Chemical Equilibrium 77.14 Free Energy of dissolution

Agenda 2/14/2020

● Unit 7 Free response questions - grade and turn in● 7.14 Free energy of dissolution

○ Including some notes on “Gibbs” free energy○ MC questions○

Unit 7 Equilibrium Free Response Questions.2.For parts of the free-response question that require calculations, clearly show the method used and the steps involved in arriving at your answers. You must show your work to receive credit for your answer. Examples and equations may be included in your answers where appropriate.

PbI2(s)⇄Pb2+(aq)+2I−(aq) Ksp=7×10−9

The dissolution of PbI2(s)is represented above.

(a) Write a mathematical expression that can be used to determine the value of S, the molar solubility of PbI2(s) (Do not do any numerical calculations.)

Let S be molar solubility of PbI2,

Unit 7 Equilibrium Free Response Questions.(a) Write a mathematical expression that can be used to determine the value of S, the molar solubility of PbI2(s) (Do not do any numerical calculations.)

Let S be molar solubility of PbI2

Ksp = 7 x 10-9 = [Pb2+] [ I- ]2 =S(2S)2 = 4 S3

4 S3= 7 x 10-9

S = 7 x 10-9

4

3

1 point

(b) If PbI2(s) is dissolved in 1.0M NaI(aq), is the maximum possible concentration of Pb2+(aq) in the solution greater than, less than, or equal to the concentration of Pb 2+(aq) in the solution in part ( a ) ? Explain.

(b) If PbI2(s) is dissolved in 1.0M NaI(aq), is the maximum possible concentration of Pb2+(aq) in the solution greater than, less than, or equal to the concentration of Pb 2+(aq) in the solution in part ( a ) ? Explain.

The amount of Pb

2+

(aq) that dissolves in 1.0M NaI solution is

less than the amount of Pb

2+

(aq) that dissolves in the original

solution. The high concentration of iodide ions already in

solution in the 1.0M NaI suppresses (lowers) the dissolution

of PbI

2

(s) because less Pb

2+

can form before the solubility

product [Pb

2+

][I

-

]

2

equals or exceeds Ksp for PbI

2

. 2 points

compound Ksp A table showing Ksp values for several lead compounds is given on the left. A saturated solution of which of the compounds has the greatest molar concentration of Pb2+(aq)? Explain. (Do not do any numerical calculations.)

PbCl2 2 x 10-5

PbI2 7 x 10 -9

Pb(IO3)2 3 x 10-13

PbCl

2

has greatest concentration of Pb

2+

(aq) because it has the

largest Ksp value and since all the compounds produce the

same number of ions, solubility can be predicted based on Ksp

values directly. 1 point

3)Can get concentration of H2 and I2 directly from table.

3)

Qc= 5.56 = [HI]2 = [HI]2

[H2] [I2] (0.030)(0.015)

5.56(0.030)(0.015) = [HI]2

0.002502 = HI = 0.05 M

3)

2 points

HI

H2

I2

After equilibrium was established, the concentration of H2 was 0.020M.

b) On the graph above, carefully draw three curves , one for each of the three gases, starting from the initial points you drew in part a). The curves must show how the concentration of each of the three gases changed as equilibrium was established.

H2 (g) + I2(g) ⇄ 2 HI (g)

I 0.03 0.015 0.05

C -x -x +2x

E 0.03 -x 0.015 -x 0.05 + 2x

Given [H2]eq = 0.02 = 0.03 - x

Therefore x = 0.01 [I2

H2 (g) + I2(g) ⇄ 2 HI (g)

E 0.03 -x 0.015 -x 0.05 + 2x

Given [H2]eq = 0.02 M = 0.03 - x

Therefore x = 0.01

[I2]eq = 0.015 - 0.01 = 0.005 M

[HI]eq = 0.05 + 2(0.01) = 0.07 M

3 points

H2(g), I2(g), and HI(g) are at equilibrium at a different temperature in a different vessel.

(c) When the temperature in the vessel is decreased, does the equilibrium shift to the right, favoring the product, or to the left, favoring the reactants? Justify your answer.

The forward reaction is exothermic according to

the information given above, therefore if

temperature is decreased the system will shift to

the right favoring products in order to restore the

higher temperature, in other words the

exothermic reaction direction will be favored

until a new equilibrium is established.

Equilibrium shifts to the right, favoring products.

1 point

(d) Does the value of Kc increase, decrease, or remain the same when the temperature is decreased? Justify your answer based on the expression for Kc and the concentrations of the product and reactants.

(d) Does the value of Kc increase, decrease, or remain the same when the temperature is decreased? Justify your answer based on the expression for Kc and the concentrations of the product and reactants.

Kc increases when temperature is decreased

Kc(new) = [HI]

2

[I

2

][H

2

]

At new equilibrium [HI] is larger than before

(and [I

2

]and[H

2

] are both less than first

equilibrium system) so the new equilibrium

constant will be larger than the original

equilibrium constant.

2 points

(e) In the following empty box, draw an appropriate number of each type of molecule to represent a possible new equilibrium at the lower temperature.

2 I2 2 H2 10 HI

(e) In the following empty box, draw an appropriate number of each type of molecule to represent a possible new equilibrium at the lower temperature.

2 I2 2 H2 10 HI 1 I2 1 H2 12 HI

2 points

Calculate your percentage (max. Was 14)

Turn in for credit for the work (not the grade - yet)

AP Ranges Score100 5

7574 46261 34342 22726 1

7.14 Free energy of dissolution AP ClassroomYou go there so can read the questions and responses, but this will be a hybrid of listening and notetaking as well as problem solving.

Readily dissolves - indicates “thermodynamically favorable” process

Chapter 16 Spontaneity, Entropy and Free Energy

Thermodynamic favorability

16.1 Thermodynamically favorable Processes and Entropy

When is a process considered to be

Spontaneous? (Thermodynamically favorable).

16.1 Spontaneous Processes and Entropy

Thermodynamically favorable process is -

When it occurs without outside intervention, can be fast or slow.

Thermodynamics can tell us the direction in which a process will occur, but nothing about the rate of reaction.

Free energy ΔG = ΔH - TΔS (see ref sheets)

16.4 Free energy, G

For a process that occurs at constant temperature

ΔG = ΔH - TΔS

And for the process to be “thermodynamically favorable” (likely to happen as far as we can tell from thermodynamics, but can’t predict kinetics, ie. how quickly) ΔG has to be negative.

16.1 Spontaneous Processes and EntropyAll our ionic solids in the hand warmer lab dissolved “spontaneously” (without outside help) - even though some of processes “required energy” from the surroundings - were endothermic processes overall.

Free energy ΔG = ΔH - TΔS Notice temperature decreases so this is an endothermic process with a +ΔH

Free energy ΔG = ΔH - TΔS If ΔG is going to be negative overall, what must ΔS be?

16.1 Spontaneous Processes and EntropyThe characteristic common to all spontaneous (thermodynamically favorable) processes is increase in

Entropy, SThe driving force for a thermodynamically favorable process is an increase in the entropy of the universe.

16.1 Spontaneous Processes and EntropyEntropy, S

A measure of molecular randomness or disorder.

Many more ways for things to not be in the right place and only one way for things to be in the right place.

Disorder more probable than order.

16.1 Spontaneous Processes and Entropy

What are we looking for in the response? Look closely at the molecular representations now.

A. The particle diagram is able to illustrate that the ΔH for the dissolution of NH4NO3 is exothermic because overcoming the attractive forces between the ions requires the absorption of energy.B. The particle diagram is able to illustrate that the ΔH for the dissolution of NH4NO3 is endothermic because forming new interactions between the ions and the water molecules releases energy.C. The particle diagram is able to illustrate that entropy increases when NH4NO3(s) dissolves in water because the ions disperse in solution.D.The particle diagram is able to illustrate that entropy decreases when NH4NO3(s) dissolves in water because many hydrogen bonds can form between the water molecules and the ions.

A. The particle diagram is able to illustrate that the ΔH for the dissolution of NH4NO3 is exothermic because overcoming the attractive forces between the ions requires the absorption of energy.B. The particle diagram is able to illustrate that the ΔH for the dissolution of NH4NO3 is endothermic because forming new interactions between the ions and the water molecules releases energy.C. The particle diagram is able to illustrate that entropy increases when NH4NO3(s) dissolves in water because the ions disperse in solution.D.The particle diagram is able to illustrate that entropy decreases when NH4NO3(s) dissolves in water because many hydrogen bonds can form between the water molecules and the ions.

Qu 2.Reaction Ksp ΔH ΔS

FeCO3(s)⇄Fe2+(aq)+CO32−(aq) 3 x 10-11 < 0 (ie.

negative)>0 ie positive

MnCO3(s)⇄Mn2+(aq)+CO32−(aq) 2 x 10-11 < 0 > 0

The table above lists the equilibrium constants and changes in thermodynamic properties for the dissolution of FeCO3 and MnCO3 at 25°C. The two particle diagrams below represent saturated solutions of each compound at equilibrium.

What can you see? (count)

7 H2O6 Fe2+

6 CO32-

7 H2O6 Fe2+

6 CO32-

7 H2O5 Mn2+

5 CO32-

Which of the following explains which of the properties listed in the table is best represented by the particle diagram?

A. The particle diagrams best represent that ΔH°<0 because the ions from both compounds are solvated by water molecules.B. The particle diagrams best represent that ΔH°<0 because both compounds produce about the same amount of CO3

2− ions from the dissolution.C. The particle diagrams best represent that ΔS°>0 because both compounds produce a very small amount of ions from the dissolution.D. The particle diagrams best represent that the molar solubility is greater for FeCO3 compared to MnCO3.

Which of the following explains which of the properties listed in the table is best represented by the particle diagram?

A. The particle diagrams best represent that ΔH°<0 because the ions from both compounds are solvated by water molecules.B. The particle diagrams best represent that ΔH°<0 because both compounds produce about the same amount of CO3

2− ions from the dissolution.C. The particle diagrams best represent that ΔS°>0 because both compounds produce a very small amount of ions from the dissolution.D. The particle diagrams best represent that the molar solubility is greater for FeCO3 compared to MnCO3.

Do 3 and 4 now.Share what you think before finalizing your decision.

Breaking = + ΔH and +ΔS(needs energy) more disorder

Breaking + ΔH +ΔS + ΔH and +ΔS(needs energy)more disorder

Breaking + ΔH and +ΔS + ΔH and +ΔS -ΔH and -ΔS (needs energy) more disorder forming more organized

Breaking + ΔH +ΔS + ΔH and +ΔS -ΔH and -ΔS(needs energy) more disorder forming more organized

A. + + + + - -

Topic 7.14 Free Energy of Dissolution

The dissolution of a salt is a reversible process that can be influenced by environmental factors such as pH or other dissolved ions (common ion effect).

You need to be able to explain the relationship between the solubility of a salt and changes in the enthalpy and entropy that occur in the dissolution process.

Essential knowledge

The free energy change ( ΔG)for dissolution of a substance reflects a number of factors:

● The breaking of intermolecular interactions that hold the solid together.

● The reorganization of the solvent around the dissolved species.

● The interaction of the dissolved species with the solvent.

It is possible to estimate the sign and relative magnitude of enthalpic and entropic contributions to each of these factors.

HOWEVER, making predictions for the total change in free energy of dissolution can be challenging as a result of the “cancellations” among the free energies associated with the 3 factors cited above.

Next Exam: 60 Multiple choice

Units 1- 7

Wednesday February 26.

AP Classroom PPC Unit 7 due before the Exam (Tuesday Feb 25 at 11:59pm?)

A Cycle of Copper ReactionsRead the lab instructions and prepare a tray with

● Index card with names of lab group members and lab station number

● Glassware and any other equipment (eg. weighing paper or boat).

Per Lab bench (2 lab groups max.)

● A 2 decimal place balance - make sure it works.

You need to answer the pre-lab questions correctly before next class meeting.