SCH4U - Unit 3 - Version C · PDF fileLesson 9: Thermochemistry ... In living organisms, the process of cellular respiration converts simple sugars into cellular energy, carbon dioxide,

SCH4U

Grade 12 University Chemistry

Lesson 9 – Thermochemistry

SCH4U – Chemistry Unit 3 - Introduction

Copyright © 2008, Durham Continuing Education Page 2 of 70

Unit 3: Rates of Reaction Energy transformations take place continually inside our bodies and in the world around us. For example, you transform the food you eat into useable cellular energy or adenosine triphosphate (ATP). Machines such as automobiles and power tools convert fuel into mechanical energy. In this unit you will learn more about energy transformations including the factors that affect reaction rate, and their kinetics. Overall Expectations • demonstrate an understanding of the energy transformations and kinetics of

chemical changes; • determine energy changes for physical and chemical processes and rates of

reaction, • demonstrate an understanding of the dependence of chemical technologies and

processes on the energetics of chemical reactions.

SCH4U – Chemistry Lesson 9


Lesson 9: Thermochemistry Heat transformations take place everyday in both living organisms and machines. Cars, for example, convert gasoline into energy to propel the car forward. Heat is also lost in the process. In living organisms, the process of cellular respiration converts simple sugars into cellular energy, carbon dioxide, and water. Again, heat is lost in the process. The study of heat transformation is called Thermochemistry, and it will be the focus of this lesson. What You Will Learn After completing this lesson, you will; • write thermochemical equations, expressing the energy change as a H value or as a

heat term in the equation; • determine heat of reaction, and use the data obtained to calculate the enthalpy

change for a reaction • compare the energy changes resulting from physical change, chemical reactions,

and nuclear reactions (fission and fusion); Thermochemistry Thermochemistry is the study of energy changes involved in chemical reactions. These changes can be physical (i.e. melting ice,) chemical (rusting iron) or nuclear (nuclear fusion reactions in the Sun). All energy transformations are governed by the 1st Law of Thermodynamics, which states that the total energy of the universe is constant. Thermodynamics Terms There are a set of terms that are necessary to understand when discussing thermodynamics. They are summarized following: Thermal energy: The energy available from a substance as a result of its motion of its molecules. For example, when you burn your foods, you break down sugars into carbon dioxide, water, ATP and thermal energy (heat). The SI unit for measuring work and all forms of energy is the Joule (J). Chemical system: The set of reactants and products undergoing energy transformations. Chemical systems can be open (energy and matter can move in or out), closed (only energy can move in or out), or isolated (neither energy or matter can move in or out)



Surroundings: All matter around the system that is capable of absorbing or releasing thermal energy. Consider the following reaction taking place in your body cells:

C6H12O6 + 6O2 6H2O +2CO2 + energy The molecules (glucose, oxygen, water, and carbon dioxide are the chemical system, while the surrounding are the extracellular fluid in your cells. Heat: Amount of heat energy transferred between substances, systems or surroundings. When a reaction occurs, heat is transferred between substances, systems and surroundings. Exothermic: A chemical system that releases energy from its surroundings Endothermic: A chemical system that absorbs energy from its surroundings Temperature: A measure of the average kinetic energy of the particles in a sample

In early work, the energy required to raise the temperature of one gram of water one Celsius degree was defined as one calorie (from the Latin word for heat, calor). The energy needed to raise the temperature of one kilogram of water one Celsius degree is 1000 times larger, so it was called a kilocalorie. The kilocalorie is the unit still used in most countries to describe the energy available in food. It was commonly referred to as one Calorie (capital C).

The calorie and the Calorie are not SI metric units. They have therefore been replaced by the joule, which is the standard unit for any form of energy. A joule (J) is the energy involved when a force of one Newton (1N) acts through a distance of one meter (1m). One calorie is equal to 4.184 J, and one Calorie (kilocalorie) is equal to 4184 J or 4.184 kJ. To raise the temperature of 1.0 kg of water by 1.0oC requires 4184 J or 4.184 kJ of energy.

Support Questions

1. Identify each of the following as physical, chemical, or nuclear change.

a) A gas stove cooking pasta b) an ice cube melting in a glass c) wax melts d) ice applied to a sore back



2. Identify the system and the surroundings in each of the examples in the previous question.

3. A cup of water at 100oC has a higher temperature than a swimming pool full of water

at 20oC, but the pool has more thermal energy. Explain. Calorimetry Calorimetry is an experimental technique used to measure heat energy changes in a chemical system. A calorimeter is an instrument for measuring the heat of a reaction. The following diagram depicts a simple calorimeter:

Figure 9.1: A simple calorimeter Any heat energy changes that take place are measured by a change (increase or decrease) by the temperature of the surroundings, which in most cases is water in the cup. However, it is important to note that we cannot just simply measure the change in temperature. There are other factors that will depend on the amount of energy transferred: 1. The mass of the water 2. Specific Heat Capacity



Specific Heat Capacity The specific heat capacity is the amount of heat energy that is needed to raise the temperature of one gram of a substance 1oC (or 1 K). Specific heat capacities vary from substance to substance, and even for different states (solid, liquid, gas) of the same substance.

• Symbol is “c”, Units of J/g●oC • Larger values indicate that more energy is required to increase its temperature. • Some specific heat capacities are summarized in table 9.1 below.

Table 9.1: Specific Heat Capacities of Various Substances

Substance Concrete Aluminum Wood Water (liquid) Specific Heat Capacity

(J/ g●oC) 0.88 0.900 1.76 4.184

Water has a very high specific heat capacity, and as such, is commonly used in calorimetry experiments. All of these factors, mass, specific heat capacity and temperature change must be considered when calculating the amount of heat energy transferred in a chemical system. This leads us to an important equation used to calculate heat transfer (Q):



Example 1: When 500 mL of water in an electric kettle is heated from 25oC to 65oC, how much heat energy flows into the water?

Solution 1: First calculate the mass, considering that the density of water is 1g/mL

500 mL × 1 g/mL = 500 g

Given: m = 500g c = 4.18J/goC

TΔ = 65oC - 25oC = 40oC

Required: Q =?

Equation: Q mc T= Δ

Solution: 500Q g= 4.18 /J g× C° 40 C× °

83600 or 83.6J kJ=

Statement: 83.6kJ of heat energy flows into an electric kettle when 500 mL of water is

heated from 25EC to 65EC.

The specific heat capacities of various substances are summarized in table 9.2.

Table 9.2: Specific Heat Capacities of Various Substances



Support Questions

4. A swimming pool contains 1000 L of water. When the water is warmed by solar

energy, its temperature increases from 15.8oC to 22.1oC. How much heat does the water absorb? (cwater(liquid)= 4.184 J/g oC)

5. A cook heats water from 20oC to 50oC. Calculate the mass of water that could be

warmed by the addition of 8.00kJ of heat. 6. A 50% ethylene glycol solution has a specific heat capacity of 3.5J/goC. What

temperature change would be observed in a solution of 4kg of ethylene glycol if it absorbs 250kJ of heat?

Enthalpy • the total internal energy of a substance at a constant pressure • chemists study the enthalpy change, ∆H, that accompanies a chemical process • ∆H of a process is equivalent to its heat change at a constant pressure • ∆H can be associated with physical changes as well as chemical reactions

Molar Enthalpies Molar enthalpy, ΔHx is the enthalpy change associated with one mole of substance. For example:

H2(g) + 1/2O2(g) H2O(g) + 242.0kJ The enthalpy change associated per mole of substance in the above reaction is -285kJ/mol. Enthapy changes for exothermic reactions are given a negative(-) sign or written or placed on the product side of the equation (with a positive (+) sign) Enthapy changes for endothermic reactions are given a positive(+) sign or placed on the reactant side of the equation Stating the molar enthalpy is a method of describing the energy changes in a variety of physical and chemical changes involving one mole of a particular reactant or product. For example consider the vaporization of water (physical change) H2O(l) + 40.8kJ H2O(g) ΔHvap = 40.8kJ/mol



These values are obtained empirically. Selected molar enthalpies are found in table 9.3. Table 9.3: Molar Enthalpies of Selected Substances Chemical Name Formula Molar enthalpy of

fusion (kJ/mol) Molar enthalpy of vaporization (kJ/mol)

Sodium Na 2.6 101 Chlorine Cl2 6.40 20.4 Sodium chloride NaCl 28 171 Water H2O 6.03 40.8 Ammonia NH3 - 1.37 Freon-12 CCl2F2 - 34.99 Methanol CH3OH - 39.23 Ethylene glycol C2H5(OH)2 - 58.8 Molar Enthalpies can also be used in heat calculations, let try a sample calculation. Example 2: Freon 12 (molar mass 120.91 g/mol) is a common refrigerant that is vaporized in tubes inside a refrigerator, releasing heat. This results in heat being released to the outside of the fridge. If 500g of the Freon is vaporized, what is the expected enthalpy change, ΔH? Solution 2: Given: Δ Hvap = 34.99kJ/mol (from table 9.3) M = 120.91g/mol m = 500g Required: ?HΔ = Equation: /m Mη = vapH HηΔ = × Δ

Solution: First calculate the moles of Freon 500 gη = / 120.91g /4.135 mol

mol=

Second calculate the enthalpy change 4.135H molΔ = 34.99 /kJ mol×144.7kJ=

Statement: The expected enthalpy change when Freon 12 is vapourized is 144.7 kJ.



Support Questions

7. Calculate the enthalpy change ΔH for the vaporization of 150.0g of water at 100oC. Using Calorimetry to Find Molar Enthalpies As mentioned above, a calorimeter is a device used to measure heat energy changes There are three assumptions used in calorimetry: • No heat is transferred between the calorimeter and the outside environment • Any heat absorbed or released by the calorimeter materials is neglible • A dilute aqueous solution is assumed to have a density and specific heat capacity

equal to that of pure water (1.00g/mL and 4.18J/goC) Since this is a correspondence course, you won’t be able to perform a calorimetry experiment, but we can practice some calculations for a typical calorimetry experiment. Example 3: In a calorimetry experiment, 7.46 g of potassium chloride is dissolved in 100.0 mL of water at an initial temperature of 24.1°C. The final temperature of the solution is 20oC. What is the molar enthalpy? Solution 3: Given: KCl 7.46m g= KCl 74.6 /M g mol= ( )water 100.0 100 100m g mL g= = water 4.18 /c J g C= ° 24.1 20.0 4.1T C C CΔ = ° − ° = ° Required: sol ?HΔ =

Equation: /m Mη = solmc THηΔ

Δ = **

Note: Recognize the Law of Conservation of Energy (in this case, P is

constant and Work = 0, therefore, ΔH = q) ΔH = q ΔH (KCl dissolving) = q (calorimeter water) Or in other words solH mc Tη × Δ = Δ



Solution: Start by calculating the moles of potassium chloride

KCl /

7.46

m M

g

η =

= / 74.6 g /0.100

molmol=

Calculate the molar enthalpy. Since, solH mc Tη × Δ = Δ , then we can rearrange to;

sol

100

mc TH

g

ηΔ

Δ =

=4.18 /J g× C° 4.1 C× °

4

0.1001.7 10 /

molJ mol= ×

Statement: The final molar enthalpy of the solution is 41.7 10 /J mol× .

Support Questions

8. The energy change in the process H2O(g) H2O(l) could be described as a molar enthalpy of condensation. State the type of molar enthalpy that would occur in each of the following reactions:

a) Br2(l) Br(g) b) CH4(g) CH4(s) c) NaCl(s) Na+

(aq) + Cl-(aq) d) KOH(aq) + HCl(aq) KCl(aq) + H2O(l)

9. In the lab, you add 50 mL of concentrated hydrochloric acid (12 mol/L), HCl to form

250mL of dilute solution. The temperature of the solution changes from 18oC to 24oC. Calculate the molar enthalpy of dilution of hydrochloric acid.

10. What mass of lithium chloride must have dissolved if the temperature of 200g of

water increased by 6.0oC? The molar enthalpy of solution of lithium chloride is 37kJ/mol.



Representing Enthalpy Changes Enthalpy changes can be represented in four ways • As part of a thermochemical equation

e.g. CH3OH(l) + 3/2 O2(g) CO2(g) + H2O(g) +726 kJ

• By writing a chemical equation and stating its enthalpy change e.g. CH3OH(l) + 3/2 O2(g) CO2(g) + H2O(g) ΔH = -726 kJ

• By stating the molar enthalpy of a specific reaction e.g. ΔHcomb = -726 kJ/mol CH3OH

• By drawing a potential energy diagram Method 1: Thermochemical Equations with Energy Terms As previously mentioned, if the reaction is exothermic, the energy is listed with the products, and if the reaction is endothermic, the energy is listed with the reactants (both with a positive (+) sign). Example 4: Write a thermochemical equation for the combustion of butane. The molar enthalpy for the combustion of butane is -2871kJ/mol Solution 4: Recall from unit two, that the products of a complete combustion reaction are carbon dioxide and water. Write out and balance your equation.

2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(l) Note that the molar enthalpy is -2871kJ per one mole, and in this reaction we have two moles, so calculate your enthalpy for two moles ΔH = 2mol x -2871kJ/mol = -5742kJ Since the enthalpy value, the reaction is exothermic, and the enthalpy must be listed with the products

2C4H10(g) + 13O2(g) 8CO2(g) + 10H2O(l) + 5742kJ Method 2: Thermochemical Equations with ΔH symbols You can also write the Thermochemical equation and then write the ΔH value beside it. Remember that the ΔH value is negative for exothermic reactions and positive for endothermic reactions.



Example 5: Sulphur dioxide and oxygen react to form sulphur trioxide. The molar enthalpy for the combustion of sulphur dioxide is -98.9kJ/mol. What is the enthalpy change for this reaction? Solution 5: Start off by writing the balanced thermochemical equation

2SO2 + O2 2SO3 Calculate your enthalpy for 2 moles of sulphur dioxide ΔH = 2mol x -98.9kJ/mol ΔH = -197.8 kJ Now state the enthalpy value with the balanced equation

2SO2 + O2 2SO3 ΔH = -197.8 kJ Method 3: Molar Enthalpy of Reaction Recall that the molar enthalpy, ΔHx is the enthalpy change associated with one mole of substance. The standard molar enthalpy, ΔHo

x is the energy change associated with one mole of substance at 100kPa and usually 25oC. Table 9.4: Molar enthalpy of Reactions Type of molar enthalpy Example of change Solution Δ( )solH + −→ +(s) (aq) (aq)NaBr Na Br Combustion Δ comb( )H 4(g) 2(g) 2(g) 2 (I)CH 2O CO 2H O+ → + Vaporization Δ( H )vap →3 (I) 3 (g)CH OH CH OH Freezing →2 (I) 2 (s)H O H O Neutralization Δ neut( ) *H + → +(aq) 2 4(aq) 2 4(aq) 2 (I)2NaOH H SO 2Na SO 2H O Neutralization Δ neut( ) *H + → +(aq) 2 4(aq) 2 4(aq) 2 (I)

1 1NaOH H SO Na SO H O2 2

Formation Δ f( ) * *H + + →(s) 2(g) 2(g) 3 (I)1C 2H O CH OH2

* Enthalpy of neutralization can be expressed per mole of either base or acid consumed.



Example 6: Write an equation whose energy change in the molar enthalpy of combustion of propanol, (C3H7OH).

Solution 6: C3H7OH(g) + 92

O2(g) 3CO2(g) + 4H2O(l)

Method 4: Potential Energy Diagrams The energy transferred can be communicated graphically using a potential energy diagram.

Figure 9.1: Potential Energy Diagrams



Key Question #9

1. Solar energy can heat cold water for household hot water tanks. What quantity of heat is obtained from solar energy if 150kg of water is heated from 15oC to 50oC? (5 marks – remember to show your work)

2. Identify each of the following processes as exothermic or endothermic; (7 marks)

a) The combustion of propane in a barbeque tank. b) Ag2O(s) + heat 2Ag(s) + 1/2O2(g) c) As solid NaNO3 dissolves in water at 25oC, the temperature changes to 22oC. d) When calcium is added to water at 22oC, the temperature changes to 27oC.

e) Mg(s) +12

O2(g) MgO(s) + bright light

f) 2H2(g) + O2(g) 2H2O(l) ∆H =-571.0kJ g) N2(g) + O2(g) 2NO(g) ∆H = 180.8kJ

3. To test the properties of a fertilizer, 15.0 g of urea, NH2CONH2(s), is dissolved in

150mL of water in a simple calorimeter. A temperature change from 20.6EC to 17.8EC is measured. Calculate the molar enthalpy of solution for the fertilizer urea. (5 marks – remember to show your work)

4. a) A laboratory technician adds 56.1mL of concentrated 12 mol/L hydrochloric acid

to water to form 500mL of dilute solution. The temperature of the solution changes from 20.2EC to 22.3EC. Calculate the molar enthalpy of dilution of hydrochloric acid. (5 marks – remember to show your work)

b) What effect would there be on the calculated value for the molar enthalpy of dilution if the technician accidently used too much water so that the volume was actually more than 500 mL. Explain. (3 marks)

5. If the molar enthalpy of combustion of ethane is -1.56 MJ/mol, how much heat is

produced in the burning of; (4 marks)

a) 5.0 mol of ethane b) 45 g of ethane 6. In a laboratory investigation into the neutralization reaction a scientist adds

potassium hydroxide to nitric acid (balanced reaction following); (5 marks)

HNO3(aq) + KOH(s) KNO3(aq) + H2O(l)

Data Collected Mass KOH = 4.6g T1 = 21oC Volume of nitric acid solution = 250mL T2 = 29.3oC Calculate the molar enthalpy of neutralization of potassium hydroxide.



Key Question #9 (continued)

7. Draw a potential energy diagram with appropriately labelled axes to represent: (4 marks)

a) The exothermic combustion of octane ∆Ho =-5.47MJ) b) The endothermic formation of diborane (B2H6) from its elements ∆Ho = +36kJ)

8. Translate each of the molar enthalpies given below into a balanced thermochemical

equation, including the enthalpy change, ∆H

a) The enthalpy change for the reaction in which solid magnesium hydroxide is formed from its elements at SATP is -925kJ/mol. (3 marks)

b) The standard molar enthalpy of combustion for pentane, C5H12, is -2018 kJ/mol (3 marks)

9. For each of the following reactions, write a thermochemical equation including the

energy term as part of the equation. (6 marks)

a) The combustion of butane in a candle lighter is -2.86MJ/mol b) The molar enthalpy of the transition of graphite to diamond is 2kJ.

SCH4U


Lesson 10 – Enthalpies of Reaction



Lesson 10: Enthalpies of Reaction In lesson 9 you learned that calorimetry is one method to determine the enthalpy of a reaction. In this lesson, you will learn about two more methods; Hess’s Law and using heats of formation. What You Will Learn After completing this lesson, you will; • explain Hess’s law, using examples; • apply Hess’s law to solve problems, including problems that involve data obtained

through experimentation (e.g., measure heats of reaction that can be combined to yield the H of combustion of magnesium);

• calculate heat of reaction using tabulated enthalpies of formation; Hess’s Law of Heat Summation Hess’s Law states that the enthalpy change of a physical or chemical process depends only on the beginning conditions (reactants) and the end conditions (products). The enthalpy change is independent of the pathway of the process and the number of intermediate steps in the process. Hess’s law allows you to determine the energy of a chemical reaction without directly measuring it. There are two ways to accomplish this:

1. by combining chemical equations algebraically 2. by using the enthalpy of formation reactions (a special class of reactions)

Combining Chemical Equations Algebraically Add equations for reactions with known enthalpy changes, so that their net result is the reaction you are interested in. This may require manipulation of the known equations by reversing or multiplying each coefficient (you need to multiply ∆Ho by the same number)

Example 1: What is the enthalpy for the formation of one mole of diamond from one mole of carbon?

Target sequence C (s, graphite) C (s, diamond) ΔH° =? kJ

(The presence of the degree sign, (°), on the enthalpy indicates that the reaction is happening under standard conditions.)



The following known equations, determined by calorimetry, are provided:

1. C (s, graphite) + O2(g) CO2(g) ΔH° = -394 kJ

2. C (s, diamond) + O2(g) CO2(g) ΔH° = -396 kJ

Solution 1: We can manipulate the above known equations, to obtain the target sequence, and thus obtain the enthalpy for the reaction.

First off, reverse the bottom equation. This will put the C (s, diamond) on the product side, where we need it. When you reverse an equation the sign on the enthalpy value is also reversed. The reason behind this is that original equation is exothermic. We know this from the negative in front of the enthalpy value. That means that the opposite, reverse equation is endothermic. Putting in enthalpy (endothermic) is the reverse, the opposite of exothermic (giving off enthalpy). Hence, we change the sign EVERY time we reverse an equation.

1. C (s, graphite) + O2(g) CO2(g) ΔH° = -394 kJ

Reverse 2. CO2(g) C (s, diamond) + O2(g) ΔH° = +396 kJ

If we add the two equations together, the oxygen and carbon dioxide will cancel out. This is, of course, what we want since those two substances are not in the final, desired equation. The enthalpy values are also added together.

CO2(g) + C (s, graphite) + O2(g) CO2(g) + C (s, diamond) + O2(g) ΔH° = (-394 kJ) + (+396 kJ)

Notice the items which are the same on both sides and cancel them:

CO2(g) + C (s, graphite) + O2(g) CO2(g) + C (s, diamond) + O2(g) ΔH° = (-394 kJ) + (+396 kJ)

C (s, graphite) C (s, diamond) ΔH° = +2 kJ



Example 2: Calculate the enthalpy for the following reaction:

Target sequence N2(g) + 2O2(g) 2NO2(g) ΔH° =? kJ

Using the following two equations:

1. N2(g) + O2(g) 2NO(g) ΔH° = +180 kJ

2. 2NO2(g) 2NO(g) + O2(g) ΔH° = +112 kJ

Solution 2: Note that in the target sequence, the nitrogen dioxide, NO2, is on the right side, so begin by reversing equation 2.

1. N2(g) + O2(g) 2NO(g) ΔH° = +180 kJ

Reverse 2. 2NO(g) + O2(g) 2NO2(g) ΔH° = -112 kJ

Notice that the sign on the enthalpy has changed from positive to negative.

Next, we add the two equations together and eliminate identical items. We also add the two enthalpies together.

N2(g) + 2O2(g) 2NO2(g) ΔH° = +68 kJ

Example 3: What is the enthalpy change for the formation of one mole of butane (C4H10) gas from its elements? The reaction is:

Target sequence: 4C(s) + 5H2(g) C4H10(g) ΔH° =? kJ

The following known equations are provided:

1. C4H10(g) + 132

O2(g) 4CO2(g) +5 H20 ΔH° = -2657.4 kJ

2. C(s) + O2(g) CO2(g) ΔH° = -393.5 kJ

3. 2H2(g) + O2(g) 2H2O(g) ΔH° = -483.6 kJ



Solution 3: Begin by reversing equation 1 and changing the enthalpy sign.

Reverse 1. 4CO2(g) +5 H2O C4H10(g) + 132

O2(g) ΔH° = +2657.4 kJ

Since we need 4 moles of carbon in the target sequence, we will then multiply equation 2 by 4. Note that the enthalpy is also multiplied by 4.

2.x 4 4C(s) + 4O2(g) 4CO2(g) ΔH° = -1574 kJ

We need 5 moles of hydrogen gas in the target sequence. In equation 3, we currently

have 2 moles. If we multiply equation 3 by 52

we will obtain 2 moles. Don’t forget to

also multiply the enthalpy by 52

.

3 x 52

5H2(g) + 25 O2(g) 5H2O(g) ΔH° = -1209kJ

Now add together all of the reactants, products, and enthalpy for your three modified equations

4CO2(g) +5 H20 + 4C(s) + 4O2(g)+ 5H2(g) + 52

O2(g) C4H10(g) + 132

O2(g) + 4CO2(g) + 5H2O(g)

ΔH° = -125.6kJ



Support Questions

11. Calculate ∆H for the reaction: C2H4(g) + H2(g) C2H6(g), from the following data.

C2H4 (g) + 3O2 (g) 2CO2 (g) + 2H2O (l) ∆HE = -1411. kJ

C2H6 (g) + 3½O2 (g) 2CO2 (g) + 3H2O (l) ∆HE = -1560. kJ

H2 (g) + ½ O2 (g) H2O (l) ∆HE = -285.8 kJ

12. Calculate ∆H for the reaction 4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g), from the following data.

N2(g) + O2(g) 2NO(g) ∆HE = -180.5 kJ

N2(g) + 3H2(g) 2NH3(g) ∆HE = -91.8 kJ

2H2(g) + O2(g) 2H2O(g) ∆HE = -483.6 kJ

13. Find ∆H° for the reaction 2H2(g) + 2C(s) + O2(g) C2H5OH(l), using the following thermochemical data.

C2H5OH(l) + 2O2(g) 2 CO2(g) + 2H2O(l) ∆HE = -875. kJ

C(s) + O2(g) CO2(g) ∆HE = -394.51 kJ

H2(g) + ½O2(g) H2O(l) ∆HE = -285.8 kJ

14. Calculate ∆H for the reaction CH4(g) + NH3(g) HCN(g) + 3H2(g), given:

N2(g) + 3H2(g) 2NH3(g) ∆HE = -91.8 kJ

C(s) + 2H2(g) CH4(g) ∆HE = -74.9 kJ

H2(g) + 2C(s) + N2(g) 2HCN(g) ∆HE = +270.3 kJ



Support Questions (continued)

15. Calculate ∆H for the reaction 2Al(s) + 3Cl2(g) 2AlCl3(s) from the data.

2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g) ∆HE = -1049. kJ

HCl(g) HCl(aq) ∆HE = -74.8 kJ

H2(g) + Cl2(g) 2HCl(g) ∆HE = -1845. kJ

AlCl3(s) AlCl3(aq) ∆HE = -323. kJ

Standard Enthalpies of Formation

Standard molar enthalpy of formation, ∆Hof, is the quantity of heat energy that is

absorbed or released when one mole of a compound is formed directly from its elements in their standard states (SATP = 25oC, 100 kPa). By definition, the enthalpy of formation of an element in its standard state is zero. What does the word formation mean?

Formation means a substance, written as the product of a chemical equation, is formed DIRECTLY from the elements involved. The substance in question is always written with a coefficient of one. Here are some examples:

C (s) + O2 (g) CO2 (g) C (s) + (1/2) O2 (g) CO (g) H2 (g) + O2 (g) H2O2 (l) H2 (g) + (1/2) O2 (g) H2O (l) C (s) + 2 H2 (g) + (1/2) O2 (g) CH3OH (l)

* Recall that some elements (Hydrogen, Oxygen, Fluorine, Bromine, Iodine, Nitrogen, and Chlorine or HOFBrINCl) are diatomic elements, meaning they exist in pairs. For example, H2, O2, etc.

Writing Formation Equations

a) Write one mole of product in the given state b) Write the reactant elements in their standard states c) Balance your equation

Example 4: Write the formation equation for liquid ethanol, C2H5OH.

Solution 4: 2C(s) + 3H2(g) + 1/2O2(g) C2H5OH(l)



Using Standard Enthalpies of Formation

The enthalpy change for the target equation equals the enthalpy of formation for the products minus the enthalpies of formation for the reactants. This can be summarized in an equation

ΔH = ΣηΔHof (products) –ΣηΔHo

f(reactants)

The standard enthalpies of formation are found in appendix A at the end of the booklet. This table will be provided on the any exams or tests that you write.

Example 5: What is the molar enthalpy of combustion of methane fuel?

Solution 5:

CH4(g) + 2O2(g) CO2(g) + 2H20(l)

ΔH = ΣηΔHof (products) –ΣηΔHo

f(reactants)

ΔH = [(1 mol x -393.5kJ

mol) + (2 mol x -285.8

kJmol

)] – [(1 mol x -74.4 kJ

mol) + 2 mol x 0

kJmol

)]

Note that there are different enthalpy values for different states, so make sure you look up the correct state.

ΔH = [-965.1kJ] – [-74.4kJ]

ΔH = -890.7 kJ

Sometimes enthalpy problems are multistep. They involve more than one step and the following relationships are used: 1. Enthalpy change in the system = heat transferred to or from the surroundings 2. ΔH =ηΔHr

ΔH = ΣηΔHo

f (products) –ΣηΔHof(reactants)



Example 6: What mass of octane is completely burned to cause the heating of 20.0g of aqueous ethylene glycol automobile coolant from 10oC to 90oC? The specific heat capacity of aqueous ethylene glycol is 3.5J/(goC). Assume water is produced as a gas and that all the heat flows into the coolant. The balanced equation is below

C8H18 (g) + 252

O2(g) 8CO2(g) + 9H2O(g)

Solution 6: First calculate the molar enthalpy of combustion of octane ΔH = [(8mol x -393.5kJ/mol) + (9mol x -241.8kJ/mol)] – [(1mol x -250.1kJ/mol) + (25/2mol x 0kJ/mol)] ΔH = -5074kJ Now we will assume that the heat produced by this reaction will be absorbed by the enthylene glycol (engine coolant). Remember: ΔHoctane = qcoolant Therefore; ηΔHc =mcΔT Solve for the moles of octane;

c

octane

20.0

mc TH

g

η

η

Δ=

Δ

=0.0035 kJ× / g C° 80 C× °

5074.1kJ /0.0011

molmol=

octane

/

1.1

m M

m mol

η=

= / 114.26 /g mol0.126g=



Support Questions

16. Write formation equations for the following:

a) benzene (C6H6) b) potassium bromate c) glucose

17. Use standard enthalpies of formation to calculate:

a) the molar enthalpy of combustion for pentane. b) the enthalpy change that accompanies the reaction between solid iron (III) oxide

and carbon monoxide gas to produce solid iron metal and carbon dioxide gas. 18. For each of the following reactions, use standard enthalpies to calculate ∆H:

a) CH4(g) + H2O(g) CO(g) + 3H2(g) b) CO(g) + H2O(g) CO2(g) + H2(g) c) N2(g) + 3H2(g) 2NH3(g)

19. Ammonium nitrate fertilizer is produced by the reaction of ammonia with nitric acid:

NH3(g) + HNO3(l) NH4NO3(s) Use standard enthalpies of formation to calculate the standard enthalpy change of the reaction used to produce ammonium nitrate.



Key Question #10

1. Use Hess’s Law to predict the enthalpy change for the reaction (5 marks)

HCl(g) + NaNO2(s) HNO2(g) + NaCl(s)

Using the following information: (1) 2NaCl + H2O 2HCl + Na2O ∆Ho

1 = -507kJ (2) NO + NO2 + Na2O 2NaNO2 ∆Ho

2 = -427kJ (3) NO + NO2 2N2O + O2 ∆Ho

3 = -43kJ (4) 2HNO2 N2O + O2 + H2O ∆Ho

4 = 34kJ 2. Use the thermochemical equations shown below to determine the enthalpy for the

reaction: (5 marks) C3H8(g) + 5O2(g) 3CO2 + 4 H2O(l)

CO2 C(graphite) + O2 HE = 221.6KJ H2(g) + 1/2O2(g) H2O(l) HE = -160.3KJ 3 C(graphite) + 4H2(g) C3H8(g) HE = -58.5KJ

3. Use the thermochemical equations shown below to determine the enthalpy for the

reaction: (5 marks) SO2(g) S(s) + O2(g)

H2S(g) + 3/2O2(g) H2SO3(l) HE = -306KJ H2SO3(l) H2O(l) +SO2(g) HE = 93KJ H2S(g) + 1/2O2(g) S(s) + H2O(l) HE = -232.5KJ

4. Write balanced equations for the formation of the following: (4 marks)

a) acetylene gas (C2H2) b) potassium chloride (KCl)




5. Use standard enthalpies of formation to calculate the enthalpy changes in each of the following equations: (6 marks)

a) Magnesium carbonate decomposes when strongly heated. b) Ethene burns in air

6. A sample of acetone (C3H6O) is burned in an insulated calorimeter to produce

carbon dioxide gas and liquid water. (9 marks)

a) Use standard enthalpies of formation to calculate the theoretical value for the molar enthalpy of combustion of acetone.

b) Suppose the experiment yielded the following results. Calculate the molar enthalpy of combustion of acetone using the experimental data.

Quantity Data Mass of water 120g Specific heat capacity of aluminum 0.91J/goC Mass of aluminum can 70g Initial temperature of calorimeter 20oC Final temperature of calorimeter 26oC Mass of acetone burned 0.087g

c) Calculate the percentage error by comparing the predicted and experimental values.

SCH4U


Lesson 11 – Energy Options



Lesson 11: Energy Options Energy changes occur in our trillions of cells that make up our body. We consume food, and then convert our food into useful energy called adenosine triphosphate or ATP. We also rely on energy to sustain our lifestyles. In this lesson, you will learn about the various ways energy can be harvested to meet these demands, including their advantages and disadvantages. What You Will Learn After completing this lesson, you will; • compare conventional and alternative sources of energy with respect to efficiency

and environmental impact (e.g., burning fossil fuels, solar energy, nuclear fission); Most of society’s energy needs are met by two sources: electricity and the combustion of fossil fuels such as natural gas and oil. In Ontario, electricity has three sources; hydroelectric, nuclear power, and the burning of fossil fuels. In all three sources, potential energy is converted into kinetic energy. All sources of energy have advantages and disadvantages. Hydroelectric power can flood large areas and displace wildlife from its natural habitat. Burning fossil fuels can cause smog and pollution such as acid rain. Nuclear power is very expensive and there are concerns about safety and waste disposal. We will discuss nuclear power in more detail below. Nuclear Fission In the generation of nuclear power, the most common reaction is the splitting of a uranium atom into two small nuclei. Nuclear fission reactions provide the energy for nuclear power generating systems. As soon as the nucleus of the uranium atom captures the neutron, it splits into two lighter atoms and throws off two or three new neutrons (the number of ejected neutrons depends on how the U-235 atom splits). The process of capturing the neutron and splitting happens very quickly, on the order of picoseconds (1x10-12 seconds).

The splitting of an atom releases an incredible amount of heat and gamma radiation, or radiation made of high-energy photons. The two atoms that result from the fission later release beta radiation (super fast electrons) and gamma radiation of their own as well.



There is controversy about use of nuclear power.

Advantages

• low costs • little air pollution • reduces dependence on fossil fuels

Disadvantages

• release of radioactive material • difficulty disposing of toxic

radioactive waste • cost of building nuclear generators • unknown health effects of low-level

radiation • thermal pollution

Key Question #11

1. There are many methods for generating electricity. These include:

Hydroelectric power Fossil Fuels Nuclear Power Geothermal Wind Solar

a) Research the following methods of generating electricity. Include the following;

advantages and disadvantages efficiency, in terms of energy per dollar spent efficiency in terms of energy output per gram of fuel used environmental impact comparison to Canadians versus other areas of the world

Your research must be summarized in a clear logical concise manner and include references that are reputable such as government agencies. Search engines such as Google, and Wikipedia are not reputable. Look for government or educational agencies for gathering information.

b) Write a report expressing the direction that you feel the government should take

on energy policy. Your report must be backed up by data from your research and should be 1-2 pages in length, in a word processed format. (25 marks – 20 for content/research, 5 for grammar/spelling)

SCH4U


Lesson 12 – Chemical Kinetics



Lesson 12: Chemical Kinetics Some reactions, such as the rusting of a nail, occur very slowly. Other reactions such as the explosion of a firework occur very rapidly. Why are some reactions so fast that they occur instantaneously, while other reactions are so slow, we don’t even know that they are occurring? This unit discusses the concept of reaction rates. It forms a branch of chemistry called chemical kinetics. What You Will Learn After completing this lesson, you will; • describe, with the aid of a graph, the rate of reaction as a function of the change of

concentration of a reactant or product with respect to time; express the rate of reaction as a rate law equation (first- or second-order reactions only); and explain the concept of half-life for a reaction;

• demonstrate understanding that most reactions occur as a series of elementary steps in a reaction mechanism.

• analyse simple potential energy diagrams of chemical reactions (e.g., potential energy diagrams showing the relative energies of reactants, products, and activated complex);

• explain, using collision theory and potential energy diagrams, how factors such as temperature, surface area, nature of reactants, catalysts, and concentration control the rate of chemical reactions;

Expressing & Measuring Reaction Rates Expressing Reaction Rate Chemical Kinetics is the study of ways to make a reaction proceed faster or slower. The pace at which the reaction occurs is termed the reaction rate. The reaction rate is the change in amount of reactants (or products) over time. Reaction rate is usually expressed as [concentration reactant] per unit of time (mol/L s), and reaction rates are always positive, by convention

cRatet

Δ=Δ



Example 1: What is the overall rate of production of nitrogen dioxide in the system shown below if the concentration of nitrogen dioxide changes from 0.34mol/L to 0.82mol/L in 4 minutes?

N2(g) + 2O2(g) 2NO2(g) Solution 1:

Given: 0.82 / 0.34 /0.48 /c mol L mol L

mol LΔ = −=

4minTΔ =

Required: Rate of production = ?

Equation: cRatet

Δ=Δ

Solution: Average & Instantaneous Reaction Rate

• reaction rates are not usually constant, they change with time

Average Rate - averages the change in [reactant] or [product] per unit of time over a given time interval (secant line)

Instantaneous Rate is the rate of a reaction at a particular time (tangent line)

Considering the graph below, the average rate of formation of oxygen gas would be calculated over a given time from (i.e. over 10 seconds), whereas the instantaneous rate would be for a given moment (i.e. at t = 10seconds). This is determined using a tangent.

0.48 /4min

0.12 / .min

cRatet

mol L

mol L

Δ=Δ

=

=



Example: Consider the following reaction: IO3(aq)

- + 5I(aq)

- + 6H+(aq) 3I2(aq) + 3H2O(l)

What are the reaction rates of the various reactant and products? The reaction rate of iodate (IO3(aq)

-) ions is 1.2 x 10-2mol/L.s In order to calculate the rate of the other reactants and products, we must consider the moles of each reactant and product. Thus:

I(aq)- = 1.2 x 10-2mol/L.s x 5 = 0.06 mol/L.s

H+

(aq) = 1.2 x 10-2mol/L.s x 6 = 0.072 mol/L.s I2(aq)

+ = 1.2 x 10-2mol/L.s x 3 = 0.036 mol/L.s H2O(l)

+ = 1.2 x 10-2mol/L.s x 3 = 0.036 mol/L.s Measuring Reaction Rates There are various ways that reaction rate can be measured in the laboratory. The method employed depends on the nature of the reaction (the reactants, how the reaction proceeds, type of reaction). The methods used are the following:

Rate of Formation of Oxygen Gas

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009

0 1000 2000 3000 4000 5000 6000 7000

Time (s)

[O2] (mol/L)



Reactions that produce a gas In reactions that produce a gas, the chemist can collect gas and measure the volume and pressure changes as a reaction occurs to get an indication of rate. Generally, the faster the reaction, the greater the change in pressure or volume in the same interval. Reactions that Involve ions Ions conduct electricity, so the more ions that are formed, the greater the conductivity. Reactions that change colour Many reactants will undergo a colour change when they form products. The intensity of the colour can be measured using a device called a spectrophotometer, which measures relative colour intensity.

Support Questions

20. Consider the following reaction:

MnO4-(aq) + 5Fe2+

(aq) + 8H+(aq) Mn2+

(aq) + 5Fe3+(aq) + 4H2O(l)

Given that the rate of reaction is 3.5 x 10-2 mol/L MnO4

-(aq), determine the rate of

reaction for all reactants and products in the reaction. 21. The reaction between ammonia and oxygen produces nitrogen monoxide and water

vapour.

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)

If the rate of consumption of ammonia is 1.6 x 10-2 mol/(L.s) calculate the rate of consumption of oxygen and the formation of water vapour.

22. Rates of reactions are generally fastest at the beginning of a reaction. Explain why

this is so.



Support Questions (continued)

23. A kinetics experiment is performed in which oxygen gas is collected. The concentration of oxygen is measured every 10 s. The table below has the data obtained from the experiment.

Time (s) Oxygen gas (mol/L) 0 0 10 0.23 20 0.40 30 0.52 40 0.60 50 0.66 60 0.70 70 0.73 80 0.75 90 0.76 100 0.76

a) Plot a graph of oxygen concentration vs. time b) Calculate the average formation of oxygen gas during the first 60 seconds c) the average rate of formation of oxygen gas between t = 20s and t = 60s



Factors Affecting Reaction Rate There are five major factors that affect the rate of a reaction. They are; • Chemical nature of the reactants • Concentration of reactants • Temperature • Presence of a catalyst • Surface Area Chemical Nature of Reactants - The rate of any reaction partially depends of the chemical nature of the reactants. For instance, metals generally will react with acid to produce hydrogen gas. However the rate of gas evolved varies from metal to metal. Understanding the important properties of metals allows chemists to choose metals that will react more slowly when trying to prevent corrosion, for example. Concentration - Generally if the concentration of the reactants is increased, the rate of the reaction increases. This is because there is a higher concentration of reactant available for collision to form product. Temperature - An increase in temperature speeds up molecule collisions and thus speeds of reaction rates and conversely cooling the temperature will slow reaction rates. Presence of a Catalyst - A catalyst is a substance that speeds up a chemical reaction without being consumed in the reaction itself. In your body, organic catalysts are called enzymes and speed up important cellular processes such as cellular respiration. When you consume food, your saliva releases an enzyme called amylase which speeds up the breakdown of starch into simpler sugars. Many chemical industrial processes rely on catalysts to speed reaction rates. Surface Area - Increasing the surface area increases the rate of reaction. You may have experienced this when cooking food. For example, chopping up potatoes into smaller allows the potatoes to cook faster.



Support Questions

24. Identify the five factors that increase the rate of a reaction and provide an example of each.

The Rate Law: Reactant Concentration and Rate Relating Reactant Concentrations and Rate Consider: In general, the rate of a reaction increases when the concentrations of the reactants increase

This relationship can be expressed in a general equation called the Rate Law Equation: • the exponents m and n must be determined by experiment • they do not necessarily correspond to the stoichiometric coefficients of their

reactants (they are usually 1 or 2, but values of 0,3, or fractions can occur) Order of the Reaction The values of the exponents establish the order of the reaction. The sum of the exponents (m + n) is known as the overall reaction order. Consider the following theoretical reaction; 2X + 2Y + 3Z products Experimental evidence suggests the following rate law equation r = k[X]1[Y]2[Z]0 The overall order of the reaction in this case is 3 (1+2+0)

[ ] [ ]m nRate A B∝

aA bB products+ →

[ ] [ ]m nRate k A B=



Rate Law Constant, k The magnitude of the rate constant, k, indicates the speed of a reaction

small k value indicates a slow reaction large k value indicates a fast reaction k remains constant throughout a reaction under constant conditions

Determining the Rate Law Exponents The exponents must be determined using experimental evidence. One method of directly measuring k, m, and n is called the method of initial rates. By measuring the initial rate (the rate near reaction time zero) for a series of reactions with varying concentrations of one reactant at a time, we can deduce to what power the rate depends on the concentration of each reactant. For example, let's use the method of initial rates to determine the rate law for the following reaction:

C3H6O + Br2 C2H5OBr + HBr

whose, rate law has the form:

rate = k[C3H6O]m[Br2]n

Using the following initial rates data, it is possible to calculate the order of the reaction for both bromine and acetone:

Experiment [BI2]o [C3H6O]0 Initial Rate (M•s-1) 1 0.1 M 0.1 M 1.64 x 10-5

2 0.2 M 0.1 M 1.65 x 10-5 3 0.1 M 0.2 M 3.29 x 10-5

To calculate the order of the reaction for bromine, notice that experiments 1 and 2 hold the concentration of acetone constant while doubling the concentration of bromine. The initial rate of the reaction is unaffected by the increase in bromine concentration, so the reaction is zero order in bromine. We can prove this mathematically by taking the ratio of the rates from experiments 1 and 2:

52 2 3 6 22

51 2 1 3 6 1

[ ] [ ] (0.2 ) (0.1 ) 1.64 10 1[ ] [ ] (0.1 ) (0.1 ) 1.64 10

m n m n

m n m n

k Br C H Orate M Mrate k Br C H O M M

−

−

×= = = =

×

As you can see in the above equations, by holding the concentrations of all but one species constant between two experiments, you can calculate the order of the reaction in a single reactant at a time. By similar reasoning, we can conclude that because the rate of reaction doubled when the concentration of acetone was doubled (cf.



experiments 1 and 3) the reaction must be first order in acetone. However, had the rate quadrupled or octupled with a doubling of the acetone concentration, the reaction would have been second or third order in acetone, respectively. In practice, you will likely never see a reaction with an order higher than 3. If you calculate an order higher than 3 for a reaction, double check your math because that is highly unusual. If you compute a fractional power for a reactant's order, do not be discouraged; they are quite common (especially half-order reactions). Therefore the rate law for this reaction is: Rate = k [Br2]0[C3H6O]1 The overall order of the reaction is 1 Example 2: Consider the following reaction 2BrO3(aq)

- + 5HSO3(aq)- Br2(g) + 5SO4(aq)

2- + H2O(l) + 3H+(aq)

The following experimental evidence was collected to determine the rate law Trial Initial [BrO3(aq)

-] (mmol/L)

Initial [HSO3(aq)-]

(mmol/L) Initial rate (mmol/L.s)

1 4.0 6.0 1.60 2 2.0 6.0 0.80 3 2.0 3.0 0.20 Solution 2: Considering the experimental data, determine the rate law. First state the rate law: Rate = k[BrO3(aq)

-]m[HSO3(aq)-]n

Begin by identifying which trials change the concentration for only one reactant. Comparing trials one and two, note how the concentration of [BrO3(aq)

-] is doubled in trial one compared to 2, while the concentration of [HSO3(aq)

-] remains the same. According to the experimental results, if the concentration of bromate ions is doubled, the rate of the reaction is also doubled. This means 2m = 2 (Read as double concentration to the rate exponent m equals double rate) Solving m = 1 or 21 = 2. Thus, the order of the reaction for [BrO3(aq)

-] ions is 1.



Comparing trials two and three, note that the concentration of [BrO3(aq)

-] ions remains constant, while the concentration of [HSO3(aq)

-] ions is doubled. According to the experimental results, if the concentration of [HSO3(aq)

-] is doubled, the rate of the reaction is quadrupled. This means 2n = 4 Solving n = 2 or 22 = 4. Thus the order of the reaction for [HSO3(aq)

-] ions is 2. Restate rate law r = k[BrO3(aq)

-]1[HSO3(aq)-]2

b) Using your rate law expression, solve for k Pick any trial to solve for k. In this case, we will use the data for concentration and rate from trial one:

2

4 2 2 1

0.00160 / .0.0040 / (0.0060 / )1.1 10 /

mol L skmol L mol LL mol s−

=×

= × •

Table 12.1: Summary of Reaction Rate Orders Order of Reaction Concentration change

0 1 2 3

X 1 10 =1 11 =1 12 = 1 13 = 1 X 2 (doubling) 20 =1 21 =2 22 = 4 23 = 8 X 3 (tripling) 30 =1 31 =3 32 = 9 33 = 27



Support Questions

25. For a particular reaction at constant temperature,

A(g) + 2 B(g) products

initial initial initial [A] [B] rate 1.00 1.00 1.00 2.00 4.00 8.00 3.00 9.00 27.00 4.00 2.00 ?

What is the value of "?" in this table?

26. What is a rate law? What is the proportionality constant called? 27. What is meant by the order of a reaction? 28. The rate law for the following reaction 2 NO + O2 2 NO2 is rate = k [NO]2[O2]. At

25oC, k = 7.1 X 109 L mol-2s-1. What is the rate of reaction when [NO] = 0.0010 mol/L and [O2] = 0.034 mol/L?

29. The initial rate of the reaction:

BrO3-(aq) + 5 Br-

(aq) + 6 H+(aq) 3 Br2(l) + 3H2O(l)

Has been measured at the reactant concentrations shown (in mol/L):

Experiment [BrO3-] [Br-] [H+] Initial rate (mol/Ls)

1 0.10 0.10 0.10 8.0 x 10-4 2 0.20 0.10 0.10 1.6 x 10-3 3 0.10 0.20 0.10 1.6 x 10-3 4 0.10 0.10 0.20 3.2 x 10-3

According to these results what would be the initial rate (in mol/Ls) if all three concentrations are:

[BrO3-] = [Br-] = [H+] = 0.20 mol/L?



Support Questions

30. The reaction of iodide ion with hypochlorite ion, OCl- (which is found in liquid bleach), follows the equation; OCl- + I- OI- + Cl- It is a rapid reaction that gives the following rate data.

Initial Concentrations Rate of Formation [OCl-] [I-] (mol L-1 s-1) (mol/L) of Cl- 1.7 X 10-3 1.7 X 10-3 1.75 X 104 3.4 X 10-3 1.7 X 10-3 3.50 X 104 1.7 X 10-3 3.4 X 10-3 3.50 X 104

What is the rate law for the reaction? Determine the value of the rate constant. Relating Reaction Rate to Time Sometime the rate of reaction in time can be measured by a visible change, by measuring the elapsed time before the visible change occurs. The average rate is inversely proportional to the elapsed time:

1avr

t∝Δ

Chemical Kinetics and Half-Life • the half-life, t1/2, of a reaction is the time that is needed for the reactant mass or

concentration to decrease by one half of its initial value • SI units for t1/2 are seconds, but it is usually expressed in whatever units of time are

appropriate to the reaction • t1/2 of any first-order reaction is always constant and it depends on k (t1/2 is independent of initial concentration of the reactant) • t1/2 for any first order reaction can be calculated using this equation:

12

0.693tk

=



Example 3: Calculating with Half-Lives Solution 3: The radioisotope lead-212 has a half-life of 10.6h. What is the rate constant for this isotope? Given:

1212PbR k −⎡ ⎤= ⎣ ⎦ 12

10.6t h=

Required: ?k =

Equation: 12

0.693tk

=

Solution:

12

12

1

0.693

0.693

0.69310.60.0654

tk

kt

hh−

=

=

=

=

Statement: The half-life rate constant for the radioisotope lead-212 is 0.0654h-1.

Support Questions

31. A radioisotope has a half-life of 24 s and an initial mass of 0.084g.

What mass of radioisotope will remain after; (i) 72 s? (ii) 192 s?

Energy Changes & Rates of Reaction

Theories of Reaction Rates Rates of reaction can be explained with collision theory. Since molecules are held together by chemical bonds, energy must be provided to break these bonds. The collision theory explains why some reactions are faster than others. Collision Theory For a reaction to occur the reactant particles must collide. Only a certain fraction of the total collisions cause chemical change; these are called successful collisions. The



successful collisions have sufficient energy (activation energy) at the moment of impact to break the existing bonds and form new bonds, resulting in the products of the reaction. Increasing the concentration of the reactants and raising the temperature bring about more collisions and therefore more successful collisions, increasing the rate of reaction.

Figure 12.1: A successful collision has sufficient energy and proper orientation. Key Points about Collision Theory • higher concentration of reactant particles will increase the number of collisions

between the particles per second • for solid reactants, greater surface area means that more collisions can occur • collisions must be effective

correct orientation of reactants/collision geometry sufficient collision energy/activation energy (Ea)



Activation Energy The activation energy is the minimum increase in potential energy of a system required for molecules to react. Figure 12.2: Potential energy diagram depicting activation energy Transition State Theory is used to explain what happens when molecules collide in a reaction. It examines the transition, or change, from reactants to products. The kinetic energy (KE) of reactants is transferred to potential energy (PE) as the reactants collide, due to the law of conservation of energy

• Transition state can be represented by a potential energy diagram “hill” illustrates the activation energy (Ea) barrier of the reaction a slow reaction has a high Ea barrier a fast reaction has a low Ea barrier transition state occurs at the top of the hill; species is an activated

complex that is neither product nor reactant Figure 12.3: Potential energy diagram showing the transition state in which the molecules are in an activated state (meaning they are neither reactant nor products)



Exothermic Reactions reactants have a higher PE than the products Endothermic Reactions reactants have a lower PE than products Note: The overall change in the potential energy is the enthalpy change (∆H). There is no way to predict the Ea of a reaction by its ∆H.

Reaction Mechanisms & Catalysts Elementary Reactions Most chemical reactions actually occur as a sequence of elementary steps. The overall sequence is called the reaction mechanism. Let’s looks at a chemical reaction involving the combustion of acetylene gas to explain this concept:

2 C2H2 + 5 O2 4 CO2 + 2 H2O

According to the equation, two molecules of acetylene (C2H2) react with 5 molecules of oxygen. The collision theory states these molecules must collide in order to react. However, it is highly unlikely that 7 molecules would collide together all at once.

Instead, the reaction most likely occurs in a series of simple steps which only required two or three molecules colliding at any one instant. Although these steps cannot always actually be observed, chemists can often make predictions about the sequence of events.

For example, nitrogen monoxide reacts with oxygen according to the equation

2 NO(g) + O2 → 2 NO2

This reaction does not occur in a single step, however, but rather through these two steps:

Step 1: 2 NO N2O2

Step 2: N2O2 + O2 2 NO2

Notice that if you add these two reactions together, you end up with the overall reaction:

Step 1: 2 NO N2O2

Step 2: N2O2 + O2 2 NO2 Overall: 2 NO(g) + O2 2 NO2



Dinitrogen dioxide (N2O2) cancels out and does not appear in our overall equation. Substances such as this are called reaction intermediates and are typically short-lived.

Given an overall reaction, it is not possible to predict what the reaction mechanism would be. However, if you are given the steps of a reaction mechanism you will need to be able to add together individual steps to end up with the overall reaction.

Rate Determining Step

Here is another reaction mechanism with some additional information concerning the relative rates of each of the individual steps:

Step 1: HBr + O2 → HOOBr Slow

Step 2: HOOBr + HBr → 2 HOBr Fast

Step 3: HOBr + HBr → H2O + Br2 Fast

Step 4: HOBr + HBr → H2O + Br2 Fast

Overall: 4 HBr + O2 → 2 H2O + 2 Br2

We don't have values for the actual rates of these individual steps, but here is a question for you to consider - would you consider the overall reaction to be fast or slow?

With three fast steps and only one slow step, many of you will predict that the reaction will be fast.

But let's make up some extreme numbers and ask the question again:

Step 1: HBr + O2 HOOBr 1 year

Step 2: HOOBr + HBr 2 HOBr 0.1 s

Step 3: HOBr + HBr H2O + Br2 0.1 min

Step 4: HOBr + HBr H2O + Br2 0.1 min

Overall: 4 HBr + O2 2 H2O + 2 Br2

Now would you consider the overall reaction to be fast or slow? Clearly it is a slow reaction, taking over a year to complete despite some fast steps.



The overall rate of any reaction depends on the rate of the slowest step. This slowest step is called the rate determining step. If you want to speed up a reaction, this is where you should focus your attention.

Proposing & Evaluating Mechanisms (Summary)

• the equations for the elementary steps must combine to give the overall equation • the proposed elementary steps must be reasonable • the mechanism must support the experimentally determined rate law

1 elementary reaction is much slower than the rest (rate-determining step) rate-determining steps can occur anywhere in a reaction mechanism

Let’s try an example: Consider the decomposition of dinitrogen pentoxide: 2N2O5(g) 2N2O4(g) + O2(g) a) What would the rate equation be if the reaction occurred in a single step? The rate equation would be the same as the co-efficient on the reactant. Therefore the rate law would be r = k[N2O5]2 b) Suppose the actual rate law is r = k[N2O5]1 What is the rate determining step? Because the coefficient on the reactant must be the same as the exponent in the rate equation, the rate-determining step must be 1N2O5 some intermediate product Suggest a possible mechanism for this reaction N2O5 N2O4 + O (slow) O + N2O5 N2O4 + O2 (fast)



Support Questions

32. Consider the overall reaction involving three elements as reactants and a compound as the product:

X + 2Y + 2Z XY2Z2

When a series of reactions is performed with different initial concentrations of reactants. The results are as follows: Doubling the concentration of X has no effect on the overall rate Doubling the concentration of Y multiplies the overall rate by 4 Doubling the concentration of Z doubles the overall rate State:

a) The rate law for this system b) The rate determining step c) A possible mechanism, indicating the slow step d) A possible reaction intermediate in your mechanism



Key Question #12

1. At 25oC a catalyzed solution of formic acid produces 44.6mL of carbon monoxide gas in 30s.

a) Calculate the rate of reaction with respect to CO(g) production. (2 marks) b) What can you state about how long you would expect the production of the same

volume to take: (4 marks)

i) at 30oC? ii) without the catalyst

2. Chlorine dioxide and hydroxide ions react to form chlorate ions, chlorite ions, and

water. 2ClO2(aq) + 2OH-

(aq) ClO3(aq) + ClO2-(aq) + H2O(l)

The reaction is found to be second order with respect to chlorine dioxide and first order with respect to hydroxide ions.

a) Write a rate equation for the reaction. (1 mark) b) What is the overall order of reaction? (1 mark) c) What would you expect the effect on rate to be of doubling the concentration

of chlorine dioxide? (2 marks)

3. Nitric oxide, NO(g) reacts with chlorine gas, Cl2(g), in the reaction

2NO(g) + Cl2(g) 2NOCl(g)

Initial rates of reaction are determined for various combinations of initial concentrations of reactants and recorded below;

Trial Initial [NO] (mol/L)

Initial [Cl2] (mol/L)

Rate of production of NOCl (mol/L.s)

1 0.10 0.10 1.8 x 10-2 2 0.10 0.20 3.6 x 10-2 3 0.20 0.20 1.43 x 10-1

a) What is the rate law equation for the reaction? (1 mark) b) What is the rate-determining step? (1 mark) c) Calculate a value for the rate constant, including units. (3 marks) d) Calculate the expected rate of reaction if the initial concentrations of NO and Cl2

gases were 0.30 and 0.40 mol/L, respectively. (3 marks)




4. Draw a sketch, roughly to scale, of the potential energy diagram for a system in which Ea = +80kJ and ∆H = -20kJ. Label the axes, reactants, products, the activation energy, the activated complex, and the enthalpy change. (5 marks)

5. An investigation is performed in which the concentration of nitrogen dioxide reacting

is measured as a function of time.

[NO2](mol/L) Time (s) 0.500 0 0.445 12 0.380 30 0.340 45 0.250 90 0.175 180

a) Plot a graph of [NO] vs. time (5 marks) b) Determine the average rate of reaction of NO2 between 10 and 60s (2 marks) c) As the [NO2] halved, what was the effect on the instantaneous rate of reaction?

(2 marks) 6. If the half life of a radioisotope is 3.5s, what percentage of the original isotope

remains after 14s? (5 marks)

SCH4U


Support Question Answers

SCH4U – Chemistry Support Question Answers


Support Question Answers 1. Identify each of the following as physical, chemical, or nuclear change.

a) A gas stove cooking pasta – chemical b) an ice cube melting in a glass – physical c) wax melts – physical d) ice applied to a sore back – physical

2. Identify the system and the surroundings in each of the examples in the previous

question.

a) gas, pasta – system, environment outside of pot – surroundings b) ice cube – system, glass, external environment – surroundings c) wax – system, external environment - surroundings d) ice cube – system, back and external environment – surroundings

3. A cup of water at 100oC has a higher temperature than a swimming pool full of water

at 20oC, but the pool has more thermal energy, Explain.

There are a greater number of water molecules, thus a greater amount of potential energy.

4. A swimming pool contains 1000 L of water. When the water is warmed by solar energy, its temperature increases from 15.8oC to 22.1oC, how much heat does the water absorb? (cwater(liquid)= 4.184 J/g oC)

Given: 61000 1000 1 10 m L kg g= = = × water/liquid 4.184 /c J g C= °

(since the density of water is 1g/mL) 22.1 15.86.3

T C CC

Δ = ° − °= °

Required: ?Q = Equation: pQ mc T= Δ

Solution: 61 10pQ mc T

g

= Δ

= × 4.184 /J g× C° 6.3 C× °

26359200 or 26359.2J kJ=

Statement: The swimming pool absorbs 26359.2 kJ of heat energy when

warmed to 22.1EC.



5. A cook heats water from 20oC to 50oC. Calculate the mass of water that could be warmed by the addition of 8.00kJ of heat.

Given: water/liquid 4.184 /c J g C= ° 8 or 8000Q kJ J=

50 2030

T C CC

Δ = ° − °= °

Required: ?m =

Equation: pQ mc T= Δ or Qmc T

=Δ

Solution: 8000p

Qmc T

J

=Δ

=4.184 J / g C° 30 C× °63.73g=

Statement: The mass of water that could be warmed 30EC with the addition of

8kJ of heat is 63.73g.

6. A 50% ethylene glycol solution has a specific heat capacity of 3.5J/goC. What temperature change would be observed in a solution of 4kg of ethylene glycol if it absorbs 250kJ of heat?

Given: ethylene glycol 50% soln 3.5 /c J g C= ° 4 or 4000m kg g= 250 or 250000Q kJ J= Required: ?TΔ =

Equation: pQ mc T= Δ or p

QTmc

Δ =

Solution: 250000p

QTmc

J

Δ =

=4000 g 3.5 J× / g17.86

CC

°

= °

Statement: The heat change that would result is 17.86EC.



7. Calculate the enthalpy change ΔH for the vaporization of 150.0g of water at 100oC. Given: vap 40.83H kJΔ = (from table 9.3) 150m g= 18.0 /M g mol=

Required: ?η = ?HΔ = Equation: m Mη = × vapH HηΔ = × Δ Solution: First calculate the moles of water.

150

m M

g

η = ×

= 18.0 g× /8.33

molmol=

Now calculate the enthalpy change.

vap

8.33

H H

mol

ηΔ = × Δ

= 40.83 /g mol×340.11kJ=

Statement: The enthalpy change for the vapourization of 150g of water at

100EC is 340.11kJ. 8. The energy in the process H2O(g) H2O(l) could be described as a molar enthalpy of

condensation. State the type of molar enthalpy that would occur in each of the following reactions:

a) Br2(l) Br(g) ans: vapourization b) CH4(g) CH4(s) ans: sublimation c) NaCl(s) Na+

(aq) + Cl-(aq) ans: solution d) KOH(aq) + HCl(aq) KCl(aq) + H2O(l) ans: neutralization

9. In the lab, you add 50 mL of concentrated hydrochloric acid (12 mol/L), HCl to form

250mL of dilute solution. The temperature of the solution changes from 18oC to 24oC. Calculate the molar enthalpy of dilution of hydrochloric acid. Start by calculating the moles of HCl (Step 1)

Given: 12c molL= 0.050V L= Required: HCl ?η =



Equation: c Vη = ×

Solution: HCl 12 molη = / 0.050L mol×

0.60mol=

Recognize and apply the Law of Conservation of Energy (Step 2) ( ) ( ) HCl dissolving calorimeter waterH qΔ = sol H mc Tη∴ ×Δ = Δ

Given: ( )water 250.0 250 250m g mL g= = water 4.18 /c J g C= °

24 186

T C CC

Δ = ° − °= °

Required: sol ?HΔ = Equation: solH mc Tη × Δ = Δ

Solution:

sol

sol

250

H mc Tmc TH

g

η

η

× Δ = Δ

ΔΔ =

=4.18 /J g× C° 6 C× °

0.6010450 /

molJ mol=

Statement: The molar enthalpy of a dilution of HCl is 10450J/mol.

10. What mass of lithium chloride must have dissolved if the temperature of 200g of

water increased by 6.0oC? The molar enthalpy of solution of lithium chloride is 37kJ/mol.

Given: water 200.0m g= 6T CΔ = ° sol 37 /H kJ solΔ = water 4.18 /c J g C= ° LiCl 42.4 /M g mol=

Required: LiCl ?m = Equation: solH mc Tη × Δ = Δ m Mη= ×



Solution:

sol

LiClsol

200

H mc Tmc T

H

g

η

η

× Δ = Δ

Δ=

Δ

=4.18 J× / g C° 6 C× °

37000 J /0.1356

molmol=

LiCl 0.14

m M

m mol

η= ×

= 42.4 /g mol×

5.7g=

Statement: It would require 5.7g of LiCl to complete this reaction.

11. Calculate ∆H for the reaction: C2H4(g) + H2(g) C2H6(g), from the following data.

C2H4 (g) + 3O2 (g) 2CO2 (g) + 2H2O (l) ∆HE = -1411. kJ

C2H6 (g) + 3½O2 (g) 2CO2 (g) + 3H2O (l) ∆HE = -1560. kJ

H2 (g) + ½ O2 (g) H2O (l) ∆HE = -285.8 kJ

Solution

1 C2H4 (g) + 3 O2 (g) 2 CO2 (g) + 2 H2O (l) ∆HE = -1411. kJ

Reverse 2

2 CO2 (g) + 3 H2O (l) C2H6 (g) + 3½ O2 (g) ∆HE = +1560. kJ

3 H2 (g) + 1/2 O2 (g) H2O (l) ∆HE = -285.8 kJ

1+R2+3 C2H4 (g) + H2 (g) C2H6 (g) ∆HE = -137. kJ

12. Calculate ∆H for the reaction 4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g), from the following data.

N2(g) + O2(g) 2NO(g) ∆HE = -180.5 kJ

N2(g) + 3H2(g) 2NH3(g) ∆HE = -91.8 kJ

2H2(g) + O2(g) 2H2O(g) ∆HE = -483.6 kJ



Solution

1 x 2 2 N2(g) + 2 O2(g) 4 NO(g) ∆HE = 2 (-180.5 kJ)

Reverse 2

4 NH3(g) 2 N2 (g) + 6 H2 (g) ∆HE = 2 (+91.8 kJ)

3 x 3 6 H2(g) + 3 O2(g) 6 H2O(g) ∆HE = 3 (-483.6 kJ)

4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) ∆HE = -1628. kJ

13. Find ∆H° for the reaction 2H2(g) + 2C(s) + O2(g) C2H5OH(l), using the following thermochemical data.

C2H5OH(l) + 2O2(g) 2 CO2(g) + 2H2O(l) ∆HE = -875. kJ

C(s) + O2(g) CO2(g) ∆HE = -394.51 kJ

H2(g) + ½O2(g) H2O(l) ∆HE = -285.8 kJ

Solution

Reverse 1

2 CO2 (g) + 2 H2O (l) C2H5OH (l) + 2 O2 (g) ∆HE = +875. kJ

2 x2 2 C (s) + 2 O2 (g) 2 CO2 (g) ∆HE = 2 (-394.51 kJ)

3 x 2 2 H2 (g) + O2 (g) 2 H2O (l) ∆HE = 2 (-285.8 kJ)

2H2(g) + 2C(s) + O2(g) C2H5OH(l) ∆H° = -486. kJ

14. Calculate ∆H for the reaction CH4(g) + NH3(g) HCN(g) + 3H2(g), given:

N2(g) + 3H2(g) 2NH3(g) ∆HE = -91.8 kJ

C(s) + 2H2(g) CH4(g) ∆HE = -74.9 kJ

H2(g) + 2C(s) + N2(g) 2HCN(g) ∆HE = +270.3 kJ



Solution

1 x½ NH3(g) 1/2 N2(g) + 3/2 1H2(g) ∆HE = ½(+91.8 kJ)

Reverse 2

CH4(g) C(s) + 2 H2(g) ∆HE = +74.9 kJ

3 x ½ 1/2 H2(g) + C(s) + 1/2 N2(g) HCN(g) ∆HE = ½(+270.3 kJ)

CH4(g) + NH3(g) HCN(g) + 3 H2(g) ∆HE = +256.0 kJ

15. Calculate ∆H for the reaction 2Al(s) + 3Cl2(g) 2AlCl3(s) from the data.

2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g) ∆HE = -1049. kJ

HCl(g) HCl(aq) ∆HE = -74.8 kJ

H2(g) + Cl2(g) 2HCl(g) ∆HE = -1845. kJ

AlCl3(s) AlCl3(aq) ∆HE = -323. kJ Solution 1 2 Al (s) + 6 HCl (aq) 2 AlCl3 (aq) + 3 H2 (g) ∆HE = -1049. kJ

2 x 6 6 HCl (g) 6 HCl (aq) ∆HE = 6 (-74.8 kJ)

3 x 3 3 H2 (g) + 3 Cl2 (g) 6 HCl (g) ∆HE = 3 (-1845. kJ)

Reverse 4 x2

2 AlCl3 (aq) 2 AlCl3 (s) ∆HE = 2 (+323. kJ)

2 Al (s) + 3 Cl2(g) 2 AlCl3(s) ∆HE = -6387. kJ

16. Write formation equations for the following:

a) benzene (C6H6) b) potassium bromate c) glucose a) 6C(s) + 3H2(g) C6H6 b) K(s) + 1/2Br2(g) + 2O2 KBrO4 c) 6C(s) + 6H2(g) + 3O2(g) C6H12O6



17. Use standard enthalpies of formation to calculate:

a) the molar enthalpy of combustion for pentane.

( ) ( ) ( )5 12 22 g 2 g l8C H O 5CO 6H O2

+ → +

Equation: ( ) ( )products reactantsf fH H Hη ηΔ = Δ ° − Δ °∑ ∑

Solution:

85 393.5 6 285.8 1 173.5 0

2

3682.30 173.5

3508.8

kJ kJ kJ kJH mol mol mol mol

mol mol mol mol

kJ kJmol mol

kJmol

Δ = × − + × − − × − + ×

= − − −

= −

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

b) the enthalpy change that accompanies the reaction between solid iron (III) oxide

and carbon monoxide gas to produce solid iron metal and carbon dioxide gas.

( ) ( ) ( ) ( )2 3 s q s 2 qFe O 3CO 2Fe 3CO+ → +


Solution:

3 393.5 2 0 1 824.2 3 110.5

1180.50 1155.7

24.8


mol mol mol mol

kJ kJmol mol

kJmol

Δ = × − + × × − + × −

= − −

= −

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤−⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦



18. For each of the following reactions, use standard enthalpies to calculate ∆H. a) ( ) ( ) ( ) ( )24 q q q 2 qCH H O CO 3H+ → +


Solution:

1 110.5 3 0 1 74.8 1 241.8

110.5 316.2

206.1


mol mol mol mol

kJ kJmol mol

kJmol

Δ = × − + × − × − + × −

= − − −

=

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

b) ( ) ( ) ( ) ( )2 2q q 2 q qCO H O CO H+ → +


Solution:

1 393.5 1 0 1 110.5 1 241.8

393.5 352.3

41.2


mol mol mol mol

kJ kJmol mol

kJmol

Δ = × − + × − × − + × −

= − − −

= −

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

c) ( ) ( ) ( )2 q 2 q 3 qN 2H 2NH+ →


Solution:

2 46.1 1 0 3 0

92.2

kJ kJ kJH mol mol mol

mol mol molkJ

mol

Δ = × − − × + ×

= −

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦



19. Ammonium nitrate fertilizer is produced by the reaction of ammonia with nitric acid:

NH3(g) + HNO3(l) NH4NO3(s)

a) Use standard enthalpies of formation to calculate the standard enthalpy change of the reaction used to produce ammonium nitrate.


Solution:

1 365.6 1 46.1 1 174.1

365.6 220.0

145.4

kJ kJ kJH mol mol mol

mol mol mol

kJ kJmol molkJ

mol

Δ = × − − × − + × −

= − − −

= −

⎡ ⎤ ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

20. Consider the following reaction:

MnO4-(aq) + 5Fe2+

(aq) + 8H+(aq) Mn2+

(aq) + 5Fe3+(aq) + 4H2O(l)

Given that the rate of reaction is 3.5 x 10-2 mol/L MnO4

-(aq), determine the rate of

reaction for all reactants and products in the reaction. MnO4

- = 3.5 x 10-2 mol/L

Fe2+ = 0.175mol/L H+

= 0.28 mol/L Mn2+ = 3.5 x 10-2 mol/L Fe3+ = 0.175mol/L H2O = 0.14mol/L

21. The reaction between ammonia and oxygen produces nitrogen monoxide and water

vapour is as follows:

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) If the rate of consumption of ammonia is 1.6 x 10-2 mol/(L.s) calculate the rate of consumption of oxygen and the formation of water vapour. O2 = 0.02mol/(L.s) H2O = 0.024mol/(L.s)



22. Rates of reactions are generally fastest at the beginning of a reaction. Explain why this is so.

There is a higher number of reactants, and hence a greater number of collisions, thus speeding up reaction rate.

23. A kinetics experiment is performed in which oxygen gas is collected. The

concentration of oxygen is measured every 10 s. The table below has the data obtained from the experiment.

Time (s) Oxygen gas (mol/L) 0 0 10 0.23 20 0.40 30 0.52 40 0.60 50 0.66 60 0.70 70 0.73 80 0.75 90 0.76 100 0.76

a) Plot a graph of oxygen concentration vs time

b) Calculate the average formation of oxygen gas during the first 60 seconds rate = 0.70mol/L/60s rate = 0.117 mol/L.s



c) the average rate of formation of oxygen gas between t=20s and t=60s rate = 0.30mol/L/40s rate = 7.5 x 10-3mol/L.s

24. Identify the five factors that increase the rate of a reaction and provide an example

of each.

-concentration of reactants -temperature -surface area -presence of a catalyst -chemical nature of reactants

25. For a particular reaction at constant temperature,

A(g) + 2 B(g) products

initial initial initial [A] [B] rate 1.00 1.00 1.00 2.00 4.00 8.00 3.00 9.00 27.00 4.00 2.00 ?

What is the value of "?" in this table?

Value is 64. There is a trend that as [A] is increased, the rate increased to the power of 3. There appears to be no trend with the concentration of [B]

26. What is a rate law? What is the proportionality constant called?

The rate law is the relationship among rate, the rate constant, the initial concentration of reactants. The proportionality constant is k, it is also called the rate constant.

27. What is meant by the order of a reaction?

The order of the reaction is the exponent value that describes the initial concentration dependence of a particular reactant.



28. The rate law for the following reaction 2 NO + O2 2 NO2 is rate= k [NO]2[O2]. At 25oC, k=7.1 X 109 L mol-2s-1. What is the rate of reaction when [NO] = 0.0010 mol/L and [O2] =0.034 mol/L?

r = 7.1 X 109 L mol-2s-1 [0.0010mol/L]2[0.034mol/L] r = 2.4 x 104L-1mols-1

29. The initial rate of the reaction:

BrO3-(aq) + 5 Br-

(aq) + 8 H+(aq) 3 Br2(l) + H2O(l)

Has been measured at the reactant concentrations shown (in mol/L):

Experiment [BrO3-] [Br-] [H+] Initial rate (mol/Ls)

1 0.10 0.10 0.10 8.0 x 10-4 2 0.20 0.10 0.10 1.6 x 10-3 3 0.10 0.20 0.10 1.6 x 10-3 4 0.10 0.10 0.20 3.2 x 10-3

According to these results what would be the initial rate (in mol/Ls) if all three concentrations are:

[BrO3-] = [Br-] = [H+] = 0.20 mol/L?

Doubling [BrO3-] doubles rate

Doubling [Br-] doubles rate

Doubling [H+] quadruples rate

r= k[BrO3-]1[Br-]1[H+]2

k = initial rate/ [BrO3-]1[Br-]1[H+]2

k = 8.0 x 10-4mol/L•s/(0.10mol/L)(0.10mol/L)(0.10mol/L)2

k = 8L3•mol-3•s-1

r = 8l3• mol-3•s-1 [0.20mol/L]1[0.20mol/L]1[0.20mol/L]2

r = 1.28 x 10-2mol/L•s



30. The reaction of iodide ion with hypochlorite ion, OCl- (which is found in liquid bleach), follows the equation;

OCl- + I- OI- + Cl-

It is a rapid reaction that gives the following rate data.

Initial Concentrations Rate of Formation [OCl-] [I-] (mol L-1 s-1) (mol/L) of Cl- 1.7 X 10-3 1.7 X 10-3 1.75 X 104 3.4 X 10-3 1.7 X 10-3 3.50 X 104 1.7 X 10-3 3.4 X 10-3 3.50 X 104

What is the rate law for the reaction? Determine the value of the rate constant. Doubling [OCl-] doubles rate 21 = 2 (1st order) Doubling [I-] doubles rate 21 =2 (1st order) Rate = k[OCl-]1[I-]1

4

1 13 3

9 1 1

1.75 10 mol/L s

1.7 10 mol/L 1.7 10 mol/L

6.06 10 L mol s

k

k

− −

− −

× •=⎡ ⎤ ⎡ ⎤× ×⎣ ⎦ ⎣ ⎦

= × • •

31. A radioisotope has a half-life of 24s and an initial mass of 0.084g.

a) What mass of radioisotope will remain after: (i) 72 s? (ii) 192 s?

i) Number of half lives in 72 s =3 m 0.084g x ½ x ½ x ½ m = 0.0105g or 10.5 mg ii) Number of half lives in 192 s =8 m = 0.084g x ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ m = 0.33mg

32. Consider the overall reaction involving three elements as reactants and a compound

as the product: X + 2Y + 2Z XY2Z2

When a series of reactions is performed with different initial concentrations of reactants. The results are as follows:



Doubling the concentration of X has no effect on the overall rate Doubling the concentration of Y multiplies the overall rate by 4 Doubling the concentration of Z doubles the overall rate State; a) The rate law for this system b) The rate determining step c) A possible mechanism, indicating the slow step d) A possible reaction intermediate in your mechanism

Solution a) r = k[Y]2[Z]1 b) 2Y + 1Z some product c) Many answers are possible, as long as it is consistent with above rules X + Y XZ (fast) 2Y + Z Y2Z (slow) XZ + Y2Z XY2Z2 (fast) d) Two possible reaction intermediates are XZ and Y2Z

SCH4U – Chemistry Appendix A


Appendix A: Thermochemical Data of Selected Elements & Compounds (at 25°C and 100.000 kPa) Substance

ΔHf° (kJ/mol)

S° (J/K mol)

ΔGf° (kJ/mol)

Substance

ΔHf° (kJ/mol)

S° (J/K mol)

ΔGf° (kJ/mol)

Al (s) 0 28.3 0 NH3 (g) -46.1 192.5 -16.5 Al2O3 (s) -1675.7 50.9 -1582.3 N2H4 (l) 50.6 121.2 149.3 Br2 (l) 0 151.6 0 NH4Cl (s) -314.4 94.6 -202.9 HBr (g) -36.4 198.7 -53.5 NH4NO3 (s) -365.6 151.1 -183.9 Ca (s) 0 41.4 0 NO (g) 90.3 210.8 86.6 CaCO3 (s) (calcite) -1206.9 92.9 -1128.8 NO2 (g) 33.2 240.1 51.3 CaCl2 (s) -795.8 104.6 -748.1 N2O (g) 82.1 219.9 104.2 C (s) (graphite) 0 5.7 0 N2O4 (g) 9.2 304.3 97.9 C (s) (diamond) 1.9 2.38 2.90 HNO3 (l) -174.1 155.6 -80.7 CCl4 (l) -135.4 216.4 -65.2 O (g) 249.2 161.1 231.7 CCl4 (g) -96.0 309.9 -60.6 O2 (g) 0 205.1 0 CHCl3 (l) -134.5 201.7 -73.7 O3 (g) 142.7 238.9 163.2 CH4 (g) -74.8 186.3 -50.7 P4 (s) (white) 0 164.4 0 C2H2 (g) 226.7 200.9 209.2 P4 (s) (red) -70.4 91.2 -48.4 C2H4 (g) 52.3 219.6 68.2 PH3 (g) 5.4 310.2 13.4 C2H6 (g) -84.7 229.6 -32.8 PCl3 (g) -287.0 311.8 -267.8 C3H8 (g) -103.8 269.9 -23.5 P4O6 (s) -2144.3 345.6 -2247.4 C6H6 (l) 49.0 172.8 124.5 P4O10 (s) -2984.0 228.9 -2697.7 CH3OH (l) -238.7 126.8 -166.3 H3PO4 (s) -1279.0 110.5 -1119.1 C2H5OH (l) -277.7 160.7 -178.8 K (s) 0 64.2 0 CH3CO2H (l) -484.5 159.8 -389.9 KCl (s) -436.7 82.6 -409.1 CO (g) -110.5 197.7 -137.2 KClO3 (s) -397.7 143.1 -296.3 CO2 (g) -393.5 213.7 -394.4 KOH (s) -428.8 78.9 -379.1 COCl2 (g) -218.8 283.5 -204.6 Ag (s) 0 42.6 0 CS2 (g) +117.4 237.8 67.1 AgCl (s) -127.1 96.2 -109.8 Cl2 (g) 0 223.1 0 AgNO3 (s) -124.4 140.9 -33.4 HCl g) -92.3 186.9 -95.3 Na (s) 0 51.2 0 Cr (s) 0 23.8 0 NaCl (s) -411.2 72.1 -384.1 CrCl3 (s) -556.5 123.0 -486.1 NaOH (s) -425.6 64.5 -379.5 Cu (s) 0 33.2 0 Na2CO3 (s) -1130.7 135.0 -1044.0 CuO (s) -157.3 42.6 -129.7 S (s)

(rhombic) 0 31.8 0

CuCl -137.2 86.2 -119.9 S (g) 278.8 167.8 238.3 CuCl2 (s) -220.1 108.1 -175.7 SF6 (g) -1209.0 291.8 -1105.3 F2 (g) 0 202.8 0 H2S (g) -20.6 205.8 -33.6 HF (g) -271.1 173.8 -273.2 SO2 (g) -296.8 248.2 -300.2 He (g) 0 126.0 0 SO3 (g) -395.7 256.8 -371.1 H2 (g) 0 130.7 0 H2SO4 (l) -814.0 156.9 -690.0 H2O (l) -285.8 69.9 -237.1 Sn (s) (white) 0 51.6 0 H2O (g) -241.8 188.8 -228.6 Sn (s) (gray) -2.1 44.1 0.1 H2O2 (l) -187.8 109.6 -120.4 SnCl2 (s) -325.1 122.6 -302.1 Fe (s) 0 27.8 0 SnCl4 (l) -155.3 258.6 -440.1 FeO (s) -272.0 57.6 245.1 Fe2O3 (s) -824.2 87.4 -742.2 Fe3O4 (s) -1118.4 146.4 -1015.4 FeCl2 (s) -341.8 118.0 -302.3 FeCl3 (s) -399.5 142.3 -344.0 FeS2 (s) -178.2 52.9 -166.9 Pb (s) 0 64.8 0 PbCl2 (s) -359.4 136.0 -314.1 Mg (s) 0 32.7 0 MgCl2 (s) -641.3 89.6 -591.8 MgO (s) -601.7 26.9 -569.4 Hg (l) 0 76.0 0 HgS (s) -58.2 82.4 -50.6 Ne (g) 0 146.2 0 N2 (g) 0 191.6 0

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SCH4U - Unit 3 - Version C · PDF fileLesson 9: Thermochemistry ... In living organisms, the process of cellular respiration converts simple sugars into cellular energy, carbon dioxide,