CHE 107 SU17 Exam 2

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Question #: 1

Given the reaction 2 S2O82–(aq) + 3 I–(aq) → 2 SO42–(aq) + I3–(aq), select the correct expression

of the rate of the reaction.

A. Rate = 3∆[I−]

∆𝑡 C. Rate =

∆[SO42−]

∆𝑡

B. Rate = −1

2

Δ[S2O82−]

Δ𝑡 D. Rate = −

Δ[I3−]

Δ𝑡

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Question #: 2

Calcium phosphide and water react to form calcium hydroxide and phosphine gas.

Ca3P2(s) + 6 H2O(l) →3 Ca(OH)2(aq) + 2 PH3(g) If calcium hydroxide is produced at a rate of 0.36 M/s, what is the rate of reaction?

A. 0.48 M/s C. 0.12 M/s

B. 0.60 M/s D. 0.96 M/s

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Question #: 3

Given the following reaction and data, what is the rate law for the reaction?

2 NO(g) + Br2(g) → 2 NOBr(g)

[NO] (M) [Br2] (M) Initial Rate [NOBr]

(M/s)

0.10 0.20 24

0.25 0.20 150

0.10 0.50 60

A. rate = k[NO]2[Br2] C. rate = k[NO][Br2]2

B. rate = k[NO][Br2] D. rate = k[NO]2[Br2]2

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Question #: 4

A patient is injected with 10.0 mg of radioactive iron-59 (t1/2 = 45 days). Knowing that radioactive decay follows first-order kinetics, and assuming all of the Fe-59 is taken up by the spleen, how much remains in that organ 65 days later?

A. 2.9 mg C. 5.4 mg

B. 3.7 mg D. 1.3 mg

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Question #: 5

Which plot shows [A] vs. time for the zero-order decomposition of A → products?

A. B. C.

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Question #: 6

Crystals of the insecticide diazinon are sprinkled around a home to repel ants. Diazinon has a half-life of 40.0 days under typical environmental conditions. Assuming first-order kinetics, it will take 1 days for the diazinon to decompose to 10.0% of its original concentration in the soil. Report your answer with three significant figures. Do NOT include units in your answer.

1.

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Question #: 7

The diagram shows the energy of a system as a reaction progresses. Fill in the letter of the

correct label for numbers 1 , 2 , 3 and 4 on the diagram. A = reactants B = products C = ΔHreaction

D = Ea

1.

2.

3.

4.

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Question #: 8

If a temperature increase from 273 K to 323 K doubles the rate constant for a reaction, what is the activation energy for the reaction?

A. 10.2 kJ/mol C. –8.92 kJ/mol

B. 2.03 kJ/mol D. 16.4 kJ/mol

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Question #: 9

Consider the following reaction mechanism:

2 NO(g) ⇌ N2O2(g) fast N2O2(g) + H2(g) →N2O(g) + H2O(g) slow N2O(g) + H2(g) →N2(g) + H2O(g) fast

2 H2(g) + 2 NO(g) →N2(g) + 2 H2O(g) overall What are the functions of the selected compounds?

A. N2O2 = catalyst, N2O = reaction intermediate

B. N2O2 = reaction intermediate, N2O = reaction intermediate

C. N2O2 = catalyst, N2O = catalyst

D. N2O2 = reaction intermediate, N2O = catalyst

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Question #: 10

Consider the following reaction mechanism:

2 NO(g) ⇌ N2O2(g) fast N2O2(g) + H2(g) →N2O(g) + H2O(g) slow N2O(g) + H2(g) →N2(g) + H2O(g) fast

2 H2(g) + 2 NO(g) →N2(g) + 2 H2O(g) overall What is the rate law for this reaction?

A. rate = k[H2]2[NO]2 C. rate = k[H2]2[NO]

B. rate = k[H2][N2O2] D. rate = k[H2][NO]2

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Question #: 11

Sucrose is broken down at a faster rate in the body when the sucrase enzyme is present. In this

example, sucrose is the 1 [active site, catalyst, substrate], the sucrase enzyme is

the 2 [active site, catalyst, substrate], and the reaction proceeds at a faster rate because the

bond is strained and weakened in the 3 [active site, catalyst, substrate] of the enzyme,

leading to a different reaction mechanism with a 4 [lower, higher, comparable] activation

energy, Ea.

1.

2.

3.

4.

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Question #: 12

Starting with pure n-butane, the reaction below reaches a state of dynamic equilibrium. Which two of the following statements about the system at equilibrium are true?

A. The concentration of n-butane is equal to the concentration of isobutane.

B. The rates of the forward and reverse reactions are equal.

C. The concentrations of n-butane and isobutane remain constant.

D. The forward and reverse reactions have stopped.

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Question #: 13

Consider the reaction below at equilibrium:

3 H2(g) + CO(g) ⇔ CH4(g) + H2O(g) Kp = 1.3 × 10–25 at 298 K. What is true about this reaction?

A. The forward reaction is favored and there will be more products than reactants at

equilibrium.

B. The reverse reaction is favored and there will be more products than reactants at

equilibrium.

C. Neither direction is favored and there will be appreciable amounts of both products and

reactants at equilibrium.

D. The forward reaction is favored and there will be more reactants than products at

equilibrium.

E. The reverse reaction is favored and there will be more reactants than products at

equilibrium.

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Question #: 14

The reactions below have the equilibrium constants shown at 298 K.

S(s) + O2(g) ⇔ SO2(g) Kc1 = 4.2 × 1052 2 S(s) + 3 O2(g) ⇔ 2 SO3(g) Kc2 = 9.8 × 10128 2 SO2(g) + O2(g) ⇔ 2 SO3(g) Kc3 =

What is the equilibrium constant, Kc3, at 298 K?

A. 5.6 × 1023 C. 5.1 ×10–52

B. 2.9 ×10–25 1.1 ×1062

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Question #: 15

Consider the reaction at equilibrium below:

3 H2(g) + CO(g) ⇔ CH4(g) + H2O(g) Kp = 1.3 × 10–25 at 298 K. What is Kc for this reaction at 298 K?

A. 2.6 ×10–25 C. 3.4 × 10–18

B. 1.3 ×1024 D. 7.8 × 10–23

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Question #: 16

Choose the correct expression for Kc given the reaction below.

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

A. 𝐾𝑐 =[NaOH]2[H2]

[Na]2[H2O]2 C. 𝐾𝑐 =

[NaOH]2[H2]

[Na]2

B. 𝐾𝑐 =[NaOH]2[H2]

[H2O]2 D. 𝐾𝑐 = [NaOH]2[H2]

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Question #: 17

At some temperature, enough S8(g) was injected into an empty, rigid reaction vessel to reach a pressure of 1.00 atm. Once equilibrium was reached at the same temperature, the pressure of S8(g) was measured at 0.250 atm. What is the value of the equilibrium constant, KP, for the reaction below?

S8(g) ⇔ 4 S2(g) KP = 1 Express your answer with three significant figures.

1.

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Question #: 18

The reaction

2 N2O(g) + 3 O2(g) ⇔ 2 N2O4(g)

has an equilibrium constant, Kp = 1.1 ×106 at 298 K. If the reaction quotient, Qp = 2.2 ×103 at the same temperature,

A. the reaction will proceed towards the reactants (to the left) to reach equilibrium.

B. the reaction will proceed towards the products (to the right) to reach equilibrium.

C. the reaction has reached equilibrium and no changes in concentrations occur.

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Question #: 19

Given

2 SO2(g) + O2(g) ⇔ 2 SO3(g) Kc = 13 at 298 K and equilibrium concentrations of [SO2] = 0.697 M and [SO3] = 1.00 M, what is the equilibrium concentration of O2 at this temperature? [O2] = 1 M Report your answer to three significant figures. Do NOT include units in your answer. Use the format 2.22E2 or 2.22E-2 for numbers in scientific notation.

1.

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Question #: 20

Initially, 0.200 M N2 and 0.320 M O2 are injected into a reaction vessel. The system reaches equilibrium according to the reaction

N2(g) + O2(g) ⇔ 2 NO(g) Kc = 4.1 × 10–4 at 298 K. At equilibrium, [N2] = 1 M [O2] = 2 M [NO] = 3 M Report your answers with three significant figures. Do NOT include units in your answer. If needed, report your answer in scientific notation with the format 2.2E2 or 2.2E‐2.

1. 2. 3.

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Question #: 21

What is the best way to maximize the concentration of products?

N2(g) + 3 H2(g) ⇔ 2 NH3(g) ΔHrxn = –46 kJ

A. Increase volume, add N2, or increase temperature.

B. Increase volume, add N2, or decrease temperature.

C. Decrease volume, add N2, or decrease temperature.

D. Decrease volume, remove N2, or decrease temperature.

E. Decrease volume, remove N2, or increase temperature.

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Question #: 22

The reaction below can only be described by the _____ theory.

NH3(g) + HBr(g) → NH4Br(s) NH3 is the _____ while HBr functions as the _____.

A. Brønsted-Lowry; base; acid

B. Brønsted-Lowry; acid; base

C. Arrhenius; base; acid

D. Arrhenius; acid; base

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Question #: 23

Select the two strong acids below.

A. HNO3 B. HF C. HC2H3O2 D. HCl E. H2CO3

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Question #: 24

1 is the strongest acid and 2 is the strongest conjugate base listed in the table below. Answer with the letter only (A–F) of each species as it appears in the table.

Acid Ka Conjugate Base

A. acetic acid 1.8 ×10–5 D. acetate ion

B. nitrous acid 4.6 ×10–4 E. nitrite ion

C. ammonium ion 2.5 ×10–5 F. aniline

1. 2.

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Question #: 25

Given the table of Kw values below, what is the pH at 100ºC if [OH–] = 1.8 × 10–7 M? Report your answer with two decimal places. pH = 1

ºC Kw

0 1.14 ×10–15

10 2.93 ×10–15

20 6.81 ×10–15

25 1.00 ×10–14

30 1.47 ×10–14

40 2.92 ×10–14

50 5.48 ×10–14

100 5.13 ×10–13

1.

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CHE 107 SU17 Exam 2 - Confidential

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Question #: 1

Given the reaction 2 S2O82–(aq) + 3 I–(aq) → 2 SO4

2–(aq) + I3–(aq), select the correct

expression of the rate of the reaction.    

A.

✓B.

C.

D.

Question #: 2

Calcium phosphide and water react to form calcium hydroxide and phosphine gas.

Ca3P2(s) + 6 H2O(l) →3 Ca(OH)2(aq) + 2 PH3(g)

If calcium hydroxide is produced at a rate of 0.36 M/s, what is the rate of reaction?

A. 0.48 M/s B. 0.60 M/s

✓C. 0.12 M/s D. 0.96 M/s

Question #: 3

Given the following reaction and data, what is the rate law for the reaction?

2 NO(g) + Br2(g) → 2 NOBr(g)

 

✓A. rate = k[NO]2[Br2]

B. rate = k[NO][Br2]

C. rate = k[NO][Br2]2

D. rate = k[NO]2[Br

2]2

[NO] (M) [Br2] (M) Initial Rate [NOBr] (M/s)

0.10 0.20 24

0.25 0.20 150

0.10 0.50 60

Question #: 4

A patient is injected with 10.0 mg of radioactive iron-59 (t1/2 = 45 days). Knowing that

radioactive decay follows first-order kinetcis, and assuming all of the Fe-59 is taken up by the

spleen, how much remains in that organ 65 days later?

A. 2.9 mg

✓B. 3.7 mg C. 5.4 mg D. 1.3 mg

Question #: 5

Which plot shows [A] vs. time for the zero-order decomposition of A → products?

✓A.

B.

C.

Question #: 6

Crystals of the insecticide diazinon are sprinkled around a home to repel ants. Diazinon has a

half-life of 40.0 days under typical environmental conditions. Assuming first-order kinetics, it

will take     1      days for the diazinon to decompose to 10.0% of its original concentration in the

soil. Report your answer with three significant figures. Do NOT include units in your answer.

1. 133|1.33E2|

Question #: 7

The diagram shows the energy of a system as a reaction progresses. Fill in the letter of the

correct label for numbers      1    ,     2    ,     3     and     4     on the diagram.

A = reactants

B = products

C = ΔHreaction

D = Ea

1. A

2. D

3. C

4. B

Question #: 8

If a temperature increase from 273 K to 323 K doubles the rate constant for a reaction, what is

the activation energy for the reaction?

✓A. 10.2 kJ/mol B. 2.03 kJ/mol C. –8.92 kJ/mol D. 16.4 kJ/mol

Question #: 9

Consider the following reaction mechanism:

What are the functions of the selected compounds?

   

A. N

2O2 = catalyst, N

2O = reaction intermediate

✓B. N2O

2 = reaction intermediate, N

2O = reaction intermediate

C. N2O2 = catalyst, N

2O = catalyst

D. N2O2 = reaction intermediate, N

2O = catalyst

Question #: 10

Consider the following reaction mechanism:

What is the rate law for this reaction?

A. rate = k[H2]2[NO]2

B. rate = k[H2][N2O2] C. rate = k[H2]2[NO]

✓D. rate = k[H2][NO]2

2 NO(g) N2O2(g)   fast

N2O2(g) + H2(g) N2O(g) + H2O(g) slow

N2O(g) + H2(g) N2(g) + H2O(g)  fast

2 H2(g) + 2 NO(g) N2(g) + 2 H2O(g) overall

2 NO(g) N2O2(g)   fast

N2O2(g) + H2(g) N2O(g) + H2O(g) slow

N2O(g) + H2(g) N2(g) + H2O(g)  fast

2 H2(g) + 2 NO(g) N2(g) + 2 H2O(g) overall

Question #: 11

Sucrose is broken down at a faster rate in the body when the sucrase enzyme is present. In this

example, sucrose is the     1     [active site, catalyst, substrate], the sucrase enzyme is the     2     

[active site, catalyst, substrate], and the reaction proceeds at a faster rate because the bond is

strained and weakened in the     3     [active site, catalyst, substrate] of the enzyme, leading to a

different reaction mechanism with a     4     [lower, higher, comparable] activation energy, Ea

.

1. substrate

2. catalyst

3. active site|active sight|

4. lower

Question #: 12

Starting with pure n-butane, the reaction below reaches a state of dynamic equilibrium. Which

two of the following statements about the system at equilibrium are true?

A. The concentration of n-butane is equal to the concentration of isobutane.

✓B. The rates of the forward and reverse reactions are equal. ✓C. The concentrations of n-butane and isobutane remain constant.

D. The forward and reverse reactions have stopped.

Question #: 13

Consider the reaction below at equilibrium:

3 H2(g) + CO(g) ⇔ CH4(g) + H2O(g)   Kp = 1.3 × 10–25 at 298 K.

What is true about this reaction?

A. The forward reaction is favored and there will be more products than reactants at

equilibrium. B. The reverse reaction is favored and there will be more products than reactants at

equilibrium. C. Neither direction is favored and there will be appreciable amounts of both products and

reactants at equilibrium. D. The forward reaction is favored and there will be more reactants than products at

equilibrium. ✓E. The reverse reaction is favored and there will be more reactants than products at

equilibrium.

Question #: 14

The reactions below have the equilbrium constants shown at 298 K.

What is the equilibrium constant, Kc3, at 298 K?

 

✓A. 5.6 × 1023

B. 2.9 ×10–25

C. 5.1 ×10–52

D. 1.1 ×1062

S(s) + O2(g)  SO2(g) Kc1 = 4.2 × 1052

2 S(s) + 3 O2(g)  2 SO3(g)  Kc2 = 9.8 × 10128

2 SO2(g) + O2(g)  2 SO3(g)  Kc3 =

Question #: 15

Consider the reaction at equilibrium below:

3 H2(g) + CO(g) ⇔ CH4(g) + H2O(g)   Kp = 1.3 × 10–25 at 298 K.

What is Kc for this reaction at 298 K?

A. 2.6 ×10–25

B. 1.3 ×1024

C. 3.4 × 10–18

✓D. 7.8 × 10–23

Question #: 16

Choose the correct expression for Kc given the reaction below.

2 Na(s) + 2 H2O(l) ⇔ 2 NaOH(aq) + H2

(aq)

A.

B.

C.

✓D.

Question #: 17

At some temperature, enough S8(g) was injected into an empty, rigid reaction vessel to reach a

pressure of 1.00 atm. Once equilibrium was reached at the same temperaure, the pressure of S8(

g) was measured at 0.250 atm. What is the value of the equilibrium constant, KP, for the

reaction below?

 S8(g) ⇔ 4 S2(g)

KP =     1    

Express your answer with three significant figures.

 

1. 324

Question #: 18

The reaction

   2 N2O(g) + 3 O2(g) ⇔ 2 N2O4(g)  

has an equilibrium constant, Kp = 1.1 ×106 at 298 K. If the reaction quotient, Qp = 2.2 ×103 at

the same temperature, 

 

A. the reaction will proceed towards the reactants (to the left) to reach equilibrium.

✓B. the reaction will proceed towards the products (to the right) to reach equilibrium. C. the reaction has reached equilibrium and no changes in concentrations occur.

Question #: 19

Given

   2 SO2(g) + O2(g) ⇔ 2 SO3(g)    Kc  = 13 at 298 K

and equilibrium concentrations of [SO2] = 0.697 M and [SO3] = 1.00 M, what is the equilibrium

concentration of O2 at this temperature?

[O2

] =     1      M

Report your answer to three significant figures. Do NOT include units in your answer. Use the

format 2.22E2 or 2.22E-2 for numbers in scientific notation.

 

1. 0.158|1.58E-1|

Question #: 20

Initially, 0.200 M N2 and 0.320 M O2 are injected into a reaction vessel. The system reaches

equilibrium according to the reaction

   N2(g) + O2(g) ⇔ 2 NO(g)    Kc = 4.1 × 10–4 at 298 K.

At equilibrium,

[N2] =     1      M

[O2] =     2      M

[NO] =     3     M

Report your answers with three significant figures. Do NOT include units in your answer. If

needed, report your answer in scientific notation with the format 2.2E2 or 2.2E‐2.

1. 0.197

2. 0.317

3. 5.12E-3

Question #: 21

What is the best way to maximize the concentration of products?

N2(g) + 3 H2(g) ⇔ 2 NH3(g)    ΔHrxn = –46 kJ

 

A. Increase volume, add N2, or increase temperature. B. Increase volume, add N

2, or decrease temperature.

✓C. Decrease volume, add N2, or decrease temperature.

D. Decrease volume, remove N2, or decrease temperature.

E. Decrease volume, remove N2, or increase temperature.

Question #: 22

The reaction below can only be described by the _____ theory. 

   NH3(g) + HBr(g) → NH4Br(s)

NH3 is the _____ while HBr functions as the _____. 

✓A. Brønsted-Lowry; base; acid

B. Brønsted-Lowry; acid; base C. Arrhenius; base; acid D. Arrhenius; acid; base

Question #: 23

Select the two strong acids below.

✓A. HNO3

B. HF C. HC2H3O2

✓D. HCl E. H2CO3

Question #: 24

    1      is the strongest acid and     2      is the strongest conjugate base listed in the table

below.

Answer with the letter only (A–F) of each species as it appears in the table.

 

 

1. B|B.|

2. D|D.|

Question #: 25

Acid Ka Conjugate Base

A. acetic acid 1.8 ×10–5 D. acetate ion

B. nitrous acid 4.6 ×10–4 E. nitrite ion

C. ammonium ion 2.5 ×10–5 F. aniline

Given the table of Kw values below, what is the pH at 100ºC if [OH–] = 1.8 × 10–7 M? Reportyour answer with two decimal places.

pH =     1      

 

1. 5.55|5.54|

ºC Kw0 1.14 ×10–15

10 2.93 ×10–15

20 6.81 ×10–15

25 1.00 ×10–14

30 1.47 ×10–14

40 2.92 ×10–14

50 5.48 ×10–14

100 5.13 ×10–13