chem hw solutions

Started on Wednesday, 6 April 2011, 09:59 PM

Completed on Wednesday, 6 April 2011, 10:00 PM

Time taken 43 secs

Marks 28/31

Grade 9.03 out of a maximum of 10 (90%)

Question1Marks: 1

Calculate ΔGº (in kJ) at 298 K for the following reaction:

2H2O(g) +2Cl2(g) → 4HCl(g) + O2(g)

Use the data in Appendix II (Tro); Appendix 4 (Zumdahl)

Answer:

CorrectMarks for this submission: 1/1.

Question2Marks: 1

Is the above reaction under standard conditions spontaneous in the forward direction?

Choose one answer.

a. YES, because ΔG is negative.

b. YES, because ΔG is positive.

c. NO, because ΔG is negative.

d. NO, because ΔG is positive.


76

Question3Marks: 1

Calculate ΔHº (in kJ) for the following reaction at 298 K:

2CO2(g) +4H2O(l) → 2CH3OH(l) + 3O2(g)


Answer:


Question4Marks: 1

Calculate ΔSº (in J/K) at 298 K for :

2CO2(g) +4H2O(l) → 2CH3OH(l) + 3O2(g)


Answer:


Question5Marks: 1

Use ΔGº =ΔHº -TΔSº to calculate ΔG (in kJ) at 298 K for :

2CO2(g) +4H2O(l) → 2CH3OH(l) + 3O2(g)

Answer:


Question6Marks: 1

1453

161.6

1404.8432

If the above reaction could be done at 3813 K, what would be your estimate for ΔGº (in kJ) at this elevated temperature? Use ΔGº =ΔHº -TΔSº and assume ΔHº and ΔSº are independent of temperature.

(The º is included because it is still for standard conditions, that is, 1 atm for gases and 1 molar for concentrations.)

Answer:


Question7Marks: 1

Is the above reaction spontaneous in the forward direction at either of the temperatures? (Pick 2 answers.)

Choose at least one answer.

a. YES, because ΔG is negative.

b. NO, because ΔG is negative.

c. YES, because ΔG is positive.

d. NO, because ΔG is positive.

e. Because ΔH is very large and positive (strongly endothermic reaction).

f. Because ΔH is very large and negative (strongly exothermic reaction).

g. Because the entropy term is more important at these temperatures.


Question8Marks: 1

What is ΔGrxno (in kJ) at 2893 K for the following reaction?

2POCl3(g) → 2PCl3(g) + O2(g)

POCl3(g): ΔHfo = -592.7 kJ/mol and Sº = 324.6 J/K mol)

836.819

PCl3(g): ΔHfo = -287.0 kJ/mol and Sº = 311.7 J/K mol)

O2(g): ΔHfo = ? kJ/mol and Sº = 205.0 J/K mol)

Hint given in feedback.

Answer:

Find ΔHº and ΔSº. ΔG =ΔHº -TΔSº.IncorrectMarks for this submission: 0/1.

Question9Marks: 1

At what temperature (in K) does the above reaction become spontaneous?


Answer:

Use ΔG =ΔHº -TΔSº. The temperature at which ΔG = 0 is the key.CorrectMarks for this submission: 1/1.

Question10Marks: 1

The above reaction is spontaneous in the forward direction(Pick 2)


a. above the temperature found in the last question

b. below the temperature found in the last question

c. because the reaction is endothermic and ΔS is positive. (In this case ΔS is more important at higher temperatures or equivalently ΔH is less important.)

d. becuase the reaction is exothermic and ΔS is negative. (In this case ΔS is less important at lower temperatures or equivalently ΔH is more important.)


Question11

281.5

3411.830357

Marks: 1

What is ΔGo (in kJ) at 428 K for the following reaction?

PbO(g) + CO2(g) → PbCO3(s)

PbO: ΔHfo = -219.0 kJ/mol and So = 66.5 J/K mol)

PbCO3(s): ΔHfo = -699.1 kJ/mol and So = 131.0 J/K mol)

CO2: ΔHfo = -393.5 kJ/mol and So = 213.6 J/K mol)

Answer:

IncorrectMarks for this submission: 0/1.

Question12Marks: 1

At what temperature (in K) does the above reaction become spontaneous?

Answer:


Question13Marks: 1

The above reaction is spontaneous in the forward direction(Pick 2)


a. above the temperature found in the last question

b. below the temperature found in the last question

c. because the reaction is endothermic and ΔS is positive. (In this case ΔS is more important at higher temperatures or equivalently ΔH is less important.)

d. becuase the reaction is exothermic and ΔS is negative. (In this case ΔS is less important at lower temperatures or equivalently ΔH is more important.)


-42.14

580.8182428

Question14Marks: 1

What is ΔGo (in kJ) at 50 ºC for the phase change of nitromethane from the liquid to the gaseous state?

CH3NO2(l) → CH3NO2(g)

CH3NO2(l): ΔHo = -113.1 kJ/mol and So = 171.8 J/K mol)CH3NO2(g): ΔHo = -74.7 kJ/mol and So = 274.4 J/K mol)


Answer:

Find ΔH and ΔS for the phase change. ΔG =ΔH -TΔS. In calculating ΔG it does not matter what kind of process it is--reaction or phase change.IncorrectMarks for this submission: 0/1.

Question15Marks: 1

Which state is more stable for nitromethane at the temperature in the previous question?

Choose one answer.

a. Liquid

b. Gas


Question16Marks: 1

What is the boiling point of nitromethane (in ºC)?

Answer:

Hint, find using K and convert to ºC.Correct

2.49

101.2690058

Marks for this submission: 1/1.

Question17Marks: 1

At 298 K, ΔGo = - 6.36 kJ for the reaction:

2N2O(g) + 3O2(g) ↔ 2N2O4(g)

Calculate ΔG (in kJ) at 298 K when PN2O = 5.71 atm, PO2 = 0.0064 atm, and PN2O4= 0.408 atm.

Help given in feedback.

Answer:

Hint, first find Q.CorrectMarks for this submission: 1/1.

Question18Marks: 1

Under the conditions in the previous question, in which direction is the reaction spontaneous?

Choose one answer.

a. to the left (more reactant)

b. to the rignt (more product)


Question19Marks: 1

At 298 K, ΔGo = + 8.68 kJ for the reaction:

ZnF2(s) ↔ Zn2+(aq) + 2F-(aq)

Under standard conditions what is the spontaneous direction (Recall, the o after ΔG means standard conditions, that is, some solid ZnF2, 1 molar concentrations for Zn2+(aq) and F-(aq), and 298 K if no other temperature is specified.)

Choose one answer.

18.1

a. to the left (more precipitate forms)

b. to the right (more solid dissolves)


Question20Marks: 1

Calculate ΔG (in kJ) at 298 K for some solid ZnF2, 0.024 M Zn2+ and 0.017 M F-(aq).Hint given in feedback.

Answer:

Hint, first find Q.CorrectMarks for this submission: 1/1.

Question21Marks: 1

Under the conditions of the previous question what is the spontaneous direction?

Choose one answer.

a. to the left (more precipitate forms)

b. to the right (more solid dissolves)


Question22Marks: 1

Endothermic reaction; decrease in entropy:Calculate the equilibrium constant at 34 K for a reaction with ΔHo = 10 kJ and ΔSo = (-100) J/K. (Don't round unil the end. Using the exponent enlarges any round-off error.)


Answer:

-20.75

2.6e-21

Hint, first calculate ΔGo at {T} K.CorrectMarks for this submission: 1/1.

Question23Marks: 1

Calculate the equilibrium constant at 114 K for the thermodynamic data in the previous question.

Notice that Keq is larger at the larger temperature for an endothermic reaction.

Answer:


Question24Marks: 1

Endothermic reaction; increase in entropy Calculate the equilibrium constant at 35 K for a reaction with ΔHo = 10 kJ and ΔSo = 100 J/K.

Answer:


Question25Marks: 1

Calculate the equilibrium constant at 122 K for the thermodynamic data in the previous question.

Notice that Keq is dramatically larger for a larger temperature when there is a substantial positive increase in entropy.

Answer:


Question26Marks: 1

1.7e-10

1.99e-10

8.75

Warm-up question Using data from the Appendix, calculate ΔGo (in kJ) at 65 oC for the reaction:

N2O(1 atm) + H2(1 atm) ↔ N2(1 atm) +H2O(l)

(Recall, all gases at 1 atm for standard conditions)

Answer:


Question27Marks: 1

Calculate ΔG (in kJ) at 65 oC for the reaction:

N2O(0.0020 atm) + H2(0.50 atm) ↔ N2(586.9 atm) +H2O(l)

Answer:


Question28Marks: 1

Assume that the ΔHo and ΔSo of vaporization do not change significantly with temperature. Calculate the vapor pressure of CH3OH at 43oC (in atm).

CH3OH (l) ↔ CH3OH(g) . . . ΔHo = 38.0 kJ and ΔSo = 112.9 J/K

(Don't round unil the end. Using the exponent enlarges any round-off error.) Hint given in feedback.

Answer:

Hint,what is the relationship between Keq and the vapor pressure.CorrectMarks for this submission: 1/1.

Question29Marks: 1

-337.2842

-305.7786194

.413

Calculate the vapor pressure of Hg at 42 oC (in atm).

Hg(l) ↔ Hg(g) . . . ΔHo = 61.32 kJ and ΔSo = 98.83 J/K

Answer:


Question30Marks: 1

Cyclopropane is a hydrocarbon that contains a ring of three carbon atoms with two hydrogen atoms bonded to each carbon (see below). (a) What are the steric number and the orbital hybridization for the carbon atoms? (b) What is the expected C—C—C bond angle? The actual bond angle is 60º. (c) Consequently, do you expect this bond to be under a lot of stress? (d) What does this imply about the expected C—C bond strength?


a. (a) SN =2, orbital hybridization = sp

b. (a) SN =3, orbital hybridization = sp2

c. (a) SN =4, orbital hybridization = sp3

d. (b) 109.5º

e. (b) 120º

f. (b) 180º

g. (c) YES--stress

h. (c) NO--stress

i. (d) Expected C-C bond strengh < 347 kJ/mol

j. (d) Expected C-C bond strengh = 347 kJ/mol

k. (d) Expected C-C bond strengh > 347 kJ/mol


9.8588e-6

Question31Marks: 1

Calculate the average C—C bond strength in cyclopropane (in kJ/mol). Its combustion and the experimental enthalpy of reaction are:

C3H6(g) + 4.5O2(g) → 3CO2(g) + 3H2O(g) ΔHºrxn = -1,957.7 kJ/mol

Since all reactants and products are in the gaseous state, bond energies may be used to estimate the enthalpy of reaction. Do not use the standard C—C bond strength, leave the C—C bond strength as an unknown and solve for its value. For review see chapter 9 p. 392 (Tro) or chapter 13 p. 608 (Zumdahl).

Answer:

Correct

Started on Saturday, 2 April 2011, 09:49 PM

Completed on Saturday, 2 April 2011, 09:50 PM

Time taken 29 secs

Marks 24/24

Grade 7 out of a maximum of 7 (100%)

Question1Marks: 1

Which of the following processes is NOT spontaneous?

Examples of spontaneous processes

Choose one answer.

a. A smoker's smokes gathers around the smoker.

b. A woman enters a room. Shortly thereafter her perfume can be smelled by

298.67

those on the other side of the room.

c. Leaves decay.

d. A lighted match burns.

e. Water evaporates from an open container on a dry day (low humidity).


Question2Marks: 1

Which of the following statements is INCORRECT?

Spontaneous processes and change in energy

Choose one answer.

a. A system tends to seek a minimum in energy. For reactions this means that exothermic reactions tend to be spontaneous.

b. Water fills a glass from the bottom up because this gives the lowest potential energy.

c. Dynamite exploding is an example of an exothermic reaction.

d. Ice forming during a reaction on a warm day is an example of an exothermic reaction.

e. Two solutions are combined and vigorous boiling ensues is an example of an exothermic reaction.


Question3Marks: 1


Spontaneous processes and change in entropy

Enthalpy versus entropy

Choose one answer.

a. The more ordered a system, the greater its entropy.

b. Perfume molecules spread throughout the room is more random or disordered than having them confined to a small portion of the room.

c. The entropy of the perfume molecules is larger when they are spread throughout the room than when they are confined to a small portion of the room.

d. The enthalpy of the system is more important in determining the outcome at a low temperature.

e. The entropy of the system is more important in determining the outcome at a high temperature.


Question4Marks: 1


Entropy

Choose one answer.

a. The entropy is larger when a disolved salt in a liquid is uniform or spread throughout the liquid than when it is highly concentration in a small portion of the liquid.

b. An iron nail at 100oC has more entropy than the same nail at 250oC

c. Entropy decreases when a gas dissolves in a solvent, mainly because the dissolved molecules are confined to a smaller volume.

d. Adding heat to a substance always increases its entropy.

e. Entropy increases in going from the solid to the liquid to the gaseous state.

f. When a solid dissolves in a liquid the entropy generally increases because the positional entropy increases. For example, dissolving a sugar cube in water.

g. The kinetic energy of the particles in a sample increases as the temperature increases. The disorder increases as the motion of the particles increases. Consequently, entropy increases as the temperature increases.


Question5Marks: 1

For which of the following reactions is there an increase in entropy (entropy of products > entropy of reactants or ΔS > 0)?

Determining the change in entropy 1


a. 2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g)

b. 2H2O2(aq) → 2H2O(l) + O2(g)

c. 2C2H2(g) + 5O2(g → 4CO2(g) + 2H2O(g)

d. 2A(g) + 3B(g) → C(g) + 2D(g)

e. 2A(g) + 3B(s) → C(g) + 2D(g)


Question6Marks: 1

Check those pairs of molecules where the one on the right has a larger entropy. (Each pair of molecules has the same conditions. For example, both temperatures are the same.)

Determining the change in entropy 2


a. CO2(g) and CO(g)

b. CH4(g) and C2H2(g)

c. CH3F(g) and CH3I(g)

d. C2H4F2(g) and C2H3F3(g)

e. C2H6(g) and C2Br6(g)


Question7Marks: 1

Calculate the increase of entropy in J/K when 8.6 mole of C2H5OH melts at (-114)oC and 1 atm. ΔHfus = 5.02 kJ/mol. (Remember to use J and K.)

Entropy definition and calculation

Answer:


Question8Marks: 1

Calculate the increase of entropy in J/K when 6.4 mole of C2H5OH vaporization at 78oC (the boiling point at 1 atm). ΔHvap = 39.30 kJ/mol.

Entropy 2nd calculation

Answer:

Did you remember to use J and K?CorrectMarks for this submission: 1/1.

Question9Marks: 1

Consider the reaction: A(g) + 2B(g) ↔ 3C(g) + 3D(s). Keq = 1,500.[A] = 0.15 M, [B] =0.31 M, [C] = 1.52 M, and some D is present. What is Q?

Review calculating Q

Answer:


Question10Marks: 1

For the above value of Q

271.52

716.58

243.62

Choose one answer.

a. the reaction is at equilibrium.

b. the reaction proceeds spontaneously to the left (making more reactant).

c. the reaction proceeds spontaneously to the right (making more product).


Question11Marks: 1

Under certain condition for the reaction A → B, ΔH = (-24.2) kJ and ΔS = (-49) J/K. What is ΔG in J at 262 K?

Gibb's energy calculation

Answer:


Question12Marks: 1

Check the correct statements for the above reaction.Reaction direction


a. The reaction is exothermic.

b. The reaction is endothermic.

c. The change in entropy is positive.

d. The change in entropy is negative.

e. The reaction is at equilibrium.

f. The reaction is spontaneous to the right.

-11362

g. The reaction is spontaneous to the left.

All answers must be correct to get credit.CorrectMarks for this submission: 1/1.

Question13Marks: 1

For the reaction A → B, ΔH = (-22.9) kJ and ΔS = (-32) J/K. What is ΔG in J at 790 K?

ΔG = ΔH - TΔS

Answer:


Question14Marks: 1

Check the correct statements for the above reaction.






e. The reaction is at equilibrium.

f. The reaction is spontaneous to the right.

g. The reaction is spontaneous to the left.


Question15Marks: 1

Under certain condition for the reaction A → B, ΔH = 22.3 kJ and ΔS = 24 J/K. At what temperature in K is the reaction at equilibrium?

2380


Answer:

Solve ΔG = ΔH - TΔS for T. Hint, what is ΔG at equilibrium?CorrectMarks for this submission: 1/1.

Question16Marks: 1







e. Above the equilibrium temperature, the reaction is spontaneous to the right.

f. Above the equilibrium temperature, the reaction is spontaneous to the left.

g. Below the equilibrium temperature, the reaction is spontaneous to the right.

h. Below the equilibrium temperature, the reaction is spontaneous to the left.

All answers must be correct to get credit.CorrectMarks for this submission: 1/1.

Question17Marks: 1

Under certain condition for the reaction A → B, ΔH = (-14.8) kJ and ΔS = (-34) J/K. At what temperature in K is the reaction at equilibrium?

Hint repeated for this problem in feedback.

Answer:

929.17

Solve ΔG = ΔH - TΔS for T. Hint, what is ΔG at equilibrium?CorrectMarks for this submission: 1/1.

Question18Marks: 1







e. Above the equilibrium temperature, the reaction is spontaneous to the right.

f. Above the equilibrium temperature, the reaction is spontaneous to the left.

g. Below the equilibrium temperature, the reaction is spontaneous to the right.

h. Below the equilibrium temperature, the reaction is spontaneous to the left.


Question19Marks: 1

Check those statements that are true.

Standard State and thermodynamic calculations


a. The standard state for H2 is 1 atm of the gas.

435.29

b. The standard state for Hg is 1 mol of the liquid.

c. The standard state of H2O is 1 mol ice.

d. The standard state of H2O is 1 mol water.

e. ΔHfo(O2(g)) = 0.

f. ΔGfo(O2(g)) = 0.

g. So(O2(g)) = 0.

Mercury is a liquid at room temperature (standard temperature).CorrectMarks for this submission: 1/1.

Question20Marks: 1

Calculate ΔSo in J/K for the following reaction:

2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g)

So(Fe2O3(s)) = 87.4 J/K/molSo(C(s)) = 5.7 J/K/molSo(Fe(s)) = 27.3 J/K/molSo(CO2(g)) = 213.7 J/K/mol

Detailed delta S Calculation

Answer:

Remember there are 3 moles of carbon dioxide, etc.CorrectMarks for this submission: 1/1.

Question21Marks: 1

Calculate ΔGo in kJ for the following reaction:

2Fe2O3(s) + 3C(s) → 4Fe(s) + 3CO2(g)

ΔGfo(Fe2O3(s)) = -742.2 kJ/mol

ΔGfo(C(s)) = ?

558.4

ΔGfo(Fe(s)) = ?

ΔGfo(CO2(g)) = -394.4 kJ/mol

Detailed delta G calculation

Answer:

Remember there are 3 moles of carbon dioxide, etc.CorrectMarks for this submission: 1/1.

Question22Marks: 1

Check the correct statement for the reduction of iron III oxide to iron.

Reminder given in general feedback.

Choose one answer.

a. The reaction is at equilibrium.

b. The reaction is spontaneous to the right.

c. The reaction is spontaneous to the left.

If ΔG. = 0 the process is at equilibrium.If ΔG. < 0 the process is spontaneous in the forward direction or to the right.CorrectMarks for this submission: 1/1.

Question23Marks: 1

For a certain reaction ΔGorxn = (-4.9) kJ/mol. What is the equilibrium constant at 157oC?

Remember to convert kJ to J to have consistent units and the temperature to K.

Calculating Keq from ΔGo

Answer:

301.2

3.94


Question24Marks: 1

Correctly match the following conditions.

Hint given in general feedback.

ΔGorxn << 0

ΔGorxn < 0

ΔGorxn > 0

ΔGorxn >> 0

The help for the previous problem tells you the answer.

However, you can check for yourself using the Gibb's energy equation. Try different values for the Gibb's energy or Keq at some temperature, say 300 K, and see what happens.CorrectMarks for this submission: 1/1.

Keq >> 1 (much much greater than 1)

Keq > 1

Keq < 1

Keq << 1 (much much less than 1)

chem hw solutions

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boiling point

carbon atoms

standard state

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