Chem 1101
A/Prof Sébastien Perrier
Room: 351
Phone: 9351-3366
Email: [email protected]
Prof Scott Kable
Unless otherwise stated, all images in this file have been reproduced from:
Room: 311
Blackman, Bottle, Schmid, Mocerino and Wille,
Chemistry, 2007 (John Wiley)
ISBN: 9 78047081 0866
Phone: 9351-2756
Email: [email protected]
A/Prof Adam Bridgeman
Room: 222
Phone: 9351-2731
Email: [email protected] 28-2
Slide 28-1
Highlights of last lecture
This lecture…
Equilibrium
The equilibrium state
The dynamic nature of eq’m (kinetic picture)
∆G=0 at equilibrium
Equilibrium constant
How to write Keq for any reaction
Work out Keq from eq’m concentrations
Work out eq’m concentrations from Keq
Different forms of the eq’m constant.
Equilibria involving liquids and solids.
Reaction Quotient
Work out Q from any set of concentrations
Work out whether a system is at eq’m
Work out the direction of chemical change.
Working out final equilibrium concentration
from any starting position.
Short-cuts and approximations you can
make.
Slide 28-3
Slide 28-4
Revision: NO2 dimerization (from L27)
N2O4 (g)
N2O4 (g)
2 NO2 (g)
Three people did an experiment on the equilibrium of this
reaction…
Expt
Stoichiometry and Units
[NO2]
(init.)
[N2O4]
(init.)
[NO2]
(final)
[N2O4]
(final)
Keq
1
0
0.1000
0.1018
0.0491
0.211 mol/L
2
0.1000
0
0.0627
0.0185
0.212 mol/L
3
0.0500
0.0500
0.0837
0.0332
0.211 mol/L
So… they were all correct, and the point [NO]2
is that the eq’m position in ALL cases is
[N2O4]
given by:
expt 1:
Keq =
[ X ]eq =
[NO2]2
[N2O4]
But, what if I wrote:
expt 1:
K’eq =
[ X ]eq =
[NO2]
[N2O4]½
Or, what if I wrote:
= 0.211 mol/L
Slide 28-5
expt 1:
K”eq =
[NO2]2
2 NO2 (g)
0.1018
mol/L
= 0.211 mol/L
½ N2O4 (g)
0.0491
=
0.1018
(0.0491)½
0.1018
=
0.1018
2 NO2 (g)
[ X ]eq =
[N2O4]
0.0491
0.0491
(0.1018)2
NO2 (g)
mol/L
= 0.459 (mol/L)½ = (Keq)½
N2O4 (g)
0.0491
= 4.74 L/mol = 1/Keq
mol/L
Slide 28-6
1
Stoichiometry and Units
Equilibria for coupled reactions
The eq’m constant must be associated with a specific
stoichiometric chemical eq’n.
If you multiply the stoichiometry by n, you change
K by a power of n
Consider the following sequential atmospheric reactions:
N2(g) + O2(g) 2NO(g)
K1 = 4.3x10-25
2NO(g) + O2(g) 2 NO2(g)
K2 = 6.4x109
We can add up these chemical reactions to provide an overall reaction:
K1 × K 2 =
Notice that the UNITS change
2
[NO2 ]
[NO 2 ]2
[NO ]2
×
=
= Ktot
2
[N2 ][O2 ] [NO ] [O2 ]
[N 2 ][O2 ]2
i.e. when you ADD chemical equations, you
MULTIPLY the K’s
In equilibrium problems, the units of K are often
omitted (see tute for explanation)
Ktot = ?
N2(g) + 2O2(g) 2NO2(g)
If you reverse the chemical equation, you take the
inverse of K
Slide 28-7
Slide 28-8
Summary from Chemistry 4th Edition, Silberberg,
Table 17.2, p.732:
Different kinds of eq’m constant
So far we have used the symbol Keq or just K to represent the
equilibrium constant.
Sometimes we want to associate the eq’m constant with a
particular type of reaction, or particular units. We then use
a different subscript on the K but really there is no
difference to the way we treat the problem.
e.g.
Ka = acid dissociation eq’m constant;
Kb = base dissociation eq’m constant;
Kw = water dissociation eq’m constant;
Kp = eq’m constant in units of pressure;
Kc = eq’m constant specifically in concentration units;
Kstab = eq’m constant for stability of a complex;
Ksp = solubility product constant.
Slide 28-9
Equilibria involving (l) and (s)
Consider the “water gas” reaction, which is used to make
combustible gases from coal:
Equilibria involving (l) and (s)
Consider two identical closed flasks containing different
amounts of water at the same temperature:
C(s) + H2O(g) CO(g) + H2(g)
The equilibrium constant can be written as:
K =
K′=
Kc = [H2O(vap)]
Kp = pH2O
[CO ][H2 ]
[H2O ][C ]
But the “concentration”, i.e. mol/L, of C(s) is constant (the “concentration”
of a solid is its density, which is independent of how much substance is
present. Therefore this constant can be incorporated into the equilibrium
constant:
[CO ][H2 ]
[H2O ]
In general: equilibrium constant expressions DO NOT contain
concentration factors for pure solids or liquids.
Slide 28-11
Slide 28-10
H2O (l) H2O (vap)
Vapour pressure does not depend on how much liquid is present.
Consider the decomposition of mercuric oxide:
2HgO (s) 2Hg (l) + O2 (g)
Kc = [O2]
Kp = pSlide
O2 28-12
2
Equilibria involving (l) and (s)
Equilibria involving (l) and (s)
Decomposition of calcium carbonate:
Decomposition of calcium carbonate:
CaCO3(s) CaO (s) + CO2 (g)
CaCO3(s) CaO (s) + CO2 (g)
The equilibrium doesn’t
depend on the amount
of solid, only on the
amount of gas.
Kc = [CO2]
Kp = pCO2
Slide 28-13
Chemistry 4th Edition, Silberberg, Fig 17.4, p.732:
Basic eq’m calculations
Example calculation (1)
We have already investigated the eq’m between NO2 and N2O4 so that we know
that NO2 and N2O4 exist in equilibrium.
Example 1: A basic calculation;
Example 2: An approximation when K is small;
Example 3: Heterogeneous equilibria (Ksp);
Example 4: Another example when K is small
Q: A 0.0240 mol sample of N2O4(g) is allowed to come into equilibrium with
NO2(g) in a 0.372 L flask at 25ºC. Calculate the amount, in moles, of NO2 and
N2O4 present at equilibrium.
N2O4 (g) Slide 28-15
2 NO2 (g)
Example calculation (1)
1. Initial concentration of N2O4
4.61 x 10 -3 =
4.
= 0.0240 / 0.372 mol/L
4X2
(0.0645-X)
= 0.0645 mol/L
4.61 x 10 -3 x (0.0645-X) = 4X2
N2O4 (g)
I
C
E
Initially:
2 NO2 (g)
0.0645
2.973x10 -4 – 4.61x10 -3 X = 4X2
0
mol/L
4X2 + 4.61x10 -3 X – 2.973x10 -4 = 0
Let x = amount of N2O4 reacted…
Change:
-X
+2X
mol/L
Equil’m
0.0645-X
2X
mol/L
x =
x =
3. K = 4.61 x 10-3 =
[NO2]2
[N2O4]
=
K = 4.61 x 10 -3 mol/L
Approach:
1. Work out concentration of N2O4 initially
2. Let x = change in concentration
3. Write down equilibrium equation in terms of x
4. Solve for x
5. Substitute x to work out final concentrations or amounts
Slide 28-16
Example calculation (1)
2.
Slide 28-14
(2X)2
(0.0645-X)
x =
− b ± b 2 − 4ac
2a
− 4.61 × 10 −3 ± ( 4.61 × 10 −3 )2 + 4 × 4 × 2.973 × 10 -4
2× 4
− 4 .61 × 10 −3
±
8
X = 8.065x10-3 or
Slide 28-17
Quadratic equation
4 .779 × 10 −3
8
-9.217x10-3
Slide 28-18
3
Example calculation (1)
N2O4 (g)
Equil’m
0.0645-X
Example calculation (2)
2 NO2 (g)
2X
mol/L
Find the equilibrium concentrations
of N2, O2 & NO at 2400 K if we
start with 0.20 mol N2 and 0.20
mol O2 in a 5.0 L vessel.
5. [N2O4] = 0.0645 – X = 0.0564 mol/L
Kc = 2.5 × 10–3 at 2400 K
[NO2] = 2X = 0.0161 mol/L
N2 (g) + O2 (g) 2NO (g)
n (N2O4) = 0.0564 mol/L x 0.372 L = 0.0210 mol
n (NO2) = 0.0161 mol/L x 0.372 L = 0.0060 mol
Kc =
Slide 28-19
[NO] 2
[N2 ][O2 ]
Example calculation (2)
Could solve as quadratic, but suspect that x is small,
⇒ 0.040 – x ≈ 0.040
=
( 2x) 2
= 2.5 × 10 −3
( 0.040 − x)( 0.040 − x)
Slide 28-20
Lead iodide is an almost insoluble salt with a dense golden yellow colour. (It is
used in ornamental work requiring a gold-like colour). The solubility product
equilibrium constant is 7.1x10 -9 mol3/L3. Calculate the solubility of PbI2 (assume
excess solid).
The balanced equation is:
= 2.5 × 10–3 × (0.040) × (0.040),
gives x = 1.0 × 10–3 M
PbI2(s) Pb2+(aq) + 2I-(aq);
Check approximation:
x indeed small: 0.040 – x = 0.039 ≈ 0.040
(solving quadratic: x = 0.98 × 10–3 M, so approx. is good)
⇒ 2x = 2.0 ×
M = [NO]
⇒ [N2] & [O2] = 0.040 – 1.0 × 10–3 = 0.039 M
Slide 28-21
Example calculation (4)
0
0
Change:
+X
+2X
mol/L
mol/L
Equilibrium
X
2X
mol/L
Solve for x and work out how many mol/L of PbI2 will dissolve.
Solution next lecture.
Slide 28-22
Example calculation (4a)
NOCl is an important transient molecule in the ozone cycle of the
stratosphere. It exists in equilibrium with NO and Cl2. At 35ºC the eq’m
constant, Kc = 1.6x10 -5 mol/L. In a laboratory experiment where the
decomposition of NOCl was being studied, 1.0 mole of NOCl was placed
into a 2.0 L flask. What are the equilibrium concentrations of NOCl, NO
and Cl2?
When gases are involved, the equilibrium constant may be written in terms of
pressure, rather than concentration. For instance, in example (4), the eq’m
constant for the decomposition of NOCl used concentrations. It might equally
well have referred to the partial pressures of NOCl, NO and Cl. For example,
at 35ºC:
2NOCl(g) 2NO(g) + Cl2(g);
The balanced equation is:
Kc = 1.6x10-5
Initial:
0.50
0
0
mol/L
Change:
-2X
+2X
+X
mol/L
Equilibrium
0.50-2X
2X
X
mol/L
Solution next lecture.
Ksp=7.1x10-9
Initial:
Ksp = [Pb2+][I-]2 = (x)(2x)2 = 7.1x10-9 mol3/L3
10–3
2NOCl(g) 2NO(g) + Cl2(g);
0
2x
2x
Example calculation (3)
⇒ 4x2 = 2.5 × 10–3 × (0.040 – x)(0.040 – x)
4x 2
0.20/5.0 0.20/5.0
-x
-x
(0.040 – x) (0.040 – x)
Initial
Change
Equilibrium
Question asked for moles, therefore
Kp = 4.1x10-4 atm
The solution to problems using Kp are exactly the same as previously, but of
course the answer comes out in pressure units, rather than concentration.
If gases are involved, and there is any likelihood of confusion, then the
equilibrium constant is usually denoted as either Kp or Kc.
Slide 28-23
Slide 28-24
4
Relationship between Kp and Kc
Relationship between Kp and Kc
Pressure and concentration are linked by the ideal gas law:
PV=nRT
or…
Using the NOCl example:
2
 PNO   PCl 

 
2
RT
  RT 
[NO ] [Cl2 ] 
Kc =
=
2
[NOCl ]2
 PNOCl 


 RT 
2
Kp
RT
=
K p = Kc (RT )
Kc=1.6x10-5 mol/L
2NOCl(g) 2NO(g) + Cl2(g)
Check :
The preceding derivation is just one example of a
general relationship:
n
P
conc. =
=
V
RT
Kp=4.1x10-4 atm
(PNO )2 (PCl
2
(PNOCl )
1
2
or
)×  RT 
= Kp ×
2
 1 


 RT 
1
RT
To avoid confusion, just remember ONE of these
−4
=
4.1 × 10
= 1.6 × 10 −5 = Kc
0.0821 × (273 + 35)
Slide 28-25
Slide 28-26
Extra homework calculation
Solid silver is added to a solution with these initial concentrations: [Ag+] =
0.200M, [Fe2+] = 0.100M and [Fe3+] = 0.300 M. The following reversible
reaction occurs:
CONCEPTS
The approach to solving eq’m problems
When to use “small x” approximation
Ag+(aq) + Fe2+(aq) Ag(s) + Fe3+(aq)
Work out K from given concentrations
Work out final eq’m concentrations from any
starting concentrations
Convert between Kc and Kp
K = 2.98 L/mol
Ag+(aq) + Fe2+(aq) Ag(s) + Fe3+(aq)
What are the ion concentrations when equilibrium is established?
CALCULATIONS
− ∆n gas
3
Example questions
Kc = K p (RT )
∆n gas
Initial:
Extra homework calculation
0.100
0.300 mol/L
Which way does the equilibrium shift?
Q=
Slide 28-27
0.200
[Fe 3+ ]
[Fe 2+ ][Ag + ]
=
Eq’m will shift to
the ________
0.300
= 15.00
(0.100) × (0.200)
Slide 28-28
More info on example (4)
Initial:
0.200
0.100
0.300 mol/L
Change:
+X
+X
-X
Equilibrium
0.200+X
0.100+X
0.300-X
K=1.6x10-5
2NOCl(g) 2NO(g) + Cl2(g);
Ag+(aq) + Fe2+(aq) Ag(s) + Fe3+(aq)
Equilibrium
0.50-2X
2X
X
mol/L
[ NO ]2 [Cl2 ]2
(2 x ) 2 ( x )
K =
=
= 1.6 × 10 − 5
[ NOCl ]2
(0.50 − 2 x ) 2
mol/L
This is a difficult cubic equation to solve!
K = 2.98 =
0.300 − x
(0.100 + x ) × (0.200 + x )
However, note that K is small, therefore the reaction will not proceed to
products to any large extent. Consequently, x is likely to be small, compared
with the original 0.50M.
The allowable solution of this quadratic equation is x = 0.11. (You
should check this.)
Therefore,
( 2x ) 2 ( x )
( 0 .50 ) 2
[Ag+] = 0.200 + x = 0.31 M;
[Fe2+] = 0.100 + x = 0.21 M;
[Fe3+] = 0.300 – x = 0.19 M.
Slide 28-29
= 1 .6 × 10 − 5
⇒
4 x 3 = 0 .25 × 1 .6 × 10 − 5
x 3 = 1 .0 × 10 − 6
x = 1 .0 × 10 − 2
Solution next lecture.
Slide 28-30
5