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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: What does this look like?

Far to the Right Reaction has gone to completion: Ex: 2H2 + O2 2H2O

Far to the Left Reaction only occurs slightly: CaO Ca(s) + O2

Heat is needed in order for it to proceed In Between 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.

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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

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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

Equilibrium Concentrations

Formula of Keq Keq

[N2]0= 1.0 M [H2]0 = 1.0 M [NH3]0 = 0 M

[N2]= .921 M [H2] = .763 M [NH3] = .157 M

Keq=[NH3]2/[N2][H2]3 6.02x10-2

[N2]0= 0 M [H2]0 = 0 M [NH3]0 = 1.0 M

[N2]= .399 M [H2] = 1.197 M [NH3] = .203 M

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

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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

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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

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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/PSO2

2PO2

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 <1Lies 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.

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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

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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

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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; Cl2O1.8 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.

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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- Ksp=[ Pb2+][ CrO4

2-]

c. Al(OH)3 Al3+ + 3OH- Ksp= [Al3+][OH-]3

d. ZnCO3 Zn2+ + CO32-

e. Zn(OH)2 Zn2+ + 2OH-

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+ + C2O42-

I 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+ + 3I-

I 0M 0M

C +x +3x

E x 3x

Ksp = x (3x)3

Ksp= 27x4 ; Since x=1.32 x 10-5M

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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+ + PO43-

I 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

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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 SO4

2- is 0.03 mol/L if the Ksp = 2.4 x 10-5?

CaSO4 Ca2+ + SO42-

Need 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.

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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+ + I-

5.3 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___

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∆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

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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

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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

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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) + SO4

2-(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:

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2C2H2 (g) + 5O2 (g) ⇌ 4 CO2 (g) + 2H2O (g) + heat

Stress Shift [C2H2] [O2] [CO2] [H2O] Keq 1. Add C2H2

R -------- Decrease Increase Increase 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:

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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

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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+ + HC6H5O7

2- K2=y

HC6H5O72-⇌ H+ + C6H5O7

3- K3=z

Which of the following expressions gives the equilibrium constant for the reaction shown below?

H3C6H5O7⇌ 3H+ + C6H5O73-

a. 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