Welcome to Grade 12 Chemistry!

Final Exam Practice - I will have a lot of questions to work on tomorrow (Wednesday) after the redox quiz.
I will also start putting questions on the "Review Questions" page of this site

Day 32 - 24 Jan 2011

homework questions from class
Practice (use table on page 805 if needed)

1. Predict whether the following oxidation-reduction reactions should occur as written:

a)      2 Ag(s) + S(s) --> Ag2S(s no

b)      2 Ag(s) + Cu2+(aq) --> 2 Ag+(aq) + Cu(sno

c)      MnO4-(aq) + 3 Fe2+(aq) + 2 H2O(l) --> MnO2(s) + 3 Fe3+(aq) + 4 OH-(aq no

d)     MnO4-(aq) + 5 Fe2+(aq)+ 8 H+(aq) --> Mn2+(aq) + 5 Fe3+(aq) + 4 H2O(lyes

2. Which of the following pairs of ions cannot exist simultaneously in aqueous solutions?

(a) Cu+ and Fe3+     (b) Fe3+ and I-     (c) Al3+ and Co2+

a) no, a redox reaction will occur
b) no, a redox reaction will occur
c) no reaction will occur so these 2 ions can exist at the same time in the same solution

3. In this reaction:   Mn²⁺ + IO₃⁻ + H₂O → MnO₄⁻ + ½I₂+  2H⁺   What is the reducing agent?

The reducing agent is oxidized in a redox reaction.  Mn is oxidized (from +2 to +7)

4. Which of the following is most likely to gain electrons during a spontaneous redox reaction?, I₂, Li, Au or Hg?

I₂ - check position in redox table

5.  In this redox reaction,  C₂O₄²⁻- + MnO₂ → Mn²⁺ + 2CO₂  what has happened to the oxidation state of the C atom?

Oxidation state of C increases by 1. 

6.  What will happen if an iron nail is put into a solution of copper sulphate (CuSO₄)?

In this case we have Fe solid in a solution with Cu2+ and (SO4)2-.   A redox reaction should occur. Cu2+ is the oxidizing agent  in this reaction so it is reduced and Fe is oxidized to Fe2+ or Fe3+  

7.  What will happen if a piece of copper is left in a solution of zinc nitrate overnight?

In this case we have Cu solid in a solution of Zn2+ and  (NO3)-   Will copper solid be oxidized?  Zn2+ is not strong enough to oxidize copper, the copper will not replace the zinc.

8. Draw, and label, a galvanic cell that has two half cells described by the equation:

C(s) ∣ Cr₂O₇²⁻(aq) H⁺  ‖ Cu²⁺ ∣ Cu(s)

answer - text, page 699

Text – page 679, # 19  
use redox tables, page 805
a) nickel metal in a solution of silver ions.  Silver ions are good oxidizing agents and nickel can act as a reducing agent.  Reaction should occur.
b) zinc metal in solution of aluminium ions.  No reaction. 
c) aqueous solution of Cu2+ and iodide ions.  No reaction
d) chlorine gas bubbled into bromide ion solution.  Cl2 gas is a strong oxidizing agent.  A redox reaction should occur.
e) aqueous solution of Cu2+ and tin(II) ions.  Tin is Sn2+ .  Cu2+ is capable of oxidizing Sn2+ to Sn4+ so a reaction can occur.
f) copper metal in nitric acid.  Cu(s) in HNO3.  Yes, copper is oxidized by nitric acid to form Cu2+ ions.  Nitric acid is a strong oxidizing agent.

Text – page 682, #24,

a) zinc metal in HCl  yes there will be a  reaction
b) gold solid in HCl   no reaction.
c) nitric acid on copper.  Same as f in previous question.

Text – page 694, #1, 2, 5, 6,

Day 32 - 21 Jan 2011

homework questions from class


Text page 659, #13 and # 16

Determine the oxidation number of each element in the following compounds:
1. Ba(NO3)2  Ba + 2, O  -2, n +5
2.   NF3  F -1, N +3
3.  (NH4)2SO4       H +1,  O -2, N -3, S + 6
4.      Al2O3  O -2, A +3
5.       XeF4   F -1, Xe +4
6.      K2Cr2O7   K +1, O -2, Cr +6

Arrange the following compounds in order of increasing oxidation state for the carbon atom.

(a) CO      
(b) CO2
(c) H2CO
(d) CH3OH
(e) CH4

e, d, c, a, b

Balance the following redox equations by inspection or by using the oxidation number method.
Al(s)  +  MnO2(s)  -->  Al2O3(s)  +  Mn(s)  
SO2(g)  +  HNO2(aq)  -->  H2SO4(aq)  +  NO(g)
HNO3(aq) + H3AsO3(aq)  -->  NO(g) + H3AsO4(aq) + H2O(l)

4 Al(s)  +  3 MnO2(s)  -->  2 Al2O3(s)  +  3 Mn(s)
SO2(g)  +  2 HNO2(aq)  -->  H2SO4(aq)  +  2 NO(g)
2 HNO3(aq) + 3 H3AsO3(aq)  -->  2 NO(g) + 3 H3AsO4(aq) + H2O(l)

Balance the following half reactions by adding electrons where needed.  Are they oxidation or reduction reactions?
·         Pb2+ → Pb
·          Cl2 → Cl-
·          Fe3+ → Fe2+
·         N2O + H2O → NO + H+

            2e- + Pb2+ → Pb   reduction
·          2e- + Cl2 → 2Cl-   reduction
·          e- + Fe3+ → Fe2+   reduction
·         N2O + H2O → 2NO + 2H+  2e-   oxidation

For each of the following, complete the half reaction.  Is it oxidation or reduction?
1.      Dinitrogen oxide (or nitrous oxide N2O) --> nitrogen gas in acidic solution

2e- + 2H+ + N2O --> N2 + H2O

2.      Nitrite ions NO2− --> to nitrate ions NO₃⁻  in basic solution
2OH- + (NO2)- --> (NO3)- + 2e- + H2O

3.      Silver (I) oxide Ag₂O  --> silver metal in basic solution
2e- + H2O + Ag2O --> 2Ag + 2OH-

4.      Nitrate ions --> nitrous acid HNO₂ in acidic solution
2e- + 3H+ + (NO3)- --> HNO2 + H2O

5.      (MnO4)- (aq) --> Mn2+(aq) in acidic solution
5e- + 8H+ + (MnO4)-  --> Mn2+ + 4H2O

6.      Zn(s) --> Zn(OH)42-(aq) in basic solution
4OH- + Zn --> Zn(OH)4 2- + 2e-

Day 30 - 19 Jan 2011
review questions for tomorrow's quiz are on the "Review Questions" page

Day 29 - 18 Jan 2011

Homework questions:   these can be found online at

The answers are there too but please try the question before you check your answers.

In class assignment questions:

Acids & Bases

1.  If you have a 0.5 M aqueous solution of HCl, what is the [H+]?     0.5M

2.  pH + pOH = _____14______  The equation for pH is ___________________.

3.  The conjugate base of benzoic acid is benzoate:  C6H5COOH -->  H+ +  C6H5COO-          

a) What is the conjugate base of ammonium NH₄⁺ ?  ___NH3_________ 

b) What is the conjugate acid of H₂PO₄⁻ ? ______H3PO4___________ 

4.   The Ka for HCO₃⁻  is 4.7 x 10-11.  What is it’s conjugate base and its Kb?   CO3 2-  and Kb = 2.1 x 10 -4

5. If the Ka of acetic acid is 1.8 x 10-5, what is the Kb for the acetate ion?  5.6 x 10 -10

6. Calculate the percent ionization of propanoic acid if a 0.050 mol/L solution has a pH of 2.78.  answer in text, page 553

7.  If a 0.15 mol/L solution of a monoprotic acid  is 1.5% ionized at SATP, what is its Ka? 

8. A 0.10 M solution of NaCN is found to have a pH of 11.15.  Using this information, calculate the Kb of the cyanide ion and the Ka of its conjugate acid, hydrocyanic acid.  Kb = 2.02 x 10-5  and Ka = 4.94 x 10-9

9. Try the example question on page 567 of the text (barbituric acid).

answer is in the text

10. Few substances are more effective in relieving intense pain than morphine.  Morphine is an alkaloid – an alkai-like compound obtained from plants – and alkaloids are all weak bases.  In 0.01 mol/L morphine, the pH is 10.10.  Calculate the Kb for morphine. 1.6 x 10-6

11. Calculate the hydroxide ion concentration, [OH-], in a 0.025 M solution of analine, C6H5NH2, a weak base with Kb = 4.3×10-10   .      What is the [H⁺]?  7.7 x 10-12

12. Calculate the pH of a 1.00 mol/L solution of sulphuric acid H₂SO₄.   0

13. Calculate the pH of a 0.100 mol/L solution of sulphuric acid H₂SO₄.  1

Day 28 - 17 Jan 2011:
Acid base notes for class are online here - see link above.

All of Chemguy's acid base youtube videos are here at one link:

Day 26 - 13 Jan 2011:  These are the problems that we did in class today:

1.  ∆G = 0  when ___at equilibrium____and when __at a phase change (= change of state)_____

2.  Entropy is a measure of __disorder, or randomness_______ in a system, or in the surroundings.

3.  A reaction is spontaneous when ∆G is _____<0_______

4.  There are many conditions that will increase the entropy of a system.  Can you name three of these conditions? pg 497

5. The Second law of thermodynamics states that all changes either directly or indirectly _____increase______ the entropy of the ____________universe_________.  

6.  If you have a reaction:  A + B --> C.  How do you determine the ∆S for the reaction? 

7. What are the units for S?    J/mol K

8.  Use the Gibbs equation to calculate the normal condensation point of hydrogen peroxide H2O2 (l) at 101.3 kPa (standard pressure).

 The ∆H formation: for H2O2 (l) is -187.8 kJ/mole and for H2O2 (g) is -136.6 kJ/mole.

∆S for H2O2 (l) is 109.6 J/molK and for H2O2 (g) it is 233 J/molK. 

temperature = 141.8 C

9.   Calculate ΔG for the following reaction at 25°C. Will the reaction occur (be spontaneous)? How do you know? 
NH3(g) + HCl(g) → NH4Cl(s). For the reaction:  ΔH = -176.0 kJ and  ΔS = -284.8 J·K-1
   -91.1 kJ, spontaneous

10. At what temperature is the following reaction spontaneous?

       Br2(l) -->  Br2(g)  whereΔH° = 30.91 kJ/mol, ΔS° = 93.2 J/mol.K    58.6 Celcius

       If the temperature is increased to 500 K, what will happen: 

·          A) the reaction will remain spontaneous

·         B) the reaction will become non-spontaneous

·         C) the reaction will become spontaneous

·         D) The question can’t be answered

11.  Determine the ΔG° for the following reaction:

C2H4(g) + H2O(l) --> C2H5OH(l)

Tabulated ΔG°f  of C2H5OH(l) = -175 kJ/mol, C2H4(g) = 68 kJ/mol, (H2O (l) = -237 kJ/mol
(tabulated means that they have been found in a book or online)
 - 6 kJ, spontaneous

12.   Use this reaction:   C(s, diamond) + O2(g) -->  CO2(g)      ΔG° = -397 kJ

And this reaction:  C(s, graphite) + O2(g) -->  CO2(g)     ΔG° = -394 kJ

To determine the ΔG° C (s, diamond) -> C (s, graphite)    - 3kJ, spontaneous

Does the answer seem right?  What is a possible explanation for the answer? 
think about reaction rates?

13.  Is the following reaction spontaneous under standard conditions?

4KClO3(s) --> 3KClO4(s) + KCl(s)  given the following information:

∆H°f (kJ/mol): KClO3(s)  -397.7,  KClO4(s)  -432.8,  KCl(s) -436.7

S° (J/mol.K):  KClO3(s) 143.1, KClO4(s) 151.0, KCl(s) 82.6

 -133 kJ, spontaneous

14. For the same reaction as in question 13, for what temperatures will this reaction be spontaneous?  [hint: what has to happen to ΔG in the equation?]
  if ΔG = 0, T = 3731 degrees K

15.  chart

16. Calculate Ksp values at SATP for the following ionic compounds from the given SATP solubilities:                   

                      a) Silver chromate (Ag2CrO4(aq)),      6.50x10-5 mol/L  1.1 x 10 -12

                      b) aqueous Iron (III) hydroxide,     1.1x10-8 g/L
       3.04 x 10-39

17.  Will a precipitate of lead (II) sulphate, PbSO4 (s), form if 255 ml of 0.00016 mol/L lead (II) nitrate, Pb(NO3)2 (aq) is poured into 456 ml of 0.00023 mol/L sodium sulphate Na2SO4 (aq)?
text page 488 - 489

Day 25 - 12 Jan 2011: Homework

youtube videos:
http://www.youtube.com/watch?v=B4SFv_2Skdc   entropy
relationship between exothermic/endothermic and

finish the vocabulary paper from yesterday [the words are below today's homework]

text, page 493, questions 1 to 11

the answers are in the text on page 817

If you want to try more equilibrium & solubility problems:

1. Calculate the [Mg2+] in a saturated solution of Mg(OH)2 at 25°C.  Ksp = 5.6 X 10 -12 

Mg(OH) <--> Mg2+ + 2 OH-
Ksp = [Mg2+] [OH¯]²  =  4x³    x = 1.1187 x 10 ¯⁴
= [Mg 2+]

2. Calculate the mass of AgIO3 which will dissolve in 2.50 L of water at 25°C.  Ksp = 3.2 x 10 -8
AgIO3 <--> Ag+  IO3-
Ksp = [Ag+] [IO3-] =
x = 1.79 x 10 ¯⁴ but this is molarity.  That is how much will dissolve in 1L.
The question asks for 2.50L, so multiply by 2.5 to get 4.47 x 10 ¯⁴ moles
1 mole AgIO3 = 282.8g, therefore, 4.47 x 10 ¯⁴ moles weighs 0.13g

3. At a certain temperature 2.2 x 10-4 grams of CuI (copper iodide) will dissolve in 1.0 L of water (to make a saturated solution). Calculate the Ksp for CuI at this temperature.

convert g to moles.  I mole CuI is 190.4 g,   1.155  x 10 ¯⁶  moles
CuI (s) <--> Cu+ +  I-
[ ] of each product is 1.155  x 10 ¯⁶  moles
Ksp = (1.155  x 10 ¯⁶  moles)²  =  1.3 x  10 ¯²

4. It is found that 1.892 x 10-13 grams of the compound cadmium (II) sulphide (CdS) will dissolve in
350.0 mL of water to form a saturated solution. Using this data, calculate the value for the Ksp of CdS

convert grams to moles, i mole CdS = 144.59g,  therefore 1.309 x 10¯ⁱ⁵ moles
volume = 0.35 L
therefore [ ] = 3.741 x 10¯ⁱ⁵ M
Ksp = (3.741 x 10¯ⁱ⁵) (3.741 x 10¯ⁱ⁵ M)  =  1.4 x 10 ¯²⁹

5. Calculate the molar solubility of calcium oxalate (CaC2O4).  Ksp = 2.3 x 10 -4
CaC2O4 <--> Ca 2+  = C2O4 2-
Ksp = [Ca2+] [C2O4 2-] = x² 
x = 4.8 x 10¯⁵ M and that is the molar solubility of CaC2O4

6. If 250.0 mL of 0.000340M Cu(NO3)2 is mixed with 350.0 mL of 3.12 x 10-4M KIO3 (potassium iodate), will a
precipitate form?

Think of possible precipitates.
Set up an equilibrium equation for that precipitate and it's ions.
You will need to know the Ksp for that solid.  check here: http://www.ktf-split.hr/periodni/en/abc/kpt.html
For the test on Friday I will give you a list of Ksp values.
Cu(IO3)2 is the most likely precipitate.  Cu(IO3)2 <--> Cu2+  + 2 IO₃⁻, Ksp for this compound is 6.9 x 10⁻⁸
calculate the concentrations of these two ions in the solution AFTER mixing.
determine Q,   Q = [Cu2+] [IO₃⁻]²  = 4.69 x 10⁻ⁱ²
Q < Ksp
so NO precipitate

7. What does "molar solubility" mean?
It is the number of moles of a substance that will dissolve in 1 L of solution  at SATP before the solution becomes saturated.

8. Which substance is probably more soluble in aqueous solution at SATP?
a) PbCl2 or KCl (potassium chloride)?    KCl
b) H2SO4 or CaSO4?   H2SO4
c) Ba(OH)2 or BaSO4?

9.  What does the Ksp for a substance tell you about that substance?
It is a measure of how soluble a substance is at a given temperature.

10. What is the relationship between trial ion product (or Q) and Ksp? 
Know the difference in meaning between Q> Ksp and Q < Ksp.

Both are calculated from the concentration of ions in the balanced equation that you are looking at.  Ksp is the value obtained if the solution is at equilibrium ( which is also the point where the solution is saturated).  Q is the value obtained at any other time during the reaction.
if Q > Ksp, there will be precipitate
if Q < Ksp, there will be no precipitate

11. If the solubility of Pb(OH)2 is 0.155 g/L, then the concentration of each ion in a saturated solution of a Pb(OH)2 is

  • a) [Pb2+] = 0.155 g/L and [OH-] = 0.155 g/L
  • b) [Pb2+] = 0.052 g/L and [OH-] = 0.103 g/L
  • c) [Pb2+] = 6.43 x 10-4 M and [OH-] 1.29 x 10-3 M
  • d) [Pb2+] = 6.43 x 10-4 M and [ 0H-]= 6.43 x 10-4 M
Pb(OH)2 <--> Pb2+ +  2OH-  so there will be twice as much OH- in moles.

12. At 25°C, the solubility of an unknown compound is 7.1 x 10-5 M. The compound is

  • a) CuI (Ksp = 1.6 x 10-16)
  • b) AgI (Ksp = 8.3 x 10-17)
  • c) CaCO3 (Ksp = 5.0 x 10-9) 
  • d) BaSO4 (Ksp = 1.5 x 10-9)
In all 4 of these compounds, Ksp = x²  where x is the concentration of one of the ions.
If molar solubility is 7.1 x 10-5 M, there will be 7.1 x 10-5 M of each ion,
if x = 7.1 x 10-5 M, Ksp = 5.04 x 10-9

13. Which of the following describes the changes in ion concentrations when 1.0 g of solid ZnS is added to a saturated solution of ZnS?   If the solution is already saturated, any added ZnS will precipitate.  The [ ] of the ions can not change.

  • a)  [Zn2+] increases and [S2-]  decreases
  • b)  [Zn2+] decreases  and [S2-] decreases
  • c)  [Zn2+] increases  and [S2-]  increases
  • d)  [Zn2+] remains constant   and [S2-] remains constant 
If the solution is already saturated, any added ZnS will precipitate.  The [ ] of the ions can not change.

14. Which of the following compounds will form a saturated solution with the greatest concentration of Ag+?

  • a) AgI (Ksp = 8.3 x 10-17)
  • b) AgBr (Ksp = 7.7 x 10-13)
  • c) AgCl (Ksp = 1.8 x 10-10)
  • d) AgCN (Ksp = 2.0 x 10-12)
Ksp = [Ag+] [anion] 
If you increase the Ksp, you increase the amount of the substance that can dissolve.  




Solubility product constant (pg. 483)


Saturated solution (pg 483)

Supersaturated solution (pg. 487)

Common ion effect (pg. 490)

Entropy, S, - measure of randomness or disorder.  (pg. 496

Bond energy (pg. 495)

Standard entropy (pg. 505)

Spontaneous reactions (pg. 494)

Free energy (Gibbs free energy) ∆G, (pg. 498, 499)

Electrolyte – any substance containing free ions that make the substance electrically conductive.

Ionic solution – a solution containing ions

Trial ion product (pg. 487)

First law of thermodynamics (pg. 494)

Second law of thermodynamics (pg. 500)

Third law of thermodynamics (pg. 505)

Day 24 - 11 Jan 2011, homework:

youtube videos on equilibrium and equilibrium shift

http://www.youtube.com/watch?v=uLA_nWvLtJc       effect of temperature on exothermic/endothermic reactions
http://www.youtube.com/watch?v=9aYe5_Xa2-I&feature=related     chemguy on calculating equilibrium constants
http://www.youtube.com/watch?v=FiHPsic2dWA&feature=related    chemguy on calculoating equilibrium concentrations
http://www.youtube.com/watch?v=4wWbfRUd-rw&feature=related      chemguy on K and Q
http://www.youtube.com/watch?v=oKRJtwMUtD4&NR=1    chemguy on Le Chateliers Principle

Day 23 - 10 Jan 2011, homework:
text page 449, #2, 3, 4, 6

1. Consider the quilibrium N2(g) + 3 H2(g) = 2 NH3(g) at a certain temperature. An equilibrium mixture in a 4.00 liter vessel contains 1.60 moles NH3, 0.800 moles N2, and 1.20 moles H2. What is the value of Kc?

Start by calculating [ ]s of reactants and products in mol/L.     0.40 moles/L  NH3, 0.200 moles/L N2 and 0.30 moles/L  H2

2.  PCl3 + Cl2 = PCl5
For this reaction - if we keep the equilibrium concentrations of the reactants at [PCl3] = 0.5 M and [Cl2]  = 0.07 M and the temperature at 400K, so that Kc = 96.2 M-1. What is the equilibrium concentration of the product, PCl5 ?

[PCl5] = (96.2)(0.5)(0.07) mol/L = 3.4 M

3. A mixture of 0.75 mol of N2 and 1.20 mol of H2 are placed in a 3.0 liter container.
When the reaction N2 (g) + 3H2 (g) <--> 2NH3 (g) reaches equilibrium,  [H2] = 0.100M.
What is the value of [N2] and [NH3] at equilibrium?

[NH3] = 0.200 M
[N2] = 0.150 M

4. A mixture of 2.5 moles H2O and 100 g of C are placed in a 50-L container and allowed to come to equilibrium subject to the following reaction:
C(s) + H2O(g) <--> CO(g) + H2(g),
The equilibrium concentration of Hydrogen is found to be [H2] = 0.040 M.

What is the equilibrium concentration of water vapor, [H2O] ?

Answer: [H2O] = 0.010 M

5. The following system is at equilibrium. In which direction (right or left) will the equilibrium position shift with the following changes?
3NO(g) <--> N2O(g) + NO2(g) + 154.9 kJ
a) increasing the temperature
b) adding more N2O
c) adding more NO
d) adding a catalyst
e) increading the volume

a) exothermic reaction, increase temp. shifts left  b) left  c) right  d) no change  e) increase volume favours side with more molecules so shift left

Day 20 - 5 Jan 2011, homework questions:

youtube videos to help you:
Collision theory -  http://www.youtube.com/user/ChemFlicks#p/u/3/SyznsEFOopg
Kinetic & Potential energy  - http://www.youtube.com/user/ChemFlicks#p/u/2/48pySSbamIg
Potential Energy diagrams:  http://www.youtube.com/user/ChemFlicks#p/u/1/D7TfmT85Cjw
Enthalpy & Enthalpy Change:  http://www.youtube.com/user/ChemFlicks#p/u/7/LjJi8nXTHqA
Rate Problems:  http://www.youtube.com/user/ChemFlicks#p/u/12/RDJrcFVJf0w

Factors Affecting Reaction Rate:  http://www.youtube.com/user/ChemFlicks#p/u/5/uUON--ay77A

Part A:

1. C2H4 (g) + 3O2(g) --> 2CO2(g) + 2H2O(g)
An experiment is done to determine the rate of the reaction.  It is found that  0.26 moles of O2 is consumed in 3.0 minutes.
What is the rate of production of CO2 in g/min ?

Combustion of ethene:
If 2 moles of CO2 are produced for every 3 moles of O2 consumed, and 0.26 moles  of O2 are consumed, then 0.173 moles of CO2 are produced in 3 minutes.
This means that 0.058 moles of CO2 produced in 1 minute.
1 mole of CO2 weighs 44.01 grams which means that 0.058 moles weighs 25.5 grams.
The rate of production of CO2 is 25.5 g/min

2. For the reaction, N2(g)  +  3 H2(g)  -->  2 NH3(g)
If the rate of formation of NH3 is 8.0 x 10-3 mol/s, calculate the rate of consumption of H2 in mol/s

3 moles of H2 are consumed for every 2 moles NH3 produced.
In one second .008 moles of NH3 are produced according to the data.
In one second 0.12 moles of H2 are consumed.   The rate of H2 consumption is 0.12 moles/sec

3. For the reaction:  Sn(s) + 2 HCl(aq)  -->  H2(g)  +  SnCl2(aq)
What could you do to try to increase the rate of this reaction?
Is this a heterogeneous or a homogeneous solution (for the reactants)?

Think about the factors that affect rate - temperature, catalysts, concentration of reactants & surface area.
The reaction is heterogeneous.

4. It is known that compounds called chlorofluorocarbons  (C.F.C.s) (eg. CFCl3) will
break up in the presence of ultraviolet radiation, such as found in the upper atmosphere, forming single chlorine atoms:

                                                       CFCl3  ->  CFCl2  +   Cl

         The Cl atoms then react with Ozone (O3) as outlined in the following mechanism.      
             Step 1:        Cl   +   O3   -->   ClO  +  O2
             Step 2:        ClO  +   O  -->  Cl  +   O2      (single "O" atoms occur naturally in the atmosphere.)
         a)  Write the equation for the overall reaction. (Using steps 1 and 2)
          b) What is the catalyst in this reaction?                 
         c)  Identify an intermediate in this reaction              
             d)  Explain how a small amount of chlorofluorocarbons can destroy a large amount of  ozone.
            e)  What breaks the bond in the CFCl3 and releases the free Cl atom?

a) O3 + O --> 2O2
b) Cl
c) ClO
d) because only a small amount of Cl is needed to catalyze a large number of reactions. 
e) heat/energy from the sun

5. What is meant by the rate determining step in a reaction mechanism?
The rate determining step is the slowest step in a multi-step chemical reaction.

6. What is meant by a reaction mechanism?

A reaction mechanism is the step by step sequence of elementary reactions in a chemical reaction.

Part B: Choose the best answer: (answers are below, in orange)

1: When chemists speak of speeds, one of the quantities that is almost always included is:
a) Mass
b) Concentration
c) Volume
d) Time

2: Which of the following is not important when considering reaction rate?
a) Molecular geometry
b) state
c) Concentration
d) Temperature

3: Generally, reaction rates will increase with _________ temperature.
a) Increasing
b) Decreasing
c) Static  (stays the same)
d) Low

4: A substance that is involved in changing the rate of a chemical reaction without undergoing change of its own is called a(n):
a) Activation Energy
b) Reactant
c) Catalyst
d) Product

1 d., 2 a., 3. a, 4. c

Day 18 - 3 Jan 2011, homework questions:

Question 1.
N₂₍g₎ + O₂₍g₎ → 2NO ₍g₎   ∆H= 180.6 kJ
a)      Rewrite the thermochemical equation, including the standard enthalpy of reaction as either a reactant or a product.
b)      Draw an enthalpy diagram for the reaction
c)      What is the enthalpy change for the formation of one mole of NO?
d)      What is the enthalpy change for the reaction of 100 g of nitrogen with sufficient oxygen?

a) N₂₍g₎ + O₂₍g₎ + 180.6 kJ → 2NO ₍g₎ 
b) endothermic potential energy diagram
c) 1/2 (180.6) kJ =
d) 1 mole of N2 = 28 g, 100g of N2 = 3.57 moles,

Question 2.
The reaction of iron with oxygen results in rust.  A large amount of heat is produced by this reaction.
4 Fe ₍s₎ + 3 O₂ ₍g₎ → 2 Fe₂O₃₍s₎ + 1.65 x 10³ kJ    (formation of iron(III) oxide)
a)      What is the enthalpy change for this reaction?
b)      Draw an enthalpy diagram for this reaction.
c)      What is the enthalpy change for the formation of 23.6g of iron(III)oxide?

a) heat is a product, so it is an exothermic reaction & ∆H is negative.    - 1.65 x 10³ kJ
b) exothermic potential energy diagram
c) 1 mole of iron(III)oxide is 159.7 g/mol,  23.6g =  0.15 moles.  Formation of 2 moles produces 1.65 x 10³ kJ of heat so 0.15 moles produces 247.5 kJ

Question 3:
25.9 kJ + ½ H₂₍g₎ + ½ I₂₍g₎ →  HI₍g₎
a) What is the enthalpy change for this reaction?
b) How much energy is needed for the reaction of 4.57 X 10²⁴ molecules of iodine, I₂, with excess hydrogen H₂?
c) Draw and label an enthalpy diagram for this reaction.

a) heat is a reactant in this equation, endothermic, ∆H is positive
b) 1 mole = 6.022 x 10²³  (this is Avogardos number).  Determine the number of moles of Iodine.  You know the enthalpy with 1/2 mole of iodine. 
c) endothermic potential energy diagram

Question 4:
Tetraphosphorous decoxide, P₄O₁₀  is an acidic oxide.  It reacts with water to produce phosphoric acid, H₃PO₄ in an exothermic reaction where ∆H = -257.2 kJ
a)      Write an equation for this reaction including the enthalpy change as a heat term in the equation.
b)      How much energy is released when 5.00 moles of  P₄O₁₀  reacts with excess water?
c)      How much energy is released when 235g of H₃PO₄ is formed?

a) exothermic - so heat is a product.  Write the equation for the reaction with heat on the product side of the equation
balanced equation:  P₄O₁₀ + 6 H₂O  -->  4H₃PO₄ + 257.2kJ
b) 1 mole releases 257.2kJ, so 5 moles releases 5(257.2 kJ) = 1286.00 kJ
c)  1 mole of H₃PO₄ is 98 g/mol so 235 g is 2.4 moles,   4 moles releases 257.2 kJ, so 2.4 moles releases 154.3 kJ

Question 5:
The molar enthalpy of vaporization of ammonia NH₃ is 1.37 kJ/mol. 
a)      How many grams of  NH₃ are there in 1 mole of NH₃?

b)      How much energy is required to vaporize 50g of NH₃?

a) 17.03 g/mol
b)  50g would be 2.94 moles, so 50g of ammonia would need 2.94 mol (1.37kJ/mol) = 4.03 kJ of energy

Question 6:
A chemist wants to determine the enthalpy of neutralization for the following reaction:
HCl aq + NaOH aq --> NaCl aq + H ₂O₍ l₎
He uses a calorimeter to neutralize completely 61.1 ml of 0.543 mol/l HCL aq with 42.6 ml of NaOH aq.  The initial temperature of both solutions was 17.8° C.  After neutralization the highest recorded temperature is 21.6° C.  Calculate the enthalpy of neutralization making the usual assumptions about density and specific heat capacity of water. 

total volume = 61.1 + 42.6 ml = 83.7 ml  so m = 83.7 g
temperature, from 17.8 to 21.6, change in temperature is 3.8 C
c = 4.184 J/g C
q = 83.7 x 3.8 x 4.184 J = 1.330.76 J for this reaction. (or 1.33 kJ) 
The temperature of the water increases which means that the reaction gives off heat and is exothermic.  q for the system = -1.33kJ
Next - determine number of moles of HCl used and then calculate the enthalpy of neutralization.

Question 7:
Given the following thermochemical data:

(1)  1/2H2(g) + 1/2Cl2(g) --> HCl(g)        ΔH = -92.3 kJmol-1
(2)  2C(s) + 3H2(g) -->  CH3CH3(g)         ΔH = -84.7 kJmol-1
(3)  2C(s) + 2H2(g) + Cl2(g) -->  ClCH2CH2Cl(l)         ΔH = -166.0 kJmol-1
Calculate the enthalpy change for the reaction
(4)  2Cl2(g) + CH3CH3(g) --> ClCH2CH2Cl(l) +  2HCl(g)       ΔH = ??? kJmol-1

ΔH -s -265.9 kj/mol
double equation 1
reverse equation 2
leave equation 3 as is.

Question 8:
Given the following standard enthalpies of combustion (298K, 1 atmos.):
C(s) -393 kJ mol-1; H2(g) -285.6 kJ mol-1; C8H18(l) -5512 kJ mol-1;
Calculate the enthalpy of formation of octane, C8H18(l)

Similar to the formation of butane question that we did in class today:
Use the equations with known enthalpies: 
1) C8H18 + O2 --> 8 CO2 + 9 H2O  ΔH = -5512 kJ mol-1
2) C + O2 --> CO2     ΔH = -393 kJ mol-1
3) H2 + O2 -->H2O   ΔH = -285.6 kJ mol-1
to find the enthalpy of this equation:
8 CO2 + 9 H2O --> C8H18  
answer:   ΔH = 5512 + 8(-393) + 9 (-285.6) kJ mol-1

Question 9:
Given the following standard Enthalpies of Formation in kJ mol-1 (298K, 1 atmos.):
CH4(g) -74.9; CH3Br(l) -36.0; HBr(g) -36.2
Calculate the enthalpy change for the reaction:
Br2(l) + CH4(g) --> CH3Br(l) + HBr(g)

Use Hess's Law: use the equation, ΔH reation = ΔH products - ΔH reactants.
ΔH formation of Br2 = 0  because the standard enthalpy of formation of an element in its standard state is 0.
ΔH products = -36.0 + (-36.2)
ΔH reactants =  0 + (-74.9)

Question 10:
The reaction that occurs when a typical fat, glycerol trioleate, is metabolized in the body is:
C₅₇ H₁₀₄O₆ ₍s₎ + 80 O₂  ₍g₎ -->  57 CO₂ (g) +  52 H₂O (l)
a)      37.8 kJ is released when 1.00g of this fat (molar mass = 884g) is metabolized. Calculate the molar enthalpy of formation of fat in kJ/mol
b)      How many kJ of energy must be in the form of heat if you want to get rid of 1 pound (454g) of this fat by combustion?

a) 1.00 g = 0.0011 moles, molar enthalpy =  34,364 kJ/mol  or 34.364 MJ/mol  [megajoules] BUT this is the enthalpy of combustion!
To determine the enthalpy of formation, you have to look at the question in a different way. 
You can use Hess's Law: the equation, ΔH reation = ΔH products - ΔH reactants. 
     ΔH products = 57 (-393) + 52 (-285.8) kJ/mol  = -37,291.1 kJ/mol    [information from question 8, or look up the ΔH formation]
     ΔH reactants = ΔH formation of C₅₇ H₁₀₄O₆ ₍s₎ + 0     [O2 gas is an element in its standard state]
     ΔH reation = 34,364 kJ/mol
If you do the math, the ΔH formation of C₅₇ H₁₀₄O₆ ₍s₎ is -2927.1 kJ/mol

b) 37.8 kJ for 1 gram, so 454 (37.8 kJ) for 454 grams 


Energy Changes and Rates of Reaction (Thermochemistry)
Vocabulary for January 3

Exothermic reaction – the reaction releases thermal energy

Endothermic reaction – the reaction absorbs thermal energy

Potential energy – the energy stored in a system. 

Kinetic energy – the energy an object has that is due to its motion

Thermal energy = kinetic energy + potential energy. 

Calorimetry – an experimental technique for measuring energy changes

Temperature – average kinetic energy of the particles in a sample of matter

Open system – one in which both matter & energy can move in or out of

Closed system – energy can move in or out, but matter can not mover

Isolated system – an ideal system in which neither matter or energy can move in or out

Specific heat capacity – the quantity of heat required to raise the temperature of a unit mass of a substance by one degree Celsius

Enthalpy – a measure of the total energy of a system

Enthalpy change in a reaction – the energy absorbed from or released to the surroundings when a reaction occurs. Or, the difference in enthalpies of the reactants and products during a change.

Physical change – a change in the form of a substance in which no chemical bonds are broken

Chemical change – Change in the chemical bonds occurs

Nuclear change – a change in the protons or neutrons of an atom resulting in a formation of new atoms.

Mole – unit of measure for the amount of a substance.  One mole of carbon has a mass of 12g.   Or, a mole of any pure substance of a mass in grams is equal to its atomic mass.

Day 15 - 10 December


1. text page 249, questions 10, 11, 13a
2. text page 250, questions 1, 2 and 3
3. text page 255-256, questions 6, 7, 8, 9, 10

Day 12 - 7 December


1. for Wednesday - read page 86 in text (lab) and learn symbols for hazardous materials
2. for Thursday - page 197, #1, 5, 6, 7, 8, 9, 10, 11, 13

Get ready for bonding!  Molecules and atoms

Valence electrons

orbital hybridization

energy level diagrams

VSPER theory

ionic and covalent bonds


hydrogen bonding

intramolecular and intermolecular forces

metallic bonds

Day 11 - 6 December

If you did not have a poster or a written report today, please bring tomorrow.
Thank-you for all your hard work on your presentations.  I know that it was not easy for anyone.

No homework tonight.  Review periodic table if you are still having troubles with the names & symbols of the elements.

Day 8 - 1 December

Homework: Review for test.  Work on lab reports & presentations.
Review Questions on polymers, proteins, polysaccharides and nucleic acids are now on the "Review Questions" page

Note:  a Lewis model, or Lewis diagram, shows the structure of a molecule. The Lewis diagram for propane:      
    H  H  H                                                                                                           
    H  H  H

If you are having trouble with the action of borate on the polymers, see this link (diagram part way down the page)

Day 7 - 30 November

Homework:  Work on lab report.  The lab report is due on Friday.

hydrogen bonding

dehydration synthesis - glucose + fructose

sickle cell anemia - how one wrong amino acid can cause problems

peptidoglycans (amino acids & sugars)

DNA structure (simple)

DNA structure (a bit more detail)

DNA & making proteins

Glucose song (Sugar, Sugar)

Day 6 - 29 November

1. What are the monomers in Kevlar (picture page 110).  Structure only.  You do NOT need to name the monomers.
2. Questions, Page 107 in text,
  #2,4,6 and 7.

making plastic cups, forks, and straws

making natural rubber

making rubber bands

making nylon

Day 5 - 26 November

Your quizzes are marked.  If you want to know your mark before Monday, you can email me.

Words to know for Monday's class (page 98 - 113 in the text book)
                  synthetic                monomer                cellulose
polymer                     petroleum                dimer
molecule           silk                              crosslinking                rubber
solvent              dissolve                       flexible                           stretch
rigid                        moulded                viscous                   process

In the addition polymerization process there are three steps:
initiation  -  to start the reaction
propagation  - to keep the reaction going
termination -  to stop the reaction

Day 4 - 25 November

youtube video on substitution reactions

youtube video on addition reactions

youtube video on combustion & elimination reactions

Day 3 - 24 November
There is a new page on this site with review questions for Friday's quiz.  The link is beside the "presentations" link.

1. Page 31, # 3 and #4

2. Page 42, #6
3. Page 44, #7, 8, & 9
4. Page 52, #5

Day 2 - 23 November

Read - text book, pages 24 - 77
Do - page 815, #1, #2

Have you signed up for your presentation yet?

youtube video on cyclical hydrocarbons & organic halides:


carboxylic acids:


ketones & aldehydes

Extra Help:

Functional Groups

The hydroxyl group -OH is found in alcohols and in carboxylic acids.
    Alcohols (or alkanols):  -OH only
    Carboxylic Acids:  Both –OH and a carbonyl group

The carbonyl group  C=O is found in aldehydes, ketones carboxylic acids, and amides. 
    Ketones:  The carbonyl group is in the carbon chain.
    Aldehydes:  C=O is part of the aldehyde group, or formyl group, at the end of the carbon chain. 
    Carboxylic Acids:   Carboxyl group = Carbonyl group  + Hydroxyl group
    Amide:  Carbonyl group links to a Nitrogen,

The carboxyl group COOH is found in carboxylic acids

The amino group  -NH2 is found in amines

The nitro group is -NO2

Double bonds and triple bonds
     Double bonds are in alkenes and aromatic hydrocarbons like benzene
     Triple bonds are in alkynes


Day 1 - 22 November

If you are having trouble with naming alkanes, check out the youtube videos: 


1. Name and draw the Lewis diagrams of the structural isomers of the following:
  • octane
  • octene
  • octyne
  • cyclooctane

2. Read text book, section 1.2 from page 11 to page 23

3. Do both questions 1 and 2 from the text book, at the bottom of page 22

4. Think about the presentations.  What topic would you like to choose?  Click on the "presentations" link on the blue bar near the top of this page to see all of the information about the presentations.