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Chemical Equilibrium
AP Chemistry
Ms. Poliner
Any reaction that occurs in a closed vessel will reach equilibrium.
Types of Chemical Equilibrium:
Far to the Right
Far to the Left
In Between
What does this look like?
Reaction has gone to completion:
Ex: 2H2 + O2 2H2O
Reaction only occurs slightly:
CaO Ca(s) + O2
Heat is needed in order for it to proceed
This is “in between” is what we are going
to be studying for this unit.
⇌
Qualitative Description of Chemical Equilibrium:
1.
-Equilibrium Pt.
(For simulation: http://www.chm.davidson.edu/vce/equilibria/BasicConcepts.html)
2.
1
Video Demonstration:
H2 (g) and I2 (g)
H2 (g) +I2 (g) ⇌ 2HI (g)
2HI (g)
Quantitative Description of Chemical Equilibrium:
jA + kB ⇌ lC + mD
1. Law of Mass Action: General description of the equilibrium condition:
K= __[C]l[D]m__
[A]j[B]k
2. Equilibrium Constant: Keq
• [C] represent the _____concentration_______ of the reactants and products when
they are in equilibrium.
• K is a constant, known as the ___________equilibrium________.
Practice Problems:
1. Determine the value of the equilibrium constant for the reaction below if
[N2O4]=1.5x10-3M and [NO2]=.571M.
N2O4(g) ⇌ 2 NO2 (g)
K = (.571M)2/(1.5 x 10-3M)
K= 217
2. For the reaction H2(g) + I2(g) = 2 HI(g), what is the concentration of HI, if the
equilibrium concentrations of H2 and I2 are both .100M? The Kc= 50.3 at 731K
K=[HI]2/.1M x .1M
[HI]=.71M
2
3. For the reaction 2SO2 (g) + O2 (g) ⇌ 2SO3 (g), [SO2]= 2.00M and [SO3]=10.0M.
What is [O2] if Kc= 800.0? K=800= [10M]2/ [O2][2M]
.031M
The equilibrium constant for a specific reaction at a certain temperature remains
____constant________ while the equilibrium concentrations vary based on the initial
concentrations of reactants and products.
Ex. N2 (g) + 3H2 (g) ⇌2NH3 (g)
Initial
Concentrations
[N2]0= 1.0 M
[H2]0 = 1.0 M
[NH3]0 = 0 M
[N2]0= 0 M
[H2]0 = 0 M
[NH3]0 = 1.0 M
Equilibrium
Concentrations
[N2]= .921 M
[H2] = .763 M
[NH3] = .157 M
[N2]= .399 M
[H2] = 1.197 M
[NH3] = .203 M
Formula of Keq
Keq
Keq=[NH3]2/[N2][H2]3
6.02x10-2
Keq=[NH3]2/[N2][H2]3
6.02x10-2
Ex. Calculate the equilibrium constant for the various examples of the Haber process
provided below. Each of the equations uses the same concentrations of reactants and
products and occurs at 127°C:
a.
N2 (g) +
3H2 (g) ⇌ 2NH3 (g)
[NH3]= 2.1 x 10-2 mol/L
[N2] = 8.5 x 10 -1 mol/L
[H2] = 3.1 x 10-3 mol/L
Ka=[NH3]2/[N2][H2]3
b. 2NH3 (g)⇌ N2 (g) +
3H2 (g)
Kb= [N2][H2]3/[NH3]2
Kb=Ka-1
c. ½ N2 (g) + ½ H2 (g) ⇌ NH3 (g) Kc =[NH3]/[N2]1/2[H2]3/2; Kc= Ka1/2
3
Based on this example, what are two conclusions that can be drawn about the
equilibrium constant?
1. The equilibrium expression for a reaction written in the reverse is the reciprocal of that
for the original reaction.
2. When the equation is multiplied by a factor of n, the new equilibrium expression can
be written as Knew=(Korig)n
3. For a reaction made up of two or more steps, the equilibrium constant for the net
reaction is the ____product___ of the equilibrium constants of the individual steps.
Ex. Calculate a value for the equilibrium constant for the reaction:
O2 (g) + O (g) ⇌ O3 (g)
Given that:
NO2 (g) ⇌ NO (g) + O (g)
K= 6.8 x 10-49
O3 (g) + NO (g) ⇌ NO2 (g) + O2 (g)
K= 5.8 x 10-34
NO (g) + O (g) ⇌ NO2 (g)
NO2 (g) + O2 (g) ⇌ O3 (g) + NO (g)
K= 2.54 x 1031
Equilibrium Constants can be expressed in terms of concentrations or partial
pressures:
Kc- Measures of __concentration_ are used in the calculation of this equilibrium constant.
Kp- _partial pressures__are used in the calculation of this equilibrium.
Relationship between Kp and Kc:
Based on the ideal gas law:
Therefore, Kp = Kc(RT)∆n
Ex. At 327°C, the equilibrium concentrations are [CH3OH]= .15M, [CO]= .24M, and
[H2]=1.1M for the reaction
CH3OH (g) ⇌ CO (g) + 2H2 (g)
Calculate Kp at this temperature.
Kc= [1.1]2[.24]/[.15]
Kc= 1.936
4
Kp= 1.936(.0821 x 600)2
Kp=4,697.8
The reactions studied thus far are examples of __homogeneous__ equilibria, or situations
that only involve one phase, the gas phase. When there is more than one phase, this is
known as ____heterogeneous_____ equilibria. When determining the equilibrium
constant in these heterogeneous reactions, pure ____liquid____ and ____gas__ are not
included.
Practice Problems:
1. Write expressions for K and Kp for the following reactions.
a. 2NH3 (g) + CO2 (g) ⇌ N2CH4O (s) + H2O (g)
Kc= [H2O]/[NH3]2[CO2]; Kp= PH2O/PNH32PCO2
b. CuO (s) + H2 (g) ⇌ Cu (l) + H2O (g)
Kc= [H2O]/[H2]; Kp= PH2O/PH2
2. Given the following equilibrium constants at 427°C,
Na2O (s) ⇌ 2Na (l) + ½ O2 (g)
K1= 2 x 10-25
NaO (g) ⇌ Na (l) + ½ O2 (g)
K2= 2 x 10-5
Na2O2(s) ⇌ 2Na (l) + O2 (g)
K3= 5 x 10-29
NaO2 (s) ⇌ Na (l) + O2 (g)
K4= 3 x 10-14
Determine the values for the equilibrium constant for the following reactions.
a. Na2O (s) + ½ O2 (g) ⇌ Na2O2 (s)
Na2O (s) ⇌ 2Na (l) + ½ O2 (g)
K1= 2 x 10-25
2Na (l) + O2 (g) ⇌Na2O2(s)
K3= 1/ 5 x 10-29
____________________________________________________
Na2O (s) + ½ O2 (g) ⇌ Na2O2 (s)
K= 4000
5
b. NaO (g) + Na2O (s) ⇌ Na2O2 (s) + Na (l)
Na2O (s) ⇌ 2Na (l) + ½ O2 (g)
K1= 2 x 10-25
NaO (g) ⇌ Na (l) + ½ O2 (g)
K2= 2 x 10-5
2Na (l) + O2 (g) ⇌Na2O2(s)
K3= 1/ 5 x 10-29
____________________________________________________
NaO (g) + Na2O (s) ⇌ Na2O2 (s) + Na (l) K=2x108
c. 2NaO (g) ⇌ Na2O2 (s)
K=8 x 1018
3. What is the equilibrium constant, Keq, for the reaction below if the following partial
pressures were measured at equilibrium: PSO2 = 0.55 atm, PO2 = 0.30 atm, and PSO3 = 0.15
atm.
2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
Kp = PSO32/PSO22PO2
K= .248
When a reaction is at equilibrium, what does the value of K tell us?
When K >1 Equilibrium lies to the right; products predominate. If large enough,
practically go to completion.
When K <1Lies to the left; reactants predominate.
But what if we’re not at equilibrium? Keq versus Q:
Q, or the ____reaction quotient___ is calculated by using the law of mass action, but
using the initial concentrations rather than the equilibrium concentrations.
Q is used to determine the___direction___the system will shift in order to reach
equilibrium:
1. Q=K At equilibrium.
2. Q > K Moves to the left.
3. Q<K Moves to the right.
6
Ex. For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 x 10-2L2/mol2.
Predict the direction in which the system will shift to reach equilibrium in each of the
following cases.
a. [NH3]0= 1.0 x 10-3M; [N2]0= 1.0x10-5; [H2]0= 2.0 x 10-3 M
N2 (g) +
3H2 (g) ⇌ 2NH3 (g)
Q= [NH3]2/[N2][H2]3
Q= 1.3 x 107- Products must decrease shift to the left.
b. [NH3]0= 2.0 x 10-4M; [N2]0= 1.5x10-5 M; [H2]0= 3.54 x 10-1 M
Q=6.01 x 102- You are at equilibrium.
c. [NH3]0= 1.0 x 10-4 M; [N2]0= 5.0 M; [H2]0= 1.0 x 10-2 M
2.0 x 10-3- Shift to the right because Q<K
Up until now we have solved for equilibrium concentrations and partial pressures
when given Keq and Kp. We can also solve for Keq and Kp when given the
equilibrium concentrations and partial pressures. But how can we solve for these
equilibrium values when these equilibrium values are not given?
Let’s use the following example to show how ICE charts can help us solve these
problems:
Ex. H2 (g) + F2 (g) ⇌ HF (g)
Keq= 1.15 x 102
In a particular experiment, 3.000 moles of each component is added to a 1.500-liter flask.
Calculate the equilibrium concentrations of all species.
Step 1: Write the balanced equation for the reaction
H2 (g) +
F2 (g) ⇌
2HF (g)
Step 2: Write the equilibrium expression using the law of mass action.
Keq= [HF]2/[H2][F2]
Step 3: List the initial concentrations: 2M
7
Step 4: Calculate Q and determine the direction of the shift to equilibrium. In this
example, Q=1, shifts to the right.
Step 5: Define the change needed to reach equilibrium and define the equilibrium
concentration by applying changes.
H2
+
F2 ⇌
2HF
I
2M
2M
2M
C
-x
-x
+2x
E
2-x
2-x
2+ 2x
Step 6: Substitute into the equilibrium equation: 1.15 x 102 = (2+2x)2/ (2-x)2
x= 1.528
Step 7:Check that equilibrium concentration gives you your K.
H2=F2= .472 M
HF= 5.056M
1.15 x 102 = (5.056)2/ (.472)(.472)
Practice Problems:
1. At 35°C the equilibrium constant for the following reaction is 1.6 x 10-5 mol/L
2NOCl (g) ⇌ 2NO (g) + Cl2 (g)
If 1.0 mol of NOCl is placed in a 2.0-liter flask, calculate the equilibrium concentrations
of all species.
2NOCl
⇌
2NO
+
Cl2
I
.5M
0
0
C
-2x
+2x
+x
E
.5-2x
2x
x
1.6x10-5= 4x3/ (5-2x)2
x=1.0 x 10-2M
5% Rule Check: To make denominator 5; .01/5<5%
NO .02M; Cl2.01M; NOCl .48M
8
2. At a particular temperature, K= 2.0 x 10-6 mol/L for the reaction
2CO2 (g) ⇌ 2CO (g) + O2 (g)
If 2.0 mol of CO2 is initially placed into a 5.0-L vessel, calculate the equilibrium
concentrations of all species.
CO .0086M; O2.0043M; CO2 .39M
3. At 25°C, K= .090 for the reaction
H2O (g) + Cl2O (g) ⇌ 2HOCL (g)
Calculate the concentrations of all species at equilibrium for each of the following cases.
a. 1.0 g of H2O and 2.0g of Cl2O are mixed in a 1.0-L flask.
H2O (g) + Cl2O (g) ⇌ 2HOCL (g)
I
.056M
.023M
0M
C
-x
-x
+2x
E
.056-x
.023-x
2x
K= [HOCl]2/[H2O][Cl2O]
.090 = (2x)2/ (.056-x)(.023-x)
To solve for x, you can’t use the 5% rule and must use the quadratic formula
.090 (.056-x)(.023-x) = 4x2
.090(.00129-.079x + x2)=4x2
1.61x10-4 - .00711x + .090x2=4x2
-3.91x2-.0071x+ 1.61x10-4=0
Now plug into quadratic formula: a=-3.91; b=-.0071; c=1.61x10-4
(-b±√b2-4ac)/2a: x cannot be – so must use the solution that gives you a positive number.
Therefore, x= .0056M.
HOCl 9.2 x 10-3; Cl2O1.8 x 10-2; H2O 5.1 x 10-2M (A tiny bit off because of
rounding)
b. 1.0 mol of pure HOCl is placed in a 2.0-L flask.
Don’t need to complete.
9
Equilibrium Constant of Solubility:
The __solubility product (Ksp)__ is the equilibrium constant between an ionic solid and
its saturated solution, while the ___solubility__ is the concentration of products or
reactants that must be dissolved in order to make a saturated solution (g/L or M).
Ex. Write the Ksp expression for these compounds.
a.
AgI Ag+ + I-
Ksp=[Ag+][I-]
b.
PbCrO4 Pb2+ + CrO42-
c.
Al(OH)3 Al3+ + 3OH-
d.
ZnCO3 Zn2+ + CO32-
e.
Zn(OH)2 Zn2+ + 2OH-
Ksp=[ Pb2+][ CrO42-]
Ksp= [Al3+][OH-]3
Practice Problems:
1. Use the following data to calculate the Ksp value for each solid:
a. The solubility of CaC2O4 is 6.1x10-3 g/L
CaC2O4 Ca2+ + C2O42I
OM
OM
C
+x
+x
E
x
x
Convert to grams: 6.1 x 10-3 g/L x (1mol/ 128g)
Solubility or x= 4.8 x 10-5M
Ksp = (4.8 x 10-5)2
Ksp = 2.3 x 10-9
b. The molar solubility of BiI3 is 1.32 x 10-5mol/L
BiI3  Bi3+ + 3II
0M
0M
C
+x
+3x
E
x
3x
Ksp = x (3x)3
Ksp= 27x4 ; Since x=1.32 x 10-5M
10
Ksp= 8.2 x 10-19
2. Calculate the solubility of each of the following compounds in moles per liter and
grams per liter.
a. Ag3PO4, Ksp= 1.8 x 10-18
Ag3PO4 3Ag+ + PO43I
0M
0M
C
+3x
+x
E
3x
x
Ksp = (3x)3x
1.8 x 10-18 = 27x4
x= 1.61 x 10-5M; Now convert to g/L 6.7 x 10-3 g/L
b. CaCO3, Ksp= 8.7 x 10-9
c. Hg2Cl2, Ksp= 1.1 x 10-18
3. For each of the following pairs of solids, determine which solid has the smallest molar
solubility.
a. CaF2 (s), Ksp= 4.0 x 10-11 or BaF2 (s), Ksp=2.4 x 10-5
11
b. Ca3(PO4)2(s), Ksp= 1.3 x 10-23 or FePO4 (s), Ksp=1.0x10-22
As before, the Q can tell us how the reaction will proceed in order to reach equilibrium.
In this case, however:
If Q> Ksp  A precipitate will form because the reaction will shift to the left.
If Q= Ksp At equilibrium
If Q< Ksp No precipitate will form.
Ex. Will a precipitate of CaSO4 form in a solution if the Ca2+ is 0.0025 mol/L and the
SO42- is 0.03 mol/L if the Ksp = 2.4 x 10-5?
CaSO4 Ca2+ + SO42Need to determine Q: Q= [Ca2+][SO42-]
Q= (.0025M)(.03M)
Q=7.5 x 10-5
Q>K  A precipitate forms.
Will a precipitate of lead (II) chloride form if 50 mL of 0.1 M lead (II) nitrate is added to
20 mL of 0.04 M sodium chloride solution?
We can further expand this concept of equilibrium constant of solubility by considering
selective precipitation, or the separation of metal ions in aqueous solution by using a
reagent whose anions forms a precipitate with only one of the metals ions in the mixture.
12
Ex. A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a source of I- is added
to this solution gradually, will PbI2 (Ksp= 1.4 x 10-8) or CuI (Ksp= 5.3x 10-12) precipitate
first? What concentrations of I- are necessary to begin precipitation?
PbI2 2Cl- + Pb2+
1.4 x 10-8 = (2.0 x 10-3) [I-]2
[I-]= 2.6 x 10-3 M At any concentration above this equilibrium concentration of I-, Q
will be larger than Ksp so a precipitate will form.
CuI Cu+ + I5.3 x 10-12 = (1.0 x 10-4) [I-]
[I-]= 5.3 x 10-8M At any concentration above this equilibrium concentration of I-, Q
will be larger than Ksp so a precipitate will form.
If we are slowly adding I-, then CuI will precipitate first because it requires a lower
concentration of I- than Pb2+.
Practice Problems:
1. If 100L of .03M Pb(NO3)2 is mixed with 200L of .09M KCl, does a precipitate (PbCl2)
form? (Ksp = 1.6 X 10-5).
Since the concentrations change as soon as you combine them, must determine the
number of moles of these ions and then calculate the new concentration.
Pb2+ .03M x .1L= .003mol/.3L .01M
Cl- .09M x .2= .018mol/.3L .06M
Q= [Pb2+][Cl-]2
Q= 3.6 x 10-5 ; Q>Ksp; Therefore a precipitate forms.
2. If 1.4M of 10L Mn2+ and .09M of 30L Fe2+ is in a solution, and S2- is slowly added
which will precipitate first? MnS has a Ksp= 2.3 X 10-13 while FeS has a Ksp=3.7 X 10-19.
…And it doesn’t end there!!! We can even use the equilibrium constant when describing
___Free Energy____.
• Remember that: ∆H is the ____enthalpy___
13
∆S is the ____entropy___
and ∆G is the ___Gibbs Free Energy___ which is the
___spontaneity of the reaction______
• Where _∆G°_ is the standard free energy change, or the change in free energy that
occurs if the ____reactants___ in their standard states are converted to the
___products___ in their standard states.
• ∆G is the ____nonstandard__ condition.
• Relationship between ∆G° and ∆G is established by the equilibrium constant:
___∆G= ∆G°+ RTlnQ__
Ex. For the ammonia synthesis reaction, N2 (g) + 3H2(g) ⇌ 2NHs(g), where ∆G°=-33.3kJ
per mole of N2 consumed at 25°C. Calculate ∆G for a reaction that has the following
mixture of reactants:
PNH3= 1.00 atm, PN2=1.47 atm, PH2= 1.00 x 10-2 atm
Q= (1)2/ (1.47)(.01)3
Q= 680, 272.1
∆G= -33 + (.008314)(298)ln(680,272.1)
• At equilibrium, ∆G=__O__and Q=__K__
Therefore, with some math, we get:
∆G= ∆G° + RTlnQ
0= ∆G° + RTlnK
∆G° = -RTlinK
14
While the _____reaction quotient______ can be calculated in order to determine the
direction the initial reaction will proceed in order to reach equilibrium, and the value of
the ___equilibrium constant____ tells us where the equilibrium lies, additional factors
can control the __position__ of a chemical equilibrium.
Le Châtelier’s Principle: “If a change in conditions is imposed on a system at
equilibrium, the equilibrium position will shift in a direction that tends to reduce that
change in conditions”
Types of Change:
1. Change in Concentration:
If _______increase_______ concentration, the system shifts away from the added
component.
If _____decrease____ concentration, the system shifts toward the removed component.
Ex. Suppose there is an equilibrium position described by the concentrations:
N2+ 3H2 ⇌ 2NH3
[N2]= .399M; [H2]= 1.197M; [NH3]= .202M
What will happen if 1.000 M of N2 is added to the system at constant volume?
Will shift to the right.
2. Change in Pressure or Volume:
Only for Gases!!!!
If _____increase___pressure, the equilibrium position will shift toward the side of the
reaction involving the smaller number of gaseous molecules (less pressure).
If ____decrease______ pressure, the equilibrium position will shift toward the side of the
reaction involving the larger number of gaseous molecules (more pressure).
If _increase_ pressure, _decrease_ volume while if _decrease__ pressure, _increase_
volume.
Ex.
a. 3 H2(g) + N2(g) ⇌ 2 NH3(g)
Stress: Increase in Pressure Shift to the right
15
Stress: Increase Volume Shift to the left.
b. 2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g)
Stress: Decrease in Pressure Shift to left
Stress: Decrease Volume Shift to right
3. Change in Temperature:
If the reaction is exothermic, or _-∆H__, then the reaction can be written as:
Reactants  Products + ∆H
Therefore, if __increase____ temperature, then will shift to favor the reactants to use up
the extra heat. If ___decrease____ temperature, then will shift to favor the products, to
produce more heat.
If the reaction is endothermic, or ___∆H ___, then the reaction can be written as:
Reactants + ∆H  Products
Therefore, if __increase___temperature, then will shift to favor the products to use up the
extra heat. If ____decrease____temperature, then will shift to favor the reactants, to
produce more heat.
Ex.
a. N2O4 (g ) ⇌ 2 NO2 (g)
ΔH° = +58.0 kJ
If increase temperature of the reaction, then the equilibrium will shift to favor the
___products____.
b. In each of the following equilibria, would you increase or decrease the temperature to
force the reaction in the forward direction?
H2 (g) + CO2 (g) ⇌ H2O(g) +CO(g)
ΔH° = +41.0 kJ
Increase
16
2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g)
ΔH° = -198 kJ
Decrease
4. Common Ion Effect:
The shift in equilibrium position that occurs because of the addition of an __ion_ already
involved in the equilibrium reaction. Its addition will result in a shift away from the side
where it was added.
Ex. BaSO4 (s) ⇌ Ba2+(aq) + SO42-(aq). If add Na2SO4, what ion will be added? As a
result, equilibrium will shift to the _left_, and the solubility of the reaction will
__decrease___.
5. Addition of a catalyst:
Increase the rate at which equilibrium is achieved. Does not change equilibrium
composition.
Ex. Based on Le Châtelier’s Principle, complete the following chart for the reaction
below:
17
2C2H2 (g) + 5O2 (g) ⇌ 4 CO2 (g) + 2H2O (g) + heat
Stress
1. Add C2H2
Shift
R
[C2H2]
--------
[O2]
Decrease
[CO2]
Increase
[H2O]
Increase
Keq
Same
2. Add O2
R
Decrease
---------
Increase
Increase
Same
3. Add CO2
L
Increase
Increase
----------
Decrease
Same
4. Add H2O
L
Increase
Increase
Decrease
----------
Same
5. Remove
C2H2
L
---------
Increase
Decrease
Decrease
Same
6. Remove O2
L
Increase
---------
Decrease
Decrease
Same
7. Remove
CO2
R
Decrease
Decrease
-----------
Increase
Same
8. Remove
H2O
R
Decrease
Decrease
Increase
---------
Same
9. Increase
Temperature
L
Increase
Increase
Decrease
Decrease
Decrease
10. Decrease
Temperature
R
Decrease
Decrease
Increase
Increase
Increase
11. Increase
Pressure
R
Decrease
Decrease
Increase
Increase
Same
12. Increase
Volume
L
Increase
Increase
Decrease
Decrease
Same
Review Questions:
18
1. For a particular salt, the solution process is endothermic. As the temperature at which
the salt is dissolved increases, which of the following will occur?
a. Ksp will increase, and the salt will become more soluble.
b. Ksp will decrease, and the salt will become more soluble.
c. Ksp will increase, and the salt will become less soluble.
d. Ksp will not change, and the salt will become more soluble.
2. 2HI (g) + Cl2⇌ 2HCl (g) + I2 (g) + energy
A gaseous reaction occurs and comes to equilibrium as shown above. Which of the
following changes to the system will serve to increase the number of moles of I2 present
at equilibrium?
a. Increasing the volume at constant temperature.
b. Decreasing the volume at constant temperature.
c. Adding a mole of inert gas at constant volume.
d. Increasing the temperature at constant volume.
e. Decreasing the temperature at constant volume.
3. A sealed isothermal container initially contained 2 moles of CO gas and 3 moles of H2
gas. The following reversible reaction occured:
CO (g) + 2H2 (g) ⇌ CH3OH (g)
At equilibrium, there was 1 mole of CH3OH in the container. What was the total number
of moles of gas present in the container at equilibrium?
a. 1
b. 2
c. 3
d. 4
e. 5
19
4. 4NH3 (g) + 3O2 (g) ⇌2N2 (g) + 6H2O (g) + energy
Which of the following changes to the system at equilibrium shown above would cause
the concentration of H2O to increase?
a. The volume of the system was decreased at constant temperature.
b. The temperature of the system was increased at constant volume.
c. NH3 was removed from the system.
d. N2 was removed from the system.
e. O2 was removed from the system.
5. A sample of solid potassium nitrate is placed in water. The solid potassium nitrate
comes to equilibrium with its dissolved ions by the endothermic process shown below.
KNO3 (s) + energy ⇌ K+ (aq) + NO3- (aq)
Which of the following changes to the system would increase the concentration of K+
ions at equilibrium?
a. The volume of the solution is increased.
b. The volume of the solution is decreased.
c. Additional solid KNO3 is added to the solution.
d. The temperature of the solution is increased.
e. The temperature of the solution is decreased.
6. Citric acid, H3C6H5O7, can give up 3 hydrogen ions in solution. The dissociation
reactions are as follows:
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H3C6H5O7⇌ H+ + H2C6H5O7-
K1 = x
H2C6H5O7-⇌ H+ + HC6H5O72-
K2=y
HC6H5O72-⇌ H+ + C6H5O73-
K3=z
Which of the following expressions gives the equilibrium constant for the reaction shown
below?
H3C6H5O7⇌ 3H+ + C6H5O73a. xyz
b. xy/z
c. x/yz
d. z/xy
e. 1/xyz
7. BaF2 (s) ⇌ Ba2+ (aq) + 2F- (aq)
The value of the solubility product, Ksp, for the reaction above is 1.0 x 10-6 at 25°C.
a. Write the Ksp expression of BaF2.
4 x 10-18
b. What is the concentration of F- ions in a saturated solution of BaF2 at 25°C?
2.0 x 10-6 M
c. 500 millileters of a .0060-molar NaF solution is added to 400 ml of a .0060-molar
Ba(NO3)2 solution. Will there be a precipitate?
2.37 x 10-8= Q; Since Q<Ksp, no precipitate will form.
d. What is the value of ∆G° for the dissociation of BaF2 at 25°
∆G°= -.008314(298)ln (2.37x 10-8)  -43.5 kJ/mol
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