unit 3 review, pages 406–413 - weebly · biocatalysts support the tenets of green chemistry...

Copyright © 2012 Nelson Education Ltd. Unit 3: Energy Changes and Rates of Reaction U3-4

Unit 3 Review, pages 406–413 Knowledge 1. (a) 2. (c) 3. (c) 4. (d) 5. (c) 6. (d) 7. (d) 8. (b) 9. (d) 10. (d) 11. (d) 12. (d) 13. False. The terms “thermal energy” and “temperature” do not mean the same thing. 14. False. Much of the energy emitted from the Sun is released during the nuclear process called fusion. 15. True 16. False. The quantity of energy required to break a chemical bond is known as bond dissociation energy. 17. True 18. True 19. False. Electricity generated by wind turbines is slightly more expensive than electricity generated by other sources. 20. True 21. False. Reactant concentration(s) can be used to express the rate law equation. Understanding 22. Kinetic energy is the energy of motion. Potential energy is energy based on position or composition. 23. (a) Combustion of methane gas to heat a home is an open system. (b) Combustion of methane gas in a sealed bomb calorimeter is a closed system. (c) Water boiling in a kettle with a closed lid is a closed system. (d) An acid-base neutralization reaction in a sealed flask is a closed system. 24. (a) Splitting a large gas molecule into smaller gas molecules is an endothermic reaction. (b) Forming a cation from an atom in the gas phase is an endothermic reaction. (c) Mixing elemental sodium and chlorine to form table salt is an exothermic reaction. (d) Nuclear fission is an exothermic reaction. 25. Nuclear fusion is the process of building large atoms by combining nuclei; nuclear fission is the process of splitting nuclei of large atoms to form smaller atoms. 26. Answers may vary. Sample answer: Three examples of endothermic reactions are cooking an egg, photosynthesis, and the synthesis of NO(g) from N2(g) and O2(g). 27. Answers may vary. Sample answer: Three examples of exothermic reactions are the combustion of methane, cellular respiration, and nuclear fission.


28. The change in enthalpy is the difference in chemical potential energy between the reactants and the products. It is consistent with the law of conservation of energy because the amount of energy released or absorbed by the reaction is exactly equal to the change in potential energy, so no energy is lost or gained. 29. Bond energies are used to find the approximate enthalpy change of a reaction. The sum of the energies required to break the bonds in the reactants plus the sum of the energies released by forming the bonds in the products (expressed as a negative quantity) equals the approximate enthalpy change of the reaction. 30. Forming any bond releases energy. Breaking a bond requires energy. So breaking a P−Cl bond releases more energy than forming a N–O bond does. 31. More energy needed to form and break triple bonds than single or double bonds because the attractions between atoms are stronger in triple bonds than they are in single or double bonds. 32. The two rules you need to apply when calculating enthalpy changes using Hess’s law are: 1) If you reverse a chemical reaction, you must also reverse the sign of ΔH; and, 2) The magnitude of ΔH is directly proportional to the number of moles of reactants and products in a reaction. If the coefficients in a balanced equation are multiplied by a factor, the value of ΔH is multiplied by the same factor. 33. The standard enthalpy of formation of a compound is the change in enthalpy that accompanies the formation of 1 mol of the compound from its elements, with all elements in their standard states. 34. The three main forms of fossil fuels used for energy are coal, petroleum, and natural gas. The primary concern with using fossil fuels as energy sources is that they are a nonrenewable resource. 35. Answers may vary. Sample answer: Three physical properties that can change during a reaction and that may be used to measure the rate of a reaction are pH of a solution, volume of gas, and colour. 36. (a) The chemical nature of the reactants (in this case the differing reactivities) is the factor that causes copper but not gold jewellery to turn green over time. (b) Temperature is the factor that affects how quickly milk left out on the counter will turn sour. (c) Papain is sometimes added to meat to make it more tender because is a catalyst (it accelerates the breaking down of peptide bonds). (d) The greater surface area of the dust in grain silos makes it explosive, whereas kernels of grain are almost non-flammable. (e) The different concentrations is what makes pure acetic acid burn skin on contact but vinegar safe to add to food and consume.


37. Chart formats and examples may vary. Sample answer: Factors That Affect Reaction Rate

Factor Effect Example nature of reactants

Reactivity of substances differs depending on state of matter, bond type, bond strength, and number of bonds (molecular size).

Sodium reacts with water but gold does not.

temperature of the reaction system

Higher temperature increases the kinetic energy of the reactant entities so entities move faster and with more energy; the probability of collisions increases and more entities have enough energy to break their bonds and form an activation complex.

Chemical changes occur more rapidly at higher temperatures when cooking.

concentration of reactants

Higher concentration increases the probability of collisions between reactant molecules, and thus of effective collisions.

Concentrated acids react with metal faster than dilute acids.

surface area of reactants

Increasing surface area increases the number of sites where reactants can collide, and thus the probability of effective collisions.

Powdered sugar dissolves faster than sugar cubes.

presence of a catalyst

A catalyst provides an alternative pathway for the reaction that has a lower activation energy, so a much greater fraction of the collisions are successful at a given temperature.

Sulfur and nitrogen oxides break down in a catalytic converter.

38. Biocatalysts support the tenets of green chemistry because biocatalysts are generally more efficient and usually less toxic than other catalysts. 39. Answers may vary. Sample answer: The order of reaction with respect to a reactant is the exponent to which the concentration of that reactant is raised in the rate law equation for the reaction. It tells us how the reaction rate is proportional to the initial concentration of the reactant. The orders of reaction in the rate law equation tell you which reactants are taking part in the rate-determining step, which can help you to work out a possible mechanism. If more than one reactant is present in a reaction, the sum of the orders of reaction of the reactants (the exponents in the rate law equation) is called the overall reaction order; for example, in the reaction A + B → C, if the reaction is first order with respect to A and second order with respect to B, the overall reaction order is 1 + 2 = 3. 40. An elementary step of a reaction mechanism is a single reaction that occurs during the overall reaction. An elementary step involves a one-, two-, or three-entity collision that cannot be explained by simpler reactions. The rate-determining step is the slowest elementary step and is the step in a reaction mechanism that determines the rate of the overall reaction.


Analysis and Application 41. Answers may vary. Sample answer: Photosynthesis is an endothermic reaction that we encounter in everyday life. The source of energy for the enthalpy change is sunlight. 42. Answers may vary. Sample answer:

Transfer of Energy when a Pan of Water Is Placed over a Hot Campfire

As the wood burns, chemical potential energy stored in the wood is transformed into thermal energy and light energy.

Thermal energy is transferred to the metal pan by conduction, convection, and radiation. The kinetic energy of the vibrating particles in the metal increases and the temperature of the metal rises.

Thermal energy is transferred from the pan to the water by conduction. The kinetic energy of the water molecules increases. When the boiling point is reached, continued heating causes the bonds between particles to begin to break. Potential energy increases as the water evaporates.

43. Given: m = 1.5 kg = 1500 g; c = 4.18 J/(g·°C); Tinitial = 20 ºC; Tfinal = 75 ºC Required: q Analysis: q = mc!T Solution:

q = mc!T

q = (1500 g) 4.18 Jg " °C

#$%

&'(

(75 °C ) 20 °C)

= (1500 g ) 4.18 Jg " °C

#

$%

&

'( (55 °C )

= 3.4 *105 Jq = 340 kJ

Statement: The amount of thermal energy that is required to heat 1.5 kg of water from 20 °C to 75 °C is 3.4 !105 J or 340 kJ. 44. Given:

!Hvap = 20.7 kJ/mol ;

mCl2

= 2.25 kg = 2250 g

Required: ΔH

Analysis: !H = n!Hvap ;

n =

mM


Solution:

!H = (2250 g Cl2 )1 mol Cl2

70.90 g Cl2

"

#$$

%

&''

20.7 kJmol Cl2

"

#$$

%

&''

!H = 657 kJ

Statement: The enthalpy change during the vaporization of 2.25 kg of elemental chlorine is 657 kJ. 45. Given:

VH2O(l) = 200.0 mL ;

dH2O(l) = 1.00 g/mL ; c = 4.18 J/(g·°C); Tinitial 1 = 22.1 °C;

Tfinal 1 = 26.8 °C; mmetal = 5.1 g; Tinitial 2 = 48.6 °C Required: c of metal Analysis: Use q = mc!T to determine the amount of thermal energy required to heat the

metal. Rearrange as c = q

m!T and substitute the calculated value for q and the given

values to determine the specific heat capacity of the metal. Solution:

q = mc!T

= (200.0 g)4.18 Jg " °C

#$%

&'(

(26.8 °C ) 22.1 °C)

= (200.0 g )4.18 Jg " °C

#

$%

&

'( (4.7 °C )

q = 3929 J (two extra digits carried)

c = qm!T

= 3929 J(5.1 g)(48.6 °C " 26.8 °C)

c = 35 J/(g # °C)

Statement: The specific heat capacity of the metal is 35 J/(g °C)⋅ . 46. Given: mNaOH(s) = 0.40 g;

VH2O(l) = 100 mL ;

dH2O(l) = 1.00 g/mL ; c = 4.18 J/(g·°C);

Tinitial = 20.02 °C; Tfinal = 21.12 °C Required: !Hsol

Analysis: q = mc!T ; n =

mM

; !H = n!Hsol


Solution:

qsurroundings = mc!T

= (100 g H2O(l)) 4.18 Jg " °C

#$%

&'(

(21.12 °C ) 20.02 °C)

= (100 g ) 4.18 Jg " °C

#

$%

&

'( (1.10 °C )

qsurroundings = 459.8 J (two extra digits carried)

Since the final temperature is higher than the initial temperature, the system transfers thermal energy to the surroundings and so the reaction is exothermic.

!Hsystem = –459.8 J Convert enthalpy change to molar enthalpy change using the molar mass, M.

NaOH(s) 0.40 g=m

MNaOH(s) = MNa + MO + MH

= (22.99 g/mol) + (16.00 g/mol) + (1.01 g/mol)MNaOH(s) = 40.00 g/mol

nNaOH(s) =mNaOH(s)

MNaOH(s)

=0.40 g

40.00 g/molnNaOH(s) = 0.010 mol

!Hsol =!Hn

= –459.8 J0.010 mol

!Hsol = 46 kJ/mol

Statement: The molar enthalpy change of the dissolution reaction of sodium hydroxide in water is –46 kJ/mol. 47. (a) Given: 60.0 mL of 0.700 mol/L NaOH(aq); 40.0 mL of H2SO4(aq); ΔT = 5.6 °C; c = 4.18 J/(g·°C) Required: !Hneut NaOH Analysis: q = mc!T ; !H = n!Hneut


Solution:


= (100.0 g )4.18 Jg " °C

#

$%

&

'( (5.6 °C )

qsurroundings = 2341 J (two extra digits carried)

Since the temperature increased, the system transferred thermal energy to the surroundings and so the reaction is exothermic.

!Hsystem = –2341 J Convert enthalpy change to molar enthalpy change.

n = (60 mL NaOH(aq) ) 0.700 mol NaOH1000 mL NaOH(aq)

!

"##

$

%&&

n = 0.042 mol NaOH

!Hneut =!Hn

= –2341 J0.042 mol

!Hneut = –56 kJ/mol

Statement: The molar enthalpy of neutralization for sodium hydroxide is –56 kJ/mol. (b) The assumptions made in the answer to (a) are: • Any thermal energy transferred from the calorimeter to the outside environment is

negligible. • Any thermal energy absorbed by the calorimeter itself is negligible. • All dilute, aqueous solutions have the same density (1.00 g/mL) and specific heat

capacity (4.18 J/(g·°C)) as water. 48. Given:

VH2O(l) = 2.00 L = 2000 mL ;

dH2O(l) = 1.00 g/mL ; c = 4.18 J/(g·°C);

Tinitial = 26.5 °C; Tfinal = 100.0 °C; Required: q Analysis: q = mc!T Solution:

q = mc!T

= (2000 g) 4.18 Jg"ºC

#$%

&'(

(100 ºC ) 26.5 ºC)

= (2000 g ) 4.18 Jg " °C

#

$%

&

'( (73.5 °C )

q = 614 kJ

Statement: The amount of energy that was transferred to the water from sunlight was 614 J.


49. Given: mH2O(l) = 255 g ;

mC3H6O(l) = 1.01 g ;

cH2O(l) = 4.18 J/(g!ºC) ;

cCu(s) = 0.385 J/(g!ºC) ; ΔT = +28.8 °C; mcalorimeter = 305 g

Required: !Hc C3H6O (l)


mM

; !H = n!Hc

Solution:


= (255 g H2O(l)) 4.18 Jg " °C

#

$%

&

'( (28.8 °C ) + (305 g ) 0.385 J

g " °C

#

$%

&

'( (28.8 °C )

qsurroundings = 34 080 J (two extra digits carried)

Since the final temperature is higher than the initial temperature, the system transfers thermal energy to the surroundings and so the reaction is exothermic.

!Hsystem = –34 080 J Convert enthalpy change to molar enthalpy change.

n = (1.01 g propanol ) 1 mol propanol58 g propanol

!

"#

$

%&

n = 0.0174 mol (two extra digits carried)

!Hc =!Hn

= –34 080 J0.0174 mol

!Hc = "1.96 #103 kJ/mol

Statement: The enthalpy of combustion of propanol is 31.96 10 kJ/mol− × . 50. Given: H2(g) + Cl2(g) → 2 HCl(g) for H2(g): nH−H = 1 mol; DH−H = 432 kJ/mol; for Cl2(g): nCl−Cl = 1 mol; DCl−Cl = 239 kJ/mol; for: HCl(g): nH−Cl = 2 mol; DH−Cl = 427 kJ/mol Required: ΔH Analysis:

!H = "n # Dbonds broken $ "n # Dbonds formed

!H = (nH$H DH$H + nCl$ClDCl$Cl ) $ nH$ClDH$Cl

Solution:

!H = (nH"H DH"H + nCl"ClDCl"Cl ) " nH"ClDH"Cl

= (1 mol # DH"H +1 mol # DCl"Cl ) " 2 mol # DH"Cl

= 1 mol # 432 kJmol

$

%&'

()+ 1 mol # 239 kJ

mol

$

%&'

()*

+,,

-

.//" 2 mol # 427 kJ

mol

$

%&'

()

!H = "183 kJ

Statement: The enthalpy change for the reaction is –183 kJ.


51. (a) Given: H2O2(g) → H2O(l) + O2(g) Required: ΔH Solution: Step 1. Balance the chemical equation.

H2O2(g) → H2O(l) +

12

O2(g)

Step 2. For each reactant and product, identify the number of bonds per mole, the amount of bonds in the reaction, and the bond energy per mole.

Substance Number of bonds per mole (nsubstance)

Amount of bonds in reaction

Bond energy per mole

reactants H2O2(g) 1 mol O–O bonds 2 mol O–H bonds

1 mol 2 mol

146 kJ/mol 467 kJ/mol

products H2O(l) 2 mol O–H bonds 2 mol 467 kJ/mol O2(g) 1 mol O=O bonds 0.5 mol 495 kJ/mol

Step 3: Calculate the enthalpy change, ΔH, of the reaction.


!H = (nDO$O + nDO$H ) $ (nDO$H + nDO=O )

Add the total energy absorbed to break the bonds in the reactants.

!n " Dbonds broken = (nDO#O + nDO#H )

= 1 mol "146 kJmol

$

%&'

()+ 2 mol " 467 kJ

mol

$

%&'

()

!n " Dbonds broken = 1080 kJ

Add the total energy released when the bonds of products form.

!n " Dbonds formed = (nDO=O + nDO#H )

= 0.5 mol " 495 kJmol

$

%&'

()+ 2 mol " 467 kJ

mol

$

%&'

()

!n " Dbonds formed = 1181.5 kJ

Subtract the energy released when the bonds of the products form from the energy absorbed to break the bonds of the reactants.


= 1080 kJ $1181.5 kJ!H = –102 kJ

The sign of the enthalpy change is negative. This is an exothermic reaction and energy is released. Statement: The enthalpy change is –102 kJ.


(b) Given: CH3OH(aq) + O2(g) → CO2(g) + H2O(l) Required: ΔH Solution: Step 1. Balance the chemical equation.

CH3OH(aq) +

32

O2(g) → CO2(g) + 2 H2O(l)

Step 2. For each reactant and product, identify the number of bonds per mole, the amount of bonds in the reaction, and the bond energy per mole.




reactants CH3OH(aq) 3 mol C–H bonds 1 mol C–O bonds 1 mol O–H bonds

3 mol 1 mol 1 mol

413 kJ/mol 358 kJ/mol 467 kJ/mol

O2(g) 1 mol O=O bonds 1.5 mol 495 kJ/mol products CO2(g) 2 mol C=O bonds 2 mol 799 kJ/mol

H2O(l) 2 mol O–H bonds 4 mol 467 kJ/mol Step 3: Calculate the enthalpy change, ΔH, of the reaction.


!H = (nDC$H + nDC$O + nDO$H + nDO=O ) $ (nDC=O + nDO$H )


!n " Dbonds broken = (nDC#H + nDC#O + nDO#H + nDO=O )

= 3 mol " 413 kJmol

$

%&'

()+ 1 mol " 358 kJ

mol

$

%&'

()

+ 1 mol " 467 kJmol

$

%&'

()+ 1.5 mol " 495 kJ

mol

$

%&'

()

!n " Dbonds broken = 2806.5 kJ


!n " Dbonds formed = (nDC=O + nDO#H )

= 2 mol " 799 kJmol

$

%&'

()+ 4 mol " 467 kJ

mol

$

%&'

()

!n " Dbonds formed = 3466 kJ


bonds broken bonds formed

2806.5 kJ 3466 kJ–660 kJ

Δ = Σ × −Σ ×= −

Δ =

H n D n D

H

The sign of the enthalpy change is negative. This is an exothermic reaction and energy is released. Statement: The enthalpy change is –660 kJ. (c) Given: CHF2CHF2(g) → CHCH(g) + 2 F2(g)


Required: ΔH Solution: Step 1. Balance the chemical equation. CHF2CHF2(g) → CHCH(g) + 2 F2(g) Step 2. For each reactant and product, identify the number of bonds per mole, the amount of bonds in the reaction, and the bond energy per mole.




reactants CHF2CHF2(g) 1 mol C–C bonds 4 mol C–F bonds 2 mol C–H bonds

1 mol 4 mol 2 mol

347 kJ/mol 485 kJ/mol 413 kJ/mol

products CHCH(g) 2 mol C–H bonds 1 mol C≡C bonds

2 mol 1 mol

413 kJ/mol 839 kJ/mol

F2(g) 1 mol F–F bonds 2 mol 154 kJ/mol Step 3: Calculate the enthalpy change, ΔH, of the reaction.


!H = (nDC$C + nDC$F + nDC$H ) $ (nDC$H + nDC%C + nDF$F )


!n " Dbonds broken = (nDC#C + nDC#F + nDC#H )

= 1 mol " 347 kJmol

$

%&'

()+ 4 mol " 485 kJ

mol

$

%&'

()+ 2 mol " 413 kJ

mol

$

%&'

()

!n " Dbonds broken = 3113 kJ


!n " Dbonds formed = (nDC–H + nDC#C + nDF$F )

= 2 mol " 413 kJmol

%

&'(

)*+ 1 mol " 839 kJ

mol

%

&'(

)*+ 2 mol "154 kJ

mol

%

&'(

)*

!n " Dbonds formed = 1973 kJ



= 3113 kJ $1973 kJ!H = 1140 kJ

The sign of the enthalpy change is positive. This is an endothermic reaction and energy is absorbed. Statement: The enthalpy change is 1140 kJ.


52. Given: 4 CH3NO2(g) + 3 O2(g) ! 4 CO2(g) + 2 N2(g) + 6 H2O(g) and

(1) C(g) + O2(g) ! CO2(g) "H = #393.5 kJ(2) 2 H2(g) + O2(g) ! 2 H2O(g) "H = #483.6 kJ(3) 2 C(g) + 3 H2(g) + 2 O2(g) + N2(g) ! 2 CH3NO2(g) "H = –226.2 kJ

Required: !H Solution: Step 1. Since this is a combustion reaction, rewrite the equation for the combustion of nitromethane so that nitromethane has a coefficient of 1:

CH3NO2(g) +

34

O2(g) ! CO2(g) +12

N2(g) +32

H2O(g)

Step 2. Reverse equation 3 and multiply it and its ΔH by

12

.

Multiply equation 2 and its ΔH by

34

.

Then, add the three equations and their changes in enthalpies.

C(g) + O2(g) ! CO2(g) "H = #393.5 kJ

32

H2(g) + 34

O2(g) ! 32

H2O(g) "H = #362.7 kJ

CH3NO2(g) ! C(g) + 32

H2(g) + O2(g) + 12

N2(g) "H = 113.1 kJ

CH3NO2(g) + 34

O2(g) ! CO2(g) + 12

N2(g) + 32

H2O(g) "H = #643.1 kJ

Statement: The enthalpy change for the combustion of nitromethane is –643.1 kJ per mole of nitromethane. 53. (a) The balanced chemical equation for the combustion of glucose is C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) (b) Given: C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l); from Table 1,

!Hf C6 H12O6 (s)

o = "1273.1 kJ/mol; !Hf CO2 (g)o = "393.5 kJ/mol;

2

of H O(l) 285.8 kJ/molΔ = −H ;

Required: !Hr o for the combustion reaction

Analysis: !Hr

! = "nproducts!H !products # "nreactants!H !

reactants Since O2(g) is in its standard state, you can rewrite the equation as

2 2 2 2 6 12 6 6 12 6

o o o or CO (g) f CO (g) H O(l) f H O(l) C H O (s) f C H O (s)H n H n H n H⎡ ⎤Δ = Δ + Δ − Δ⎣ ⎦


Solution: Glucose has a coefficient of 1 in the balanced equation so you can use the equation as is. Insert the appropriate values into the equation for standard enthalpy of formation and solve.

2 2 2 2 6 12 6 6 12 6

o o o or CO (g) f CO (g) H O(l) f H O(l) C H O (s) f C H O (s)

or

6( 393.5 kJ) 6( 285.8 kJ) ( 1273.1 kJ)= 2802.7 kJ/mol

H n H n H n H

H

⎡ ⎤Δ = Δ + Δ − Δ⎣ ⎦= − + − − −

Δ −

Statement: The enthalpy change of the combustion of glucose is –2802.7 kJ. (c) Given: C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) Required: ΔH Solution: Step 1. For each reactant and product in the balanced equation, identify the number of bonds per mole, the amount of bonds in the reaction, and the bond energy per mole.



Bond energy per

mole reactants C6H12O6(s) 5 mol C–C bonds

7 mol C–O bonds 5 mol O–H bonds 7 mol C–H bonds

5 mol 7 mol 5 mol 7 mol

347 kJ/mol 358 kJ/mol 467 kJ/mol 413 kJ/mol

O2(g) 1 mol O=O bonds 6 mol 495 kJ/mol products CO2(g) 2 mol C=O bonds 12 mol 799 kJ/mol

H2O(g) 2 mol O–H bonds 12 mol 467 kJ/mol Step 2: Calculate the enthalpy change, ΔH, of the reaction.


!H = (nDC$C + nDC$O + nDO$H + nDC$H + nDO%O ) $ (nDC=O + nDO$H )


!n " Dbonds broken = (nDC#C + nDC#O + nDO#H + nDC#H + nDC=O + nDO$O )

= 5 mol " 347 kJmol

%

&'(

)*+ 7 mol " 358 kJ

mol

%

&'(

)*+ 5 mol " 467 kJ

mol

%

&'(

)*

+ 7 mol " 413 kJmol

%

&'(

)*+ 6 mol " 495 kJ

mol

%

&'(

)*

!n " Dbonds broken = 12 437 kJ


!n " Dbonds formed = (nDC=O + nDO#H )

= 12 mol " 799 kJmol

$

%&'

()+ 12 mol " 467 kJ

mol

$

%&'

()

!n " Dbonds formed = 15 192 kJ




= 12 437 kJ $15 192 kJ!H = –2755 kJ

The sign of the enthalpy change is negative. This is an exothermic reaction and energy is released. Statement: The enthalpy change of the combustion of glucose is –2755 kJ. (d) Given:

VH2O(l) = 100.0 mL ; c = 4.18 J/(g·°C); ΔT = 37.0 °C;

mC6 H12O6 (s) = 1.00 g

Required: ΔH


mM

; !H = n!Hsol

Solution:


= (100.0 g H2O(l) ) 4.18 Jg " °C

#

$%

&

'( (37.0 °C )

qsurroundings = 15.466 kJ (two extra digits carried)

Since the temperature increased, the system transfers thermal energy to the surroundings and so the reaction is exothermic.

!Hsystem = –15.466 kJ Convert enthalpy change to molar enthalpy change using the molar mass, M.

mC6 H12O6 (s) = 1.00 g

MC6 H12O6 (s) = 6MC +12MH + 6MO

= 6(12.01 g/mol) +12(1.01 g/mol) + 6(16.00 g/mol)MC6 H12O6 (s) = 180.18 g/mol

nC6 H12O6 (s) =mC6 H12O6 (s)

MC6 H12O6 (s)

=1.00 g

180.18 g/molnC6 H12O6 (s) = 0.00555 mol (two extra digits carried)

!Hsol =!Hn

= –15.466 kJ0.00555 mol

!Hsol = "2790 kJ/mol

Statement: The enthalpy change of the combustion of glucose is !2790 kJ/mol .


(e) The answers to (b), (c), and (d) are different because the calculations use average values and because there is some experimental error in the calorimetry experiment. 54. Three different ways to express the standard molar enthalpy of formation for vinyl chloride gas are: 2C(s) + 3/2 H2(g) + 1/2 Cl2(g) → C2H3Cl(g) + 37.3 kJ 2 C(s) + 3/2 H2(g) + 1/2 Cl2(g) → C2H3Cl(g) !H o = –37.3 kJ

55. (a) A thermochemical equation for the formation of sulfur dioxide gas is S(s) + O2(g) → SO2(g) ΔH = –298.6 kJ (b)

(c) Given: !Hfo = "296.8 kJ/mol ;

mSO2 (g) = 9.63 g

Required: ΔH

Analysis: !H = n!Hfo ;

n =

mM

Solution:

!H = (9.63 g SO2(g) )1 mol SO2(g)

64.07 g SO2(g)

"

#$$

%

&''

(296.8 kJmol SO2(g)

"

#$$

%

&''

!H = 44.6 kJ

Statement: The reaction releases 44.6 kJ of thermal energy.


56. (a) An equation for the combustion of 1 mol of pentane gas is C5H12(g) + 8 O2(g) → 5 CO2(g) + 6 H2O(l) (b) Given: C5H12(g) + 8 O2 → 5 CO2(g) + 6 H2O(l); from Table 1,

!Hf C5H12 (g)

o = "146 kJ/mol; !Hf CO2 (g)o = "393.5 kJ/mol;

!Hf H2O(l)

o = "285.8 kJ/mol ;

Required: !Hr o for the combustion reaction

Analysis: !Hr

! = "nproducts!H !products # "nreactants!H !

reactants Since O2(g) is in its standard state, you can rewrite the equation as

!Hr

o = nCO2 (g)!Hf CO2 (g)o + nH2O(l)!Hf H2O(l)

o"#

$% & nC5H12 (g)!Hf C5H12 (g)

o

Solution: Pentane gas has a coefficient of 1 in the balanced equation so you can use the equation as is. Insert the appropriate values into the equation for standard enthalpy of formation and solve.

!Hro = nCO2 (g)!Hf CO2 (g)

o + nH2O(l)!Hf H2O(l)o"

#$% & nC5H12 (g)!Hf C5H12 (g)

o

= 5(&393.5 kJ) + 6(&285.8 kJ) & (&146 kJ)!Hr

o = & 3536.3 kJ/mol

Statement: The enthalpy change of the combustion of pentane is –3536.3 kJ. (c) Given: !Hc = "3534.5 kJ/mol ;

mSO2 (g) = 2.0 g

Required: ΔH

Analysis: !H = n!Hc ; n =

mM

Solution:

!H = (20 g C5H12(g) )1 mol C5H12(g)

72.17 g C5H12(g)

"

#$$

%

&''

(3530 kJmol C5H12(g)

"

#$$

%

&''

!H = 1000 kJ

Statement: The combustion of 20 g of pentane would release 1000 kJ of thermal energy.


57. Answers may vary. Sample answer: Comparison of Five Different Uses of Fossil Fuels

Fossil fuel Use

Possible alternative fuel

Advantage(s) of alternative

Disadvantage(s) of alternative

petroleum gasoline for transportation, small motors

syngas more available than petroleum

emits greenhouse gases

coal electricity wind energy no emissions limited availability natural gas

electricity hydro power no emissions; renewable

limited availability; environmental concerns

natural gas

heating and cooling

solar energy renewable limited availability at times

petroleum diesel fuel for transportation

biodiesel renewable; reduced greenhouse gases

impact on food supply

58. Answers may vary. Sample answer: Advantages and Disadvantages of Wind Energy

Advantages Disadvantages Social factors: – can offer energy independence to individuals or small towns Economic factors: – can be located close to where it is needed – in the long term, may cost less than fossil fuels because damage to environment may be less Environmental factors: – renewable – no greenhouse gases are created – no pollution

Social factors: – distracting noise and motion Economic factors: – less efficient than fossil fuels – in the short term, more expensive per unit of energy than fossil fuels – requires large land or water areas – needs consistent wind Environmental factors: – can interfere with flying or swimming animals

59. For the reaction represented by the balanced equation CO(g) + NO2(g) → CO2(g) + NO(g) (a) The sum of the two concentrations must equal 0.10 mol/L, since CO(g) becomes CO2(g) in the reaction, so the missing values are:

Time (s) [CO(g)] (mol/L) [CO2(g)] (mol/L) 0 0.100 0.000

20 0.050 0.050 40 0.033 0.067 60 0.026 0.074 80 0.020 0.080

100 0.017 0.083


(b) The rate of NO2(g) consumption is the same as the rate of CO(g) consumption. If the initial concentration of nitrogen dioxide gas is 0.250 mol/L, [NO2(g)] after 80 s is 0.250 mol/L – 0.080 mol/L = 0.170 mol/L. (c)

(d) Yes, the graph reflects the stoichiometry of the balanced equation, because [CO(g)] decreases at the same rate that [CO2(g)] increases. 60. For the reaction represented by the balanced equation 2 X2O5(g) → 4 XO2(g) + O2(g): (a)


(b)

Time (h) [X2O5(g)] (mol/L)

[XO2(g)] (mol/L)

[O2(g)] (mol/L)

0.0 1.20 0 0 2.0 0.80 0.80 0.20 4.0 0.55 1.30 0.325 7.0 0.30 1.68 0.42

12.0 0.10 2.20 0.55

(c) (i) Rate of X2O5(g) consumption:

![X2O5(g)]!t

=[X2O5(g)]t=12.0 h " [X2O5(g)]t=0 h

!t

= 0.10 mol/L "1.20 mol/L12.0 h " 0 h

![X2O5(g)]!t

= "0.092 mol/(L # h)

(ii) Rate of O2(g) formation:

![O2(g)]!t

=[O2(g)]t=12.0 h " [O2(g)]t=0 h

!t

= 0.55 mol/L " 0.00 mol/L12.0 h " 0 h

![O2(g)]!t

= 0.046 mol/(L # h)

(iii) Rate of XO2(g) formation:

![XO2(g)]!t

=[XO2(g)]t=12.0 h " [XO2(g)]t=0 h

!t

= 2.20 mol/L " 0.00 mol/L12.0 h " 0 h

![XO2(g)]!t

= 0.183 mol/(L # h)


(d) Answers may vary, as graphs in (a) may vary. Sample answer:

To calculate the instantaneous rate of consumption of X2O5(g) at t = 2.0 h, use the points (0 h, 1.08 mol/L) and (4.0 h, 0.52 mol/L) on the tangent to the [X2O5(g)] curve at 2.0 h.

rateinstantaneous at t=2.0 h = slope of the tangent at 2.0 h

= !y!x

at 2.0 h

= 1.08 mol/L " 0.52 mol/L0 h " 4.0 h

rateinstantaneous at t=2.0 h = "0.14 mol/(L # h)

The instantaneous rate of consumption of X2O5(g) at t = 2.0 h is 0.14 mol/L h⋅ . To calculate the instantaneous rate of consumption of X2O5(g) at t = 7.0 h, use the points (3.0 h, 0.55 mol/L) and (10.0 h, 0.12 mol/L) on the tangent to the [X2O5(g)] curve at 7.0 h.

rateinstantaneous at t=7.0 h = slope of the tangent at 7.0 h

= !y!x

at 7.0 h

= 0.55 mol/L " 0.12 mol/L3.0 h "10.0 h

rateinstantaneous at t=7.0 h = "0.061 mol/(L # h)

The instantaneous rate of consumption of X2O5(g) at t = 7.0 h is 0.061 mol/L ! h . (e) The observed trend is that the rate of consumption of X2O5(g) decreases over time. The rate of consumption of a reactant decreases as more and more of the reactant is converted into products.


61. Given: N2(g) + 2 O2(g) → 2 NO2(g); [NO2(g)]initial = 0.32 mol/L; [NO2(g)]t = 3 min = 0.80 mol/L Required: rate of NO2(g) production

Solution: 2 3 min 2 0 h2

2

[NO (g)] [NO (g)][NO (g)]

0.80 mol/L 0.32 mol/L3 min 0 min

[NO (g)] 0.2 mol/(L min)

t t

t t

t

= =−Δ =Δ Δ

−=−

Δ = ⋅Δ

Statement: The overall rate of production of nitrogen dioxide is 0.2 mol/(L !min) . 62. Given: 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g); instantaneous rate of consumption of NH3(g) is 2.0 × 10−2 mol/(L·s) Required: (a) instantaneous rate of consumption of oxygen gas (b) instantaneous rate of formation of water vapour Analysis: Scale the rates by the inverse of their coefficients in the balanced chemical equation. Then, substitute the given rate and solve for the required rates.

Solution: (a)

14

!"[NH3(g)]

"t#$%

&'(= 1

5!"[O2(g)]

"t#$%

&'(

"[O2(g)]"t

= 54

(2.0 )10!2 mol/(L * s))

"[O2(g)]"t

= 2.5)10!2 mol/(L * s)

(b)

14

!"[NH3(g)]

"t#$%

&'(= 1

6"[H2O(g)]

"t#$%

&'(

"[H2O(g)]"t

= 64

(2.0 )10!2 mol/(L * s))

"[H2O(g)]"t

= 3.0 )10!2 mol/(L * s)

Statement: (a) The instantaneous rate of consumption of oxygen gas is

2.5!10"2 mol/(L # s) . (b) The instantaneous rate of formation of water vapour is 3.0 !10"2 mol/(L # s) . 63. (a) According to the “Methane Gas Volume versus Time” graph in Figure 2,

![CH4(g)]!t

=[CH4(g)]t=7 s " [CH4(g)]t=3 s

!t

= 18 mL " 35 mL7.0 s " 3.0 s

![CH4(g)]!t

= "4.3 mL/s

The reaction rate of the decomposition of methane between 3 s and 7 s is 4.3 mL/s.


(b) According to the “Oxygen Gas Volume versus Time” graph in Figure 2,

![O2(g)]!t

=[O2(g)]t=6 s " [O2(g)]t=2 s

!t

= 36 mL "13 mL6.0 s " 2.0 s

![O2(g)]!t

= 5.8 mL/s

The reaction rate of the production of oxygen gas between 2 s and 6 s is 5.8 mL/s. (c) Methane is a reactant, because the graph shows that its concentration is decreasing; oxygen is a product, because the graph shows that its concentration is increasing. 64. The reaction Pb2+(aq) + 2 Cl–(aq) → PbCl2(s) would have a higher reaction rate at room temperature than the reaction Pb(s) + Cl2(g) → PbCl2(s) because the reactants are dissolved ions which are free to move about in the solution and collide. A reaction involving a solid like Pb(s) can only occur at the exposed surface, so the number of collisions that can occur are limited. 65. The reason why containers of powdered aluminum metal must carry a warning that the contents are dangerously combustible while many everyday objects made of aluminum metal are not required to carry the same warning is that the powder is highly combustible because the small particle size forms a very large surface area. Objects made of bulk aluminum do not react quickly with oxygen. 66. (a) If 44.2 mL of carbon monoxide gas forms in 30.0 s, the rate of reaction with respect to carbon monoxide gas production is

![CO(g)]!t

= 44.2 mL30.0 s

![CO(g)]!t

= 1.47 mL/s

(b) (i) The temperature of run 1 was 25 ºC. If the temperature is increased to 30 ºC, the reaction rate will increase because higher temperature increases the kinetic energy of the reactant entities so entities move faster and with more energy; the probability of collisions increases and more entities have enough energy to break their bonds and form an activation complex. (ii) In run 1, the solution of formic acid was catalyzed. If the reaction is performed without the catalyst, the reaction rate will be slower because the activation energy will be higher, so a smaller fraction of the collisions will be successful at a given temperature (iii) If the reaction is performed using formic acid that is half as concentrated as in run 1, the reaction rate will be slower because the probability of collisions between reactant molecules, and thus of effective collisions, will decrease. 67. Hydrogen peroxide can be stored for months on the shelf but will bubble strongly when applied to an open cut because there is a compound in blood that acts as a catalyst for the decomposition of hydrogen peroxide.


68. Answers may vary. Sample answer: A collision between a hydrogen and a bromine molecule that is likely to be an effective collision

69. When a magnesium strip is heated in a flame, the reaction is exothermic, and once the strip is sufficiently heated, the reaction itself provides activation energy to entities, which speeds up the reaction rate. A strip of magnesium metal can be safely stored at room temperature for long periods of time but magnesium granules stored at room temperature will react slowly with the oxygen in the air to form magnesium oxide because the granules have a large surface area, so all of the magnesium reacts over time. 70. Answers may vary. Sample answer: Advantages of Biocatalysts over Traditional Catalysts

Biocatalysts Traditional catalysts naturally occurring must be synthesized no toxic by-products generally toxic metals or salts inexpensive often expensive elements homogeneous generally heterogeneous very efficient less efficient

71. A highly exothermic chemical reaction will affect the rate of this reaction as follows: The reaction rate will increase after the reaction starts because the exothermic reaction provides thermal energy that acts as activation energy. 72. In the reaction of calcium oxide and acetic acid, two ways of increasing the rate at which calcium oxide dissolves are 1) increasing the concentration of acetic acid to increase the probability of collisions between reactant molecules, and thus the probability of effective collisions, and 2) heating the acetic acid to increase the kinetic energy so the entities move faster and with more energy, increasing the probability of collisions and causing more entities to have enough energy to form an activation complex.


73. A clock reaction involving the hypochlorite ion, ClO-(aq), is represented by ClO–(aq) + I–(aq) → Cl–(aq) + IO–(aq) If the rate is first order with respect to each of the reactants: (a) (i) when the initial [ClO–(aq)] is doubled, the initial rate will double. (ii) when the initial [I–(aq)] is halved, the initial rate will halve. (iii) when the same initial numbers of moles of reactants are placed in a container with half the volume of water, the initial rate will quadruple because both concentrations double. (b) Answers may vary. Sample experiment to study the effect of change in temperature using the iodine clock reaction in Investigation 6.5.1: Question: How does temperature affect the iodine clock reaction? Prediction: The iodine clock reaction rate will increase with temperature. Experimental design: In the iodine clock reaction, two colourless solutions are combined and the combined solution initially stays colourless. After a certain amount of time, the combined solution suddenly changes colour. By changing the reaction temperature, the effects of temperature on reaction rate will be determined. The time from mixing to the appearance of the blue-black product will be measured and then graphed. Equipment and materials: • chemical safety goggles • lab apron • 2 beakers, 250 mL • 2 graduated cylinders, 10 mL • 2 large test tubes • thermometer • water baths • kettle • ice cubes • timer • Solution A (contains iodate ions, IO3

–(aq) • Solution B (contains HSO3

–(aq) and starch) Procedure: 1. Put on your safety goggles and lab apron. 2. Prepare a water bath for the solutions. Fill two 250 mL beakers about two-thirds full

with water of a given temperature. For water baths below room temperature, use ice to chill the water. For warmer baths, use warm water from the tap or a kettle. There should be enough water in the beakers so that the solutions in the test tubes are well beneath the water level in the baths.

3. Measure equal volumes of the two solutions, pouring each into its own test tube. Place the two test tubes in the water baths, allowing them to remain for about 10 min to allow the solution temperatures to reach the temperature of the water baths.

4. Use the thermometer to monitor water bath temperatures. Try to keep the water bath temperature within 0.5 °C of the desired temperature. Add more ice or warm water as necessary. Record the water bath temperature just before mixing the solutions.


5. Once the solutions are at the desired temperatures, prepare to record time and mix the two solutions. Pour the mixture back and forth between the two test tubes several times to ensure mixing. Then, place the test tube containing the mixed solutions back in the water bath. Record the time it takes from initial mixing until the blue-black colour appears.

6. Follow steps 2 to 5 for at least 4 different temperatures; for example, 5 ºC, 15º C, 25 ºC, and 35 ºC.

7. Plot a graph of temperature versus reaction time. Safety and disposal precautions: Wear safety glasses and apron. Discard chemicals following school policy. 74. (a) Given: CO(g) + Cl2(g) → COCl2(g); reaction is first order with respect to chlorine; k = 1.3!10"2 L/(mol # s) Required: the order of the reaction with respect to carbon monoxide Solution: The units of k indicate that the reaction is a second order reaction overall, so the order of the reaction with respect to carbon monoxide is 1. (b) If you wanted to increase the reaction rate as much as you could by doubling the initial concentration of one of the reactants, you could choose either of the reactants, because the order of reaction is the same with respect to both reactants. 75. Given: zero-order reaction; [A]initial = 1.5 mol/L ; [A]t=120 s = 0.75 mol/L Required: the rate constant, k Analysis: In a zero-order reaction, the rate constant equals the rate, since rate = k[A]0[B]0 = k. Solution:

k = rate

= ![A]!t

= 0.75 mol/L "1.5 mol/L120 s – 0 s

k = 6.3#10"3 mol/(L $ s)

Statement: The rate constant, k, is 6.3!10"3 mol/(L # s) . 76. Given: H3O+(aq) + OH–(aq) → 2 H2O(l); k = 1.0 !1011 L/(mol " s) Required: the rate of the reaction when equal volumes of 0.10 mol/L solutions of hydrochloric acid and sodium hydroxide are mixed. Analysis: Since hydrochloric acid is a strong acid, the solutions will ionize completely, and so the concentration of the hydronium ion, H3O+, and of HCl are actually the same. Similarly, sodium hydroxide is a strong base and will ionize fully, so

[H3O+ (aq)] = [HCl(aq)] = 0.10 mol/L

[OH! (aq)] = [NaOH(aq)] = 0.10 mol/L

The units of the rate constant are L/(mol ! s) , so the overall reaction order is 2. Hence, the reaction is first order with respect to both H3O

+ (aq) and OH! (aq) . Thus, the rate law

equation is r = k[H3O+ (aq)][OH! (aq)] . Substitute the given values in the equation.


Solution:

rate = k[H3O+ (aq)][OH! (aq)]

= [1.0 "1011 L/(mol # s)](0.10 mol/L)(0.10 mol/L)rate = 1.0 "109 mol/(L # s)

Statement: The rate of the reaction is 1.0 !109 mol/(L " s) . 77. (a) The reaction C3H8(g) + 5 O2(g) = 3 CO2(g) + 4 H2O(g) is likely to occur in a series of steps rather than in a single step because a collision involving 6 molecules is extremely unlikely.

(b) Given: C3H8(g) + 5 O2(g) = 3 CO2(g) + 4 H2O(g); ![C3H3(g)]

!t= –4 "10#2 mol/(L $ s)

Required: the rate of reaction with respect to oxygen gas and carbon dioxide gas Analysis: Scale the rates by the inverse of their coefficients in the balanced chemical equation. Then, substitute the given rate and solve for the required rates.

Solution:

!"[C3H3(g)]

"t= 1

5!"[O2(g)]

"t#$%

&'(

"[O2(g)]"t

= 5(–4 )10!2 mol/(L * s))

"[O2(g)]"t

= !2 )10!1 mol/(L * s)

!"[C3H3(g)]

"t= 1

3"[CO2(g)]

"t#$%

&'(

"[CO2(g)]"t

= !3(–4 )10!2 mol/(L * s))

"[CO2(g)]"t

= 1)10!1 mol/(L * s)

Statement: The rate of reaction with respect to oxygen gas is 2 !10"1 mol/(L # s) , and the rate of reaction with respect to carbon dioxide gas is 1!10"1 mol/(L # s) . 78. (a) Given: proposed reaction mechanism

2 NO(g) ! N2O2(g)N2O2(g) + H2(g) ! N2O(g) + H2O(g)N2O(g) + H2(g) ! N2(g) + H2O(g)

Required: overall balanced equation


Solution:

2 NO(g) ! N2O2(g)N2O2(g) + H2(g) ! N2O(g) + H2O(g)N2O(g) + H2(g) ! N2(g) + H2O(g)

2 NO(g) + N2O2(g) + H2(g) + N2O(g) + H2(g)

! N2O2(g) + N2O(g) + H2O(g) + N2(g) + H2O(g)

Overall balanced equation: 2 NO(g) + 2 H2(g) ! N2(g) + 2 H2O(g) (b) The reaction intermediates are N2O2(g) and N2O(g). (c) If Step 1 is the slow step, the rate law equation for this reaction is r = k[NO(g)]2. 79. (a) Given: 2 NO2Cl(g) → 2 NO2(g) + Cl2(g); experimental data provided in Table 4 Required: the rate law equation Solution: The rate doubles when [NO2Cl(g)] doubles so the reaction is first order with respect to NO2Cl(g), and the rate equation is r = k[NO2Cl(g)]. (b) Given: proposed reaction mechanisms; rate = k[NO2Cl(g)] (i)

NO2Cl ! NO2 + Cl (slow)NO2Cl + Cl ! NO2 + Cl2 (fast)

(ii)

2 NO2Cl ! N2O2Cl2 (slow)N2O2Cl2 ! 2 NO2 + Cl2 (fast)

(iii)

NO2Cl ! NO2 + Cl (fast)NO2Cl + Cl ! NO2 + Cl2 (slow)

Required: the incorrect mechanisms, using the rate law Solution: (i) The first step is the rate-determining step, so the rate law equation is r = k[NO2Cl(g)], which agrees with the experimentally determined rate law equation. (ii) The first step is the rate-determining step, so the rate law equation is r = k[NO2Cl(g)]2, which does not agree with the experimentally determined rate law equation, so this mechanism is incorrect. (iii) The second step is the rate-determining step, so the rate law equation is r = k[NO2Cl(g)][Cl], which does not agree with the experimentally determined rate law equation, so this mechanism is incorrect. Statement: Proposed mechanisms (ii) and (iii) are incorrect because they do not agree with the experimentally determined rate law equation. (c) No, it is not appropriate to ask which mechanism is correct, because the data only shows whether the proposed mechanism is consistent with the data. Other mechanisms could also be consistent. 80. (a) Between run 1 and run 3, the rate quadruples when [NO(g)] doubles and [O2(g)] remains constant, so the order of reaction with respect to NO(g) is 2. (b) Between run 1 and run 2, the rate doubles when [O2(g)] doubles and [NO(g)] remains constant, so the order of reaction with respect to O2(g) is 1. (c) The total order of the reaction is 1 + 2 = 3.


(d) Given: proposed reaction mechanism; orders of reaction: NO(g) = 2, O2(g) = 1

NO(g) + O2(g) ! NO2(g) + O(g) (fast)NO(g) + O(g) ! NO2(g) (slow)

Required: agreement of mechanism with the experimentally determined rate law Solution: The experimentally determined rate law is r = k[NO(g)]2[O2(g)]. The second step in the proposed mechanism is the rate-determining step, so the rate law equation is r = k[NO(g)][O(g)], which does not agree with the experimentally determined rate law equation. Statement: The mechanism does not agree with the proposed rate law (ii) and (iii) are incorrect because they do not agree with the experimentally determined rate law equation. Evaluation 81. Answers may vary. Sample answers: (a) Some reasons why the per capita energy use of Canadians is one of the highest of any country are that most of Canada has very cold winters so we use a lot of energy for heat, the distance between populated places is large so we use a lot of energy to transport goods and people from one place to another, and we are a relatively wealthy country, so people tend to use purchase new clothing and other items (which take fuel to create) and to use fuel for recreational purposes. (b) To decrease our energy use, Canadians could keep buildings slightly less warm in winter and slightly warmer in summer, increase efficiency of heating systems, improve public transportation systems, add high taxes to fuels to discourage their use, carpool, and walk or ride bicycles for short errands. (c) The cold winters and the large distances between population centres in Canada affect our energy use, but cannot be changed. (d) I agree that the per capita energy use in Canada will always be one of the highest in the world, because Canada has a fairly small population and very cold winters. We can adopt some of the measures mentioned above to conserve some energy, but the cold winters and great distances will always mean we have to use more energy than others for basic needs. 82. Answers may vary. Sample answer: (to the Minister of Natural Resources) Dear Minister, At current rates of energy use, Canadians are using traditional energy sources faster than they can be renewed. We desperately need to develop sustainable, large scale alternatives. I encourage your ministry to promote research into ways of harnessing tidal energy for use across the country. We are surrounded by three oceans, the force behind tidal energy is immense, and tidal energy is renewable and predictable. Canada is well placed to become a leader in this field as we develop the capacity to supply our own energy needs for the foreseeable future. We need to break our reliance on fossil fuels. Although this research could be costly in the short term, I believe that Canadian taxpayers would recognize it as a very worthwhile long-term investment.


83. Answers may vary. Sample answer: No, I would not be concerned if a wind farm was in my area, because wind farms produce renewable energy that is important to reducing our global footprint and they have very few risks. We all have to bear some of the responsibility for creating the energy that we use as a society and I would rather have a wind farm in my area than an oil well or a nuclear plant, which carry far more environmental risks. 84. No, fire departments do not need to warn people about storing lumber near potential sources of sparks the way they do for large masses of paper, because the surface area of lumber is very small compared to its mass, so a single spark is unlikely to ignite a pile of lumber. 85. Answers may vary. Sample answers: Industrial processes are likely to increase reaction rates by using catalysts, increasing temperature, or increasing concentration of reactants. Increasing concentration is relatively inexpensive but generally works best in homogeneous reactions. Increasing temperature is effective but higher temperatures can introduce hazards to material handling and extra fuel costs. Catalysts are safer but some catalysts are expensive. 86. Answers may vary. Sample answers: (a) The advantages to society of producing sugars using a biocatalyst such as glucose isomerase are that the biocatalyst does not use any toxic chemicals and production is less expensive. (b) The main disadvantage to society of the use of high fructose corn syrup as a sweetener is overconsumption of sugar leading to health problems. (c) I think glucose isomerase has had a positive effect on society because it has reduced food costs. 87. Answers may vary. Sample answer: I agree with the scientist who proposed mechanism 1 because it is consistent with the rate law. (Note that mechanism 3 is also consistent with the rate law.) 88. From the experimental data in Table 6, the reaction is first order with respect to H2(g) and second order with respect to NO(g). Therefore, step 2 in the proposed mechanism is the rate-determining step because a fast first step is first order in both reactants, producing the intermediate for step 2, which is first order with respect to NO(g). 89. (a) It is important to know the rate of decomposition of greenhouses gases in the atmosphere because greenhouse gases increase the retention of thermal energy until they decompose. (b) The lifespan of a molecule is considered in determining the global warming potential (GWP) of an atmospheric gas because the warming effect is cumulative over time.


Reflect on Your Learning 90. Answers may vary. Sample answer: Ways to Determine the Enthalpy of a Reaction

Method Summary of method Strengths Weaknesses calorimetry Reaction run in a

calorimeter and temperature change is recorded.

– real data – may be difficult to run the reaction in calorimeter

bond energy

Enthalpy change calculated from energies of reactant and product chemical bonds

– fast and simple calculation

– based on average bond energy, not that of specific compounds

heats of formation

Calculate enthalpy based on Hess’s law and the standard enthalpies of formation.

– accurate data available

– simple calculation

– heats of formation not available for every compound

– result is an estimate 91. Answers may vary depending on individual student experience. 92. Answers may vary depending on individual student experience. 93. Answers may vary. Sample answer: Because our universe is infinite, and because the human brain has an infinite capacity to wonder about things it does not understand, I do not believe science will ever reach a point where there are no theoretical constructs. As we have seen, scientific inquiry extends in many directions, for example: near and far; large and small; identities, descriptions, processes, and causes. There will always be something we have not gathered enough direct evidence to prove. Every answer leads to new questions. 94. Answers may vary. Sample answer: While research that deals directly with societal issues is crucial, research that is driven solely by the desire to understand is also important. Much of the research being carried out today with definite societal goals is based on previous research that was carried out to learn. For example, research into alternative fuel technologies makes use of centuries of knowledge gained about various chemicals and about geophysical properties and processes. It could not proceed without this scaffolding. Research 95. Answers may vary. Students’ answers should include a method for comparing the efficiency of the two fuels, evaluating the availability of resources to produce sufficient ethanol, and the costs and benefits to consumers and to food production. 96. Answers may vary. Sample answer: Coal is used to generate approximately 13 % of all electrical power in Canada, but its use is not uniform. In some provinces, coal is the main source of energy, and in other provinces it is not used at all. From the largest amount of coal used for power generation to the smallest, Canadian provinces rank as follows: Ontario, Alberta, Saskatchewan, Nova Scotia, New Brunswick, and Manitoba, with other provinces using insignificant amounts.


It is important to realize that British Columbia and Alberta produce over 80 % of Canada’s coal. Nova Scotia is another coal-producing province. Alberta likely uses a lot of coal to generate power because it produces a lot of coal and does not incur shipping costs to acquire it. Ontario’s high use of coal to generate power is a result of its great need for power due to its high population and its historic (but changing) role as the industrial engine of Canada. British Columbia, Quebec, and Newfoundland and Labrador have access to large rivers that can be used for hydroelectric power generation. While Alberta is actively developing new coal-fired power generation plants, provinces that are not coal producers are looking for ways to reduce their reliance on coal as a result of concerns about the greenhouse gases emitted when it is burned. Ontario, for example, is planning to decrease the amount of power generated by burning coal, and is researching the use of nuclear power instead. 97. Answers may vary. Answers should include the following information: • While wind turbines in North America and Europe can be huge and grouped in large “farms”, wind turbines in developing areas of China and India tend to be small and generate power that will be used locally. This eliminates the need to build infrastructure to bring electrical power to remote areas. • Turbines for offshore power generation tend to be larger than those used on land, because installation is so challenging, the turbines must be as efficient as possible. Since offshore winds are often of higher speeds, countries with access to ocean or lakes that can be used in this way tend to construct offshore wind turbines. • Other than size differences, modern wind turbines have been becoming more and more similar, using three blades, and turning the blades to catch the wind, for example. These turbines can be used in both low and high wind conditions, and at both low and high temperatures. • In some countries, such as Germany, economic incentives have been provided to providers of wind power. This has led directly to Germany becoming one of the word’s leaders in wind-power generation. 98. Answers may vary. Sample answer: The major greenhouse gases are carbon dioxide, methane, nitrous oxide, and ozone. Since fossil-fuel combustion causes a large proportion of carbon dioxide emissions, methods for capturing the carbon dioxide before it enters the atmosphere have focused on power generating plants. In some cases, a solvent is added to trap carbon dioxide from the flue emissions and prevent its release. The carbon dioxide trapped is transported to be stored underground (for example, in sandstone). In other cases, a chemical process can be used to capture carbon dioxide before it enters the air. This provides a higher yield, but can only be used where the facility was built specifically to accommodate it. There is some research ongoing into the development of a hydrogen-powered car that would trap carbon dioxide emissions and store them in liquid form to prevent their release into the environment. Nitrous oxide is also produced during combustion at high temperatures. Many companies use a selective catalytic reduction process to remove nitrous oxide from the atmosphere. This involves combining ammonia with the nitrous oxide, then allowing a catalyst to absorb it, and separating it into nitrogen and oxygen—both harmless components of the atmosphere.


Significant amounts of methane are liberated from coal seams during mining. Additional methane enters the atmosphere as a result of raising livestock. Still more methane may be released as large areas of permafrost thaw with global warming. Research is being done into regenerative thermal oxidation strategies to reduce the amount of methane produced during mining. Citizen groups advocate less reliance on meat, which would reduce farming emissions. All attempts to stop global warming will reduce emissions from thawing permafrost. Ozone is formed when nitrous oxides and volatile organic compounds in the atmosphere interact. Steps to curb nitrous oxide emissions will help reduce ozone levels. In addition, paints and cleaners have been formulated with lower levels of volatile organic compounds, to reduce the formation of ozone in the atmosphere. 99. Because biological catalysts are designed to work with molecules of living things, they may be particularly effective for producing chemicals useful as medicines. In addition, because they are natural substances, they may not have side effects as severe as some manufactured catalysts may have. 100. Answers may vary. Answers should include information such as the following: Adsorption catalysts work by binding reactant molecules temporarily on their surface. They are used as heterogeneous catalysts; for example, the metal catalyzes the decomposition of nitrogen oxides in a catalytic converter. Catalysts that form intermediate compounds react with other chemicals to make an unstable compound that breaks down to release a new molecule and the catalyst. Enzyme reactions in cells use intermediate catalysts.

unit 3 review, pages 406–413 - weebly · biocatalysts support the tenets of green chemistry...

Documents