Transcript Equilibrium

Equilibrium
Phase
Solution
Chemical
Reversible Reactions

a number of chemical reactions have
a ΔH and ΔS that are both positive or
both negative; one force favours the
reaction, while the other opposes it:
energy + H2O(l)
ΔH = +
 H2O(g)
ΔS = +

for the opposite reaction the ΔH and ΔS
are both negative:
H2O(g)

→
energy + H2O(l)
ΔH = ΔS = as a result this reaction is reversible
Equilibrium



occurs when a reaction can go in
either direction.
the majority of reactions can be
equilibrium situations.
there are 3 main types of equilibria.
Phase Equilibrium

occurs between the phases of a substance:
energy + H2O(l)


H2O(g)
even at room temperature water evaporates
until the air above is saturated.
(relative humidity is the amount of water
vapour in the air, in %, relative to the
amount the air can hold at that temperature.
The air is saturated if the relative humidity is
100%.)

as a water molecule evaporates, another
molecule of water vapour condenses. The
rate of the evaporation reaction equals the
rate of the condensation reaction; the
reactions are at equilibrium:
energy + H2O(l)

H2O(g)
Solution Equilibrium



Occurs when a soluble non-volative
substance is dissolved in a solvent.
If the solution is saturated and there is
still solid solute at the bottom of the
beaker the system is at equilibrium.
For example:
NaCl(s)  Na1+(aq) + Cl1-(aq)
C12H22O11 (s)  C12H22O11 (aq)
Chemical Equilibrium


occurs in a chemical reaction system
(the other two were examples of
physical reactions)
Examples:
2 NO2 (g)  N2O4 (g)
PCl5 (g)  PCl3 (g) + Cl2 (g)
Equilibrium vs Steady State


Equilibrium is a closed system; no
inputs or outputs.
Steady state is an open system;
inputs and outputs.
Dynamic Equilibrium

5 conditions are necessary for
equilibrium to occur:
• the reaction must be reversible
• the system must be closed
• both reactant and product must be
present
• concentrations of reactant and product
do not change
• conditions of temperature and pressure
must be held constant
Dynamic Equilibrium



the forward reaction continues
the reverse reaction continues
the rate of the forward reaction
equals the rate of the reverse
reaction.
Le Chatelier’s Principle


reactions continue in both directions
during equilibrium
any factor which changes the rate of
one or both reactions will alter the
relative amounts of reactants and
products.
Le Chatelier’s Principle

we can predict how an equilibrium
system will respond to a given
change by using Le Chatelier’s
Principle:
“When a system at equilibrium is
disturbed by application of a stress,
it attains a new equilibrium position
that minimizes the stress.”
Le Chatelier’s Principle

in other words, for every action there
is an equal and opposite reaction;
whatever you do to a reaction
system, the system will respond in
the opposite direction.
Effect of Changing Concentraton
on a Chemical System

Increasing the concentration of any
component will cause the reaction
system to try to decrease it, by
favouring the opposite side.
Effect of Changing Concentraton
on a Chemical System
3 CH4 (g) + 6 H2O(g) + 4 N2 (g)  8 NH3 (g) + 3 CO2 (g)


Increase [CH4]; system responds by trying
decrease the [CH4] by making more product.
The concentration of reactants will go down
while the concentration of products goes up.
In other words, increasing [CH4] in this
reaction favours the creation of product.
Effect of Changing Temperature on
a Chemical System


Increasing the temperature of a
system will favour the endothermic
direction.
Decreasing the temperature will
favour the exothermic direction.
Effect of Changing Temperature on
a Chemical System
3 H2(g) + N2 (g)  2 NH3 (g) + 92 kJ


adding energy causes the system to try to
absorb energy; the endothermic direction
is favoured. More reactant is made and
the amount of product decreases.
taking energy away favours the
exothermic reaction; in this case products
are favoured.
Effect of Changing Pressure on a
Chemical System


equal numbers of moles at the same
temperature and pressure have the same
volume.
increasing pressure has the effect of
favouring the side of the equation with
fewer particles.
Effect of Changing Pressure on a
Chemical System
3 CH4 (g) + 6 H2O(g) + 4 N2 (g)  8 NH3 (g) + 3 CO2 (g)



in this reaction the reactant side has
13 particles; the product side has 11.
increasing the pressure will favour
the products.
decreasing the pressure will favour
the reactants.
Effect of Changing Volume on a
Chemical System


in a closed system reducing the
volume has the effect of increasing
the pressure.
increasing the volume reduces the
pressure.
Catalysts and Inhibitors


catalysts reduce activation energy in
both directions; they increase the
rate of reaction in both directions but
does not change the equilibrium.
inhibitors have the opposite effect,
but also do not affect the
equilibrium.
Common Ion Effect


is related to solubility equilibria.
adding an ionic substance with an ion
in common with an aqueous
equilibrium will affect the equilibrium
in the same way as changing
concentration
Common Ion Effect



NaCl(s)  Na1+(aq) + Cl1-(aq)
if NaNO3(aq) is added to this system,
it breaks into Na1+(aq) and NO31-(aq).
It is just like adding Na1+ ions to the
NaCl system. That system responds
by favouring the formation of solid
NaCl; the concentration of Cl1- ions
decreases.
Common Ion Effect
NaCl(s)  Na1+(aq) + Cl1-(aq)



add AgNO3
Ag1+ precipitates chloride(Cl1-)
removing it from solution.
system responds by producing more
chloride, also increases sodium ions
and reduces the NaCl solid
Equilibrium Constant

For the reaction:
aA(g) + bB(g)  cC(g) + dD(g)


the lower-case letters represent
number of moles (the numbers used
to balance the equation)
the upper-case letters represent the
chemical species.
Equilibrium Constant
aA(g) + bB(g)  cC(g) + dD(g)



At equilibrium a mathematical relationship
exists where a constant results from the
equation:
Keq = [Products] =
[Reactants]
[C]c[D]d
[A]a[B]b
where [ ] represents the concentration.
Equilibrium Constant



this a useful mathematical relationship.
no matter what concentrations of reactant
or product you start with we know that
the final ratio of products over reactants
will equal the value of the Keq.
This applies only if conditions of
temperature and pressure stay constant.
Writing Equilibrium Expressions

Consider the reaction:
N2 (g)


+
3 H2 (g)  2 NH3 (g)
Keq =
+
92 kJ
[NH3]2
[N2] [H2]3
energy is not included in the equilibrium
constant expression
Writing Equilibrium Expressions
2 CaCO3 (S)  2 Ca(s) + 2 CO2 (g) + O2 (g)


Keq =
[CO2]2[O2]
solids and liquids are not included in
the equilibrium expression.
Writing Equilibrium Expressions
2 H2 (g) + O2 (g)  2 H2O (l)


Keq =
1
[H2]2[O2]
since the liquid is not included in the
equilibrium expression the
numerator becomes a 1.
Equilibrium Calculations


the equilibrium expression is a
mathematical relationship of several
variables.
given the values of all but one of the
variables allows you to calculate the
unknown, whether it be the Keq or
any of the concentrations.
Equilibrium Calculations

consider the following case:
N2 (g) + 3 H2 (g)  2 NH3 (g)


+
92 kJ
given this equation we can write the
equilibrium expression:
Keq =
[NH3]2
[N2] [H2]3
Equilibrium Calculations


at equilibrium the concentration of
N2 = 0.142 mol/L, the
[H2] = 0.341 mol/L and the
[NH3] = 0.220 mol/L.
calculate the value of the Keq
Equilibrium Calculations
[N2] = 0.142 mol/L
[H2] = 0.341 mol/L
[NH3] = 0.220 mol/L

Keq =

Keq = 8.60

[NH3]2 =
(0.220 mol/L)2
[N2] [H2]3
(0.142 mol/L)(0.341 mol/L)3
This is slightly larger than 1; products are
favoured over reactants
ICE box problems



This type of problem gives you
information about the initial
situation, before equilibrium is
established.
Then you get 1 or more pieces of
information concerning the situation
at equilibrium.
You are required to fill in the blanks
by figuring out the change.
ICE box problems
2 A(g) + B(g)  C(g) + 3 D(g)



0.100 mol/L each of A and B are
placed in a flask and allowed to come
to equilibrium.
At equilibrium 0.022 mol/L of C is
present.
Calculate the value of the Keq.
2 A(g) + B(g)  C(g) + 3 D(g)
[Initial]
0.100 M
0.100 M
0.0
0.0
[Change]
[Equilibrium]


0.022 M
this is what you know.
Nothing is said in the problem about the amount of C and
D initially, so we assume they start at zero.
2 A(g) + B(g)  C(g) + 3 D(g)
[I]
[C]
[E]

0.100 M
0.100 M
0.0
0.0
+0.022 M
0.022 M
since C started with nothing and ended with 0.022 mol/L
we can see that it changed by +0.022 mol/L.
2 A(g) + B(g)  C(g) + 3 D(g)
[I]
[C]
[E]


0.100 M
0.100 M
0.0
0.0
+0.022 M 3/1(+0.022 M)
0.022 M
if C goes up, so does D.
the stoichiometry is 1 C : 3 D. For every 1 that C goes up,
D goes up 3.
2 A(g) + B(g)  C(g) + 3 D(g)
[I]
0.100 M
0.100 M
[C] 2/1(-0.022M) 1/1(-0.022M)
[E]


0.0
0.0
+0.022 M 3/1(+0.022 M)
0.022 M
the change to A and B are negative (teeter-totter)
always take the stoichiometry into account.
2 A(g) + B(g)  C(g) + 3 D(g)
[I]
0.100 M
0.100 M
[C] 2/1(-0.022M) 1/1(-0.022M)
= -0.044 M
[E] 0.056 M


= -0.022 M
0.078 M
0.0
0.0
+0.022 M 3/1(+0.022 M)
0.022 M
= +0.066 M
0.066 M
adding or subtracting the changes from the initial
concentration yields the equilibrium concentration of
each species.
all that remains is to write the equilibrium
expression and solve for Keq
Keq =
[C][D]3
[A]2 [B]
= (0.022 mol/L)(0.066 mol/L)3
(0.056 mol/L)2( 0.078 mol/L)
Keq = 0.026
Another ICE box
N2 (g)


+
3 H2 (g)  2 NH3 (g)
If 1.65 mol of ammonia is placed in a
3.00 L flask at 142 °C and allowed to
come to equilibrium the concentration of
hydrogen gas is 0.771 mol/L
Calculate the value of the Keq
N2 (g)
[I]
0.00 mol/L
[C]
1/3
[E]
3 H2 (g)
0.00 mol/L
(+0.771 M) +0.771 mol/L
= 0.257 mol/L
0.257 mol/L
0.771 mol/L
2 NH3 (g)
1.65 mol/3.00L
= 0.550 mol/L
2/3
(-0.771 mol/L)
= -0.514 mol/L
0.036 mol/L
N2 (g) + 3 H2 (g)  2 NH3 (g)

Keq =
=
[NH3]2
[N2] [H2]3
(0.036 mol/L)2
(0.257 mol/L)(0.771 mol/L)3
= 0.011
A(g) + B(g)  2 C(g)

If the initial concentration of C is
0.350 mol/L, calculate the
equilibrium concentrations of all
three species if the value of the Keq
is 145.
A(g) + B(g)  2 C(g)
[I]
0.0
0.0
0.350 mol/L
[C]
[E]


this is what we know, other than the Keq value.
where do we go from here ?
A(g) + B(g) 
[I]
0.0
0.0
[C]
+x
+x
2 C(g)
0.350 mol/L
- 2x
[E]

We don’t know the changes in concentration. When we
don’t know something in math we call it x.
A(g) + B(g) 
[I]
0.0
0.0
[C]
+x
+x
[E]
x
x


2 C(g)
0.350 mol/L
- 2x
0.350 mol/L - 2x
The equilibrium concentrations are expressed in terms of
x.
Now we write the Keq expression
Keq =
[C]2
[A] [B]
145 =
(0.350 mol/L – 2x)2
(x)(x)
145 =
(0.350 mol/L – 2x)2
(x)2
√145
=
0.350 mol/L – 2x
x
x = 0.0249 mol/L
A(g) + B(g) 
[I]
0.0
0.0
[C]
+x
+x
[E]
x
x


2 C(g)
0.350 mol/L
- 2x
0.350 mol/L - 2x
[A] = [B] = x = 0.0249 mol/L
[C] = 0.350 M – 2x = 0.350 M – 2(0.0249 M)
= 0.300 mol/L
A(g)

+
B(g) 
C(g)
If the initial concentration of C is
0.350 mol/L, calculate the
equilibrium concentrations of all
three species if the value of the Keq
is 145.
A(g)
+
B(g) 
[I]
0.0
0.0
[C]
+x
+x
[E]
x
x
C(g)
0.350 mol/L
-x
0.350 mol/L - x
Keq =
[C]
[A] [B]
145 =
(0.350 mol/L – x)
(x)(x)
145 =
0.350 mol/L – x
x2
145x2
=
0.350 mol/L – x
145x2 + x - 0.350 mol/L = 0

now we apply the quadratic equation:
145x2 + x - 0.350 mol/L = 0
a
b
c
x
= -b ± √ b2 – 4ac
2a
x
= 0.0458 mol/L
A(g) + B(g) 
[I]
0.0
0.0
[C]
+x
+x
[E]
x
x


C(g)
0.350 mol/L
-x
0.350 mol/L - x
[A] = [B] = 0.0458 mol/L
[C] = 0.350 M – x = 0.350 M – 0.0458 M
= 0.304 mol/L
Calculations using the Ksp


A saturated solution of CuBr has a concentration of
2.0 x 10-4 mol/L. Calculate the Ksp.
Write the dissociation equation and determine the
concentration of each ion in solution:
CuBr(s)

Cu1+
(aq)
+
2.0 x 10-4 mol/L 1(2.0 x 10-4 mol/L)
= 2.0 x 10-4 mol/L
Br1-
(aq)
1(2.0 x 10-4 mol/L)
= 2.0 x 10-4 mol/L
Write the Ksp expression, insert the numbers and
solve:
Ksp
= [Cu1+][Br1-]
= (2.0 x 10-4 mol/L)( 2.0 x 10-4 mol/L)
= 4.0 x 10-8

Calculate the Ksp for Bi2S3, which has a solubility
of 1.36 x 10-15 mol/L at 25°C.
Bi2S3
(s)
1.36 x 10-15 mol/L


2 Bi3+(aq)
+
3S
2-
(aq)
2(1.36 x 10-15 mol/L) 3(1.36 x 10-15 mol/L)
= 2.72 x 10-15 mol/L = 4.08 x 10-15 mol/L
Ksp = [Bi3+]2[S2-]3
= (2.72 x 10-15 mol/L)2(4.08 x 10-15 mol/L)3
= 5.02 x 10-73
The Ksp value for Cu(IO3)2 is 1.4 x 10-7 at 25°C.
Calculate its solubility at this temperature.

Like the ICE box, express concentrations in terms
of x:
Cu(IO3)2
(s)

x
Cu2+(aq) +
1(x)
Ksp
2 IO31-(aq)
2(x)
= [Cu2+][IO31-]2
= (x)(2x)2
1.4 x 10-7

x
=
= 4x3
3√1.4
x 10-7
4
= 3.3 x 10-3 mol/L
Predicting a Precipitate


Ksp values can also be used to predict
if mixing solutions will produce a
precipitate.
e.g.
• A solution is prepared by adding 750.0
mL of 4.00 x 10-3 mol/L Ce(NO3)3 to
300.0 mL of 2.00 x 10-2 mol/L KIO3.
• The Ksp of Ce(IO3)3 is 1.9 x 10-10.
• Will Ce(IO3)3 precipitate from this
solution ?
Step 1. Write a dissociation equation for each
substance and determine the concentration of
each ion in solution.
Ce(NO3)3
4.00 x 10-3 mol/L
KIO3
(aq)

(aq)
Ce3+(aq)
+
1(4.00 x 10-3 mol/L)
= 4.00 x 10-3 mol/L

K1+(aq)
2.00 x 10-2 mol/L 1(2.00 x 10-2 mol/L)
= 2.00 x 10-2 mol/L
3 NO31-(aq)
3(4.00 x 10-3 mol/L)
= 1.20 x 10-2 mol/L
+
IO31-(aq)
1(2.00 x 10-2 mol/L)
= 2.00 x 10-2 mol/L
Step 2. Use the dilution equation to determine the
concentration of each of the significant species. In
this case they are Ce3+(aq) and IO31-(aq)
[Ce3+] =
csvs
vd
=
(4.00 x 10-3 mol/L)(750.0 mL)
750.0 mL + 300.0 mL
= 2.86 x 10-3 mol/L
[IO31-] =
csvs
vd
=
(2.00 x 10-2 mol/L)(300.0 mL)
750.0 mL + 300.0 mL
= 5.71 x 10-3 mol/L
Step 3. Write the Ksp expression for the substance you
are testing and solve for the Ksp using the numbers you
just calculated. This is a trial Ksp

Substance: Ce(IO3)3

Ce(IO3)3 (s)

Ksp

Ce3+(aq) + 3 IO31-(aq)
= [Ce3+][IO31-]3
Trial Ksp = (2.86 x 10-3 mol/L)( 5.71 x 10-3 mol/L)3
= 5.52 x 10-10
Step 4. Compare the trial Ksp with the actual
Ksp from the question:




if the trial Ksp is larger than the actual Ksp, there
is more solute than can dissolve; there will be a
precipitate.
if the trial Ksp is smaller than the actual Ksp,
there is less solute than can dissolve; there will
not be a precipitate.
In this case 5.52 x 10-10 is larger than
1.9 x 10-10
there will be a precipitate