1

Chem. 1B Final Practice Test 2 Solutions

Name__________________________________________ Print Neatly. You will lose 1 point if I cannot read your name or perm number.

Student Number _________________________________

If you are sitting next to someone with the same version of the test, you will be moved and lose 5

points.

All work must be shown on the exam for partial credit. Points will be taken off for incorrect or

no units. Calculators are allowed. Cell phones may not be used for calculators. On fundamental

and challenge problems you must show your work in order to receive credit for the problem. If

your cell phone goes off during the exam you will have your exam removed from you.

Fundamentals (of 72 possible)

Problem 1 (of 13 possible)

Problem 2 (of 17 possible)

Problem 3 (of 18 possible)

Problem 4 (of 20 possible)

Multiple Choice (of 60 possible)

Extra Credit (of 5 possible)

Final Total (of 200 possible)

First letter

of last name

2

Fundamental Questions Each of these fundamental chemistry questions is worth 6 points. You must show work to get

credit. Little to no partial credit will be rewarded. Make sure your answer includes the proper

units.

1) 6 pts The equilibrium constant K for the following reaction at 900°C is 0.0028.

What is KP at this temperature?

CS2(g) + 4H2(g) ⇌ CH4(g) + 2H2S(g)

𝐾 =[𝐶𝐻4][𝐻2𝑆]2

[𝐶𝑆2][𝐻2]4 𝐾𝑃 =

𝑃𝐶𝐻4𝑃𝐻2𝑆

2

𝑃𝐶𝑆2𝑃𝐻2

4

𝐶 =𝑛

𝑉=

𝑃

𝑅𝑇

𝐾 =

𝑃𝐶𝐻4

𝑅𝑇𝑃𝐻2𝑆

2

𝑅2𝑇2

𝑃𝐶𝑆2

𝑅𝑇

𝑃𝐻2

4

𝑅4𝑇4

= 𝑅2𝑇2𝑃𝐶𝐻4

𝑃𝐻2𝑆2

𝑃𝐶𝑆2𝑃𝐻2

4 = 𝑅2𝑇2𝐾𝑃

𝐾𝑃 =1

𝑅2𝑇2𝐾

𝐾𝑃 =1

(0.08206 𝐿∙𝑎𝑡𝑚𝑚𝑜𝑙∙𝐾

)2

(1173𝐾)2(0.0028)

𝐾𝑃 = 3.02 × 10−7

2) 6 pts When 1 mol of a fuel burns at constant pressure it produces 3452 kJ of heat

and does 11 kJ of work. What are the values of ΔE and ΔH for the

combustion of the fuel? Report your answer in 𝑘𝐽

𝑚𝑜𝑙

𝑞 = −3452 𝑘𝐽 𝑤 = −11 𝑘𝐽

∆𝐸 = 𝑞 + 𝑤 = −3452 𝑘𝐽 + −11 𝑘𝐽 = −3463 𝑘𝐽 ∆𝐻 = 𝑞 at constant temperature ∆𝐻 = −3452 𝑘𝐽

3)

6 pts

If ΔH is -12.4 𝑘𝐽

𝑚𝑜𝑙 and ΔS is 0.91 𝐽

𝑚𝑜𝑙∙𝐾 what is ΔG at 75°C and is the reaction

spontaneous.

∆𝐺 = ∆𝐻 − 𝑇∆𝑆 = −12.4 𝑘𝐽

𝑚𝑜𝑙− (348𝐾) (0.00091

𝑘𝐽

𝑚𝑜𝑙∙𝐾) = −12.7

𝑘𝐽

𝑚𝑜𝑙

Since ΔG is negative the reaction is spontaneous.

3

4) 6 pts What is the molar solubility of CaF2 in 0.010 M LiF?

CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)

𝐾𝑠𝑝 = [𝐶𝑎2+] [𝐹−]2 = 𝑆(0.010 + 2𝑠)2 = 4.0 × 10−11

Since Ksp is small assume 0.010+2S = 0.010

𝑆(0.010)2 = 4.0 × 10−11 𝑆 = 4.0 × 10−7

Check Assumption 2(4.0×10−7)

0.010100% = 0.0080% Good

Molar Solubility S=4.0×10-7

5) 6 pts

ΔG°rxn for 2SO2(g) + O2(g) 2SO3(g) is -141.74 𝑘𝐽

𝑚𝑜𝑙 at 25.00°C. What is

ΔGrxn when the partial pressure of each gas is 100. atm? What is the

spontaneous direction of the reaction under these conditions?

∆𝐺 = ∆𝐺° + 𝑅𝑇𝑙𝑛(𝑄)

∆𝐺 = ∆𝐺° + 𝑅𝑇𝑙𝑛 (𝑃𝑆𝑂3

2

𝑃𝑆𝑂2

2 𝑃𝑂2

)

∆𝐺 = −141.74 𝑘𝐽

𝑚𝑜𝑙+ (0.0083145

𝑘𝐽

𝑚𝑜𝑙∙𝐾) (298.15𝐾)𝑙𝑛 (

100.2

100.2 100.)

∆𝐺 = −153.16 𝑘𝐽

𝑚𝑜𝑙

Since ΔG is negative the reaction is spontaneous in the forward direction.

6) 6 pts What is K for the following reaction at 25°C?

Sn4+(aq) + 2Ag(s) Sn2+(aq) + 2Ag+(aq) E°=0.65 V

Are there more reactants or products at equilibrium?

Sn4+ + 2e- Sn2+

2(AgAg+ + e-)

𝐸° =𝑅𝑇

𝑛𝐹𝑙𝑛(𝐾)

0.65 𝑉 =(8.3145 𝐽

𝑚𝑜𝑙∙𝐾)(298𝐾)

(2)(96,485 𝐶𝑚𝑜𝑙

)𝑙𝑛(𝐾)

𝑙𝑛(𝐾) = 51

𝐾 = 9.7 × 1021 More products

Ca2+ F-

Initial 0 0.010 M

Change +S +2S

Equilibrium S 0.010+2S

4

7) 6 pts The preparations of two aqueous solutions are described in the table below.

For each solution, write the chemical formulas of the major species present

at equilibrium. You can leave out water itself.

Write the chemical formulas of the species that will act as acids in the 'acids'

row, the formulas of the species that will act as bases in the 'bases' row, and

the formulas of the species that will act as neither acids nor bases in the

'other' row.

You will find it useful to keep in mind that NH3 is a weak base.

0.5 mol of HNO3 is added to 1.0 L of a 0.5

NH3 solution.

acids: NH4+

bases: none

other: NO3-

0.1 mol of NaOH is added to 1.0 L of a

solution that is 0.3 M in both NH3 and

NH4Br.

acids: NH4+

bases: NH3

other: Na+, Br-

8) 6 pts A solution has a pH of 4.90, what is the pOH, [H+], and [OH-]?

𝑝𝑂𝐻 = 14.00 − 𝑝𝐻 = 14.00 − 4.90 = 9.10

[𝐻+] = 10−𝑝𝐻 = 10−4.90 = 1.3 × 10−5 𝑀

[𝑂𝐻−] =1.0 × 10−14

1.3 × 10−5= 7.7 × 10−10𝑀

9) 6 pts What is Ecell for the following reaction at 25°C?

I2(s)+ Fe(s)Fe2+(aq, 0.20 M) + 2I-(aq, 0.30 M)

𝐸 = 𝐸° −𝑅𝑇

𝑛𝐹𝑙𝑛(𝑄)

Fe Fe2+ + 2e- E° =0.44 V

I2 + 2e- 2I- E° = 0.54 V

Fe + I2 Fe2+ + 2I- E°=0.98 V

𝑄 = [𝐹𝑒2+][𝐼−]2

𝐸 = 0.98𝑉 −(8.3145 𝐽

𝑚𝑜𝑙∙𝐾)(298𝐾)

(2)(96,485 𝐶

𝑚𝑜𝑙)

𝑙𝑛((0.20)(0.30)2) = 1.03 𝑉

5

10) 6 pts A piece of iron of mass 20.0 g at 100.°C is placed in a vessel of negligible

heat capacity but containing 50.7 g of water at 22.0°C. Calculate the final

temperature of the water. Assume that there is no energy lost to the

surroundings.

𝑞𝑚𝑒𝑡𝑎𝑙 = −𝑞𝑤𝑎𝑡𝑒𝑟

𝑞 = 𝑚𝐶∆𝑇

(20.0 𝑔 𝐹𝑒) (0.45 𝐽

℃∙𝑔) (𝑇𝑓 − 100. ℃)

= −(50.7 𝑔 𝐻2𝑂) (4.18 𝐽

℃∙𝑔) (𝑇𝑓 − 22.0℃)

𝑇𝑓 = 25℃

11)

6 pts

Consider the following exothermic reaction at 25°C and 1 atm.

2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

Circle the correct answer for each of the following quantities: w, ΔH, ΔE, ΔS, and ΔG.

w<0 w>0 w=0

ΔH<0 ΔH>0 ΔH=0

ΔE<0 ΔE>0 ΔE=0

ΔS<0 ΔS>0 ΔS=0

ΔG<0 ΔG>0 ΔG=0

12) 6 pts Identify if the following reactions are redox reactions or not. If they are

redox reactions, specify what is oxidized and what is reduced.

Cu(OH)2(s) + 2HNO3(aq) Cu(NO3)2(aq) + 2H2O(l)

Not a redox reaction

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

Redox Reaction

Fe +30 gained electrons

Fe is reduced

C +2 +4 lost electrons

C is oxidized

6

Challenge Questions

Each of the following short answer questions are worth the noted points. Partial credit will be

given. Make sure to show work, report answers to the correct number of significant figures and

use the proper units.

1) 13 pts A metal forms the salt MCl3. Electrolysis of the molten salt with a current

of 0.700 A for 6.63 h produced 3.00 g of the metal. What is the molar mass

of the metal?

Determine the moles of electrons

𝑛 =𝐼𝑡

𝐹

𝑡 = 6.63ℎ (60 𝑚𝑖𝑛

1 ℎ) (

60 𝑠

1 𝑚𝑖𝑛) = 23,900 𝑠

𝑛 =𝐼𝑡

𝐹=

(0.700 𝐴)(23,900 𝑠)

96,485 𝐶𝑚𝑜𝑙

= 0.173 𝑚𝑜𝑙

Because M forms the compound MCl3 the ion that M forms is M3+

M3+ + 3e- M

Determine the moles of M

0.173 𝑚𝑜𝑙 𝑒− (1 𝑚𝑜𝑙 𝑀

3 𝑚𝑜𝑙 𝑒−) = 0.0577 𝑚𝑜𝑙 𝑀

Determine molar mass M

3.00 𝑔

0.0577 𝑚𝑜𝑙= 52.0 𝑔

𝑚𝑜𝑙

7

2) 17 pts Consider the process

A(l, 75°C) A(g, 155°C)

which is carried out at constant pressure. The total ΔS for this process is

known to be 75.0 𝐽

𝐾∙𝑚𝑜𝑙. For A(l) and A(g), CP values are 75.0 𝐽

𝐾∙𝑚𝑜𝑙, and

29.0 𝐽

𝐾∙𝑚𝑜𝑙, respectively, and are not dependent on temperature. Calculate

ΔHvap for A(l) at 125°C (its boiling point).

Break this process into 3 steps

A(l) (75°C to 125°C)A(l) to Al(g) 125°C A(g) (125°C to 155°C)

Step 1:

∆𝑆 = 𝐶𝑃𝑙𝑛 (𝑇2

𝑇1) = (75.0 𝐽

𝑚𝑜𝑙∙𝐾)𝑙𝑛 (

398𝐾

348𝐾) = 10.1 𝐽

𝐾∙𝑚𝑜𝑙

Step 2:

Phase change, therefore, T is constant. The problem indicates that

P is also constant.

∆𝑆 =𝑞

𝑇=

∆𝐻𝑣𝑎𝑝

𝑇=

∆𝐻𝑣𝑎𝑝

398𝐾

Step 3:

∆𝑆 = 𝐶𝑃𝑙𝑛 (𝑇2

𝑇1) = (29.0

𝐽

𝑚𝑜𝑙∙𝐾) 𝑙𝑛 (

428𝐾

398𝐾) = 2.11

𝐽

𝐾∙𝑚𝑜𝑙

∆𝑆𝑡𝑜𝑡 = 10.1𝐽

𝑚𝑜𝑙∙𝐾+

∆𝐻𝑣𝑎𝑝

398 𝐾+ 10.1

𝐽

𝐾∙𝑚𝑜𝑙= 75.0

𝐽

𝐾∙𝑚𝑜𝑙

∆𝐻𝑣𝑎𝑝 = 2.50 × 104 𝐽

𝑚𝑜𝑙= 25.0

𝑘𝐽

𝑚𝑜𝑙

8

3) Calculate the number of moles of HCl(g) that must be added to 1.0 L of

1.0 M NaC2H3O2 to produce a solution buffered at each pH.

3a) 6 pts pH = pKa

Na+, C2H3O2-, H

+, and Cl-

Yes a reaction will go to completion

H+ + C2H3O2- HC2H3O2

H+ C2H3O2- HC3H5O2

Initial

(mol) x 1.0 0

Final (mol) 0 1.0-x x

Major Species: C2H3O2- , HC2H3O2, Na+, and Cl-

Buffer

When the pH = pKa [HC2H3O2] = [C2H3O2-]

[𝐴−]

[𝐻𝐴]= 1 =

1.0 𝑚𝑜𝑙−𝑥

𝑉𝑥

𝑉

𝑥 = 0.5 𝑚𝑜𝑙 0.5 moles of HCl is needed

3b) 6 pts pH = 4.20

Still have a buffer

𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 ([𝐴−]

[𝐻𝐴])

𝑝𝐾𝑎 = −𝑙𝑜𝑔(1.8 × 10−5) = 4.74

4.20 = 4.74 + 𝑙𝑜𝑔 (

1.0 𝑚𝑜𝑙 − 𝑥𝑉𝑥𝑉

)

0.29 =1.0 𝑚𝑜𝑙 − 𝑥

𝑥

𝑥 = 0.78 𝑀 0.78 moles of HCl is needed.

3c) 6 pts pH = 5.00

Still have a buffer

𝑝𝐻 = 𝑝𝐾𝑎 + 𝑙𝑜𝑔 ([𝐴−]

[𝐻𝐴])

5.00 = 4.74 + 𝑙𝑜𝑔 (

1.0 𝑚𝑜𝑙 − 𝑥𝑉𝑥𝑉

)

1.8 =1.0 𝑚𝑜𝑙 − 𝑥

𝑥

𝑥 = 0.36 𝑀

0.36 moles of HCl is needed.

9

4a) 10 pts A technician carries out the reaction 2SO2(g) + O2(g) 2SO3(g) at 25˚C

and 1.00 atm in a constant-pressure cylinder fitted with a piston. Initially,

0.0300 mol SO2 and 0.0300 mol O2 are present in the cylinder. The

technician then adds a catalyst to initiate the reaction. How much work

takes place, and is it done by the system or on the system. Assume that the

reaction goes to completion and the temperature of the system is constant.

𝑤 = −𝑃∆𝑉 Determine the change in volume

Determine initial volume

𝑃𝑉 = 𝑛𝑅𝑇

𝑉 =𝑛𝑅𝑇

𝑃=

(0.0600 𝑚𝑜𝑙)(0.08206 𝐿∙𝑎𝑡𝑚𝑚𝑜𝑙∙𝐾)(298𝐾)

1.00 𝑎𝑡𝑚= 1.47 𝐿

Determine the limiting reagent and final moles of substances

Total moles of gas in the container after the reaction comes to completion

0.0150 𝑚𝑜𝑙 + 0.0300 𝑚𝑜𝑙 = 0.0450 𝑚𝑜𝑙 Determine final volume

𝑉 =𝑛𝑅𝑇

𝑃=

(0.0450 𝑚𝑜𝑙)(0.08206 𝐿∙𝑎𝑡𝑚𝑚𝑜𝑙∙𝐾

)(298𝐾)

1.00 𝑎𝑡𝑚= 1.10 𝐿

Calculate work

∆𝑉 = 1.10𝐿 − 1.47𝐿 = −0.37𝐿

𝑃 = 1.00𝑎𝑡𝑚 (101325 𝑃𝑎

1 𝑎𝑡𝑚) = 101325 𝑃𝑎

𝑤 = −𝑃∆𝑉 = −(1.00 𝑎𝑡𝑚)(−0.37 𝐿) = 0.37 𝐿 ∙ 𝑎𝑡𝑚

0.37 𝐿 ∙ 𝑎𝑡𝑚 (101.325 𝐽

1 𝐿∙𝑎𝑡𝑚) = 37𝐽

Work is done on the system

4b) 6 pts What is the change in enthalpy for the reaction in J?

∆𝐻𝑟𝑥𝑛° = ∑ ∆𝐻𝑓

°(𝑝𝑟𝑜𝑑) − ∑ ∆𝐻𝑓°(𝑟𝑒𝑎𝑐𝑡)

∆𝐻𝑟𝑥𝑛° = 2∆𝐻𝑓

°(𝑆𝑂3) − 2∆𝐻𝑓°(𝑆𝑂2) − ∆𝐻𝑓

°(𝑂2)

∆𝐻𝑟𝑥𝑛° = 2 (−396

𝑘𝐽

𝑚𝑜𝑙) − 2 (−297

𝑘𝐽

𝑚𝑜𝑙) − (0

𝑘𝐽

𝑚𝑜𝑙) = −198

𝑘𝐽

𝑚𝑜𝑙

This is the amount of heat released when 1 mole of O2 reacts with 2 mol of SO2

Calculate ΔH for 0.015 mol O2

0.15 𝑚𝑜𝑙 𝑂2 (−198 𝑘𝐽

1 𝑚𝑜𝑙 𝑂2

) = −3.0 𝑘𝐽 = −3.0 × 104𝐽

4c) 4 pts What is the change in internal energy for the reaction in J? Since the reaction takes place at constant pressure ΔH = q

∆𝐸 = 𝑞 + 𝑤 = −3.0 × 104 𝐽 + 37 𝐽 = −3.0 × 104 𝐽

SO2 (L.R) O2 SO3

I 0.0300

mol

0.0300

mol

0

F 0 mol 0.0150

mol

0.0300

mol

10

Multiple Choice Questions Each of the following multiple choice questions are worth 5 points. Your

answers need to be filled in on the Scantron provided. Note: Your

Scantrons will not be returned to you, therefore, for your records, you may

want to mark your answers on this sheet. On the Scantron you need to fill

in your perm number, test version, and name. Failure to do any of these

things will result in the loss of 1 point. Your perm number is placed and

bubbled in under the “ID number”. Do not skip boxes or put in a hyphen.

In addition, leave bubbles blank under any unused boxes. The version

number (A) is bubbled in under the “test form.”

1.The pH of a 0.10 M solution of a weak base is 9.82. What is the Kb for this base?

A) 6.6 10-4

B) 4.3 10-8

C) 2.0 10-5

D) 2.1 10-4

E) None of the above

2. For the vaporization of a liquid at a given pressure,

A) G is negative at all temperatures.

B) G is negative at low temperatures but positive at high temperatures (and zero at

some temperature).

C) G is positive at all temperatures.

D) G is positive at low temperatures but negative at high temperatures (and zero at

some temperature).

3. Predict which of the following reactions has a negative entropy change.

I. CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

II. NH3(g) + HCl(g) NH4Cl(s)

III. 2KClO4(s) 2KClO3(s) + O2(g)

A) II

B) I and II

C) I

D) III

E) II and III

11

4.Arrange the following 0.10 M solutions from lowest to highest pH: NaF, NaC2H3O2,

C5H5NHCl, KOH, HCN. (Ka for HCN is 6.2 10–10; Ka for HF is 7.2 10–4; Ka for HC2H3O2

is 1.8 10–5; and Kb for C5H5N is 1.7 10–9)

A) HCN, C5H5NHCl, NaF, NaC2H3O2, KOH

B) C5H5NHCl, HCN, NaF, NaC2H3O2, KOH

C) NaF, NaC2H3O2, HCN, C5H5NHCl, KOH

D) KOH, NaC2H3O2, NaF, HCN, C5H5NHCl

E) None of the above

5. For a reaction in a voltaic cell, both H° and S° are positive. Which of the following

statements is true?

A) G° > 0 for all temperatures.

B) E°cell will increase with an increase in temperature.

C) E°cell will decrease with an increase in temperature.

D) E°cell will not change when the temperature increases.

E) None of the above statements is true.

Use the following to answer questions 4-5:

Consider the equation 2A(g) 2B(g) + C(g). At a particular temperature, K = 1.6 104.

6. Raising the pressure by lowering the volume of the container will

A) cannot be determined

B) none of the these

C) cause [B] to increase.

D) have no effect.

E) cause [A] to increase.

7. If you mixed 5.0 mol B, 0.10 mol C, and 0.0010 mol A in a 1-L container, in which direction

would the reaction initially proceed?

A) The above mixture is the equilibrium mixture.

B) To the left.

C) We cannot tell from the information given.

D) To the right.

8. A concentration cell is constructed using two Ni electrodes with Ni2+ concentrations of 1.26 M

(cathode) and 4.95 10–4 M (anode) in the two half-cells. The reduction potential of Ni2+ is –

0.23 V. Calculate the potential of the cell at 25°C.

A) +0.101 V

B) +0.331 V

C) –0.201 V

D) –0.0287 V

E) –0.232 V

12

9. At 25°C, the following heats of reaction are known:

2C2H2 + 5O2 4CO2 + 2H2O H = –2600.0 kJ

C + O2 CO2 H = –394 kJ

2H2 + O2 2H2O H = –572 kJ

At the same temperature, calculate ΔH for the following reaction:

2C + H2 C2H2 H = ?

10. Of the following five ions or molecules, which is the strongest reducing agent?

A) Cl2

B) Fe2+

C) Cr2+

D) F-

E) H2

11. Ammonium metavanadate reacts with sulfur dioxide in acidic solution as follows (hydrogen

ions and H2O omitted):

xVO3– + ySO2 xVO2+ + ySO4

2–

The ratio x : y is

A) 2 : 1

B) 1 : 1

C) 1 : 3

D) 1 : 2

E) None of the Above

12. The following reaction has a G° value of 42.6 kJ/mol at 25°C.

HB(aq) + H2O(l) H3O+(aq) + B–(aq)

Calculate Ka for the acid HB.

A) 3.41 10–8

B) 14.0

C) 42,600

D) –17.2

E) 1.63

B, D, B, B, B, E, B, A, B, C, A, A

A) 2422 kJ

B) 226 kJ

C) -2422kJ

D) -226 kJ

E) none of these