©Prep101 Chem209 Final Booklet Chem 209 Final Booklet ...learnfaster.ca/.../Chem209-Final-Booklet-Solutions.pdf · ©Prep101 Chem203 Final Booklet 9 of 38 Problem 22. Solution: pH

©Prep101 Chem209 Final Booklet

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Chem 209 Final Booklet Solutions


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Solutions to Equilibrium Practice Problems

Problem 3. Solution:

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

22

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 5. 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 7. Solution: Equation 2 is equal the double and reverse of equation1 therefore

Kp2 = Kp1−2 = (

1

0.157) 2 = 40.6

Problem 8. 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 9. 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 12. 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 13. 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 14. 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 16. 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 17. 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 18. 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 19. 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 20. 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 21. 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 22. Solution: pH = 8.91

pH = pKa + HCN

CN

log

= 9.20 + 0.2

0.1log

= 8.91

Problem 23. 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 25. 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 28. 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 30. 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 31. 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 33. 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 34. 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 35. Answers:

Cr = +3, O = -2

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

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

Problem 36. 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 37. 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.


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Problem 39. 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 40. 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.


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Problem 41. 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 42. 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 43. 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


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Problem 45. 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 46. 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 47. 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


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Problem 48. 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


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Problem 50. (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


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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


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Problem 53. 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 54. 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 55. 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


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Problem 56. 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 57. Answer

Problem 58. Solution: Since Hrxn > 0, then Hproducts > Hreactants

1/[A]

1/[A]

0

slope = +k

t


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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 8.314 369 298

as E

s JK mol K K

and solve for Ea, Ea = 12.8 kJ/mol

Problem 60. 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 61. 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


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Problem 62. 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]

© Prep101 Chem201 Final Booklet

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Periodic Properties Practice Problems

Problem 63. Answer: a) V5+ < Ti4+ < Sr2+ < Br-

The electron configurations are...

# of protons

Sr2+: [Kr] 38

Br-: [Kr] 35

V5+: [Ar] 23

Ti4+: [Ar] 22

Sr2+ and Br- has the same number of electrons; however, Zeff is greater for Sr2+ due to a greater # of protons, resulting in a greater + charge that pulls the electrons closer to the nucleus. Thus, Sr2+ is smaller than Br-.

V5+ and Ti4+ has the same number of electrons; however, Zeff is greater for V5+ due to a greater # of protons. Thus V5+ is smaller than Ti4+.

Sr2+ and Br- are larger than V5+ and Ti4+ because there are more electrons held in a larger subshell.

(smallest) V5+ < Ti4+ < Sr2+ < Br- (largest)

b) Cl > Br > I

Problem 64. Solutions

a) Electron affinity becomes more negative from left to right because Cl has higher Zeff than S.

b) For k, valence electron is in 4s orbital while valence electron in Li is in 2s orbital. 4s orbital is larger than 2s.

c) Ionization of Xe removes 5p electron while ionization of Kr removes 4p electron. Ionization energy

2

1

n

d) Zeff is larger in O than in B.

Problem 65. a) Mg, Ionization energy increase as you move up a group and it also increases as you move right across a period.

b) Mg, Radius increase as you move left along a period and as you move down a group.


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c) Ca<Be<P<Cl<O

Electronegativity increases as you up a group and as you move right across a period. F is the most electronegative element and Cs is the least electronegative element.

Problem 66. Answer:

a) [Ar] or [Ne]3s23p6 b) S2- c) S2- d) Ca2+ e) S2- < Ar < Ca2+

Solution:

a) [Ar] or [Ne]3s23p6 b) Ar, Ca2+ and S2- all have the same number of electrons; however, Ar has 18 protons, Ca2+ has

20, and S2- has 16. Thus S2- has the least Zeff since it has the smallest charge pulling on the electrons.

c) The species with the least favorable electron affinity is S2- because it has the smallest Zeff (a smaller positive charge to pull electrons towards the nucleus).

d) Ca2+ has the greatest Zeff because it has the greatest number of protons pulling on the 18 electrons, resulting in the electrons being closer to the nucleus.

Ionization energy is the energy required to remove an electron. It is most difficult to remove an electron from Ca2+, as this will involve the removal of a core electron instead of a valence electron, and Ca2+ has the greatest Zeff. It is easier to remove an electron from S2- versus Ar because S2- has a smaller Zeff. S2- < Ar < Ca2+


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Problem 67.

has a noble gas electron configuration? Kr

has the smallest ionization energy? Cs

has an atomic number Z = 13? Al

has a half-filled sub-shell with l = 2? Cr

is a hydrogen-like species? He+

has only one 4s electron? Cr

have two unpaired electrons? O, C, V3+

has only two d electrons with n = 3? V3+

is diamagnetic? Kr

has the largest radius? Cs

has only one electron with l = 1? Al

has the largest number of unpaired electrons?

Cr

are transition metal species? V3+, Cr


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Problem 68. Answer:

a) IE1 for K > IE1 for Ca because Zeff increases from left to right across a period, so Ca has a higher Zeff and it is therefore harder to remove an electron. IE2 for K > IE2 for Ca because the ionization process for Ca leads to the formation of the stable noble gas configuration (Ca2+) while the ionization reaction for K requires the destruction of a stable noble gas configuration. The former will always be lower in energy.

b)

K(g)+ → K(g)

2+ + e

Ca(g)+ → Ca(g)

2+ + e

c) The larger the negative charge, the greater the repulsion between electrons, the larger the radius. These are all atoms that have the configuration of Kr, so only charge influences radius. Therefore, Rb+

< Br < Se2

The ionization energy decreases as you go down Group 1 because as n increases it is easier to remove the outer electron. Radius increases as you go down Group A because as n increases, the radius increases.


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Problem 69. a)Answer:

Ca2+ < K+ < Cl < S2

b) Answer:

B < Be < O < N

c) Answer:

Sr < Mg < S < F

Problem 70. Answer:

a) When AB is put into aqueous solution, it will dissociate into A- and B+ ions rather than A+ and B-. A has a greater ionization energy, and thus is less able to form A+ than B forming B+. Further, electron affinity for A is larger, and thus the reaction A A- is more favorable than B B-.

b) A has a greater electronegativity; A has a larger electron affinity value and thus releases more energy when it gains an electron, making the reaction more favorable than B gaining an electron.

c) A will be more to the right on the periodic table than B, since it has a higher ionization energy and larger electron affinity. Because A is further to the right, it has a greater Zeff and thus will be smaller than B.

d) Since nonmetals are on the right hand side of the periodic table, and thus have more favorable electron affinity due to a greater Zeff, element A is the nonmetal.


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Chemical Bonding Solutions

Problem 71. Answer: B

Formal Charge = Valence electrons – lone pair electrons – bonds

FCF = 7 − 6 − 1 = 0

FCP = 5 − 2 − 3 = 0

FCS = 6 − 4 − 2 = 0

Problem 72. Answer: X is Carbon (C).

FCX = x − 0 − 4 = 0,

Therefore x = 4 and that is the number of valence electrons in the element. Giving us carbon.


There are 4 lone pairs of electrons.


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VSEPR Solutions


Solution: A triple bond corresponds to 2 bonds and 1 bond. The two single bonds consist of one

bond each. Therefore there is a total of three bonds and two bonds.

Problem 75. Answer:

a) N(1): sp3 N(2): sp2 O(1): sp2 O(2): sp3

b) i) 109 I ii) 120 iii) 6 lone pairs

Solution:

N(1) has 3 bonding pairs and 1 lone pairs of electrons sp3 hybridized

N(2) has 3 bonding pairs of electrons sp2 hybridized

O(1) has 1 bonding pair and 2 lone pairs of electrons sp2 hybridized

O(2) has 1 bonding pair and 3 lone pairs of electrons sp3 hybridized

i) N(1) is sp3 hybridized; tetrahedral structure; 109 bond angle

ii) N(2) is sp2 hybridized; trigonal planar; 120 bond angle

iii) By the Lewis Structure, there are 6 lone pairs of electrons.

Problem 76. Answer: D

Solution: From the Lewis diagram of the molecule we can see that the shape of the molecule is T-Shaped.


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Problem 77. Which of the following statements is/are correct for the formate ion HCO2 ?

Answer: C

Solution: the oxidation number of the C atom is +2, there are only 2 plausible contributing structures, and HCO2

= 18 e- = AX3 = trigonal planar

Problem 78. a) Answer:

b) Answer:

Sigma bonds = 4

Pi bonds = 1

c) Answer: A single bond contains one σ bond, whereas a double bond consists of one σ and one п bond. σ bonds are covalent bonds between electron pairs in the area between two atoms. п bonds are a covalent bond between parallel p orbitals and the electron pairs shared are below & above the line joining the 2 atoms.

C N

d) Answer: There are 3 electron pairs around the central C atom, and thus it is sp2 hybridized. Thus

the bond angle is 120.

п bond

σ bond

2p orbitals

sp2 orbitals


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Problem 79. Predict the geometric shape of ClO2- ion.

Answer: Bent

Solution:

Total valence electrons: 7 + 2(6) + 1 = 20 electrons

Center Cl has 2 bonding pairs and 2 lone pairs of electrons sp3 hybridized & tetrahedral formation. Since there are 2 lone pairs of electrons, the structure is bent-shaped.

Problem 80. Draw the Lewis structure for the peroxymonosulfate ion, H-O-O-SO3-, and estimate the H-O-O bond angle.

Answer: < 109

Solution:

The center oxygen in H-O-O has 4 pairs of electrons (2 lone pairs of electron and 2 bonding pairs of electrons); as such, this part of the molecule has a tetrahedral structure. The bond angle is less than

109.5 due to the 2 lone pairs of electrons that bend the structure more than bonding pairs of electrons. The tetrahedral formation means the electrons are distributed in sp3 orbitals.

-1


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Problem 81. a) Applying your knowledge of VSEPR, indicate the geometry of the three atoms indicated in acetic acid.

b) What is the O=C-O bond angle?

c) Indicate the orbital hybridization of the three atoms with the arrows.

Answer:

a) H3C- : 4 bonding pairs of electrons, thus it is tetrahedral

3 bonding pairs of electrons; thus it is trigonal planar

4 bonding pairs of electrons; thus it is tetrahedral.

b) The geometry around the central C carbon is trigonal planar structure. Thus, the bond angle is 120.

c)

H3C - C - OH Tetrahedral structure = sp3hybridization

H3C - C - OH Trigonal planar structure = sp2 planar

H3C - C - OH Tetrahedral structure = sp3 hybridization

CH3

OH

O

CH3

OH

O

CH3

OH

O


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Problem 82. a) Cl2CS (thiophosgene) (Carbon is central atom, Cl are equivalent).

Lewis:

*All formal charges are zero. Total valence electrons = 24

VSEPR

AX3

Electronic geometry: trigonal planar

Molecular geometry: trigonal planar

b) PF6

Lewis Structure

*Formal Charge on F = 0, Formal Charge on P = -1. Total valence electrons = 48

AX6

Electronic geometry: octahedral

Molecular geometry: octahedral


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Problem 83. a) HNO3

a) Total # of electrons = 1 + 5 + 3(6) = 24

Formal charge:

On O(1) 6 – [4 + ½ (4)] = 0

On O(2) 6 – [6 + ½ (2)] = -1

On O(3) 6 – [4 + ½ (4)] = 0

On N 5 – [0 + ½ (8)] = +1

On H 1 – [0 + ½ (2)] = 0

Molecular drawing:

b) O3

Total # of valence electrons: 3(6) = 18

Lewis structure:

Formal charge on:

O(1) 6 – [6 + ½ (2)] = -1

O(2) 6 – [2 + ½ (6)] = + 1

O(3) 6 – [4 + ½ (4)] = 0

Molecular structure around central O = bent

(1) (2) (3)

(2) (1)

(3)

Structure around central N: 3 electron pairs = trigonal planar

Structure around central O: 4 electron pairs = bent

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