exam iii - napa valley college homepage 121/exam iii... · exam iii thermodynamics 1st, 2nd, ......

Exam IIIThermodynamics

1st, 2nd, and 3rd LawsEnthalpy, Entropy, and Gibbs Energy

Heats of FormationHess LawSpontaniety

Gibbs EquationElectrochemical Cells

Balancing Redox ReactionsNernst Equation

BatteriesElectroplating and Hydrolysis

Thermodynamics

1) A 100 gram block of copper was heated with a bunsen burner and then thrown into a cup ofwater (250 mL) initially at 25C. When the system reached thermal equilibrium the finaltemperature of the water and the block was 61C. What was the initial temperature of the blockof copper? Cp(Cu)=24.5 J/mol-K, Cp(H2O)=75.2 J/mol-K, FWT(Cu) = 63.55 g/mol, FWT(H2O)= 18 g/mol, density of water = 1 g/ml

2) Given the following information calculate the heat of formation of C2H4.

C2H4 + 3 O2 2 CO2 + 2 H2O H = -414 kJ/molC + O2 CO2 H = -393.5 kJ/molH2 + O2 H2O H = -241.8 kJ/mol

3) When ethanol (C2H5OH) burns in oxygen it forms CO2 and water. If the heat of combustionis -1237.7 kJ/mol, the heat of formation of water is -241.8 kJ/mol, and the heat of formation ofcarbon dioxide is -393.5 kJ/mol, what is the heat of formation of C2H5OH?

4) Given a block of ice at 0C how much of the ice will melt if you put a 100 gram block ofaluminum on it that is initially at 100C? Cp(Al)=24.1 J/mol-K, Cp(H2O)=75.2 J/mol-K, Heat ofFusion = 6.01 kJ/mol for ice, FWT(Al) = 26.98 g/mol, FWT (water) = 18 g/mol

5) 10 grams of diethyl ether was placed into a bomb calorimeter and then combusted in excessoxygen. The balanced combustion reaction was as follows;

C4H10O + 6 O2 -----> 4 CO2 + 5 H2O

The heat released by this reaction heated 2500 mL of water 8.6C. What is the heat of formationof the diethyl ether? Cp(H2O) = 75.2 J/mol-K, Hf(CO2) = -393.5 kJ/mol, Hf(H2O) = -241.8kJ/mol, FWT (Diethyl ether) = 74 g/mol, FWT (water) = 18 g/mol.

6) If a 25 gram block of copper at 45C is added to 50 grams of liquid ammonia at -60C,calculate the final temperature of the system. For NH3: FWT(NH3) = 17 g/mol, Hvap = 23.4kJ/mol, Cp(liq)= 35.1 J/mol.K and for Cu : Cp = 24.4 J/mol.K, FWT(Cu) = 63.55 g/mol

7) You have 50 grams of ice at -2.5C sitting in a dixie cup. If you heat 5 nickels and throwthem into the cup and 8 grams of the ice melts, how hot were the nickels? Each nickel weighs 5grams and has a heat capacity of 24.3 J/mol-K. Assume that nickels are pure nickel with anatomic mass of 57.81 g/mol.

8) What is the G and S for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25C? Theenthalpy for this reaction is - 210 kJ/mol.

9) Calculate the G for the making of 6M NaOH from solid NaOH. Saturated solutions of NaOH have a concentration of 19.25 M.

10) Mathematically we write the first law of thermodynamics as U = q + w. The work, w, ispressure-volume work. What is the sign on this work (+ or -) and why does it have this sign?

11) Please give me a definition of a heat capacity.

12) What is the G for the following reaction at 25C?

BaBr2(aq) + CuSO4(aq) ----> BaSO4(s) + CuBr2(s)

H S

BaBr2 -186.47 kJ/mol 42.0 J/mol KCuSO4 -201.51 kJ/mol 19.5 J/mol KBaSO4 -345.57 kJ/mol 7.0 J/mol KCuBr2 -25.1 kJ/mol 21.9 J/mol K

Circle all that apply about this reaction,

spontaneous not spontaneousexothermic endothermicentropy increases entropy decreasesthe reaction occurs quickly the reaction occurs slowly

What is the equilibrium constant for this reaction?

13) Given the following set of thermodynamic data calculate the G, H, S, and Keq for thefollowing reaction at 25C (Enthalpies are in kJ/mol and entropies are in J/mol-k)

CoCl2(s) Co2+ + 2 Cl-

H SCoCl2 -325.5 106.3Co2+ -67.36 -155.2Cl- -167.4 55.1

Electrochemical Cells

14) Please balance each of the following redox reactions;

MnO4- + Fe2+ > Fe3+ + Mn2+

CrO42- + I- > I2 + Cr

3+

Co2+ + NO3- > Co2O3 + NO

I3- + S2O3

2- > I- + S4O62-

H2S + Cr2O72- > Cr3+ + S

Fe2+ + NiOOH > Fe3+ + Ni(OH)2

15) Consider the following electrochemical reaction,

PbI2 + 2 e- ---> Pb + 2 I-

Pb ---> Pb2+ + 2 e-

PbI2 ----> Pb2+ + 2 I-

How would the voltage change if you,

a) Add KI(s) to the oxidation cell Inc. N.C. Dec.b) Add AgNO3 to the reduction cell Inc. N.C. Dec.c) Add heat Inc. N.C. Dec.d) Add water to the reduction cell? Inc. N.C. Dec.e) Add NaNO3(s) to the oxidation cell Inc. N.C. Dec.f) Enlarge the oxidation electrode? Inc. N.C. Dec.


Au ---> Au3+ + 3e-

AuCl4- + 3e- ---> Au + 4 Cl-

------------------------------AuCl4

- ---> Au3+ + 4 Cl-


a) Enlarge the Au electrodes? Inc. N.C. Dec.

b) Add KCl(s)? Inc. N.C. Dec.

c) Add AgNO3? Inc. N.C. Dec.

d) Add water to the reduction cell? Inc. N.C. Dec.

e) Add NaNO3(s) Inc. N.C. Dec.

f) Cool the reaction vessel? Inc. N.C. Dec.

17) Given the following half-reactions draw an electrochemical cell, calculate the voltage of thecell and lable the anode and cathode. You may use a carbon electrode for the I- solution.

Fe3+ + e- ----> Fe2+ +0.771 VI2 ----> 2 I

- +0.535 V

18) What is the value of the half reaction,

Cu2+ + e- ----> Cu+

given that,

Cu2+ + 2e- ----> Cu = +0.3402Cu+ + e- ------> Cu = +0.522

19) Given the following half-reactions draw an electrochemical cell that would work. Calculatethe voltage of the cell and label the anode and cathode, tell which electrode is positive and whichis negative, and where the oxidation and reduction reactions are occurring. In addition indicatethe direction of electron flow, and the concentration of any ionic species in solution.

Co3+ + e- ---> Co2+ +1.842 VPb2+ + 2e- ---> Pb -0.126 V

20) Given the following half-reactions, properly set up a WORKING electrochemical cell.

Ag + I- AgI + e- = 0.1519 VPbI2 + 2e

----> Pb + 2 I- = 0.3580 V

Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of allionic species. Indicate what the electrodes are made of.

20b) Draw the cell diagram for the above cell.

20c) What is the Keq for the above cell?

20d) Look very carefully at the overall reaction, and explain whether your answer for 20c) isreasonable or not.

21) Balance the following half-reactions in a BASE and then use these reactions to set up aWORKING electrochemical cell.

Bi(s) Bi2O3(s) 0.460 VoltsHg2O(s) Hg(l) -0.123 Volts

Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of allionic species. Indicate the composition of the electrodes.



22) How long will it take to deposit 1 ounce of gold (31.1 g) at a constant current flow of 10amps? MW (Au) = 197.8

23) How long will it take to deposit 1 lb. of Palladium (453.6 g) from a solution of PdCl4 at aconstant current of 25 amps? MW (Pd) = 106.4

Thermodynamics

1) A 100 gram block of copper was heated with a bunsen burner and then thrown into a cup ofwater (250 mL) initially at 25C. When the system reached thermal equilibrium the finaltemperature of the water and the block was 61 C. What was the initial temperature of the blockof copper? Cp(Cu)=24.5 J/mol-K, Cp(H2O)=75.2 J/mol-K, FWT(Cu) = 63.55 g/mol, FWT(H2O)= 18 g/mol, density of water = 1 g/ml

-nCpTCu = nCpTH2O

-100g

63.5524.5

J

molK61C =

250g

1875.2

J

molK(25C - 61C)

Ti = 1036.3C

2) Given the following information calculate the heat of formation of C2H4.

C2H4 + 3 O2 ------> 2 CO2 + 2 H2O H = -414 kJ/molC + O2 -----> CO2 H = -393.5 kJ/molH2 + O2 -----> H2O H = -241.8 kJ/mol

2 CO2 + 2 H2O C2H4 + 3 O2 H = 414 kJ/mol2 (C + O2 CO2) H = -393.5 kJ/mol x22 (H2 + O2 H2O) H = -241.8 kJ/mol x22 C + 2 H2 C2H4 H = 414 kJ/mol + 2 (-393.5 kJ/mol )+ 2(-241.8 kJ/mol)

H = -856.6 kJ/mol

3) When ethanol (C2H5OH) burns in oxygen it forms CO2 and water. If the heat of combustionis -1237.7 kJ/mol, the heat of formation of water is -241.8 kJ/mol, and the heat of formation ofcarbon dioxide is -393.5 kJ/mol, what is the heat of formation of C2H5OH?

C2H5OH + 3 O2 2 CO2 + 3 H2O

H = Hprod - Hreact -1237.7 kJ/mol= [2(-393.5kJ/mol) + 3(-241.8kJ/mol)] [HC2H5OH + 3(0kJ/mol)]HC2H5OH = -274.7 kJ/mol

4) Given a block of ice at 0C how much of the ice will melt if you put a 100 gram block ofaluminum on it that is initially at 100C? Cp(Al)=24.1 J/mol-K, Cp(H2O)=75.2 J/mol-K, Heat ofFusion = 6.01 kJ/mol for ice, FWT(Al) = 26.98 g/mol, FWT (water) = 18 g/mol

-nCpTAl = nfus

-100g

26.9824.1

J

molK100C 0C=n6010

J

molK

n = 1.4863 mol ice x 18g/mol = 26.75 g ice will melt

5) 10 grams of diethyl ether was placed into a bomb calorimeter and then combusted in excessoxygen. The balanced combustion reaction was as follows;

C4H10O + 6 O2 -----> 4 CO2 + 5 H2O

The heat released by this reaction heated 2500 mL of water 8.6 C. What is the heat of formationof the diethyl ether? Cp(H2O) = 75.2 J/mol-K, Hf(CO2) = -393.5 kJ/mol, Hf(H2O) = -241.8kJ/mol, FWT (Diethyl ether) = 74 g/mol, FWT (water) = 18 g/mol.

-nCpTH2O = rxn note: a negative sign was added because energy was given off.

-2500g

1875.2

J

molK(8.6C )= -89,822 J for 10 grams

-89,822 J

= -664,683 J/mol

H = Hprod - Hreact

-664,683 kJ/mol= [4(-393.5kJ/mol) + 5(-241.8kJ/mol)] [HEther + 6(0kJ/mol)]

HEther = -2118.3 kJ/mol

6) If a 25 gram block of copper at 45 C is added to 50 grams of liquid ammonia at -60C,calculate the final temperature of the system. For NH3: FWT(NH3) = 17 g/mol, Hvap = 23.4 kJ/mol, Cp(liq)= 35.1 J/mol.K and for Cu : Cp = 24.4 J/mol.K, FWT(Cu) = 63.55 g/mol

-nCpTCu = nCpTNH3

-25g

63.5524.4

J

molKTf 45C=

50g

1735.1

J

molK(Tf (-60C))

-9.9587(Tf 45C) = 103.235(Tf + 60C)-Tf + 45 = 10.3663(Tf + 60) = 10.3663Tf + 621.98-11.3663Tf = 666.98 Tf = -50.76C

7) You have 50 grams of ice at -2.5 C sitting in a dixie cup. If you heat 5 nickels and throw theminto the cup and 8 grams of the ice melts, how hot were the nickels? Each nickel weighs 5 gramsand has a heat capacity of 24.3 J/mol-K. Cpice = 37.7 J/molK. Assume that nickels are purenickel with an atomic mass of 57.81 g/mol.

Note: You must heat all of the ice but melt only 8 grams of it.

nCpTNi = nCpTice + nfus

25 g

57.8124.3

J

molKTf 0C=

6010

J

molK+

50g

1837.7

J

molK(2.5C)

Tf 0C = 279.1C so Tf = 279.1C

8) What is the G and S for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25C? Theenthalpy for this reaction is - 210 kJ/mol.

G = G + RT ln Q = 0 + (8.314 J/molK)(298K) ln (0.03 M/ 4.5M)

G = -12,414 J/mol

G = H - TS =>

= =

,

,

= 663 /

9) Calculate the G for the making of 6M NaOH from solid NaOH. Saturated solutions of NaOH have a concentration of 19.25 M.

There are two processes going on,

1. Solid NaOH Satd NaOH2. Satd NaOH 6 M NaOH

The first process is an equilibrium so G = 0 for the first process.

For the second process, since all solutions have a standard state of 1M, the standard statefor Satd NaOH and 6 M NaOH are both 1 M and the G = 0 since there is no differencebetween the Gs for these two solutions. So,

G1 = 0G2 = G + RT ln QG2 = 0 + (8.314 J/molK)(298K) ln (6 M/ 19.25M) = -2888.2 J/mol

G1+ G2 = 0 + (- 2888.2 J/mol) = - 2888.2 J/mol

10) Mathematically we write the first law of thermodynamics as U = q + w. The work, w, ispressure-volume work. What is the sign on this work (+ or -) and why does it have this sign?

Work = PV

A chemist thinks of himself as being inside the system pushing out (like expanding aballoon). To a chemist, pushing out (expanding the balloon) is negative work. Negativework is work that you must do as opposed positive work with is work being done for you.In mathematics, a change in volume is always the Final Volume Initial Volume. In thecase of an expanding balloon, the final volume will be larger than the initial volume soFinal Volume Initial Volume will produce a positive value (Final Volume InitialVolume = + change in volume). Since PV = negative for an expanding balloon, andboth P and V are positive, we must add a negative sign to the work expression to makethis work overall negative. So, chemist define pressure-volume work as,

Work = - PV (Chemist)

It should be noted that engineers see the world opposite from chemists. Engineers viewthemselves as being outside of the system pushing in. Work for an engineer iscompressing the balloon, making it smaller. Since V = Vfinal V initial, and the Vfinalis smaller than Vinitial, the V is negative. So, for an engineer,

Work = PV (Engineer)

This value is still negative, like the chemist, but we see the world differently from anengineer.

11) Please give me a definition of a heat capacity.

It is the amount of energy needed to raise the temperature of one mole of substance 1C.

12) What is the G for the following reaction at 25C?

BaBr2(aq) + CuSO4(aq) ----> BaSO4(s) + CuBr2(s)

H S

BaBr2 -186.47 kJ/mol 42.0 J/mol KCuSO4 -201.51 kJ/mol 19.5 J/mol KBaSO4 -345.57 kJ/mol 7.0 J/mol KCuBr2 -25.1 kJ/mol 21.9 J/mol K

H = Hprod - Hreact and S = Sprod - Sreact

H = Hprod - Hreact H = [(-345.57 kJ/mol) + (-25.1 kJ/mol)] [(-186.47 kJ/mol) + (-201.51 kj/mol)]H = 17.31 kJ/mol

S = Sprod - SreactS = [(7.0 J/molK) + (21.9 J/molK)] [(42.0 J/molK) + (21.9 J/molK)]S = -32.6 kJ/mol

G = H - TSG = 17,310 J/mol + (298 K) (-32.6 J/molK)G = 27,024.8 J/mol

Circle all that apply about this reaction,

spontaneous not spontaneousexothermic endothermicentropy increases entropy decreasesthe reaction occurs quickly the reaction occurs slowly

What is the equilibrium constant for this reaction?

G = - RT ln Keq

Keq = e

= e

,.

(.)() = 1.832x10

13) Given the following set of thermodynamic data calculate the G, H, S, and Keq for thefollowing reaction at 25C (Enthalpies are in kJ/mol and entropies are in J/mol-k)

CoCl2(s) Co2+ + 2 Cl-

H SCoCl2 -325.5 106.3Co2+ -67.36 -155.2Cl- -167.4 55.1

H = Hprod - Hreact and S = Sprod - Sreact

H = Hprod - Hreact H = [(-67.36 kJ/mol) + 2(-167.4 kJ/mol)] [(-325.5 kj/mol)]H = -76.66 kJ/mol (exothermic)

S = Sprod - SreactS = [(-155.2 J/molK) + 2(55.1 J/molK)] [(106.3 J/molK)]S = -151.3 J/molK (entropy decreases)

G = H - TSG = -76,660 J/mol - (298 K) (-151.3 J/molK)G = -31,572.6 J/mol (reaction is spontaneous as written)

G = - RT ln Keq

Keq = e

= e

,.

(.)() = 3.423x10

Electrochemical Cells

14) Please balance each of the following redox reactions;

MnO4- + Fe2+ > Fe3+ + Mn2+

Fe2+ Fe3+ + e-

5e- + 8 H+ + MnO4- Mn2+ + 4 H2O

CrO42- + I- > I2 + Cr

3+

3 e- + 8 H+ + CrO42- Cr3+ + 4 H2O

2 I- I2 + 2 e-

Co2+ + NO3- > Co2O3 + NO

3 H2O + 2 Co2+ Co2O3 + 6 H

+ + 2e-

3 e- + 4 H+ + NO3- NO + 2 H2O

I3- + S2O3

2- > I- + S4O62-

2 e- + I3- 3 I-

S2O32- S4O6

2- + 2 e-

H2S + Cr2O72- > Cr3+ + S

H2S S + 2 H+ + 2 e-

6 e- + 14 H+ + Cr2O72- 2 Cr3+ + 7 H2O

Fe2+ + NiOOH > Fe3+ + Ni(OH)2

Fe2+ Fe3+ + e-

e- + H+ + NiOOH Ni(OH)2


PbI2 + 2 e- Pb + 2 I-

Pb Pb2+ + 2 e-

PbI2 Pb2+ + 2 I-


a) Add KI(s) to the oxidation cell Inc. N.C. Dec.b) Add AgNO3 to the reduction cell Inc. N.C. Dec.c) Add heat Inc. N.C. Dec.d) Add water to the reduction cell? Inc. N.C. Dec.e) Add NaNO3(s) to the oxidation cell Inc. N.C. Dec.f) Enlarge the oxidation electrode? Inc. N.C. Dec.

In electrochemical cells all concentrations are 1 M. If all concentrations are 1 M, the reversereaction is going to occur. Therefore, the real reaction going on is,

Pb + 2 I- PbI2 + 2 e- (oxidation)

Pb2+ + 2 e- Pb (reduction)Pb2+ + 2 I- PbI2

To understand what would happen, you need to write down the Nernst Equation,

= - RT/n ln 1/[Pb2+][I-]2

a) When you put KI in the oxidation cell you are increasing the I- concentration. Thiscauses 1/[Pb2+][I-]2 to become smaller. The natural log of a smaller number is a negativenumber. The negative of a negative is a positive so the voltage will increase.

b) Nothing happens. There is nothing for the silver to react with.

c) Adding heat increases T in the equation. If everything else remains the same, the- RT/n ln 1/ [Pb2+][I-]2 becomes more negative and the voltage decreases.

d) Adding water is a dilution. Diluting the reduction cell lowers the concentration of I-.If I- gets smaller, the 1/[Pb2+][I-]2 gets larger. The natural log will become larger butthere is a negative so it becomes more negative, so the voltage goes down.

e) NaNO3 does not participate in this reaction. No change.

f) Changing the size of the electrodes does not change the voltage. The electrodes arenot a part of the reaction. They are not in the Nernst Equation.


Au Au3+ + 3e-

AuCl4- + 3e- Au + 4 Cl-

------------------------------AuCl4

- Au3+ + 4 Cl-


a) Enlarge the Au electrodes? Inc. N.C. Dec.b) Add KCl(s)? Inc. N.C. Dec.c) Add AgNO3? Inc. N.C. Dec.d) Add water to the reduction cell? Inc. N.C. Dec.e) Add NaNO3(s) Inc. N.C. Dec.f) Cool the reaction vessel? Inc. N.C. Dec.

Note: This is NOT a Ksp problem since AuCl4- is not a solid (solids are not charged) so

we will leave the reaction as is.

= - RT/n ln [Au3+][Cl-]4/[ AuCl4-]

a) No change. Changing the size of the electrode makes no difference.

b) Adding Cl- makes the natural log large which decreases the voltage

c) The Ag+ will react with the Cl- to make AgCl. This removes Cl-, making the naturallog smaller (more negative). So you will be subtracting a smaller number so the voltagewill increase.

d) Diluting the reduction cell lowers the concentration of both AuCl4- and Cl- but the Cl-

is raised to the 4th power so it reduces faster. Reducing Cl- makes the natural log smaller(more negative so you are subtracting a smaller (more negative) number so voltageincreases.

e) No change.

f) If T is smaller, then you are subtracting a smaller number so voltage increases.

17) Given the following half-reactions draw an electrochemical cell, calculate the voltage of thecell and label the anode and cathode. You may use a carbon electrode for the I- solution.

Fe3+ + e- Fe2+ +0.771 eV2 I- I2 + 2 e

- -0.535 eV2 Fe3+ + 2 I- 2 Fe2+ + I2 0.236 eV

18) What is the value of the half reaction,

Cu2+ + e- ----> Cu+

given that,

Cu2+ + 2e- Cu = +0.3402Cu+ + e- Cu = +0.522

Cu2+ + 2e- Cu = +0.3402Cu Cu+ + e- = -0.522

Cu2+ + e- Cu+

= 2(0.3402) + 1(-0.522) = 0.1584 volts1

C Pt

1M KI1M FeCl1M FeCl

3

2

e-

AnodeNegativeOxidation

CathodePositiveReduction

I2

19) Given the following half-reactions draw an electrochemical cell that would work. Calculatethe voltage of the cell and label the anode and cathode, tell which electrode is positive and whichis negative, and where the oxidation and reduction reactions are occurring. In addition indicatethe direction of electron flow, and the concentration of any ionic species in solution.

Co3+ + e- Co2+ = + 1.842 eVPb2+ + 2e- Pb = - 0.126 eV

Co3+ + e- Co2+ = + 1.842 eVPb Pb2+ + 2e- = + 0.126 eV3 Pb + 2 Co3+ 3 Pb2+ + 2 Co = 1.968 V

Pb Pt

1MPb(NO )3 2

1M CoCl1M CoCl

3

2

e-



20) Given the following half-reactions, properly set up a WORKING electrochemical cell.

AgI + e----> Ag + I- = -0.1519 VPbI2 + 2e

----> Pb + 2 I- = 0.3580 V



Ag/AgI(s)/1 M KI/PbI2(s)/Pb


= 0.5099 volts = 0.0592/2 log KeqKeq = 1.684x1017

20d) Look very carefully at the overall reaction, and explain whether your answer for 20c) isreasonable or not.

The problem is that everything is a solid except for the 1 M KI. The equilibrium constant must be,

So the voltage is completely dependent on the difference between the amount of I- in equilibrium with thePb and the I- in equilibrium with the Ag. This makes what is known as a concentration cell, that is, a cellwhose voltage is due only to differences in concentration on the two sides of the cell.

Ag Pb

PbI (s)2



AgI(s)

e-

1M KI

Ag

Pb

I

IKeq

][

][

21) Balance the following half-reactions in a BASE and then use these reactions to set up aWORKING electrochemical cell.

6 OH- + 2 Bi(s) Bi2O3(s) + 3 H2O + 6 e- 0.460 Volts

2 e- + H2O + Hg2O(s) 2 Hg(l) + 2 OH- -0.123 Volts



Bi / Bi2O3(s) / 1 M NaOH / Hg2O(s) / Hg / Pt


= 0.337 volts = 0.0592/6 log KeqKeq = 1.43x1034

Hg O2

Pt

e-



Bi

Bi O (s)2 3

1M NaOH

22) How long will it take to deposit 1 ounce of gold (31.1 g) at a constant current flow of 10amps? MW (Au) = 197.8

31.1 g/197.8 g/mol = 0.1572 mol Au has a 3+ charge, so multiply by 3.

3 mole e-/mol Au x 0.15723 mol Au = 0.4717 mol e-

0.4717 mol e- x 96,487 C/mol e- = 45,511.8 C

Amp x sec = Coulombs (C)

(10 amps)(sec) = 45,511.9 C

sec = 4,551.18 sec => 75.85 minutes

23) How long will it take to deposit 1 lb. of Palladium (453.6 g) from a solution of PdCl4 at aconstant current of 25 amps? MW (Pd) = 106.4

453.6 g/106.4 g/mol = 4.263 mol Pd, but Pd has a 4+ charge, so multiply by 4.

4 mole e-/mol Pd x 4.263 mol Pd = 17.053 mol e-

17.053 mol e- x 96,487 C/mol e- = 1,645,357 C

Amp x sec = Coulombs (C)

(25 amps)(sec) = 1,645,357 C

sec = 65,814 sec => 18.28 hours

Chemistry 121 Name____________________Third Exam May 15, 2007

CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for fullcredit. You may use a calculator.

Question Credit

1 ( 18)

2 (40)

3(12)

4(15)

5(15)

Total

R = 8.314 J/mol-K = 96486 C/mol e- C = amp x sec = - 0.0592/n log Q G = G + RT ln Q G= -RTlnKeqG = -n PV = nRT R = 0.08205 L-atm/mol-K

1) Balance each of the following oxidation-reduction half-reactions;

IO3- > I3

- (in acid)

Mn(OH)2 > Mn2O3 (in base)

2) Given the following set of half-reactions, set up a WORKING electrochemical cell.

Hg2Cl2(s) + 2 e- > 2Hg + 2Cl- = 0.27 V

2 H+ + 2 e- > H2 = 0.000 V



2c) What is the voltage of the cell when all ionic species have a concentration of 0.10 M?

3) How long will it take to galvanize (coat with Zinc) a nail with 2 grams of zinc using a current(amperage) of 0.8 amps? FWT of Zn2+ = 65.41 g/mol

4) You are given 100 mL of water at 25 C. You add a 20 gram ice cube at 0C and 50 mL ofboiling water at 100 C. What is the final temperature of the system?

5) What is the G and S for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25 C? The H = -210 kJ/mol.for this reaction.

Chemistry 121 Name__Answer Key____Third Exam May 15, 2007

CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for fullcredit. You may use a calculator.

Question Credit

1 ( 18)

2 (40)

3(12)

4(15)

5(15)

Total


1) Balance each of the following oxidation-reduction half-reactions;

IO3- > I3

- (in acid)

16 e- + 18 H+ + 3 IO3- > I3

- + 9 H2O

Mn(OH)2 > Mn2O3 (in base)

2 Mn(OH)2 > Mn2O3 + H2O + 2 H+ + 2e-

2 OH- > 2 OH-

2 Mn(OH)2 + 2 OH- > Mn2O3 + 3 H2O + 2e

-

2) Given the following set of half-reactions, set up a WORKING electrochemical cell.

Hg2Cl2(s) + 2 e- > 2Hg + 2Cl- = 0.27 V

2 H+ + 2 e- > H2 = 0.000 V


2b) Draw the cell diagram for the above cell. (Note: The solution in both cells is HCl so no saltbridge is required).

Pt/H2 (1atm) / 1M HCl / Hg2Cl2(s) / Hg(l)/ Pt

2c) What is the voltage of the cell when all ionic species have a concentration of 0.10 M?

The overall reaction is,

Hg2Cl2(s) + 2 e- > 2Hg + 2Cl- = 0.27 V

H2 > 2 H+ + 2 e- = 0.000 V

Hg2Cl2(s) + H2 > 2 Hg + 2 H+ + 2 Cl- = 0.27 V

So the voltage is,

= 0.27 - 0.0592/2 log [H+]2 [Cl-]2 = 0.27 - 0.0591/2 log [0.10]2 [0.10]2 = 0.3882 Volts

H2 1atm

1M HCl

Pt

Hg Cl2 2

1M HCl

Pt

e-



3) How long will it take to galvanize (coat with Zinc) a nail with 2 grams of zinc using a current(amperage) of 0.8 amps? FWT of Zn2+ = 65.41 g/mol

2 grams/65.41g/mol = 0.03058 mole of Zn

0.03058 mole of Zn x 2 mole e-/mol Zn = 0.06115 mole e-

0.06115 mole e- x 96,486 C/mol e- = 5900.4 Coulombs

1 C = 1 amp x sec

5900.4 C = 0.8 amps x sec

5900.4 C/0.8 amps = 7375.47 seconds

4) You are given 100 mL of water at 25C. You add a 20 gram ice cube at 0C and 50 mL ofboiling water at 100C. What is the final temperature of the system? Hfus = 6.01 kJ/mol andCp(H2O) = 75.2 J/mol K, FWT H2O = 18g/mol (These were not given in the original problem).

The easiest way of doing this is by mixing the hot and cold water and then adding the ice cube.

First add 100 mL of 25C water to 50 mL of 100C water,

nCpT25C = - nCpT100C100g/18g/mol (75.2 J/molK)(Tf - 25) = -50g/18g/mol(75.2 J/molK)(Tf-100)The 18g/mol and the 75.2 J/molK cancel on both sides so,100(Tf-25) = -50(Tf-100)Tf = 50C

So now we have 150 mL of water at 50 C and we add the 20 gram ice cube. I assume that the icecube will melt and then waters will mix to get to some final temperature so,

nHfus + nCpT0C = - nCpT50C

20g/18g/mol(6010 J/mol) + 20g/18g/mol (75.2)(Tf - 0) = - 150g/18g/mol(75.2)(Tf - 50)

All the 18 g/mol cancel so,

120,200 + 1504Tf = - 11280Tf + 56400012784Tf = 443,800Tf = 34.71C

5) What is the G and S for the dilution of 4.5 M H2SO4 to 0.03 M H2SO4 at 25C? The H = - 210 kJ/mol.for this reaction.

G = G + RT ln Q

But G = 0 since both concentrations have the same standard state of 1M. So,

G = 0 + RT ln 0.03M/4.5M

G = -5391.42 J/mol

Now, since G = H - TS, we can solve for S,

-5391.42 J/mol = -210,000 J/mol - (298)S

S = (-5391.42 J/mol + 210,000 J/mol )/298 = 686.61 J/mol K

Chemistry 121 Name____________________Third Exam May 14, 2009

CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.You may use a calculator.

Question Credit

1 (18)

2 (28)

3(10)

4(20)

Total


1a) Please balance the following half-reaction in a base.

BH3 > B2O3

1b) What is the voltage of the following half-cell reactions when added together?

Cu2+ + e- > Cu+ = 1.29Cu+ + e- > Cu = 1.68Cu2+ + 2 e- > Cu = ?

2) In lab, you were asked to experimentally determine the Ksp of PbI2. Given the informationbelow (found in the CRC), what voltage should you have obtained for your cell, and what shouldthe Ksp have been?

PbI2(s) Pb2+ + 2 I-

H SPbI2 -174.1 kJ/mol 176.98 J/mol-KPb2+ 1.63 kJ/mol 21.34 J/mol-KI- -55.94 kJ/mol 109.37 J/mol-K

The reaction is (Circle all that apply).

Spontaneous Not SpontaneousExothermic EndothermicEntropy increases Entropy DecreasesOccur Fast Occurs Slow

2c) Like you did in lab, draw the electrochemical cell that was used to determine the Ksp of PbI2.Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of allionic species. Indicate the composition of the electrodes.

2c) Draw the cell diagram for the above cell.

3) It took 35 minutes and 10 amps to electroplate 10.615 grams of platinum (FWT = 195.08g/mol). What was the charge on the platinum?

4) A 20 gram block of ice initally at -10.0C was dropped into 250 mL of water at 25C. What isthe final temperature of the system? Hfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/mol-K, Cp(ice) =35.2 J/mol-K, FWT (H2O)= 18 g/mol

Chemistry 121 Name__Answer Key___Third Exam May 14, 2009

CLOSED BOOK EXAM - No books or notes allowed. ALL work must be shown for full credit.You may use a calculator.

Question Credit

1 (18)

2 (28)

3(10)

4(20)

Total


1a) Please balance the following half-reaction in a base.

BH3 > B2O3

3 H2O + 2 BH3 > B2O3 + 12 H+ + 12 e-

12 OH- > 12 OH-

12 OH- + 2 BH3 > B2O3 + 9 H2O + 12 e-

1b) What is the voltage of the following half-cell reactions when added together?

Cu2+ + e- > Cu+ = 1.29Cu+ + e- > Cu = 1.68Cu2+ + 2 e- > Cu = (1.29(1) + 1.68 (1))/2 = 1.485 Volts

Pb Pb

1MPb(NO )3 2

1M KI

PbI (s)2

e-



2) In lab, you were asked to experimentally determine the Ksp of PbI2. Given the informationbelow (found in the CRC), what voltage should you have obtained for your cell, and what shouldthe Ksp have been?

PbI2(s) Pb2+ + 2 I-

H SPbI2 -174.1 kJ/mol 176.98 J/mol-KPb2+ 1.63 kJ/mol 21.34 J/mol-KI- -55.94 kJ/mol 109.37 J/mol-K

H = [1.63 + 2(-55.94)] - [-174.1] = 63.85 kJ/mol S = [21.34 + 2(109.37)] - [176.98] = 63.1 J/molKG = 63,850 - 298(63.1) = 45,046.2 J/mol45,046.2 J/mol = -8.314(298)ln(Ksp)Ksp = 1.27x10-8

The reaction is (Circle all that apply).

Spontaneous Not SpontaneousExothermic EndothermicEntropy increases Entropy DecreasesOccurs Fast Occurs Slow

2c) Like you did in lab, draw the electrochemical cell that was used to determine the Ksp of PbI2.Label the anode, the cathode, the direction of electron flow, and indicate the concentrations of allionic species. Indicate the composition of the electrodes.

2c) Draw the cell diagram for the above cell.

Pb / PbI2(s) / 1 M KI // 1 M Pb(NO3)2 / Pb

3) It took 35 minutes and 10 amps to electroplate 10.615 grams of platinum (FWT = 195.08g/mol). What was the charge on the platinum?

10.615 g/195.08 g/mol = 0.05441 mole of Pt

0.05441 mole of Pt x Charge on Pt = moles of e-

Also,

amp x sec/ 96,486 C/mol e- = mole of e-

So,

0.05441 mole of Pt x Charge on Pt = amp x sec/ 96,486 C/mol e-

0.05441 mole of Pt x Charge on Pt = 10 amps x 35 min x 60 sec/min / 96,486 C/mol e-

Charge on Pt = [10 amps x 35 min x 60 sec/min / 96,486 C/mol e-]/0.05441 mole of Pt

Charge on Pt = +4

4) A 20 gram block of ice initally at -10.0C was dropped into 250 mL of water at 25C. What isthe final temperature of the system? Hfus = 6.01 kJ/mol, Cp(H2O) = 75.2 J/mol-K, Cp(ice) =35.2 J/mol-K, FWT (H2O)= 18 g/mol

nCpTice + nHfus + nCpT0C H2O = -nCpT25C H2O

(20/18)35.2(10) + (20/18)6010 + (20/18)75.2(Tf-0) = -(250/18)75.2(Tf-25)

All the 18's cancel so,

7040 + 120200 + 1504Tf = -18800Tf + 470000

20304Tf = 342760

Tf = 16.88C

exam iii - napa valley college homepage 121/exam iii... · exam iii thermodynamics 1st, 2nd, ......

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