CHAPTER 9
Describing Chemical Reactions
BLM 9.0.1OVERHEAD
A balanced equation gives you many details about chemical reactions. Some of these
details are listed here:
• The formula of each reactant and product tells you exactly how many atoms of each
element comprise one molecule or formula unit of the compound.
• The symbol in brackets tells you the state of the compound when it is undergoing the
reaction.
• The coefficient of each reactant and product tells you the mole ratios of the
compounds in the reaction. For example, in the second equation you know that one
mole of ethene gas (C
2H
4(g)) reacts with three moles of oxygen gas to produce two
moles of carbon dioxide gas and two moles of water vapour.
2Na(s) + Cl
2(g) → 2NaCl(s)
C
2H
4(g) + 3O
2(g) → 2CO
2(g) + 2H
2O(g
)Another way to describe a chemical reaction involves energy. As you know, when
some reactions occur in a test tube, the test tube feels warmer than it did before you
added the reactants. The reaction generated heat, thus raising the temperature of the
solution. Such a reaction is called
exothermic
. Conversely, some reactions make the
solution cooler by absorbing heat. These reactions are called
endothermic
.
NAME: CLASS:
relative number of molecules relative number of atoms relative number of
formula units
CHAPTER 9
Energy and Change PreQuiz
BLM 9.1.1ASSESSMENT
1. Consider the following equation:
2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) + energy
(a) Does the equation represent an endothermic reaction or an exothermic reaction? Justify your answer.
(b)What effect would this reaction have on the temperature of its surroundings? Compare the temperature of the surroundings before the reaction takes place to the temperature of the surroundings after the reaction takes place.
(c) What does the equation communicate about the relative energy of the chemical bonds in the reactants and products?
Energy and Change PreQuiz
(continued) ASSESSMENT
3. Use the following information to answer the questions below:
Q = m × c × Dt Q = heat (J) m = mass (g)
c = specific heat capacity (J/g • °C) Dt = tf –ti (°C)
For water, c = 4.184 J/g • °C
(a) A 50.0 g sample of water at 80.0 °C is mixed in a calorimeter with a 50.0 g sample of water at 5.0 °C. What is the final temperature of the water?
(b) A reaction raises the temperature of 100.0 g of water by 3.20 °C. How much heat does the reaction release?
CHAPTER 9
Energy and Change PreQuiz
Answer Key
BLM 9.1.1A
ANSWER KEY
1. (a)You can tell the reaction is exothermic because the energy term is written with the products, meaning heat is produced.
(b) The reaction would increase the temperature of the surroundings; for the surroundings, Tf > Ti.
(c)The energy released by forming all chemical bonds in the products is greater than the energy required to break all chemical bonds in the reactants.
2. Thermal energy is the total kinetic energy of the particles in a substance. Heat is the transfer of thermal energy from a region of higher to lower temperature. Temperature is a measure of the average thermal energy of the particles in a substance.
3. (a)42.5 °C
(b)1.34 J
CHAPTER 9
Energy Changes in Chemical
Reactions
BLM 9.1.2
OVERHEAD
This illustration shows bonds being broken and formed during a chemical
reaction.
The difference between the amount of energy needed to break bonds
(energy
1) and the energy released when bonds form (energy
2) determines
whether a reaction is endothermic or exothermic.
CHAPTER 9
Endothermic and Exothermic
Reactions Quiz
BLM 9.1.3
ASSESSMENT
Read the following summary of exothermic and endothermic reactions. Then answer the questions on the next page.
A chemical change is always accompanied by a change in energy because the atoms or ions that make up the reactants are rearranged.
During a reaction, chemical bonds that hold the reactant atoms or ions together must be broken and new chemical bonds must be formed within the product substance(s). The breaking of chemical bonds requires the input of energy and is defined as an
endothermic process.
The formation of chemical bonds releases energy and is defined as an exothermic process. Since any chemical change involves both the breaking and formation of chemical bonds,
there are two possible outcomes of any chemical change.
The overall change is exothermic if more energy is released to form the product chemical bonds than is required to break the reactant chemical bonds.
The overall change is endothermic if less energy is released to form the product chemical bonds than is required to break the reactant chemical bonds.
The energy that is released or absorbed is related to the external environment of the chemical reaction. Endothermic reactions absorb thermal energy from the surrounding environment and result in a decrease in temperature. Exothermic reactions, on the other hand, release thermal energy to the surrounding environment and result in an increase in temperature.
Type of
reaction Breaking chemicalbonds (reactants) Forming chemicalbonds (products) Overall energychange
exothermic energy released
endothermic energy absorbed
1. (a) What is the source of the energy that is released in an exothermic reaction? What absorbs the energy that is released?
CHAPTER 9
Endothermic and Exothermic
Reactions Quiz
(continued)BLM 9.1.3
ASSESSMENT
2. Classify each reaction as either endothermic or exothermic, and briefly explain your answer.
Description of chemical reaction Endothermic or
exothermic? Explanation A piece of paper is ignited and
burns with a bright flame.
Pentaborane (a colourless liquid), B5H9, reacts violently with oxygen
gas to form solid boric oxide, B2O3, and water, typically bursting
into flame and often exploding. Pure iron metal is formed and carbon dioxide is released when iron(III) oxide ore is heated to a very high temperature in the presence of solid carbon.
Sodium hydroxide solution and hydrochloric acid solution are mixed. The temperature of the mixture increases.
Mixing ammonium thiocyanate and barium hydroxide octahydrate in a beaker causes water on the outside of the beaker to freeze. The high temperature in an oven causes baking soda (sodium hydrogencarbonate) to break down into carbon dioxide, water, and sodium carbonate.
DATE: NAME: CLASS:
CHAPTER 9
Endothermic and Exothermic
Reactions Quiz Answer Key
BLM 9.1.3A
ANSWER KEY
1. (a) The energy that is released in an exothermic reaction results from the difference between the energy absorbed when reactant bonds break and the energy released when product bonds form. In an exothermic reaction, the energy that is required to break bonds is less than the energy that is released to form bonds. The energy that is released is absorbed by the surroundings.
(b) The energy that is absorbed by an endothermic reaction comes from the surroundings. Energy is required by the reaction because the energy that is required to break bonds is greater than the energy that is released to form bonds.
2.
Description of chemical reaction Endothermic orexothermic? Explanation
A piece of paper is ignited and burns with a bright flame.
exothermic Heat and light are emitted.
Pentaborane (a colourless liquid), B5H9,
reacts violently with oxygen gas to form solid diborane, B2O3, and water,
typically bursting into flame and often exploding.
exothermic Bursting into flame and explosion are both energy emissions.
Pure iron metal is formed and carbon dioxide is released when iron(III) oxide ore is heated to a very high temperature in the presence of solid carbon.
endothermic High temperatures are required to form the new substances.
Sodium hydroxide solution and hydrochloric acid solution are mixed. The temperature of the mixture increases.
exothermic Heat is emitted.
Mixing ammonium thiocyanate and barium hydroxide octahydrate in a beaker causes water on the outside of the beaker to freeze.
endothermic Heat is absorbed from the surroundings.
The high temperature in an oven causes baking soda (sodium
hydrogencarbonate) to break down into carbon dioxide, water, and sodium carbonate.
endothermic Heat is absorbed from the surroundings to cause the reaction to occur.
DATE: NAME: CLASS:
CHAPTER 9
Specific Heat Capacity
Problems
BLM 9.1.5
ASSESSMENT
1. How much energy is needed to heat enough water to make a cup of tea (250 mL), if the water is initially at 20.0 ºC and you want to increase the temperature to 85.0 ºC? (Assume that 1.00 mL of water has a mass of 1.00 g.)
2. As the tea in Question 1 steeps, it cools from 85.0 ºC to 75.0 ºC. How much energy is lost by the tea as it steeps?
CHAPTER 9
Specific Heat Capacity
Problems
(continued)BLM 9.1.5
ASSESSMENT
4. A 1.00 kg block of ice, at 25.0 ºC, is warmed by 35 kJ of energy. What is the final temperature of the ice?
5. Which substance100 g of moist air, 100 g of water, or 100 g of icewill experience the greatest temperature change if 100 J of energy is used to warm it? Why?
6. Imagine that you have a 500 g iron pot (c = 0.440 ), a 500 g copper pot (c = 0.385 ),
CHAPTER 9
Specific Heat Capacity
Problems
(continued)BLM 9.1.5
ASSESSMENT
7. A 5.0 g sample of an unidentified metal absorbs 71 J of energy as its temperature increases from 125 ºC to 162 ºC. What metal is the sample? (Hint: See the data in Question 6.)
8. What mass of seawater is needed to provide 300.0 MJ as it cools from 75 ºC to 33 ºC?
9. How much water can be heated from its melting to its boiling point by adding 2.75 106 J of
CHAPTER 9
Specific Heat Capacity
Problems
(continued)BLM 9.1.5
ASSESSMENT
10.If 100.0 g of a substance releases 45 kJ of energy as it cools from 13.0 ºC to –15.0 ºC, what is the specific heat capacity of the substance?
CHAPTER 9
Specific Heat Capacity
Problems Answer Key
BLM 9.1.5A
ANSWER KEY
1. Mass of water, m = 250 mL
= 250 = 250 g
Specific heat capacity of water, c = Final temperature, T2 = 85.0 °C
Initial temperature, T1 = 20.0 °C
Change of temperature, DT = 65.0 °C
Q = mcDT
=
= 68 087.5 J 6.8 104 J
It would take 6.8 104 J of energy to heat 250 mL of water from 20.0 ºC to 85.0 ºC.
2. Mass of water, m = 250 mL
= 250
= 250 g
Change of temperature, DT = –10.0 °C
Q = mcDT
=
= –10 475 J –1.05 104 J
The 250 mL of tea loses 1.05 104 J as it cools from 85.0 ºC to 75.0 ºC.
3. Q = 100.0 kJ = 1.000 105 J
m = 500.0 g c =
CHAPTER 9
Specific Heat Capacity
Problems Answer Key
(cont’d)BLM 9.1.5A
ANSWER KEY
Q = MCDT
= 47.7327 °C 47.8 °C
4. Q = 35 kJ = 3.5 104 J
m = 1.00 kg = 1.00 103 g
c =
T1= –25.0 °C
Q = mcDT DT = Q/mc
=
= 17.5 °C DT = T2 – T1
T2 = T1 +DT
5. The moist air will experience the greatest temperature change because it has the lowest specific heat capacity. (Less energy is needed to heat it the same amount, or the same number of
degrees).
6. The aluminium pot will keep the water warm the longest. It will store more energy while heating to 100 °C because it has the highest specific heat capacity.
7. m = 5.0 g Q = 71 J T2 = 162 °C
T1 = 125 °C
DT = T2 – T1
DT = 162 °C – 125 °C = 37 °C
CHAPTER 9
Specific Heat Capacity
Problems Answer Key
(cont’d)BLM 9.1.5A
ANSWER KEY
Q = mcDT
The specific heat capacity of the metal is similar to the specific heat capacity of copper, so the metal is most likely copper.
8. Q = –300.0 MJ = –3.000 108 J
c = T2 = 33 °C
T1 = 75 °C
DT = 33 °C –75 °C = –44 °C
1.8 106 g
9. Q = 2.75 106 J
c =
T2 = 100.00 °C
T1 = 0.00 °C
DT = 100.00 °C
CHAPTER 9
Specific Heat Capacity
Problems Answer Key
(cont’d)BLM 9.1.5A
ANSWER KEY
10.m = 100.0 g
Q = 45 kJ = –4.5 104 J
T1 = 13.0 °C
T2 = –15.0 °C
DT = –28.0 °C
Q = MCDT
= 16.071
CHAPTER 10
Hess’s Law: Two Pathways, One
Change
BLM 10.1.1
OVERHEAD
Hess’s Law
The enthalpy change of a physical or chemical process depends only on the
initial and final states. The enthalpy change of the overall process is the
sum of the enthalpy changes of its individual steps.
Carbon dioxide can be formed by the reaction of carbon with oxygen via
two different pathways. The enthalpy change of the overall reaction is the
same regardless of the method or pathway that is used.
CHAPTER 10
Using Hess’s Law to Determine
Enthalpy Change
BLM 10.1.2
Sample Problem
One of the methods that the steel industry uses to obtain metallic iron is to react iron(III) oxide, Fe2O3(s), with carbon monoxide, CO(g), as shown in the balanced equation below:
Fe2O3(s) + 3CO(g) → 3CO2(g) + 2Fe(s)
Determine the enthalpy change of this reaction, given the following equations and their enthalpy changes.
(1) CO(g) + O2(g) → CO2(g) ΔH° = –283.0 kJ
(2) 2Fe(s) + O2(g) → Fe2O3(s) ΔH°= –824.2 kJ
What Is Required?
You need to manipulate equations (1) and (2), along with their enthalpy changes, so they add up to the overall equation:
Fe2O3(s) + 3CO(g) → 3CO2(g) + 2Fe(s)
What Is Given?
You know the chemical equations for reactions (1) and (2), and you know their corresponding enthalpy changes.
Plan Your Strategy
Step 1Examine equations (1) and (2) to see how they compare with the overall equation. Decide how you need to manipulate equations (1) and (2) so that they add to the overall equation. (Reverse the equation, multiply the equation by a coefficient, do both, or do neither.) Remember to adjust ΔH° accordingly for each equation.
Step 2Write the manipulated equations so that their equation arrows line up. Add the reactants and products on each side, and cancel substances that appear on both sides in equal amounts.
Step 3Ensure that you have obtained the overall equation. Add ΔH° for the combined equation.
CHAPTER 10
Using Hess’s Law to Determine
Enthalpy Change
(continued)BLM 10.1.2
Act on Your Strategy Step 1
Equation (1) has CO(g) as a reactant and CO2(g) as a product, as does the overall equation. The
stoichiometric coefficients do not match the coefficients of the overall equation, however. To achieve the same coefficients, you must multiply equation (1) by 3.
Equation (2) has the required stoichiometric coefficients, but Fe(s) and Fe2O3(s) are on the wrong
sides of the equation. You need to reverse equation (2) and, therefore, change the sign of the ΔH°. Note: Since oxygen gas, O2(g), is present in both equations that you are manipulating, do not use
O2(g) to decide on how to manipulate the equations. Always start with a chemical species that is
present in only one of the equations and is also present in the overall equation.
Step 2
3 × (1) 3CO(g) + → 3CO2(g) ΔH° = 3(–283.0 kJ)
–1 × (2) Fe2O3(s) → 2Fe(s) + ΔH° = –1(–824.2 kJ)
Fe2O3(s) + 3CO(g) → 3CO2(g) + 2Fe(s)
Step 3
Fe2O3(s) + 3CO(g) → 3CO2(g) + 2Fe(s)
ΔH° = 3(–283.0 kJ) + (–1(–824.2 kJ)) = –24.8 kJ
The manipulated equations add to the overall equation. Therefore, the sum of the manipulated enthalpy changes is the enthalpy change of the overall equation:
Fe2O3(s) + 3CO(g) → 3CO2(g) + 2Fe(s) ΔH° = –24.8 kJ
Check Your Solution
The equations added correctly to the overall equation. Check to ensure that you adjusted ΔH°
accordingly for each equation. Because you added the ΔH° values, the final answer will be as precise as the least precise number used in the calculation. The final answer has one digit after the decimal point, which is correct.
CHAPTER 10
Using Hess’s Law to Determine
Enthalpy Change for Formation
Reactions
BLM 10.1.3
OVERHEAD
Sample Problem
From the following information, calculate the enthalpy of formation of solid phosphorus pentachloride as shown in the equation below:
P4(s) + Cl2(g) → PCl5(s) ΔH° = ?
(1) P4(s) + 6Cl2(g) → 4PCl3(ℓ) ΔH° = –1272 kJ
You need to manipulate and add equations (1) and (2) along with their enthalpy changes to obtain the overall equation.
What Is Given?
You know the overall equation and the thermochemical equations for reactions (1) and (2).
Plan Your Strategy
The overall equation contains fractions. To simplify your manipulations of equations (1) and (2), you may find it easier to first find the enthalpy change for the overall equation with whole-number coefficients and then divide to obtain the true overall equation at the end of the problem. First, balance the overall equation with whole numbers by multiplying the coefficients by 4:
P4(s) + 10Cl2(g) → 4PCl5(s) ΔH° = ?
Second, manipulate equations (1) and (2).
Equation (1) can be used as written to obtain P4(s) as a reactant.
Equation (2) has PCl5(s) on the correct side of the equation but not in the correct stoichiometric
quantities. Multiply equation (2) by 4.
Act on Your Strategy
(1) P4(s) + 6Cl2(g) → ΔH° = –1272 kJ
(2) × 4 + 4Cl2(g) → 4PCl5(s) ΔH° = 4(–125 kJ)
P4(s) + 10Cl2(g) → 4PCl5(s) ΔH° = –1272 kJ + 4(–125 kJ)
= –1772 kJ
The desired overall equation is for the formation of only 1 mol of PCl5(s). Therefore, divide the
equation and corresponding ΔH° by 4, thus giving the following solution.
P4(s) + Cl2(g) → PCl5(s) ΔH° =
= –443 kJ
Therefore, the enthalpy of formation of phosphorus pentachloride is ΔH° = –443 kJ/mol.
Check Your Solution
