Equilibrium
Equilibrium
•
•
•
•
Products can not escape reaction vessel
Reaction becomes reversible
Rate of forward equals rate of reverse
Concentrations are constant
Equilibrium Constant
• Equilibrium constants, Keq, are an important concept in chemistry.
Equilibrium constants tell us the progress of a chemical reaction by
relating the concentrations of both reactants and products at
equilibrium to a numerical constant. If the equilibrium constant is
greater than one, then there is more product formed at equilibrium
than reactants remaining. If the equilibrium constant is less than
one, there is less product than reactant. A good example of this
concept is a special type of equilibrium constant, the ionization or
acidity constant, Ki or Ka. It is used to compare the strengths of
acids. The larger the ionization constant, the more ions and hence,
the stronger the acid. Consider the ionization constant for acetic
acid at 1.8 x 10-5. The magnitude of this Ki tells us that there are
very few ions in a 1 M solution of acetic acid, about 0.4% at room
temperature.
HCH3COO(aq)
H + (aq) + CH3COO- (aq), Ki = 1.8 x 10-5
Ca(OH)2 (s)
Ca 2+ (aq) + 2OH - (aq), K sp = 5.5 x 10-6
• Another type of equilibrium constant is solubility product
constant, Ksp. As the name implies, solubility product
constants are used to compare the solubilities of similar
substances. Again, the larger the constant, the greater
to solubility at a given temperature. If we compare the
Ksp of the Group II hydroxides in the adjacent table, we
will note that the solubility of Mg(OH)2 is considerably
less than the other Group II hydroxides. The solubility
product constant can also be used to determine the
actual ion concentration in a solution at a given
temperature. Note that the Ksp of calcium hydroxide is
1.3 x 10-6 at 25 C. Using this value, we can predict the
pH of a saturated calcium hydroxide solution to be 11.8
at 25 C.
How is the equilibrium constant determined? Given the
expression for a general reaction where the brackets “[]”
represent the concentration of aqueous solutions in
moles/liter, M, we can write an equilibrium expression:
aA + bB
cC + dD
C  D 
product concentrations

=
=
reactant concentrations  A a  Bb
c
equilibrium expression, K eq
d
Rules for expression
• There are some rules to following when writing
an equilibrium expression and calculating its
numerical value.
• 1. The concentrations of the products are
always in the numerator and the concentrations
of the reactants are always in the denominator.
• 2. The concentrations are raised to the power
of the coefficients in the balanced equation of
the reaction.
Rules
• 3. Pure solids and liquids have a constant
concentration and are not included in the
equilibrium expression.
• 4. Equilibrium constants are temperature
dependent. Therefore, equilibrium
constants are always stated at a given
temperature, often 25 C.
2NaF(aq) + Mg(OH)2 (s)
MgF2 (s) + 2NaOH(aq)
• In one experiment, a researcher mixed 25
ml of 0.20 M sodium fluoride solution with
some solid magnesium hydroxide at
25EC. After two days, it was found that
0.40 ml of 0.10 M HCl was need to
neutralize 0.85 ml of the solution in the
flask. Assuming that the mixture is at
equilibrium, we can determine the
equilibrium constant from the reaction:
Since one mole of NaOH reacts with one mole of HCl by means of the
relationship, NaOH(aq) + HCl(aq) = NaCl + H2O, then we can easily
calculate the equilibrium concentration of sodium hydroxide. Where
Vacid x M acid = Vbase x M base
0.40 ml HCl x 0.10 M HCl = 0.85 ml x Molarity of the NaOH solution
Molarity of the NaOH solution = 0.047 M NaOH
2NaF(aq) + Mg(OH)2 (s)
Initial
0.2 M
Change
-x
Equilibrium 0.2 - x
MgF2 (s) + 2NaOH(aq)
Constant Constant 0
+x
+x
• Now that we know the concentration of the
NaOH produced, we can calculate the
concentration of the fluoride ions remaining at
equilibrium. If “x” represents to concentration of
NaOH at equilibrium, then 0.2 - x would
represent the equilibrium concentration of
sodium fluoride. The [F-] is 0.2 - x because we
started with a 0.2 M sodium fluoride solution.
We can make this statement as one mole of
NaOH is formed when each mole of NaF reacts.
The ratio is actually 2:2 but simplifies to 1:1.
• Since the equilibrium concentration of
NaOH is 0.047 M, then the equilibrium
concentration of sodium fluoride would
calculate to be 0.2 M - 0.047 M or 0.153
M.
• Now that we know the concentrations of both the
reactants and products, we are ready to
calculate the value for the equilibrium constant,
Keq, for this reaction. It is not necessary to
know the concentrations for magnesium fluoride,
MgF2, and magnesium hydroxide, Mg(OH)2, as
they are solids and deleted from the equilibrium
expression. Therefore, the equilibrium
expression and constant become
 NaOH   MgF2  , but simplifies to
product of the products
=
=
product of the reactants  NaF2  Mg(OH) 2 
2
equilibrium constant, K eq
equilibrium constant, K eq =
 NaOH 2
 NaF2
=
 0.047 M 2
 0.153 M 2
= 0.094
• What is the expected equilibrium constant for
this reaction? The theoretical equilibrium
constant for the reaction can be determined from
the change in the free energy, ΔG reaction. It is
calculated from the free energies of formation,
ΔGfo, for the substances reacting and produced
in the balanced chemical equation. Given the
free energy values in Table #1, we can find the
free energy change for the overall reaction.
2NaF(aq) + Mg(OH) 2 (s)
MgF2 (s) + 2NaOH(aq)
G oreaction =  nG fo products -  nG fo reactants
G oreaction =  2 mol(G fo NaOH(aq)) + (G fo MgF2 (s))  - 2 mol(G fo NaF(aq)) + (G fo Mg(OH) 2 (s)) 

 

G oreaction =  2 mol(-419.2 kJ) + 1 mol (-1071.1 kJ)
 -  2 mol(-540.7 kJ) + 1 mol(-833.9 kJ)
G oreaction = (-1909.5 kJ) - (-1915.3 kJ) = +5.8 kJ
The equilibrium constant for the reaction can now be
determined from the change in free energy of +5.8 kJ
(+5800 joules) according to the relationship:
G oreaction = -RTlnK eq , or lnK eq
ln K eq 
G oreaction
=
 RT
+5800 joules
-(8.31
joules
)(298K)
mol K
K eq = e-2.3 = 0.096
 -2.3
The opposite relationship is also true. That is,
once the equilibrium constant is determined, the
change in free energy for the reaction can be
found. In our example, the equilibrium constant
was found to be 0.094. Therefore the change in
free energy of the reaction is also +5800 joules or
+5.8 kJ.
G oreaction = -RTlnK eq
G oreaction = -8.31
joules
mol K
x 298K x ln 0.094 = +5800 joules (+5.8 kJ)
Another useful quantity is the enthalpy or heat of the reaction. The
enthalpy or heat of the reaction is calculated in the same fashion as the
change in the free energy, by finding the difference in the heats of
formation between the reactants and products as given on Table #1.
2NaF(aq) + Mg(OH) 2 (s)
MgF2 (s) + 2NaOH(aq)
H oreaction =  nH fo products -  nH fo reactants
H oreaction =  2 mol(H fo NaOH(aq)) + (H fo MgF2 (s))  - 2 mol(H fo NaF(aq)) + (H fo Mg(OH)2 (s)) 

 

H oreaction =  2 mol(-469.2 kJ) + 1 mol (-1124.2 kJ)
 -  2 mol(-572.7 kJ) + 1 mol(-924.7 kJ)
H oreaction = (-2062.6 kJ) - (-2070.1 kJ) = +7.5 kJ
• What does the enthalpy or heat of the reaction tell us?
The energy change in a reaction tells us whether or not
the reaction is endothermic or exothermic. An
endothermic reaction has a positive ΔH and requires
energy for the reaction to proceed. Since endothermic
reactions required energy, they are favored by an
increase in the temperature of the reaction. If ΔH is
negative, the reaction is exothermic. Exothermic
reaction liberate heat as the reaction proceeds.
Exothermic reactions are favored by a lower temperature
than an exothermic reaction. In this case, ΔH is positive.
Therefore, the reaction should be heated for it to
proceed at a reasonable rate.
Indirect Method
Mg 2+ ( aq) + 2OH - ( aq), K sp = 5.6 x 10-11
Mg(OH) 2 ( s)
and
Mg 2+ (aq) + 2F- (aq), K sp = 5.8 x 10-10
MgF2 ( s)
Mg(OH) 2 ( s)
2+
-
Mg (aq) + 2F (aq)
2F- (aq) + Mg(OH) 2 ( s)
Mg 2+ (aq) + 2OH - ( aq), K1 = 5.6 x 10-11
-10
1
MgF2 ( s), K 2 = (5.8 x 10 ) = 1.7 x 109
MgF2 ( s) + 2OH - ( aq), K new = K1 x K 2 = 0.095
Lab Procedure
• Cut the very top of a jumbo polyethylene transfer pipet
and pack it with about 2 cm (1") of cotton. This can be
achieved by pushing small pieces of cotton into the
bottom of pipet with a glass rod. Add dry calcium
carbonate to the pipet in small portions, tapping the desk
to pack the solid after each addition. When the level of
solid has reached the bottom of the pipet bulb, insert the
pipet into a small test tube. Wrap the open area
between the test tube and pipet with plastic wrap to hold
the pipet in place and inhibit the exchange of air with the
filtrate. Fill the pipet bulb with 0.1 M NaF. Cover the top
of the pipet with a small piece of plastic wrap and place
the reaction system your lab drawer or another assigned
area until the next laboratory period
Procedure
• Remove the pipet filter from the test tube containing the equilibrium
mixture from your lab drawer or designated area. While setting up
the experiment, measure the temperature of the solution in the test
tube.
• If necessary, attach tip extenders to both hypodermic syringes. It is
important to rinse and prepare the syringes as directed by your
instructor. Be certain that you fill the syringe designated for acids
with standardized 0.1 M hydrochloric acid solution and fill the “base”
syringe with the filtrate. Note that both syringes are color coded.
Record the initial volume of the solution in both syringes to the
nearest 0.01 ml in the space provided on your data table. Partly fill
a polyethylene transfer pipet with bromocresol green/methyl red
solution, the acid-base indicator in this experiment.
Procedure
• Add roughly 0.80 ml - 0.85 ml of the equilibrium mixture
to a small beaker or plastic cup followed by one drop of
the bromocresol green/methyl red indicator. Carefully
add standardized 0.1 M HCl dropwise with constant with
swirling until one drop of acid just turns the solution
orange. Always wait between drops near the end point
as the reaction slows. You may inadvertently add too
much acid and miss the endpoint. The solution will turn
from blue to red at the endpoint. When you are satisfied
that the titration is complete, record the final volume of
the solutions in both syringes to the nearest 0.01 ml in
the space provided on your data table .
Procedure
• Refill both the syringes with their respective
solutions, rinse the beaker or plastic cup
throughly with deionized or distilled water, and
repeat the experiment as time permits. When
you have completed the required number of
trials, throughly rinse all glassware and the
syringes. Return all common items to their
storage area, pour the unused solutions into the
sink, and flush with plenty of water.
Write the net ionic equation for the reaction studied in this experiment.
-
2F (aq) + CaCO3 (s)
CaF2 (s) +
2CO3 (aq)
Calculate the concentration (molarity) of the sodium carbonate
solution produced from the volumes of both solutions reacted and the
concentration of the hydrochloric acid, 0.10 M HCl, titrated.
1 mol CO320.61ml x 0.10 M HCl = 0.061mmol HCl x
= 0.030 mmol CO322 mol HCl
0.030 mmol CO32= 0.033M
0.92 ml
According the balanced equation for this reaction, what is the
relationship between the change in the fluoride ion concentration and
the change in the carbonate ion concentration?
-
Δ[F ] =
22Δ[CO3 ]
2F- (aq) + CaCO3 (s)
CaF2 (s) + CO32- (aq)
Initial
.1M
Constant Constant 0
Change
-2x
+x
Equilibrium
.1M-2x
+x
-
[F ] = 0.1 -
22[CO3 ]
K eq
K eq
= 0.1 M - 2(0.033M) = 0.034 M
[CO32- ]
=
[F- ]2
[CO32- ] (0.033 M)
=
=
= 28
- 2
2
[F ]
(0.034 M)
G oreaction = -RTlnK eq
G oreaction = -8.31
joules
mol K
x 298K x ln 29 = - 8300 joules (-8.3 kJ)
2F- (aq) + CaCO3 (s)
CaF2 (s) + CO32- (aq)
G oreaction =  nG fo products -  nG fo reactants
G oreaction = (G of CO32- (aq)) + (G of CaF2 (s))  -  2 mol(G fo F- (aq)) + (G fo CaCO3 (s)) 

 

G oreaction =  (-528.10 kJ) + 1 mol (-1161.90kJ)  -  2 mol(-276.48 kJ) + 1 mol(-1128.76 kJ) 
G oreaction = (-1690.00 kJ) - (-1681.72 kJ) = = -8.28 kJ
Calculate the theoretical equilibrium
constant from the free energy of the reaction
determined in part (a) above
G oreaction = -RTlnK eq
G oreaction = -8.28 kJ = -8280 J = -8.31
lnK eq =
joules
mol K
-8280 J
-8.31
joules
mol K
x 298K
K = 28.3
x 298K x ln K eq
= 3.34
Using the theoretical equilibrium constant,
calculate the expected concentrations of each
reactant and product.
2F- (aq) + CaCO3 (s)
CaF2 (s) + CO32- (aq)
Initial
.1M
Constant Constant 0
Change
(-2x)
+x
Equilibrium
.1M-2x
+x
[CO32- ]
K eq =
= 28.3
- 2
[F ]
[CO32- ] = 0.033 M, [F- ] = 0.1 - (0.066) = 0.034 M
The solubility product of CaF2 is 1.6 x 10-10 and that of CaCO3 is 4.5 x
10-9 at 25 C. Given that information, calculate the equilibrium constant
for the reaction studied in this experiment from the solubility product
data. How does this value compare with the value obtained from the
thermodynamic and your empirical data?
CaCO3 (s)
Ca 2+ (aq) + 2F- (aq)
2F- (aq) + CaCO3 (s)
Ca 2+ (aq) + CO32- (aq), K1 = 4.5 x 10-9
-1
CaF2 (s), K 2 = (1.6 x 10-10 ) = 6.3 x 109
CaF2 (s) + CO32- (aq), K new = K1 x K 2 = 28
Given the standard enthalpies (heat) of formation, Hfo, for both the
reactants and products in the Table #2 below, find the theoretical heat
or enthalpy of the reaction
2F- (aq) + CaCO3 (s)
CaF2 (s) + CO32- (aq)
H oreaction =  nH fo products -  nH fo reactants
H oreaction = (H of CO32- (aq)) + (H of CaF2 (s))  -  2 mol(H of F- (aq)) + (Hof CaCO3 (s)) 

 

H oreaction = 1 mol(-676.26 kJ) + 1 mol (-1214.62 kJ)  -  2 mol(-329.11 kJ) + 1 mol(-1206.87 kJ) 
H oreaction = (-1890.88 kJ) - (-1865.09 kJ) = -25.79 kJ