Download A matter of Equilibrium

Introduction to equilibrium
Imagine injecting water into an evacuated, sealed container:
A matter of
Equilibrium
Oxtoby Chapter 7
Reversible Reactions
H2O (g)
H2O (l)
H2O (l)
H2O (l)
&
H2O (g)
H2O (l)
H2O (l)
2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g)
Simply mixing CO2 and H2O does not give butane back!
In each case the system will eventually reach equilibrium in
which both reactions occur at equal rates. We can write:
N2 (g) + 3 H2 (g)
Such reactions are called irreversible
However many reactions can proceed in either direction:
If we mix pure N2 and H2 we form ammonia:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
But if we take pure ammonia we form N2 and H2:
2 NH3 (g) → N2 (g) + 3 H2 (g)
Such reactions are called reversible
2 NH3 (g)
Such a state is a dynamical equilibrium – both the forward and
backward reactions are proceeding simultaneously but the
rates of each balance one another.
In other words if we imagine following a particular H atom we
would see that it spends some of its time bound in ammonia
molecules and some time within H2 molecules.
Quantifying Equilibrium
But how is this dimensionless?
We can quantify any general reaction;
If we apply this to the ammonia reaction N2 + 3 H2
a A(g) + b B(g)
c C(g) + d D(g)
at equilibrium in terms of a dimensionless quantity, K, the
equilibrium Constant where
K=
[C]c[D]d
[A]a[B]b
For gases Px ∝ [x]
so we can write:
H2O (g)
Equilibrium
Some reactions go (essentially) to completion:
e.g.,
vacuum
• Initially the liquid evaporates
and the vapour pressure
rises.
• Once vapour is formed some
starts to condense.
• Eventually the rate of
condensation equals the rate
of evaporation, and
• an equilibrium is established:
K=
(PC)c(PD)d
(PA)a(PB)b
In which [X] is the concentration of X and the integers a, b,.. are
the Stoichiometric coefficients of the balanced equation.
(n.b., this can be generalised to all reactions – see solutions)
K=
(PNH3)2
(PH2)3(PN2)
2 NH3,
which has dimensions of atm-2.
Although rarely include it in the expression for K, instead of
using the partial pressure Px of a component we really should
use Px/Pref, so that K is then dimensionless.
Pref is the standard pressure. This should be 1 bar, but older
books (and some new ones) will have used 1 atm. This can
make a difference of a few percent to the values of K
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Examples of Equilibrium
Uses of K
K=?
N2 (g) +
O2 (g)
2 NO (g)
2 H2 (g) + O2 (g)
2 H2O (g)
2 H2O (g)
There are two main uses of K;
1 If a reaction has reached equilibrium we can use K to
determine the pressures (concentrations) of reactants and
products.
At 623 K for the equilibrium N2 + 3H2
2 NH3, K = is 9.0
Calculate the partial pressure of NH3 given that the equilibrium
N2 and H2 partial pressures are 0.3 and 0.9 atm, respectively.
2 H2 (g) + O2 (g)
9.0 =
Use of K (cont.)
(PNH3)2
Hence PNH3 = 1.40 atm.
(0.9)3(0.3)
Reaction Quotient, Q
2 If a system has not yet reached equilibrium we can make
use of K to determine which reaction (forward or backward)
will dominate in order to achieve equilibrium.
The reaction quotient is useful because by calculating Q we
can work out in which direction the reaction must proceed
preferentially to achieve equilibrium.
In cases of non-equilibrium we can define a Reaction
Quotient, Q. This has the same form as for K, but uses the
non-equilibrium pressures rather than the equilibrium
values.
A non equilibrium system will always move in such a
direction as to make Q tend towards K.
i.e., for a non-equilibrated N2 + 3 H2
In other words the rate of one or other of the reactions will
exceed that of the other until equilibrium is achieved.
Q=
2 NH3 system
(PNH3)2
This is most easily seen by example:
(PH2)3(PN2)
Example of Q → K
The reaction N2 + 3 H2
2 NH3 has en equilibrium constant
at 400 °C of 1.9 x 10-4.
Suppose 1 mol N2, 0.2 mol H2 and 0.4 mol NH3 are sealed
in a 1 dm3 flask at 400 °C. In which direction will the
reaction proceed?
PV = nRT, hence; PN2 = 55 atm; PH2 = 11 atm; PNH3 = 22 atm
(22)2
(PNH3)2
= 6.61 x 10-3
=
Calculate Q =
(11)3(55)
(PH2)3(PN2)
Q > K and thus the reaction will move preferentially in a such
a way as to reduce Q, in this case right to left. In other words
the ammonia decomposes until Q = K (equilibrium achieved).
Calculations with K
The equilibrium constant for the reaction H2(g) + I2(g)
2HI(g)
at 720 K is 50. If an initial mixture is prepared of 2 atm H2
and 5 atm I2, calculate the equilibrium pressures of each
gas at 720 K.
Think what happens with time:
H2
initially
at Eqm
Change
2
2–x
–x
+
I2
2 HI
5
5–x
–x
0
2x
+2x
(Avagadro)
2
Pressures or Concentrations?
So, at equilibrium
(PHI)2
K = 50 =
Now a bit of algebra:
(2x)2
=
(PH2)(PI2)
(2-x)(5-x)
50 (10 - 7x + x2) = 4x2
or 46x2 – 350x + 500= 0
Gave two definitions of K earlier — which to use?
Concentrations
K=
Pressures
[C]c[D]d
K=
[A]a[B]b
(PC)c(PD)d
(PA)a(PB)b
Solving for using the quadratic formula (see Oxtoby App. C-3)
gives two possibilities: x = 5.703 atm or 1.906 atm.
• Use with liquids or gases
• Use only with gases
The first solution is unfeasible since it implies negative H2, I2
pressures at equilibrium. Hence x = 1.906 atm.
• Standard (reference) state
is taken to be a 1 M
solution
• Standard (reference) state
is 1 bar (1 atm pre 1982)
So at equilibrium PH2 = 0.094 atm, PI2 = 3.094 atm, PHI = 3.812 atm
(take care with significant figures!)
• Denote KP
• Denote Kc
Kc and KP: does it Matter?
Example
• The ideal gas law relates pressure (P) and concentration (c):
KP = 11.5 at 573 K for the reaction
nRT
n
P =
=
RT = cRT
V
V
• Hence for the equilibrium
a A(g) + b B(g)
c C(g) + d D(g)
( PC ) ( PD ) = ( cCRT ) ( cDRT )
a
b
a
b
( PA ) ( PB ) ( c A RT ) ( cBRT )
c
d
(c ) (c )
(c + d −a − b )
(c +d −a −b )
= C a D b × ( RT )
= K c ( RT )
( c A ) ( cB )
c
KP =
Strictly:
d
c
d
K p = K c ( RT )
( c + d −a − b )
⎛c ⎞
× ⎜ ref ⎟
⎝ Pref ⎠
PCl3 (g) + Cl2 (g)
PCl5 (g)
What is the value of Kc?
( c + d −a − b )
(1+1−1)
⎛ P
⎞
K c = K p ⎜ ref ⎟
⎝ cref RT ⎠
⎛
⎞
1.00 bar
= 11.5 × ⎜
⎟
-3
-1
-1
⎝ 1.00 mol dm × 8.314 J K mol × 573 K ⎠
⎛ 105 Pa ⎞
= 2.41× 10−3 × ⎜ 3 -3
⎟
⎝ 10 m × J ⎠
= 0.241
Heterogeneous equilibria
Reactions often involve different states of matter e.g., the
decomposition of calcium carbonate
CaO (s) +
CaCO3 (s)
Work with concentrations:
K´ =
[CaO] [CO2]
[CaCO3]
CO2 (g)
But CaO and CaCO3 are solids!!
flconcentrations (densities) do not change (unless we
change temperature of pressure)
It is conventional to include the concentration of all solids, and
solvent if that is present in excess, as part of the constant on
the left hand side of the equation
K = K′
[CaCO3 ]
[CaO]
= [CO2 ]
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States can change!!!
H2 + I2
2HI
So how would we write the equilibrium constants for the
following reactions?
C (s) + CO2 (g)
2 CO (g)
At low temperatures, I2 is a solid, and so
K=
[HI]2
[H2]
CaO (s) + 3 C (s)
CaC2 (s) + CO (g)
At high temperatures, I2 is a gas, and so
K=
H2O (l)
[HI]2
H2O (g)
[H2][I2]
Always remember to specify the states of your compounds!
3 C2H2 (g) + 3 H2 (g)
C6H12 (l)
Disturbing Equilibrium
Disturbing
Equilibrium
What happens if we take a system at equilibrium and disturb it?
vacuum
Placing solid CaCO3 into a closed, evacuated
container will lead to the equilibrium
CaCO3
Oxtoby Chapters 7 – 8.2
Now remove the gaseous CO
The reaction quotient, Q, falls below
the value of the equilibrium constant,
K.
CaCO3 (s)
CaO (s) + CO2 (g)
K = [CO2] = PCO2
Le Châtelier’s Principle
The system is no longer in equilibrium
pump
CaCO3 (s)
CaO (s) + CO2 (g)
•To re-establish equilibrium, more CaCO3 must decompose to
replace (some of) the CO2 removed.
This type of response is quite general, and
was encapsulated by Henri Louis le Châtelier
in 1884:
“A system at equilibrium subjected to an
external stress reacts in such a way as to
counteract the stress.”
•Continuous would eventually decompose all of the calcium
carbonate: the reaction would have been driven the reaction to
completion
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Examples of Le Châtelier’s Principle
N2 + 3 H2
for which K =
[NH3]2
[N2][H2]3
2 NH3
ª
(PNH3)2
A note on Reaction Energetics
Thermodynamics is the study of energy and energy changes
and we will study it explicitly next week, and week 7.
(PN2)(PH2)3
If we….
1 increase the pressure externally….
…. then the reaction will be driven to the __________
2 pump in extra H2 ….
…. then the reaction will be driven to the __________
3 remove some of the ammonia….
…. then the reaction will be driven to the __________
4 Increase the volume of the container….
…. then the reaction will be driven to the __________
At this point it is useful to know that the heat gained or lost in
a reaction is known as the enthalpy change, ∆H (kJ mol-1).
∆H and Reversible Reactions
∆H and Le Châtelier
Energy must be conserved overall in a chemical reaction.
If a forward reaction is exothermic (∆H < 0) then the back
reaction must be endothermic (∆H > 0) to the same degree.
N2 + 3 H2
2 NH3
∆H = -92.4 kJ
mol-1
We should read this as “when 1 mole of gaseous N2 react with
3 moles of gaseous H2 to produce 2 moles of ammonia gas,
92.4 kJ of energy are released” — and vice versa
The equilibrium can be thought of as:
N2 + 3 H2
2 NH3 + 92.4 kJ mol–1
So, let’s imagine…
2 NH3 (g)
If ∆H<0, then heat is released in the reaction and the reaction
is said to be exothermic.
N2 + 3 H2
2 NH3
∆H = -92.4 kJ mol-1
If we raise the temperature of the equilibrium mixture, then
the reaction moves in the direction that counteracts the
stress applied.
This means that the reaction proceeds in the endothermic
direction — from right to left — so as to absorb some of the
heat we put in.
(warning in passing: in cases where ∆H ≠ 0, the equilibrium
constant itself changes with temperature dependent)
Equilibria in Solution
… you set yourself up in business making ammonia (which is very
important in the fertiliser industry) from N2 and H2 (the so-called
Haber-Bosch process. We know the basic chemistry:
N2 (g) + 3 H2 (g)
If ∆H>0, then during the reaction heat is absorbed (i.e., must
be provided) and the reaction is said to be endothermic.
∆H = -92.4 kJ mol-1
Equilibrium constants and Le Châtelier’s principle apply to all
chemical reactions, including equilibria in solution
One of the most important examples is the autoionization of
water.
What reaction conditions (pressures, temperatures, etc.,)
should we use to make the ammonia most efficiently?
Pure liquid water always exists as the following equilibrium:
What other factors are important?
or, more correctly 2 H2O (l)
H2O (l)
OH-(aq) + H+(aq)
OH-(aq) + H3O+(aq)
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2 H2O (l)
These are small numbers…..
OH-(aq) + H3O+(aq)
Kw = 1.01 x 10-14 at 25 °C. Hence at 25 °C the actual H3O+
concentration is 1 x 10-7 mol dm-3 ([OH-] is the same).
[H3O+][OH-]
K=
The equilibrium constant is
[H2O]2
It is more convenient to introduce a logarithmic scale, the pH
scale, where:
However, since the concentration of water is not changed by
this reaction, we include it in the constant, Kw. (Oxtoby 7.6)
pH = -log10[H3O+]
Kw = [H3O+][OH-]
•Kw is small, but immensely
important.
•As with most equilibrium
contants, its value does
change with temperature.
T / °C
0
25
40
60
Kw
0.114 x 10-14
1.01 x 10-14
2.92 x 10-14
9.61 x 10-14
(you may see it in some books as pH = -log10[H+])
The pH of pure water at 25 °C is
pH = -log10(1 x10-7) = 7.00
Anyone unfamiliar with logarithms needs
to study Oxtoby appendix C-1
pH of a strong acid
Strong bases
The [H3O+] of aqueous solutions can vary from 10-15 – 20 M
NaOH is a strong base:
HCl is a strong acid:
– it fully dissociates in solution
– The equilibrium lies at the extreme right
– it fully dissociates in solution
– The equilibrium lies at the extreme right
H3O+(aq)
HCl(aq) + H2O(l)
NaOH(aq) + H2O(l)
+
Cl-(aq)
OH–(aq) + Na+(aq)
– K is huge
A 0.1 M NaOH solution we will give [OH–] = 0.1 M (pOH = 1).
– K is huge
A 0.1 M HCl solution we will give [H3O+] = 0.1 M
The corresponding [H3O+] at equilibrium is
(pH = 1).
[H3O+] = Kw / [OH–] = 1.0 x 10–13 M
The corresponding [OH-] at equilibrium is
[OH–] = Kw / [H3O+] = 1.0 x 10–13 M
(pOH = 13)
(pH = 13)
At 25 °C it is always true that pH + pOH = 14.00 = pKw.
pH ranges
A solution with pH < 7 then [H3O+] > [OH-] and is called acidic
A solution with pH = 7 then [H3O+] = [OH-] and is called neutral
A solution with pH > 7 then [H3O+] < [OH-] and is called basic
[H3O+]
1
10-1
0
1
10-2
10-3
10-4
10-5
10-6
2
3
4
5
6
pH
vinegar
cola
OJ
10-7
7
H2O
10-8
8
blood
10-9
9
10-10 10-11 10-12 10-13 10-14
10
11
12
13
14
toothpaste
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