CHM 1046 FINAL REVIEW...CHM 1046 FINAL REVIEW Prepared & Presented By: Marian Ayoub PART II Chapter...

CHM 1046 FINAL REVIEW Prepared & Presented By: Marian Ayoub

PART II

Chapter Description

14 Chemical Equilibrium

15 Acids and Bases

16 Acid-Base Equilibrium

17 Solubility and Complex-Ion Equilibrium

19 Electrochemistry

20 Nuclear Chemistry

CH. 14 CHEMICAL EQUILIBRIUM

• DYNAMIC EQUILIBRIUM: WHEN FORWARD AND REVERSE REACTION RATES ARE THE SAME.

*PURE SOLIDS AND LIQUIDS DO NOT APPEAR IN THE EQUILIBRIUM CONSTANT EXPRESSION (ANY K OR Q)

• THE EQUILIBRIUM CONSTANT KC: THE VALUE OBTAINED FOR THE EQUILIBRIUM-CONSTANT

EXPRESSION WHEN EQUILIBRIUM CONCENTRATIONS ARE SUBSTITUTED.

• THE EQUILIBRIUM CONSTANT KP: THE EQUILIBRIUM CONSTANT EXPRESSION FOR A GASEOUS

REACTION IN TERMS OF PARTIAL PRESSURES.

FOR GASES: KP = K

C (RT)ΔN , WHERE ΔN = DIFFERENCE OF COEFFICIENT.

Ch. 14 Chemical Equilibrium

CALCULATING VALUES OF K

• IF 2 OR MORE REACTIONS ARE ADDED TO ACHIEVE A GIVEN REACTION, THE

EQUILIBRIUM CONSTANT FOR THE GIVEN EQUATION EQUALS THE PRODUCT

OF THE EQUILIBRIUM CONSTANTS (K) OF THE ADDED EQUATIONS.

• IF REACTION IS REVERSED YOU TAKE THE INVERSE OF ORIGINAL K OF REACTION.

• IF MULTIPLIED OR DIVIDED, RAISE K TO THAT POWER.

EX: MULTIPLY BY 2

EX: DIVIDE BY 3

1/K

K1xK2

K2

K1/3


USING QC TO DETERMINE THE DIRECTION OF EQUILIBRIUM

• REACTION QUOTIENT (QC): THE INITIAL REACTION RATE OF A REACTION.

ITS NOT A CONSTANT BUT DYNAMIC.

• THE RELATIONSHIP BETWEEN QC AND KC GIVES THE DIRECTION OF THE

EQUILIBRIUM.

• IF KC > QC EQUILIBRIUM IS IN THE FORWARD DIRECTION.

• IF KC = QC REACTION IS IN EQUILIBRIUM.

• IF KC < QC EQUILIBRIUM IS IN THE REVERSE DIRECTION.

• IF QC = 0 ONLY REACTANTS ARE PRESENT.

• IF QC = ∞ ONLY PRODUCTS ARE PRESENT.


1. Consider the equilibrium between Dinitrogen Tetroxide and Nitrogen Dioxide:

N2O4(g) ↔ 2NO2(g) Kp = 0.660 at 319 K

a) What is the value of Kc for this reaction?

b) What is value of Kp for the reaction 2NO2 (g) ↔ N2O4(g)

c) If the equilibrium partial pressure of NO2 (g) is 0.332 atm, what is the equilibrium partial pressure of N2O4(g)?


1. Consider the equilibrium between Dinitrogen Tetroxide and Nitrogen Dioxide:

N2O4(g) ↔ 2NO2(g) Kp = 0.660 at 319 K

a) What is the value of Kc for this reaction?

b) What is value of Kp for the reaction 2NO2 (g) ↔ N2O4(g)

c) If the equilibrium partial pressure of NO2 (g) is 0.332 atm, what is the equilibrium partial pressure of N2O4(g)?


a) Kp = K

c (RT)Δn

Kp

Kc = _______

(RT)Δn

0.660

Kc = ________________

(0.0821 x 319 )2-1

b) Reaction reversed then take the

inverse of Kp

1/ Kp = 1/ 0.660

c) (PNO2)2

Kp = _________________

(PN2O4 )

(0.223) 2

0.660 = ___________

(PN2O4 )

2. If a 2.50 L vessel at 1000˚C containes 0.525 mol CO2, 1.25 mol CF4, and 0.75 mol COF2, in what

direction will a net reaction occur to reach the equilibrium?

CO2 (g) + CF4 (g) ↔ 2COF2 (g) Kc = 0.50 @ 1000 K


[CO2] = 0.525 mol / 2.5 L = 0.21 M

[CF4] = 1.25 mol / 2.5 L = 0.5 M

[COF2] = 0.75 mol / 2.5 L = 0.3 M

[COF2]2

Qc = ___________ [CO2] [CF4]

[0.3]2

Qc = ___________

[0.21] [0.5]

Kc = 0.50

Solution:

How do you determine the direction of a reaction in order to reach equilibrium?

3. Starting with 0.100 mol each of CO and H2O in a 5.00 L flask, equilibrium is established in the

following reaction at 600K:

CO(g) + H2O(g) ↔ CO2 (g) + H2 (g) Kc = 23.2 at 600K

What is the concentration of hydrogen at equilibrium?



[CO] = 0.100 mol/ 5 L = 0.02 M

[H2O] = 0.100 mol / 5 L = 0.02 M

[X] [X]

23.2 = ______________

[0.02-X] [0.02-X]

[X]2

23.2 = _________________

[0.02-X]2

*Take the square root of both sides:

[X]

4.82 = _________________

[0.02-X]

0.096 -4.82X = X

0.096 = X + 4.82X

5.82X = 0.096

X = 0.096/ 5.82

X = 0.0165

[CO2] [H2]

Kc = _________________

[CO] [H2O]

SO [H2]= X at equilibrium

Balanced Equation

CO(g) + H2O(g) ↔

CO2 (g) H2 (g)

Initial

0.02

0.02

0

0

Change

- x

-x

+x

+x

Equilibrium

0.02-x

0.02-x

x

x

Solution

4. The Keq of a reaction is 4x10-7. At equilibrium_____.

a) The products are favored

b) The reactants are favored

c) The reactants and products are present in equal amounts

d) The rate of the forward and reverse reaction are the same






Explanation: Here Keq= 0.0000004

Keq<< 1 Therefore Reactants are favored!!

What if Keq >> 1?????






Explanation: Here Keq= 0.0000004

Keq<< 1 Therefore Reactants are favored!!

What if Keq >> 1?????

Products would be favored


5. For the reaction, PCl5 (g) ↔ PCl3 (g) + Cl2 (g) ∆Hrxn= +111 kJ. Fill

in the following table:

Changes Shifts RX'N which way?

Add PCl5

Remove Cl2

Add Ar

Decrease V (or increase P)

Increase T

Add Catalyst





Add PCl5 Right

Remove Cl2

Add Ar


Increase T

Add Catalyst





Add PCl5 Right

Remove Cl2 Right

Add Ar


Increase T

Add Catalyst





Add PCl5 Right

Remove Cl2 Right

Add Ar No effect


Increase T

Add Catalyst





Add PCl5 Right

Remove Cl2 Right

Add Ar No effect

Decrease V (or increase P) Left

High to Low #of moles of Gases

1 mol ← 2 mol

Increase T

Add Catalyst





Add PCl5 Right

Remove Cl2 Right

Add Ar No effect

Decrease V (or increase P) Left High to Low #of moles of Gases

1 mol ← 2 mol

Increase T Right

Add Catalyst





Add PCl5 Right

Remove Cl2 Right

Add Ar No effect


1 mol ← 2 mol

Increase T Right

Add Catalyst No effect on equilibrium

*Catalysts Lowers Activation Energy only





Add PCl5 Right

Remove Cl2 Right

Add Ar No effect


1 mol ← 2 mol

Increase T Right

Add Catalyst No effect on equilibrium

*Catalysts Lowers Activation Energy only


Which of the top changes would cause a change in K?

• ONLY a change in Temperature can cause a change in K

CH. 15 ACIDS AND BASES

Arrhenius Definitions:

1) Acid: substance that when dissolved in water increases [H+]

2) Base: substance that when dissolved in water increases [OH-]

Example: Identify following as Arrhenius Acid/Base

HCl NaOH

Ch. 15 Acids and Bases





Example: HCl (Acid) NaOH (Base)

Brønsted-Lowry Definitions:

1) Acid: substance that donates a proton, H+, in a reaction

2) Base: substance that accepts a proton, H+, in a reaction

Example: Identify following as Bronsted-Lowry Acid/Base

:NH3 HCl










Example: HCl (Acid) :NH3 (Base)

Lewis Definitions:

1) Acid: is an electron pair acceptor.

2) Base: is an electron pair donor.

Example: Identify following as Lewis Acid/Base

Ag+ :NH3










Example: HCl (Acid) :NH3 (Base)

Lewis Definitions:

1) Acid: is an electron pair acceptor.

2) Base: is an electron pair donor.

Example: Ag+ (Acid) :NH3(Base)


STRONG ACIDS AND BASES

• HNO3

• H2SO4

• HClO4

• HCl

• HBr

• HI

• H3O+ or H +

• GROUP 1A & 2A HYDROXIDE

• EX: NaOH, Ca(OH)2

• EXCEPT Be(OH)2


COMPARING THE STRENGTHS OF GIVEN ACIDS

• Binary Acids (H-NONMETAL):

A. Electronegativity if same period (↑ EN = ↑ strength)

EN increases Left to Right

B. Atomic size if same group (↑ atomic size = ↑ strength)

Atomic size increases Top to Bottom

• Oxacids (H-O-nonmetal): • Electronegativity (↑ EN = ↑ Strength)

EN increases Left to Right and Bottom to Top

• Other Groups (H-O-NM-On): • More Oxygen = Stronger Acid

• Carboxylic Acids (RCOOH): A. Resonance (EN)

B. Inductive Effect: distance of EN element from COOH group.

(Closer = More Acidic)


EX: H2Te vs HI

EX: HI vs HF

EX: HOCl vs HOBr

EX: H3AsO4 vs H3AsO3

EX: FCH3COOH

vs

FCH3CH2CH2COOH

Strong base gives weak CA and weak base gives strong CA

Strong acid gives weak CB and weak acid gives strong CB

Reaction is favored in the direction which leads to the formation

of weaker acid or base (or CB and CA).

Ex: Which side of the following reaction is favored?

CH3COOH + H2O ↔ CH3COO- + H3O+

Strength of Conjugate Acid and Conjugate Bases








Answer: Weak Acid Strong C.B.









Answer: Weak Acid Strong C.B.

Weak Base Strong C.A.

Therefore the Reactant side (left) is Favored




1. Identify the acid and base on each side of the following equation:

H2S + NH3 ↔ NH4+ + HS-


1. Identify the acid and base on each side of the following equation:

H2S + NH3 ↔ NH4+ + HS-

ANSWER: Acid + Base → C.A. + C.B.

2. Which of the following is the strongest Acid?

A) HOI B) HOCl C) HOBr


2. Which of the following is the strongest Acid?

A) HOI C) HOBr

Answer: HOCl

With Oxacids (H-O-nonmetal)

We use Electronegativity trend (increases bottom to top in a group)

↑EN= ↑Strength of Acid.


↑EN

3. What is the pH of a solution prepared by dissolving

0.025 mol Ba(OH)2 in water to give 455 mL of solution?





Solution: There is more than one way to solve this..

0.025 mol x 2 = 0.05 mol OH-

0.05 mol/0.455 L = 0.1099 M OH-

pOH = -log [OH-]

pOH = -log [0.1099]

pOH = 0.959

pOH + pH = 14

pH = 14 - pOH

pH = 14 – 0.959





0.025 mol x 2 = 0.05 mol OH-

0.05 mol/0.455 L = 0.1099 M OH-

pOH = -log [OH-]

pOH = -log [0.1099]

pOH = 0.959

pOH + pH = 14

pH = 14 - pOH

pH = 14 – 0.959

OR

0.1099 M OH-

Kw= [H+] [OH-] = 1x10-14

1x10-14

[H+] = __________ = 9.1x10-14 M H+

0.1099

pH = -log [H+]

pH = -log [9.1x10-14 ]

4. A sample of milk is found to have a pH of 6.25. What is the

[OH-] in this milk?


4. A sample of milk is found to have a pH of 6.25. What is the

[OH-] in this milk?



pH + pOH = 14

pOH = 14 - 6.25

pOH = 7.75

[OH-] = 10-pOH

[OH-] = 10-7.75

OR

[H+] = 10-pH

[H+] = 10-6.25

[H+] = 5.62x10-7

Kw= [H+] [OH-] = 1x10-14

1x10-14

[OH-] = __________ =

5.62x10-7

CH. 16 ACID-BASE EQUILIBRIUM

Acid ionization constant:

*Lower value of pKa = strong acid

*Higher value of pKa = weak acid

Base ionization constant:

*Lower value of pKb = strong Base

*Higher value of pKb = weak Base

Polyprotic acids: more than one ionizable proton e.g. H2SO4 is diprotic and H3PO4 is triprotic.

In a solution there will be more than just the one conjugate acid/base pair.

Ch. 16 Acid-Bases Equilibrium

• Strong Acid + Strong Base → Neutral (ex. NaCl, KNO3)

• Strong Acid + Weak Base → Acidic Solution (ex. NH4Cl )

• Weak Acid + Strong Base → Basic Solution (ex. Na2CO3)

• Weak Acid + Weak Base → Depends on K and Kb

Hydrolysis


The common ion in an equilibrium can shift the equilibrium to the opposite

side.

Ex: adding sodium acetate CH3COONa to the following reaction will

increase the concentration of acetate ion thus moving the equilibrium to the

left.

Common Ion Effect


• Buffers are solutions containing weak acid and its conjugate base

OR weak base and its conjugate acid.

• The pH changes slightly with the addition of a little acid or base.

•Acid Base Indicators: chemicals that change color with pH.

• pH of buffers can be calculated using

The traditional ICE method

or by

Henderson-Hasselbalch equation

Buffers


Buffer Calculation:

1. What is the [H3O

+] for a buffer solution that is 0.250 M in acid and 0.600

M in the corresponding salt if the weak acid Ka = 5.80 x 10

−7?


Buffer Calculation:

1. What is the [H3O



−7?

Solution:

HA(aq) + H2O(l) ↔ H3O

+ (aq) + A

− (aq)

pKa = −log(5.80 x 10−7

) = 6.237

[base] = 0.600 M and [acid] = 0.250 M


Buffer Calculation:

1. What is the [H3O



−7?

Solution:


+ (aq) + A

− (aq)

pKa = −log(5.80 x 10−7

) = 6.237

[base] = 0.600 M and [acid] = 0.250 M


6.237 [0.600]

[0.250]

Buffer Calculation:

1. What is the [H3O



−7?

Solution:


+ (aq) + A

− (aq)

pKa = −log(5.80 x 10−7

) = 6.237

[base] = 0.600 M and [acid] = 0.250 M


6.237 [0.600]

[0.250]

pH = 6.237 + log 2.40

pH = 6.237 + 0.380 = 6.617

[H3O

+] = 10

–pH

[H3O

+] = 10

–6.617

• The reactant in a solubility equilibrium is a slightly soluble salt and the

equilibrium constant for the reaction is the Solubility Product Constant, Ksp.

• Solubility problems are equilibrium problems.

Ex: the solubility equilibrium and Ksp for the salt SrF2(s) are

SrF2(s) Sr2+(aq) + 2 F – (aq) Ksp = [Sr2+][F –]2 = 2.0 x 10-10

*Note that since the reactant is a Solid, its concentration Does Not appear in the Ksp expression.

•The Molar Solubility of a salt in water can be formed by setting up an

equilibrium table and solving for x.

•The Solubility is the same quantity expressed in g/L (rather than M = mol/L).

CH. 17 Solubility and Complex-Ion Equilibrium

Ch. 17 Solubility Equilibrium

Solubility Quotient (Qip): determines if precipitation will occur

Qip > Ksp precipitation occurs (Supersaturated)

Qip = Ksp saturation of solution (Saturated)

Qip < Ksp precipitation will not occur (Unsaturated)

*In general very small value of Ksp indicates complete precipitation)

Solubility Quotient (Qip)


• Common ion effect: Addition of an ion “common” to a solubility equilibrium will reduce

solubility. *Le Châtelier’s Principle.

Ex: SrF2(s) Sr2+(aq) + 2 F –

(aq)

Adding fluoride ion, F–(aq), to the SrF2(s) equilibrium above will shift it left.

The shift will increase the amount of SrF2(s) in solid form, and thus decrease solubility.

• Effect of pH on solubility: pH of a solution will affect solubility if the conjugate ion (acid

or base) is acidic or basic.

Ex: Cl- is very weak conjugate base (CB) and is not considered basic,

whereas HCO3 - is a weak CB whose solubility is affected by the pH.

*An Anion (CB) of a weak Acid is more soluble in Acidic solutions.

Factors Affecting Solubility


• If two precipitates are possible, the least soluble salt will precipitate first (the one

with the smaller Ksp).

Ex: If you add NaOH(aq) to a solution containing equal amounts of Ca2+and Mg2+

*Mg(OH)2(s) will precipitate before Ca(OH)2(s) since

Ksp(Mg(OH)2) = 1.2 x 10-11 < Ksp(Ca(OH)2) = 8.0 x 10-6

Comparing Precipitation Of Given Salts


Common Ion Effect:

1. What would be the molar solubility of SrF2(s) in a 0.10 M NaF(aq) solution?

Ksp = 2.0 x 10-10


We set up an equilibrium (ICE) table, but now we have an initial concentration of fluoride ion:


Common Ion Effect:


Ksp = 2.0 x 10-10

Balanced Equation

SrF2(s)

Sr2+(aq) +

2 F – (aq)

Initial (M)

0

0.10

Change (M)

+x

+2x

Equilibrium (M)

x

(0.10 + 2x)


Common Ion Effect:


Ksp = 2.0 x 10-10

Balanced Equation

SrF2(s)

Sr2+(aq) +

2 F – (aq)

Initial (M)

0

0.10

Change (M)

+x

+2x

Equilibrium (M)

x

(0.10 + 2x)

We next use the equilibrium concentrations in the table and the Ksp value given above, and solve for x.

[0.10/ 2.0 x 10-10] = 500,000,000 >>>>1000

Note that since x is small, (0.10 + 2x) ~ 0.10


Common Ion Effect:


Ksp = 2.0 x 10-10

Balanced Equation

SrF2(s)

Sr2+(aq) +

2 F – (aq)

Initial (M)

0

0.10

Change (M)

+x

+2x

Equilibrium (M)

x

(0.10 + 2x)

We next use the equilibrium concentrations in the table and the Ksp value given above, and solve for x.

[0.10/ 2.0 x 10-10] = 500,000,000 >>>>1000

Note that since x is small, (0.10 + 2x) ~ 0.10

Ksp = [Sr2+][F –]2

2.0 x 10-10 = (x)( 0.10) 2

2.0 x 10-10 = (x)( 0.10)2

2.0 x 10-10 = 0.010 x


Common Ion Effect:


Ksp = 2.0 x 10-10

2. Which of the following salts is most soluble in pure water?

A) HgS, Ksp = 2.0 x 10-53

B) AgI, Ksp = 8.5 x 10-17

C) PbI2, Ksp = 7.1 x 10-9

D) CuS, Ksp = 8.7 x 10-36

E) ZnS, Ksp =1.6 x 10-24



A) HgS, Ksp = 2.0 x 10-53

B) AgI, Ksp = 8.5 x 10-17

C) PbI2, Ksp = 7.1 x 10-9

D) CuS, Ksp = 8.7 x 10-36

E) ZnS, Ksp =1.6 x 10-24

The higher the Ksp , the more soluble it would be.



A) HgS, Ksp = 2.0 x 10-53

B) AgI, Ksp = 8.5 x 10-17

D) CuS, Ksp = 8.7 x 10-36

E) ZnS, Ksp =1.6 x 10-24

The higher the Ksp , the more soluble it would be.


CH. 19 Electrochemistry

Ch. 19 Electrochemistry

Galvanic cell is Spontaneous

Ecell > 0

∆G < 0

Galvanic/Voltaic Cells: A cell that has a spontaneous redox reaction.

The terminology of a cell are:

Electrode : a metal piece at which the electrochemical reaction takes place.

Anode: where oxidation (loss of e- ) occurs. (AN OX)

Cathode: where reduction (gain of e- ) occurs. (RED CAT)

Half-cell: the reduction or the oxidation part of the cell.

Coulomb (C): the unit of electric charge.

Volt (V): one joule per coulomb.

Voltmeter: measures volts.

Cell potential (Ecell): electro-potential difference that moves the electrons from the

anode to cathode.

*Electrons flow from Anode to Cathode (Mnemonic: A to C in alphabetical order)

Galvanic/Voltaic Cells


Galvanic cell is Spontaneous

Ecell > 0

∆G < 0

Standard Electrode Potential: the tendency for reduction to occur at an

electrode under the conditions of SHE.

More Reduction Potential = Oxidizing Agent = Reduction

More Reduction Potential = Reducing Agent = Oxidation

+

-

Standard Cell Potential: E

o cell = E

o (cathode) – E

o (anode)

Eo cell = E

o (right) – E

o (left)

*Reduction potential = potential of the reduction

half reaction.

*Oxidation potential = potential of the oxidation

half reaction = reverse the sign of the reduction

potential.


Oxidation-Reduction

Oxidation

• Losing Electrons

• Reducing Agent

• Increases In Charge

Reduction

• Gaining Electrons

• Oxidizing Agent

• Decreases In Charge


1. Identify the following two reactions:

A) 1 is oxidation, 2 is reduction

B) 1 is reduction, 2 is oxidation

C) Neither

D) Both oxidation

E) Both reduction


1) Cu(s) → Cu2+ (aq) + 2e-

2) 2 Ag+(aq) + 2e- → 2 Ag(s)

1. Identify the following two reactions:

B) 1 is reduction, 2 is oxidation

C) Neither

D) Both oxidation

E) Both reduction

LeO GeR


1) Cu(s) → Cu2+ (aq) + 2e-

2) 2 Ag+(aq) + 2e- → 2 Ag(s)

2. The following questions are about this cell:

Al|Al+3

||Pb+2

|Pb

a. Identify the anode and the cathode.

b. Write the balanced overall reaction.

c. What is the potential of this cell under standard conditions?


Al|Al+3

||Pb+2

|Pb




Solution: a. An Ox, LEO (Anode Oxidation, Lose e- Oxidation)

Al loses e- and it is on the left, therefore it is the Anode


Al|Al+3

||Pb+2

|Pb






Red Cat, GeR (Reduction Cathode, Gain e- Reduction)

Pb gain e- and it is on the right, therefore it is the cathode


Al|Al+3

||Pb+2

|Pb








b. 2Al + 3Pb+2

→ 2Al+3

+ 3Pb


Al|Al+3

||Pb+2

|Pb








b. 2Al + 3Pb+2

→ 2Al+3

+ 3Pb

c.

Eo cell = E

o (cathode) – E

o (anode)

Eo cell = E

o (right) – E

o (left)

Eo cell = (-0.13) – (-1.66)


Balance the following redox reactions • (a) give the balanced half-reactions; identify the oxidation half-reaction and the reduction half-reaction. • (b) give the balanced net reaction. • (c) identify the oxidizing agent and the reducing agent.

Cl2(g) + S

2O

3

-2 (aq) Cl

-(aq) + SO

4-2 (aq) in acid solution.

Oxidation- Reduction Reactions in

Acidic/Basic Solutions


Balance the following redox reactions • (a) give the balanced half-reactions; identify the oxidation half-reaction and the reduction half-reaction. • (b) give the balanced net reaction. • (c) identify the oxidizing agent and the reducing agent.

Cl2(g) + S

2O

3

-2 (aq) Cl

-(aq) + SO

4-2 (aq) in acid solution.

1. Assign Oxidation numbers and determine which is oxidized and which is reduced Cl2(g) + S

2O

3

-2 (aq) Cl (aq) + SO4(aq)

Cl : 0 → -1 (reduced)

S : +2→ +6 (Oxidized)

S2O

3

-2

2S + 3O = -2

2S + 3(-2) = -2

2S - 6 = -2

2S = +4

S = +2

SO4

-2

S + 4O = -2

S + 4(-2) = -2

S - 8 = -2

S = +6

S = +6

2. Split into two half reactions (Oxidation and Reduction) S

2O

3

-2(aq) SO4-2 (aq)

Cl2(g) Cl

-(aq)

Oxidation- Reduction Reactions in

Acidic/Basic Solutions

3. Balance both half Reactions

1) Balance all atoms except H and O S

2O

3

-2(aq) 2 SO4-2 (aq)

Cl2(g) 2 Cl-(aq)

2) Balance O atoms by adding H20 to one side of the equation

S2O

3

-2(aq) + 5 H20 2 SO4-2 (aq)

3) Balance H atoms by adding H+ ions to one side of the equation

S2O

3

-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+

4) Balance electric charge by adding electrons (e-) to the more

positive side

S

2O

3

-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-

Cl2(g) + 2 e- 2 Cl

-(aq)




2O

3

-2(aq) 2 SO4-2 (aq)

Cl2(g) 2 Cl-(aq)


S2O

3

-2(aq) + 5 H20 2 SO4-2 (aq)


S2O

3

-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+


positive side

S

2O

3

-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-

Cl2(g) + 2 e- 2 Cl

-(aq)

4. Combine the two half reactions and obtain a final balanced equation

a. Multiply each half-reaction by a factor to make the number of electrons equal in both reactions so when added electrons cancel out

S2O

3

-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-

x4 (Cl2(g) + 2 e- 2 Cl-(aq))

4 Cl2(g) + 8 e- 8 Cl-(aq)

b. Simplify the balanced equation

S2O

3

-2(aq) + 5 H20 + 4 Cl2(g) 2 SO4-2 (aq) + 10 H+(aq) + 8 Cl

-(aq)




2O

3

-2(aq) 2 SO4-2 (aq)

Cl2(g) 2 Cl-(aq)


S2O

3

-2(aq) + 5 H20 2 SO4-2 (aq)


S2O

3

-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+


positive side

S

2O

3

-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-

Cl2(g) + 2 e- 2 Cl

-(aq)



S2O

3

-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-

x4 (Cl2(g) + 2 e- 2 Cl-(aq))

4 Cl2(g) + 8 e- 8 Cl-(aq)


S2O

3


-(aq)

Oxidizing Agent?

Reducing Agent?




2O

3

-2(aq) 2 SO4-2 (aq)

Cl2(g) 2 Cl-(aq)


S2O

3

-2(aq) + 5 H20 2 SO4-2 (aq)


S2O

3

-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+


positive side

S

2O

3

-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-

Cl2(g) + 2 e- 2 Cl

-(aq)



S2O

3

-2(aq) + 5 H20 2 SO4-2 (aq) + 10 H+ + 8 e-

x4 (Cl2(g) + 2 e- 2 Cl-(aq))

4 Cl2(g) + 8 e- 8 Cl-(aq)


S2O

3


-(aq)

Oxidizing Agent? Cl2(g)

Reducing Agent? S2O

3

-2(aq)


If this was asking to balance in Basic solution, then there are two more steps.

5. Add as many OH- ions to both sides of the equation as there H

+ ions.

S2O

3


-(aq) + 10 OH

-(aq)

6. each added OH- will react with H

+ to form H

2O

S

2O

3

-2(aq) + 5 H

20 + 4 Cl2(g) 2 SO4-2 (aq) + 10 H

20 + 8 Cl

-(aq)

*simplify the equation

S2O

3

-2(aq) + 4 Cl2(g) 2 SO4-2 (aq) + 5 H

20 + 8 Cl

-(aq)


Bibliography

Ebbing, D. D., & Gammon, S. D. (2013). General Chemistry. Cengage Learning.

General Chemistry Course Information. (2014). Retrieved from Dr. Sapna Gupta: http://drsapnag.manusadventures.com/index.php/general-chemistry

CHM 1046 FINAL REVIEW...CHM 1046 FINAL REVIEW Prepared & Presented By: Marian Ayoub PART II Chapter...

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