, then...
• the initial concentrations of all the species are not in
equilibrium. To reach equilibrium, the reactants must
convert to products. The reaction will travel from left to
right.
– if Kc = Qc, then the reaction is in equilibrium.
Nothing will happen. There will be no reaction.
• 2.9375 > 2.2
• Qc > Kc
• For the reaction to reach equilibrium, more
products must converted into reactants
• The reaction will move from right to left
BELLRINGER
f
PRACTICE PROBLEM
LET'S LOOK AT HOW
TO SOLVE THIS...
g
LET'S CHECK THIS ANSWER BY
SUBSTITUTION
• WRITE OUT THE Kc
EXPRESSION FOR
THE 1ST STEP
REACTION:
• WRITE OUT THE Kc
EXPRESSION FOR
THE 2ND STEP
REACTION:
A( g )  B( g )  Kc  2.00
[ B]1
 2.00 d Let' s say we make B  2mol/L
1
[ A]
[2]1
 2.00
1
[ A]
That would make A  1mol/L
d
B( g )  C ( g )  Kc  0.01
[C ]1
[C ]1
 0.01  1
1
[ B]
[2]
dThat would make C  0.02mol/L
A = 1 mol/L
B = 2 mol/L
C = 0.02 mol/L
Now let's change directions and
add the coefficients
• 2nd STEP
REACTION IN
REVERSE:
C(g) 
→ B( g ) 
1
0.01
The Kc in one direction is the inverse
in the opposite direction
1
[B]
2
1
s


 100
1
s
[C] 0.02 0.01
s
• 2nd STEP
REACTION WITH
COEFFICIENTS:
[ B ]2
[2]2
2


10
,
000

100
[C ]2 [0.02]2
We
s just proved that when we square each
species
in the expression, the quotient is the
s
s
original
value to the power of 2
Therefore, in the future, we can just raise the Kc to a power equal
to that of the stoichiometric coefficients belonging to the species.
Now change the direction and add
coefficients to Reaction 1
• RAISE THE 1st STEP
Kc TO THE POWER
EQUAL TO THAT OF
THE STOICHIOMETRIC
COEFFICIENTS OF
THE SPECIES:
1
B( g )  A( g ) 
 0.5
2.00
[ A]2 [1]2
2
2


Kc

(
0
.
5
)
 0.25
2
2
[ B] [2]
Multiply the Kc' and Kc" to find the
net Kc
Kc'Kc" = Kc = 0.25 x 10000 = 2500 (answer B)