Transcript Equilibrium - University of Illinois at Urbana–Champaign

Equilibrium
March 28, 2007
Chemical Reactions
Reactants  Products
A + 3B  2C
Forward Reaction
But the reverse can also happen
2C  3B + A
Reverse Reaction
So we can describe the reaction as equilibrium
A + 3B
2C
Equilibrium
rate of forward reaction = rate of reverse reaction
Equilibrium - Concentration
At equilibrium, the concentration of product
and reactants stays constant
Equilibrium Visualization
http://www.dlt.ncssm.edu/TIGER/chem5.htm
Law of Mass Action
• Once a reaction has reached equilibrium, the
relative concentration of products remains
constant
• We call this constant K, the equilibrium constant
concentration of products at equilibrium
K
concentration of reactants at equilibrium
Example: N2 + 3H2  2NH3
2
[NH
]
3
K
3
[N
H
2][
2]
Equilibrium Expression
• Things that appear in the equilibrium
expression
– Concentration of solutions
– Pressure of gases
– Reaction coefficient
• Things that do NOT appear
– Pure liquids
– Pure solids
– Units
Learning Check
Does this graph represent a K > 1, K< 1, or K =1 ?
Greater number of reactants – K < 1
Learning Check
Determine the Equilibrium Expression (K) for
each of the following reactions
CaCO3 (s)
CaO (s) + CO2(g)
K  [CO2 ]
2NO2(g)
H2CO3(aq)
N2O4(g)
[ N 2O4 ]
K
[ NO2 ]2
CO2(g) + H2O(l)
[CO2 ]
K
[ H 2CO3 ]
Calculating the Equilibrium Constant
Calculate the equilbrium constant if the
equilibrium concentrations of NO2 and
N2O4 are 2.0 mol/L.
2NO2(g)
N2O4(g)
[products] [ N 2O4 ] 2
K

 2  0.50
2
[reactants] [ NO2 ]
2
Equilibrium Position
• At constant temperature….reaction can only have
one equilibrium constant but many equilibrium
positions
N2 + 3H2
2NH3
K = 640 (25 °C)
• Equilibrium concentration of each product can be…
N2 (mol/L)
H2 (mol/L)
NH3 (mol/L)
30
30
900
10
40
800
5
45
675
Value of Equilibrium Constant
A + 2B
2C + D
[C]2[D]
K
2
[A][B]
• If K >>>>1, forward reaction is favored
– Large concentration of products
• If K <<<<1, reverse reaction predominates
– Large concentration of reactants
• If K = 1, reverse reaction and forward reaction equal
– Equal concentration of reactants
Changing the Equilibrium Constant
• Change the temperature
• Change the reaction coefficients
• N2 + 3H2
2N2 + 6H2
2NH3
4NH3
[ NH 3 ]2
K1 
 640
3
[ N 2 ][ H 2 ]
[ NH 3 ]4
K2 
?
2
6
[N2 ] [H 2 ]
Relationship between K1 and K2
K2 = K12
K2 = (640)2 = 4.096 x 105
Changing the Equilibrium Constant
• Change the temperature
• Change the reaction coefficients
• N2 + 3H2
2NH3
2NH3
N2 + 3H2
[ NH 3 ]2
K1 
 640
3
[ N 2 ][ H 2 ]
[ N 2 ][ H 2 ]
K3 
?
2
[ NH 3 ]
Relationship between K1 and K3
K3 = 1/K1
3
K3 =1/ 640 = 0.00156
Learning Check
• Determine the value of the equilibrium
constant for the following reaction
• 2NO2
½ N2O4
N2O4
NO2
K  0.25
K ?
[ NO2 ]
1
1
K2 


2
1/ 2
[ N 2O4 ]
K1
0.25
Disturbing the Equilibrium
Heat as a Reactant/ Product
• UO2(s) + 4HF(g)  UF4(g) + 2H2O(g
FeSCN2+ Equilibrium
• KSCN + Fe(NO3)3  FeSCN2+ + KNO3
• SCN- + Fe3+  FeSCN2+
• Fe3+ + HPO42-  FeHPO4+
• Ag+ + SCN-  AgSCN
• Ag+ + Cl-  AgCl
Spectator ions….ignore