February 18, 2002
Lab announcements – 2nd lab quiz week before spring break
photocopy before turning in or study before turning in
-same as 101/other 102 sections
be extra careful with dilutions this week (Expt. 21)
Most chemical reactions do not go to completion.
chemical equilibrium – two opposing reactions occur
simultaneously at the same rate
‘equilibrium’ doesn’t necessarily mean ‘equal’ amounts of
reactants and products – in fact, it usually doesn’t.
Equilibrium constant – measure of this balance
aA +
Kc
=
bB ?
[C]c[D]d
[A]a[B]b
Kc
cC +
dD
(*at equilibrium*)
- is a measure of the extent to which a reaction occurs
- is determined experimentally
- changes with Temp
- not affected by initial concentration
- equation specific
no (l) or (s) in the expression, just substances in state (aq) or (g) (in
mol/L). (We’ll discuss other ways to express gas later.)
Kc can be related to thermodynamics (end of chapter)
K has no units
Kc >> 1 mostly products
Kc << 1 mostly reactants
February 18, 2002
Consider the following system in a 1.0 L flask at a certain temp:
PCl5 (g)
?
[PCl3][Cl2] = Kc
[PCl5]
Initially 1.0 mol PCl5 is added to the empty flask. At
equilibrium, 0.60 mol each of PCl3 (g) and Cl2 (g) are present.
What is Kc?
init
eq.
1.0 mol
0.60 mol PCl3 x
PCl3 (g) + Cl2 (g)
0
0.60 mol
0
0.60 mol
1 mol PCl5 = 0.60 mol PCl5 used
1 mol PCl3
1.0 mol PCl5 init - 0.60 mol PCl5 used = 0.40 mol PCl5 remains
Kc =
[0.60/1.0][0.60/1.0]
[0.40/1.0]
= 0.90
Consider this system in a 2.00 L flask at a certain temp:
SO2 (g) + NO2 (g) ?
SO3(g) + NO (g)
Kc = 3.00
Initially 1.00 mole each of SO 2 and NO2 were added to the flask.
What are the equilibrium concentrations of all the species in the
reaction?
can go ‘backwards’ – calculate []eq from Kc
February 18, 2002
2.0 L flask
SO2 (g) + NO2 (g)
init. 1.0 mol
?
1.0 mol
SO3(g) + NO (g)
0
Kc = 3.00
0
what are concentrations of species at equilibrium?
eq.
1.0 mol
- x mol
Kc =
1.0 mol
- x mol
[SO3][NO]
[SO2][NO2]
x mol
x mol
= [x/2.0][x/2.0]
[1.0-x/2.0][1.0-x/2.0]
it might have been easier to work in molarity to begin with:
eq.
0.5 M
-yM
Kc =
Kc =
0.5 M
-yM
[SO3][NO]
[SO2][NO2]
[y]2
[0.5 – y]2
yM
=
yM
[y][y]
[0.5 - y][ 0.5 - y]
= 3.00
take square root of both sides
Kc =
[y]
[0.5 – y]
= 1.73
y = 1.73(0.5 – y)
y = 0.866 – 1.73y
2.73y = 0.866
y = 0.317 M (conc. products)
0.5 M - 0.317 M = 0.183 M (conc. reactants)
February 18, 2002
A more difficult example:
A
init 0
init 0
eq. x
Kc
=
+
B
0
0
?
x
[C][D]
[A][B]
C
0.600 mol
0.300 M
0.300-x
=
2.00 L container
+ D
0.200 mol
0.100 M
Kc = 24.0
0.100-x
[0.300 – x][0.100 – x]
[x][x]
0.0300 - 0.400x + x2
x2
24.0x2 = x2 + 0.400x – 0.0300
23.0x2 + 0.400x – 0.0300 = 0 use quadratic equation
24.0 =
− b ± b 2 − 4ac
2a
x=
− 0.40 ± ( 0.40) 2 − 4( 23.0)( −0.030)
2(23.0)
x = 0.028 or
-0.046 – this one doesn’t make any sense.
[A],[B] = 0.028 M
[C] = 0.300 – 0.028 M = 0.272 M
[D] = 0.100 – 0.028 M = 0.072 M
February 18, 2002
When system is not in equilibrium, we can calculate a similar
term called the ‘reaction quotient’, Q. This helps us diagnose
the direction of a particular reaction.
aA +
Q
=
bB ?
[C]c[D]d
[A]a[B]b
cC +
dD
(*not at equilibrium*)
If Q = Kc system is at equilibrium
If Q > Kc system moving toward reactants (too much product)
If Q< Kc system moving toward products (too much reactant)
Ex.
?
H2 (g) + I2 (g)
0.22M 0.22M
?
2HI (g)
0.66M
Kc = 49.0
is the system at equilibrium?
Kc
=
[HI]2 =
[H2][I2]
=
9.0 system is not at equilibrium. Moving toward
products.
H2 (g) +
init
eq.
0.22M
0.22-x
I2 (g)
0.22M
0.22-x
[0.66]2 .
[0.22][0.22]
?
2HI (g)
Kc = 49.0
0.66M
0.66+2x
work out on your own what the equilibrium concentrations would be.