©Prep101 Chem203 Final Booklet Solutions to Thermodynamics ...learnfaster.ca/.../Final/...Chem-203-Final-Booklet.pdf · ©Prep101 Chem203 Final Booklet 1 of 36 Solutions to Thermodynamics

©Prep101 Chem203 Final Booklet

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Solutions to Thermodynamics Problems

Problem 1. Solution:

Moles of AgNO3 = 0.050 L x 0.100 M = 0.05 moles AgNO3 = 0.05 moles

Moles of HCl = 0.050 L x 0.100 M = 0.05 moles HCl = 0.05 moles Cl(aq)−

Since the reaction runs to completion, and all species are in a 1:1 ratio, 0.05 moles of AgCl are produced (you can prove this to yourself using an ICE table)

Heat lost by reaction = heat gained by solution

Heat gain = m s T

= 100.0 g H2O x 4.18J

g °C− 1 × (23.40 °C– 22.60 °C)

= 330 J = Heat Loss, this is the heat evolved from the formation of 0.050 moles of AgCl

Therefore, the heat produced per mol of AgCl =

1330 16.60

0.050 1000

J kJkJ mol

mol AgCl J

Problem 2. Answer: -1370 kJ mol-1

Heat capacity of calorimeter = C

q=(360 s) (4000 J/s) = 1440 kJ

q = C ∆T = C (50.3 - 25.0) rearrange and solve C = 56.92 kJC-1

mol / kJ 1370mol) (0.100

C) 25.0)(27.4C kJ (56.92Q

1

rxn

Problem 3. Solution: -14.2 J

Absorbing heat is an endothermic process (positive)

Work done by the system on the surroundings is negative by convention, therefore

E = q + w = 1.0 J - 15.2 J= -14.2 J


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Problem 4. Solution: q = n × smol × T

= (20.8 J J

mol °C) × 39.1 mol × (38.0 – 0.0)°C x

J

kJ

1000

1 = 30.9 kJ

Pressure-Volume work can be calculated by:

w = -PΔV = -1.00 atm x (998 – 876 L) = -122 L atm x atmL

J3.101x

J

kJ

1000

1= -12.4 kJ

So, the total internal energy, ΔE = q + w = 30.9 kJ + (−12.4 kJ) = 18.5 kJ

Problem 5. Solution: NO(g) + O3(g) → NO2(g) + O2(g) ΔH = − 199 kJ

3

2O2(g) → O3(g) ΔH = −½ (− 427 kJ)

O(g) → ½ O2(g) ΔH = −½ (+ 495 kJ)

______________________________________

NO(g) + O(g) → NO2(g) ΔH = − 233 kJ

Problem 6. Solution: 1367 kJ

C2H5OH(l) → C2H4(g) + H2O(l) ΔH = +44 kJ

C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l) ΔH = −1411 kJ

_________________________________________________

C2H5OH(l) + 3 O2(g) → 3 H2O(l) + 2 CO2(g) ΔH = −1367 kJ

Problem 7. Answer: ΔH° = 2(33.82) − 2(90.29) − 0 = −112.94kJ

mol


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Problem 8. Solution: CH3CH2OH(g) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)

The energy lost from breaking bonds and energy gained in bond formation must be determined.

Bonds Broken Bonds Formed

5 C-H Bonds (413 kJ mol-1)

1 C-C Bond (347 kJ mol-1)

1 C-O Bond (358 kJ mol-1)

1 O-H Bond (467 kJ mol-1)

3 O=O Bonds (498 kJ mol-1)

4 C=O bonds (CO2) (799 kJ mol-1)

6 O-H bonds (467 kJ mol-1)

Total = 4731 kJ Total = 5998 kJ

H = BE(Broken) BE(Formed)

H = 4731 kJ 5998 kJ = -1267 kJ

Problem 9. Answer: a) Water at 105 oC b) Na(aq)

+ + Cl(aq)−

c) H2O/MeOH mixture d) Reactants e) F2(g)


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Sreaction coeffpS ( products) coeffpS (reactants)

= CH3ClS + ½

H2S - (CH4S + ½

Cl2S )

= (245.2 J mol-1K-1) + ½(130.6 J mol-1K-1) – [(186.2 J mol-1K-1) + ½ (223 J mol-1K-1)]

= +12.8 J mol-1 K-1

Hreaction coeffpH ( products) coeffpH (reactants)

= CH3ClfH + ½

H2fH - (CH4fH + ½

Cl2fH )

= (+90 kJ mol-1) + ½(0 kJ mol-1) – [(-74.85 kJ mol-1) + ½(0 kJ mol-1)] = +164.8 kJ mol-1

At 298 K, ΔGo = ΔHo -TΔSo = 164.8 kJ – (298K)(12.8 J mol-1 K) = 161 kJ mol-1 so non-spontaneaous

Since both ΔHo and ΔSo are positive, the reaction will be spontaneous at high temperatures.

To determine at which temperature the reaction becomes spontaneous, find when ΔG < 0

Solve for ΔG = 0 = ΔHo -TΔSo

T = o

o

S

=

11

1

8.12

16480

molKJ

molJ= 12875 °K


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Problem 11. Calculate ΔHfo (C2H4(g)) at 25 0C.

Solution:

ΔH0 = ΔHf0(C2H6(g)) − ΔHf

0(C2H4(g)) − ΔHf0(H2(g))

−137.0 = −84.7 − ΔHf0(C2H4(g)) − 0

ΔHf0(C2H4(g)) = +52.3

kJ

mol

Calculate ΔS0 for the reaction at 25 0C.

Solution:

ΔG0= ΔH0-TΔS0

−101.1 = −137.0 − 298 × ΔS0

ΔS0 = −120.5J

K

Calculate S0 (C2H4(g)) at 25 0C.

Solution:

0 0 0

0

2 4

0 1

2 4

120.5 229.5 ( ( )) 130.6

( ( )) 219.4

prod reactS S S

S C H g

S C H g JK

Give a brief rationalization in physical terms for the algebraic sign of ΔS0

calculated in part (ii) above.

Solution: The product side contains fewer gas molecules than the reactant side→ more ordered.

ΔS0 is negative


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Problem 12. Answer: ΔG0=-69.9kJ, ΔS0=-94.9kJ, ΔH0=98.4kJ, ΔE0=-97.1kJ

Solution:

0 0 0

3( ) 2( )

0 1

3( ) 2( ) 2( )

0 0 0

0 0 0

0 0 0

( ) ( ) 370.3 300.4 69.9

1( ) ( ) ( ) 256.1 248.5 205.0 94.9

2

94.9 30069.9 98.4

1000

f g f g

g g g

g g

G G SO G SO kJ

S S SO S SO S O JK

G H T S

KH G T S kJ kJ

H E P V E n RT where n

0 0

1

2

8.314 198.4 300 97.1

1000 2gE H n RT kJ kJ kJ


(a)

Greaction coeffpGf( products)

coeffpGf(reactants)

= (51 kJ mol-1) – [(½(0 kJ mol-1) + 86.60 kJ mol-1) = -35.6 kJ mol-1

eqln KRTG

)298)(314.8(

6.35ln

11

1

eqKKmolJ

molkJ

RT

GK

eqln K 14.4

6

eq 108.1 K

(b) Same method. Answer: 14

eq 105.4 K


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Problem 14. Step 1: Solve for Keq at 273 K:

2

1 2 1

1 1ln

K H

K R T T

1 1

273 298273 298

H

R K KK KK K e

1

1

( 198.4 ) 1 1

24 273 2988.314273 7.0 10

kJmol

K KJKmolKK e

28

273 1.1 10KK

Step 2: Solve for 273

o

KG :

273

o

KG eqlnRT K

273

o

KG = -(8.314 J mol-1 K-1) (298 K) ln(1.1 x 1028) = -146 kJ mol-1

Step 3: Solve for 273KG with given concentrations

Q = 2

3

2

2 2

[ ]

[ ] [ ]

SO

SO O=

2

2

(0.100)

(0.500) (0.0100)= 4.0

273 273 lno o

K KG G RT Q

1 1 1

273 146 (8.314 )(298 )ln4.0o

KG kJ mol J mol K K

273

o

KG = -143 kJ mol-1


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Problem 15. (a) Calculate ΔH° for the above reaction at 25°C.

Answer: ΔH° = 467.9kJ

H0 npH f

0 ( prod ) nrH f

0 (react )

= (4x0+3(-393.5))-(2(-824.2)+3(0)) = +467.9kJ

(b) Calculate ΔS° for the above reaction at 25°C.

Answer: ΔS°= 558.4 JK-1

ΔS°= )()( 00 reactSnprodSn rp

= (4(27.3)+3(213.7))-(2(87.4)+3(5.7)) = 558.4 JK-1

(c) Account for the algebraic sign of ΔS° for the above reaction.

Answer: Entropy (disorder) increases (ie. ΔS° is positive) because 5 moles of solid are connected to 4 moles of solid and 3 moles of gas.

(d) Calculate ΔG° for the above reaction at 25 °C.

Answer: 301.5 kJ

Solution: ΔG°= ΔH°- TΔS°=467.9-298x558.4/1000=301.5kJ

(e) Calculate the equilibrium constant, K, for the above reaction at 25°C.

Answer: K=1.4x10-53

53

0

104.1

7.121298314.8

500,301ln

K

RT

GK

(f) Determine the temperature above which this reaction would become spontaneous under standard state conditions.

Answer: 838K

Solution: ie. We require ΔG°≤ 0

ie ΔG°=0 when T= KJK

J

S

H838

4.558

900,46710

0


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Solutions to Equilibrium Practice Problems


The expression for

4 10

5

4 2

3eq

eq eq

POK

P O

In (a)

4 10

5 5

4 2

1 1 3

1 1

init

init init

PO MQ

P O M M

, the reaction proceeds to the right.

In (b)

4 10

5 5

4 2

5.0 3.7 3

8.0 0.70

init

init init

PO MQ

P O M M

, the reaction proceeds to the left.


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To determine the final concentrations, the first thing needed are the initial reactant concentrations and an expression for the reaction coefficient Q.

[A2(g)] = 1.00

0.250

mol

L= 4.00 M, [AB(g)] = [B2(g)] =

2.00

0.250

mol

L= 8.00 M,

and

2 2

2 2

(8.00)2.0

(4.00) (8.00)init

init init

ABQ

A B

The direction of the reaction needs to be determined. To do this, we compare Q vs K.

Since Q = 2.0 > K = 0.5, the reaction proceeds to the left.

To determine what final concentrations will be from initial concentrations, a handy tool – the Initial, Change, Equilibrium (ICE) table can be used.

All compounds involved in the reaction are included in an ICE table as follows, with the species that will be consumed on the left side, and the species that will be produced on the right side. Since the reaction proceeds to the left, A2 and B2 will be formed and AB will be consumed.

Concentration (M)

2 AB(g) A2(g) B2(g)

Initial 8.00 M 4.00 M 8.00 M

Change -2x +x +x

Equilibrium 8.00 – 2x 4.00 + x 8.00 + x

As the reaction proceeds, x moles of A2(g) and B2(g) are formed as 2x moles of AB(g) are consumed.

Make sure to consider stoichiometric coefficients appropriately.

The equilibrium values are simply the sum of the initial + change concentrations.

Substitute the equilibrium concentrations into K, and solve for x (remember that K is defined for the reaction in the way that it was initially described):

22

2 2

(8.00 2 )0.5

(4.00 ) (8.00 )

eq

eq eq

AB xK

x xA B

so,


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2

2 2

2 2

2

(8.00 2 ) 0.5(4.00 )(8.00 )

64 32 4 0.5(32 12 )

64 32 4 16 6 0.5

3.5 38 48 0

x x x

x x x x

x x x x

x x

To solve for x, the quadratic formula must be used,

22 38 ( 38) 4(3.5)(48)41.46

2 2(3.5)

b b acx

a

or 9.39,

Since 2 x 9.39 = 18.78 > 8.00 (which would leave the equilibrium concentration of AB at equilibrium negative), then the x = 1.46

So, final concentrations are:

[A2] = 4.00 + 1.46 = 5.46 M

[B2] = 8.00 + 1.46 = 9.46 M

[AB] = 8.00 – 2(1.46) = 5.08 M

To check, substitute these concentrations into the Equilibrium constant expression,

22

2 2

(5.08)0.5

(5.46) (9.46)

eq

eq eq

ABK

A B , matches up.


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Problem 18. Answer: 4

4 4

3

[ ]

[ ] [ ]c

NOK

H NO

Solution: Pure solids and pure liquids and solvents are not included in the expression. Products go over Reactants, and coefficients in the equation are written as superscripts.


Kc = [CO2(g)]

Remember: Solids and liquids are not included in the expression.

Problem 20. Solution: Equation 2 is equal the double and reverse of equation1 therefore

Kp2 = Kp1−2 = (

1

0.157) 2 = 40.6

Problem 21. Answer: Equilibrium Constant = K1−1/2

=1

√K1

Solution: The second equation if reversed (1/K) and halved (K1/2). Combining these two gives 1/K1/2.

Problem 22. Answer: The reaction equation is:

2 CO(g) + O2(g) ⇌ 2 CO2(g) Kc = 3.3 × 1091

The relationship between Kc and Kp is:

Kp = Kc(RT)∆n gasIn this case there are 3 moles of gas in the reactants and 2 moles of gas in the

products, so ∆n = -1

So solving for Kp:

Kp = Kc(RT)∆n gas = (3.3 × 1091) [(0.0821 L atm

mol K) (298 K)]

−1

= 1.35 × 1090


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N2(g) + C2H2(g) ⇌ 2 HCN(g)

i: 1.00 1.00 1.00

c: +x +x -2x

e: 1.00+ x 1.00 + x 1.00-2x

Q=1>K so then rxn goes to the left

2

2

(1.00 2 ) (1.00 2 )

(1.00 ) (1.00 )c c

x xK K

x x

1

0.4882

c

c

Kx

K


COCl2(g) ⇌ CO(g) +Cl2(g)

initial: 0.04 0 0

change: -x +x +x

equil: 0.04-x x x

Kc = 2

0.04

x

x

x2 + Kcx – 0.04Kc = 0

23(4)(0.04)( )

5.29 102

Kc Kc Kcx


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Problem 25. Solution: 2 2

2

[ ][ ]5.10

[ ][ ]

CO HKc

CO H O

CO + H2O ⇌ CO2 +H2

I 0.1M 0.1 0.1 0.1

C -x -x +x +x

E 0.1-x 0.1-x 0.1+x 0.1+x

2 2

2 2

2 2

(0.1 ) 0.01 0.25.10

(0.1 ) 0.01 0.2

0.01 0.2 0.051 1.02

0 0.041 1.22

1.22 0.041

0.034

x x xKc

x x x

x x x x

x

x

x M

Therefore, at equilibrium [H2] = 0.1 + x = 0.1 + 0.034M = 0.134M

Problem 26. Solution: Only C is true

An increase in volume will shift the equilibrium to the right thus causing an increase in the total moles of CO at equilibrium.

Problem 27. Solution: a) shift to the left b) no effect c) shift to the right d) shift to the right


The reaction will shift left forming Ni(CO)4(g) to reach equilibrium

Problem 29. Answer: C

Solution: Q =[NO]2[Cl2]

[NOCl]2=

(1.2)2(0.56)

(1.3)2= 0.51. Q=K, therefore we are already at equilibrium.


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Solutions to Acids and Bases Practice Problems

Problem 30. Solution: Equal volumes of 0.1 M NaF and 0.1 M HF

Adding an acid and the salt of its conjugate base can form a buffer, this is the case in (b).

Problem 31. Solution: 1 and 3.

A buffer normally consists of a weak acid and its conjugate base in roughly equal amounts. The ratio should be no greater than 0.1 to 10 of weak base to weak acid. In 1, the ratio of acid to base is 3:1. In 3 the ratio is 4:1. In 2, there is a greater amount of strong base than acid, so all of the acetic acid is consumed so it is not a buffer. 4 is not a buffer solution since all of the weak acid is consumed with strong base, 5 is not a buffer because sodium acetate is a weak base, so you have a weak base with a strong base in solution.

Problem 32. Solution: B

This is a buffer solution, therefore:

2[ ] 0.400

log log(1.20 10 ) log 1.74[ ] 0.600

eq

eq

ApH pKa

HA

Problem 33. Solution: This is a buffer solution, therefore:

[ ]log

[ ]

eq

eq

ApH pKa

HA

= 3[ ]

4.75 log0.100

CH COONa

=4.75

Therefore 3[ ]log

0.100

CH COONa

=0 and 3[ ]

0.100

CH COONa

=1

[CH3COONa] = 0.10M

# moles CH3COONa = 0.10 moles

Problem 34. Solution: For this equilibrium [H+] = [In-] = 10-8 = 1 x 10-8

Ka = 1 x 10-6 = ][

]][[

HIn

InH

= ][

)10 x 1( 2-8

HIn

[HIn] = 1 x 10-10, therefore [HIn]/[In- ] = 1 x 10-10/1 x 10-8 =0.01 = 1/100


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Problem 35. Solution: pH = 8.91

pH = pKa + HCN

CN

log

= 9.20 + 0.2

0.1log

= 8.91

Problem 36. Solution: C

This is a buffer system as there are equal amounts of conjugate acid and base, but the addition of H+ can change the pH slightly. The original pH of the buffer is approximately equal to pKa. When H+ ions are added from the strong acid HCl, A- is converted into HA. Therefore the addition of 0.01 moles H+ produces 0.01 moles HA and consumes 0.01 moles A-.

nHA= 0.5 + 0.01 = 0.51 mol

nA- = 0.5 – 0.01 = 0.49 mol

[H+]=Ka x nHA/nA- = (1.8 x 10-4) (0.51/0.49) = 1.873 x 10-4

pH = -log (1.873 x 10-4) = 3.727


There are more moles of carbonate buffer in solution c than any of the other solutions

Problem 38. Solution: E

The solution is a buffer with pH above 7. A buffer is resistant to both addition of strong acid and strong base and the concentration of the hydronium ion is not more than the hydroxide ion (pH >7).


HCN + H2O ⇌ CN− + H3O+

I 0.5 𝑦 0

C −1 × 10−7 +1 × 10−7 +1 × 10−7

E 0.5 𝑦 + 1 × 10−7 +1 × 10−7

Ka = 6.2 x 10-10 = [(1x10-7)( Y+1x10-7)]/ 0.5

Y=0.003M x 1L = 0.003moles

Mass = 49.0075g/molNaCN x 0.003 moles = 0.15g


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A buffer normally consists of a weak acid and its conjugate base in roughly equal amounts. The ratio should be no greater than 0.1 to 10 of weak base to weak acid. In 1, the ratio of acid to base is 3:1. In 3 the ratio is 4:1. In 2, there is a greater amount of strong base than acid.

Problem 41. Solution: C or E

In order to determine which solutions are able to act as buffer solutions, determine what ions will be found in the solution and whether those ions are acidic, basic or spectator ions. If acidic and/or basic ions are found, calculate the amount of ions that are present. For solution A) all ions present (H+, NO3-, and Na+) are spectator ions. No buffer abilities possible. For solution B) all ions present (Na+, OH- and Cl-) are spectator ions. No buffer abilities possible. For solution C) This is a 1:1 ratio of conjugate acid to its conjugate base. This IS a buffer. For solution D) the HCl completely dissociates to form H+ and Cl- ions. NH3 can form an equilibrium where NH3 + H+ ↔ NH4+. However, HCl and its dissociated ions are present in 0.01mol amounts—the same as the amount of NH3. Therefore, all the NH3 is “used up” in reacting with the HCl, so no buffering abilities are possible. For solution E) the NaOH completely dissociates to form Na+ and OH- ions. The 0.004mol of OH- will react with the 0.01mol of HF to form 0.004mol of F- and have 0.006mol left of HF since the OH- is the limiting reagent. This means we have HA and A- both present in a ratio of 3:2. This is a buffer.


B + H2O ↔ BH+ + OH-

Calculate from the given pH, the concentration of OH- ions that dissociate form from the reaction of the base with water.

pOH = 14 – pH = 14 – (8.88) = 5.12

[OH-] = 10-pOH = 10-5.12 = 7.585 x 10-6M

Assume 1 L of solution, therefore the [B] = 0.40mol/L and [BH+] = 0.250mol/L.

K =[BH +][OH −]

[B]=

(0.250)( 7.585 x 10 − 6)

0.40= 4.74 x 10 − 6


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Problem 43. Solution: D

Because equal volumes of the acid and weak base are being mixed, all concentrations (M) can be treated as moles (mol). HNO3 is a strong acid so it completely dissociates

HNO3 H+ + NO3-

0.1 0.1 0.1

NH3 will react with the H+ released by the HNO3 to form NH4+. Initially there is 0.3M or 0.3moles of NH3. Upon addition of 0.1moles H+ (from the HNO3), 0.1mol of NH3 will react to form 0.1mol of NH4+, leaving 0.2mol NH3 unreacted.

Therefore,

NH3 + H+ ↔ NH4+

Using the following equation, solve for [H+].

[H+] = Ka x [HA]

[A-]

Kw = Ka x Kb so that Ka = Kw = 1.0 x 10-14 = 5.56 x 10-10

Kb 1.8 x 10-5

[H+] = 5.56 x 10-10 x [0.1] = 2.78 x 10-10M

[0.2]

pH = -log[H+] = -log[2.78 x 10-10] = 9.56


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Problem 44. Solution: B

Write out the equations that are occurring in the solution described above. Because both solutes are being added to 1L of water, all concentrations (M) can be treated as moles (mol).

C2H5COONa is a soluble salt so it completely dissociates

C2H5COONa C2H5COO- + Na+

0.1 0.1 0.1

HCl is a strong acid so it completely dissociates

HCl H+ + Cl-

0.01 0.01 0.01

C2H5COO- will react will all the H+ to form C2H5COOH. Initially there is 0.1 mole of C2H5COO-. When 0.01mol of H+ is added, the C2H5COO- reacts leaving 0.09mol. There is also 0.1mol of C2H5COOH to start, but when the C2H5COO- reacts with the H+, it forms 0.01mol more C2H5COOH so that the total amount of C2H5COOH is 0.11mol.

C2H5COOH ↔ C2H5COO- + H+

Before HCl is added 0.1 0.1

After HCl is added 0.11 0.09

Calculate [H+] using the following equation. (n = moles)

[H+] = Ka x nHA = 1.41 x 10-5 x (0.11) = 1.72 x 10-5M

nA- (0.09)

Calculate pH from the [H+].

pH = -log[H+] = -log[1.72 x 10-5] = 4.76


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Solutions to Electrochemistry Practice Problems

Problem 45. Solution: List the oxidation and reduction steps:

Reduction: Br2(g) + 2 e ⇌ 2 Br(aq)−

cathodeE = 1.080 V

Oxidation: Ag(aq)+ + e ⇌ Ag(s)

anodeE = 0.799 V

cellE =

cathodeE -

anodeE

= 1.080 V – 0.799 V = 0.281 V

eqcell Kn

VE ln

0257.0

or eqcell K

n

VE log

0592.0

V

nEK cell

eq0592.0

log

49.90592.0

)281.0(2log

V

VKeq

9101.3 eqK

Problem 46. Solution: The half-cell reactions are as follows (note: these are not at standard state or Ecell= 0 V):

Reduction: Pb(aq)2+ + 2 e ⇌ Pb(s) cathodeE = ?

Oxidation: Pb(aq)2+ + 2 e ⇌ Pb(s) anodeE = ?

cellE = cathodeE - anodeE

cathodeE = M

V

100.0

1log

2

0592.0125.0 = - 0.1546 V

anodeE = ][

1log

2

0592.0125.0

2

Pb

V

cellE 0.0700 V =

][

1log

2

0592.0125.01546.0

2Pb

V

[Pb(aq)2+ ] = 4.3 × 10−4

Solubility PbSO4 Pb(aq)2+ SO4

2−

Initial Some 0 0

Change -s +s +s

Equilibrium Some-s S S

Ksp = [Pb(aq)2+ ][SO4

2−]

Ksp = (4.3 × 10−4)2 = 1.9 × 10−7


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Problem 47. Answers:

Cr = +3, O = -2

Ca = +2, C = +4, H = +1, O = -2

Fe = +3, C = +4, O = -2

Problem 48. a) Answer: 0.15 V

Pb2+ + 2 e ⇌ Pb Eo = 0.13 V

Co ⇌ Co2+ + 2 e Eo= +0.28 V

Ecello = 0.28 𝑉 – 0.13𝑉 = 0.15 𝑉

(b) Answer: B

Reduction occurs at the cathode; therefore, the lead electrode is the cathode.

Problem 49. Answer: MnO4

(aq) > Zn(s) > I2(aq) > I(aq) > Zn(aq)

2+ > MnO2(aq)

An oxidizing agent gets reduced therefore MnO4

(aq) is the strongest oxidizing agent as it has the

largest Eo.


Br2(aq) + 2 e ⇌ 2 Br(aq) E = 1.09 V

2 I(aq) ⇌ I2(aq) + 2 e E = −0.54 V

Br2(aq) will be reduced and I(aq) will be oxidized.


24 of 36

Problem 51. What is the value of Ecell?

Answer: 0.55 V

Br2(aq) + 2 e ⇌ 2 Br(aq) E = 1.09 V

2 I(aq) ⇌ I2(aq) + 2 e E = −0.54 V

Br2(aq) + 2 I(aq) ⇌ 2 Br(aq)

+ I2(aq) Ecell = 0. 55V

Then put the Ecell into the Nerst:

Ecell = 0.55 − (0.0592

2) (log (

[0.2]2[0.1]

[0.2][10]2) = 0.66V

Problem 52. Answer: D>B>A>C

Solution: Reducing agent undergoes oxidation (most easily oxidized = strongest reducing agent)

From A+ + B ⇌ A + B+

B can oxidize to B+, B is a stronger reducing agent than A.

From A+ + C ⇌ no reaction

C cannot oxidize to C+ A is a stronger reducing agent than C.

From 2 B+ + D ⇌ 2 B + D2+

D can oxidize to D2+ D is a stronger reducing agent than B.

The decreasing order of reactivity (most easily oxidized to least easily oxidized) is:

D>B>A>C, where D is the strongest reducing agent.


25 of 36

Problem 53. Answer: B

Given: 2 Hg(aq)+ + 2 Br(aq)

− ⇌ 2 Hg2Br2(s)

To Find: anode reaction

ANODE allows oxidation to take place, which is the loss of electrons.

- Answers A, C and E are incorrect as they are reduction reactions.

- Answer D is incorrect as the charges on both sides are NOT balanced.

ANODE ∶ 2 Hg(l) + 2 Br(aq)− ⇌ Hg2Br2(s) + 2 e−

CATHODE: 2 Hg(aq)+ + 2 e− ⇌ 2 Hg(l)

2 Hg(l)acts as an intermediate

Problem 54. Answer: Ag+ only Cu ⇌ Cu2+ + 2 e E = −0.34V

So addition of this to the cathode reaction should be a positive number for the reaction to proceed.

The only one that this works for is Ag+

Problem 55. Answer: 0.25 V

3 × (I2(s) + 6 H2O(l) ⇌ 2 HIO3(aq) + 10 H+ + 10 e) E = −1.20V

5 × (ClO3(aq) + 6 H+ + 6 e ⇌ Claq

+ 3 H2O(l)) E = +1.45V

3 I2(s) + 5 ClO3(aq) + 3 H2O(l) ⇌ 6 HIO3(aq) + 5 Cl(aq)

E = 0.25V


In the reaction above, Sn(aq)2+ is being oxidized to Sn(aq)

4+ (it lost 2 e−) and Fe(aq)3+ is being reduced to

Fe(aq)2+ (gaining 1 e−). Ecell = Ereduction + Eoxidation. Since all E values are given as Ereduction values,

the Ered value for Fe in the reaction is +0.77 V. The E value for Sn in the reaction must be reversed because Sn is undergoing oxidation, therefore Eox = −0.15V.

Ecell = Ereduction + Eoxidation = (+0.77V) + (−0.15V) = 0.62V


26 of 36

Problem 57. Answer: A only

Calculate the Ecell for all three reactions. In statement I), Ecell= -1.86V (Cd(aq)2+ is being reduced and

Cl2(g) is being oxidized). In statement II), Ecell= -0.29V (Sn(aq)2+ is being oxidized and reduced into

Sn(aq)4+ and Sn respectively). In statement II) Ecell = 0.91V (Sn is being oxidized and Fe3+ is being

reduced). Ecell values that are positive means that the redox reactions will occur spontaneously, while a negative Ecell value means the redox reaction is not spontaneous.

Problem 58. Answer: I, II, and III

Solution: Statement I) is true becauseCu2+ has a more positive Ereduction value than Cr3+, which means that Cu2+ is a better oxidizing agent than Cr3+ (remember that an oxidizing agent oxidizes other substances and becomes reduced in the process).

Problem 59. Answer: -0.26

In the reaction given above, Ag(aq)+ is being reduced to Ag while Ni is being oxidized to Ni(aq)

2+ . The

overall Ecello for the reaction is calculated as

Ecello = Ereduction + Eoxidation

The reduction half of the reaction is given as + 0.80V. In order to calculate the E value for the oxidation half, rearrange the Ecell

o equation to solve for Eoxidation.

Eoxidation = Ecello − Ereduction = (+1.06V)– (+0.80V) = +0.26V. The Eoxidation value is the opposite

sign from the Ereduction value, so in order to solve for the Ereductionn value as asked, the sign on the

Eoxidation value must be reversed. Therefore, Ereduction for Ni(aq)2+ = −0.26V


27 of 36

Problem 60. Answer: n = 4

2 H2 + O2 ⇌ 2 H2O can be broken down into its half-cell reactions:

2 H2 ⇌ 4 H+ + 4 e−

O2 + 4 H+ + 4 e− ⇌ 2 H2O

Since there is an exchange of four electrons, n = 4 in the Nernst Equation.


Solution: E = Eo – n

V0257.0lnQ = 0.15V –

][

][ln

2

0257.0

2

2

Pb

CoV= 0.21V


28 of 36

Problem 62. (a) When current is allowed to flow, which species is oxidized?

Solution:

MnO4 + 8 H+ + 5 e ⇌ Mn2+ + 4 H2O Eo = 1.51 V

2 Cr3+ + 7 H2O ⇌ Cr2O72 + 14 H+ + 5 e Eo = −1.33 V

Oxidation is a loss of electrons. Cr3+ is being oxidized to Cr2O72

(b) When current is allowed to flow, which species is reduced?

Solution:

MnO4 is being reduced to Mn2+

(c) What is the value of cellE ?

Solution:

MnO4 + 8 H+ + 5 e ⇌ Mn2+ + 4 H2O Eo = 1.51 V

2 Cr3+ + 7 H2O ⇌ Cr2O72 + 14 H+ + 5 e Eo = −1.33 V

MnO4 + 2 Cr3+ + 3 H2O ⇌ Cr2O7

2 + 6 H+ + Mn2+ Eo = 0.18V

(d) What is the oxidation state of Cr in Cr2O72?

Solution: oxid. # = [7(-2) + 2]/2 = +6


29 of 36

Solutions to Kinetics Practice Problems


2

1 1 2 8 1

2

2 2 2 8 2

5

6

5

1

6

3

1 2 1

2 8

[ ] [ ]

[ ] [ ]

1.25 10 (0.080) (0.040)2 2 1

6.25 10 (0.040) (0.040)

1.25 10 (0.080) (0.040)2 2 1

6.25 10 (0.080) (0.020)

[ ] [ ]

[ ][

m n

m n

m nm

m n

m nn

m n

v k I S O

v k I S O

km

k

v kn

v k

rate k I S O

ratek

I S

2

2 8

53 1 1

1

]

1.25 103.91 10

(0.080) (0.040)

O

k M s


Rate = k[H2O]x[CH3Cl]y

4

4

1 (0.0100) 3.6 10

2 (0.0200) 1.44 10

x

x

Rate k

Rate k

0.5x = .25, therefore x = 2 and the reaction is second-order in H2O

4

4

1 (0.0100) 3.6 10

2 (0.0200) 1.44 10

x

x

Rate k

Rate k

5.54 (2

3)

𝑦

= 3.69

(2

3)

y

=2

3, therefore y = 1 and the reaction is first-order in CH3Cl

Rate = k[H2O]2[CH3Cl]1


30 of 36

Problem 65. Solution

1/2

1/2

1/2

1/2

00 1/2 1/2

01/2

0 01/2

0

1/2

[ ]ln[ ] ln[ ] , ln

[ ]

[ ]for [ ]

2

[ ] [ ]ln ln ln 2

[ ] [ ] 2

ln 2

t

t

t

t

AA A kt kt

A

At A

A Akt

A A

tk

Problem 66. Solution: If 20% decomposes, then 80% of the sample remains.

ktAA 0]ln[]ln[ rearrange

ln0.8[A]o

[A]o

k(50s)

k 4.463103 sec1

t1/ 2 ln 2

k155 sec

Problem 67. Solution: (a)

1 3 3 1

2 3 3 2

6 2

5 1

[ ]

[ ]

2.8 10 (5.13 10 )0.25 0.25 1

1.1 10 (2.05 10 )

n

n

nn

n

v k CH NNCH

v k CH NNCH

n

Therefore, the reaction is the first order.

Rate = k [CH3NNCH3(g)], therefore k = 6

2

2.8 10 /

5.13 10

M s

M

= 5.46 x 10-5 s-1

(b) ln[A]t = ln[A]0– kt

ln[0.1 − 5.13102] = ln[5.13102] − (5.4610−5) × t

-5.27 = -2.97 - (5.4610-5)t

(5.4610-5)t = 2.30

t = 4.21104 s = 11.7 hours

(c) [A]10% = 0.15.13102 = 5.1210-3M

Rate = k[A] = (5.4610-5 s-1)( 5.1210-3 M) = 2.8010-7 M/s


31 of 36

Problem 68. Solution: Because this reaction is second order 1 1

[ ] [ ]o

ktA A

if 1/[A] was plotted vs. t

then the y-intercept would be 1/[A]0 and the slope would be k

Problem 69. Answer

Problem 70. Solution: Since Hrxn > 0, then Hproducts > Hreactants

1/[A]

1/[A]

0

slope = +k

t


32 of 36

Thus, Ea rev = 66 – 41 = 25 kJ/mol

(draw an energy diagram to see this better)


Let T1 = 96 C = 369 K, and T2 = 25 C = 298 K.

Using the Arrhenius Equation 𝑙𝑛 (𝑘1

𝑘2) =

𝐸𝑎

𝑅(

1

𝑇2−

1

𝑇1)

10 1

10 1 1 1

3.55 x 10 1 1ln

1.2 x 10 8.314 369 298as E

s JK mol K K

and solve for Ea, Ea = 12.8 kJ/mol

Problem 72. Solution: from Arrhenius equation

121

2 11ln

TTR

E

k

k a

since vk, then

lnk2

k1

ln

v2

v1

Ea

R

1

T2

1

T1

ln(3) Ea

8.314

1

325

1

313

Ea 77433.3 J / mol 77.4 kJ / mol

Problem 73. Solution: Since we have an elementary process, then rate of the reaction is

rate k[A]2,

This reaction is second order therefore ktAA o

][

1

][

1

1

0.1M

1

0.2M k(35.2 min)

5 M-1 = k(35.2 min)

k = 1.42 x 10-1 M-1min-1


33 of 36

Problem 74. The steps are elementary, and the first step is the rate determining step. So the overall rate law only depends on the first step:

rate = k[O3][NO]


34 of 36

Solutions to Gases Practice Problems


1 1 2 2 2 22 1

1 2 1 1

2

2.00 1200273 7.00 1120

1.00 600

1120 273 847

PVPV nRT nR

T

atm mLPV PV PVT T C K K

T T PV atm mL

T K C


Increasing the temperature increases the kinetic energy of the molecules in the liquid causing the molecules to move more quickly and increase the vapor pressure.


7751 760 ; 775 1 1.02

760

273 310 583

mmHgatm mmHg P atm mmHg P atm atm atm

mmHg

T K C K

1 1

4

0.08206 5832.64 / 123.8 /

1.02

123.8 /3.99 4

31.0 /

x x x x

x

x

P P P P

P

P

P

P RTd Mr Mr d

RT P

L atm mol K KMr g L g mol

atm

Mr g molx P

Mr g mol


2 2

2

2

1 1

4 8

7351 760 ; 735 1 0.967

760

273 110 383

0.08206 3830.9720 56.5 /

0.967 0.559

56.5 /4.04 4

14.0 /

x x

x

CH CH

CH

CH

mmHgatm torr P atm torr P atm atm atm

mmHg

T K C K

L atm mol K KRTM m g g mol

PV atm L

Mr g molx C H

Mr g mol


35 of 36

Problem 79. Solutions

6 12 6 2 2 26 6 6C H O O CO H O

2 2

6 12 62

2 6 12 6

6 12 6 6 12 6

6 12 6

2

6 12 6

1 1

66 6

1

0.08206 2731.506 6 1.12

180.0 / 1.00

O O

C H OO

O C H O

C H O C H O

C H O

O

C H O

RTPV nRT V n

P

mnn n

n Mr

atm L K mol Km gRTV L

Mr P g mol atm


T K 273 30.0 C 303.0 K

MX mX RT

PV 0.225 g

8.3145 J mol1 K1 303.0 K 101.3 kPa 70.4 103 L

79.5 g/mol


Solution: atm

mmHgatmO

1

76025.0 2 = 190 mmHg O2

Total pressure = 190mmHg O2 + 440mmHg N2 = 630mmHg


Solution: PV = nRT

(1.00atm)(22.4L) = n(0.0821atmL/molK)(303K)

n = 0.90 moles helium gas


Solution:2

100.3125

32

0.3125 8.314 37332.3

30

gnO

nRTP kPa

V


36 of 36

Problem 84. Answer: A

Solution: To determine which gas has the highest density, consider the equation

Density =molar mass (

gmol

) × pressure(kPa)

R(gas constant) × temperature(K)

If measured at the same pressure and temperature, the gas that will have the highest density will have the largest molar mass. The molar masses of the gases are as follows: F2, 38g/mol; C2H6, 30.08g/mol; H2S, 34.086g/mol; NO, 30.01g/mol and SiH4, 32.13g/mol. Therefore, F2 will have the highest density because it has the largest molar mass.

Problem 85. Answer: D

Solution: PV = nRT 1 1 2 2

1 2

PV PV

T T

2

(1.00 )(600 ) (2.00 )(1200 )

280

atm mL atm mL

K T

T2 = 847C


We need to determine what mass is in the gas phase, and the rest must be in the liquid phase. We make the assumption that the liquid takes up a negligible volume of the container.

The pressure of the water vapour will be 233.7mmHg

P=1 atm / 760mmHg * 233.7 mmHg = 0.3075 atm

Now calculate the moles of gas:

n =PV

RT =

0.3075 atm ∗ 0.5 L

0.08206 L atm mol − 1 K − 1 ∗ (273.15 + 70)K= 0.0055 𝑚𝑜𝑙𝑒

mass of gaseous water is:

18g

mol× 0.0055 mol = 0.098 g

So the mass of liquid water is 0.5-0.098 = 0.402 g.

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