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Chapter 5: Thermochemistry. Problems: 5.1-5.95, 5.97-98, 5.100, 5.106, 5.108, 5.118-5.119, 5.121, 5.126. Energy: Basic Concepts and Definitions. energy: capacity to do work or to produce heat thermodynamics : study of energy and its transformations from one form to another

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chapter 5 thermochemistry

Chapter 5: Thermochemistry

Problems: 5.1-5.95, 5.97-98, 5.100, 5.106, 5.108, 5.118-5.119, 5.121, 5.126

energy basic concepts and definitions
Energy: Basic Concepts and Definitions

energy:capacity to do work or to produce heat

thermodynamics: study of energy and its transformations from one form to another

thermochemistry: study of heat flow accompanying chemical reactions

heat: energy that is transferred from a body at a higher temperature to one at a lower temperature heat always transfers from the hotter to the cooler object!

  • "heat flow" means heat transfer
heat transfer and temperature
Heat Transfer and Temperature

One becomes hotter by gaining heat.

One becomes colder by losing heat—i.e., when you “feel cold”, you are actually losing heat!


a. You burn your hand on a hot frying pan.

________________ loses heat, and ______________ gains the heat.

b. Your tongue feels cold when you eat ice cream.

_______________ loses heat, and ______________ gains the heat.

heat transfer and temperature1
Heat Transfer and Temperature

Example: A student puts a few drops of rubbing alcohol on her palm then spreads it out with her finger and notices that the area with the alcohol felt cooler. A minute later her palm is dry. Explain what physical change has occurred and why her palm felt cooler.

types of energy
Types of Energy

work (w): w = F  d

Work is done whenever a force (F) moves an object a specified distance (d)

kinetic energy (KE):energy associated with an object’s motion

Example: a car moving at 75 mph has much greater KE than the same car moving at 15 mph

Greater damage if the car crashes at 75 mph than at 15 mph

KE = ½ mv2m=mass, v=velocity

types of energy1
Types of Energy

potential energy (PE): energy due to position or its composition (chemical bonds)

A 10-lb bowling ball has higher PE when it is 10 feet off the ground compared to 10 inches off the ground

 Greater damage on your foot after falling 10 feet compared to falling only 10 inches

positional P.E. = mass  force of gravity  height

In terms of chemical bonds, the stronger the bond,

more energy is required to break the bond,

 the higher the potential energy of the bond

energy is a state function
Energy is a State Function

state function: a property that is based only on the physical state and chemical composition of a substance, so it is independent of the path followed to achieve that state or composition.

For example:

Potential energy is a state function because the potential energy of two skiers at the top of a given hill would be the same whether they climbed that hill or took a ski lift to the top.

units of energy
Units of Energy

joule (J): 1 J =

  • SI (SystèmeInternationale or standard) unit of energy
    • To recognize the size of a joule, note that 1 watt = 1 J/s
    • So a 100-watt light bulb uses 100 J every second.
  • Heat is also often reported in kilojoules (kJ), where 1 kJ = 1000 J
  • 1 J = 0.239 calories; 1000 calories = 1 kcal = 1 food calorie
  • How many Joules in a banana?
    • A medium banana has about 105 kcal = 439,330 J
conservation of energy
Conservation of Energy

Energy is neither created nor destroyed but converted from one form to another.

For example, the kinetic energy of a car can cause considerable damage if the car is stopped suddenly in a crash.

  • Example: Calculate the kinetic energy (in joules) for a 3.63103 kg Hummer H2 traveling at 25 miles per hour. (Use 1 mph = 0.4469 m/s and 1 J = kg m2/s2).
conservation of energy1
Conservation of Energy

Example (cont’d): At what speed (in miles per hour) must a 7.3102 kg Smart Car move to have the same kinetic energy as the Hummer H2?

at the molecular level
At the Molecular Level
  • Temperature governs the motion of particles at the molecular level.
    • At higher temperature, particles move faster and have higher kinetic energy.
  • Temperature is a measure of the average kinetic energy for a substance.
  • Thermal energy is the kinetic energy associated with the motion of particles.
    • Proportional to the temperature for any given substance
    • Increases with the size of a sample
  • The particles in a cup of boiling water at 100°C and those in a pot of boiling water (also at 100°C) have the same average kinetic energy, but the pot of boiling water has more thermal energy than the cup of boiling water because it contains more water molecules.
at the molecular level1
At the Molecular Level
  • Electrostatic potential energy is due to the electrostatic interactions due to the charges or dipoles in atoms, ions, and molecules.
    • Coulomb\'s law is used to determine the strength of ionic bonds.
    • Thus, at the molecular level, potential energy is determined in terms of the strength of the bonds holding atoms, ions, and molecules together in various substances.
systems surroundings and energy transfer
Systems, Surroundings and Energy Transfer

system: that part of the universe being studied

surroundings: the rest of the universe outside the system

Systems can be isolated, closedor open.

isolated closed and open systems
Isolated, Closed and Open Systems
  • isolated system: exchanges neither energy nor matter with the surroundings
    • Hot soup in a perfectly insulated thermos that does not allow any heat to escape.
  • closed system: exchanges energy but no matter with the surroundings
    • Hot soup in a cup with a lid allows heat to escape to the surroundings but no matter.
  • open system: exchanges energy and matter with the surroundings
    • Hot soup in an open cup allows both heat and water vapor (steam) to escape to the surroundings
direction and sign of heat flow
Direction and Sign of Heat Flow

Let q = heat flow,

  • q is + when heat flows into the system from the surroundings
  • q is – when heat flows out of the system into the surroundings

endothermic change: a physical or chemical change that requires energy or heat to occur

Boiling water requires energy:

H2O(l) + heat H2O(g)

Electrolysis of water requires energy:

2 H2O(l) + electrical energy  2 H2(g) + O2(g)

direction and sign of heat flow1
Direction and Sign of Heat Flow

exothermic change: a physical or chemical change that releases energy or heat

Water condensing releases energy:

H2O(g) H2O(l) + heat

Hydrogen burning releases energy:

2 H2(g) + O2(g) 2 H2O(g) + heat

For physical changes, consider whether the reactants or products have more kinetic energy.

  • If the reactants have greater kinetic energy than the products exothermicprocess.
  • If the products have greater kinetic energy than the reactants endothermic process.
heat of reaction
Heat of Reaction

What causes the Heat of Reaction?

Bond Energy

  • energy required to break a particular bond in 1 mol of gaseous molecules
    • always positive since breaking a bond always requires energy
    • a quantitative measure of the strength of a bond (i.e. stability of compound)

Breaking and Forming Bonds

  • Energy is absorbed by reactants when their bonds are broken, and energy is released by products when their bonds are formed.

heat of reaction (qreaction):heat associated with a chemical reaction

heat of reaction1
Heat of Reaction

If  Endothermic Reaction

If  ExothermicReaction

heat of reaction2
Heat of Reaction

For chemical changes, observe if the surroundings (including you) feel hotter or colder after the reaction has occurred.

  • If the surroundings are hotter, the reaction released heat exothermicreaction.
  • If the surroundings are colder, the reaction absorbed heat endothermicreaction.

Which of the following are endothermic changes:

freezing vaporizing sublimation

deposition melting condensation

When a student dissolves ammonium chloride in a large test tube, he notices the test tube feels colder. Explain what is releasing heat and what is gaining heat.

1st law of thermodynamics
1st Law of Thermodynamics

The energy of the universe is constant. The energy gained or lost by a system must equal the energy lost or gained by the surroundings.

Essentially, the Law of Conservation of Energy: Energy can neither be created or destroyed but converted from one form to another.

A system’s Internal Energy (E) = kinetic energy (KE) + potential energy (PE) of all the particles in the system.

  • While the values of KE and PE at a given instant are difficult to determine, changes in KE and/or PE can be determined by measuring any temperature changes for the system.
heat and work
Heat and Work

A system’s internal energy (E) can be changed using heat (q), work (w), or both:

  • The total increase in the energy of a system is the sum of heat flowing into it and the work done to it.

∆E = q + w

In this course, we will focus on work that involves the expansion or compression of gases.

  • For example, consider the work done during the combustion of fuel in an engine…

Consider the following sign conventions for the change in volume, ∆V, for the system:

  • If a gas is created, the system’s volume expands. ∆V is positive
  • The system does work to expand into the surroundings. w is negative
  • If a gas is compressed, the system’s volume is compressed. ∆V is negative

 Work is done to the system. w is positive


Atmospheric pressure, P, is positive

work is defined as w = –P∆V

Thus, for the expansion or compression of a gas, the change in internal energy (∆E) can be shown as:

∆E = q + w or ∆E = q – P∆V


Consider the combustion of octane (C8H18), a primary component of gasoline.

2 C8H18(l) + 25 O2(g) 18 H2O(g) + 16 CO2(g)

a. Calculate the change in volume (in L) due to the total number of moles of gases produced when 1.000 gal (~3.784 L) of octane undergoes combustion at 1.00 atm and 25.00°C. The density of octane at 25.00°C is 0.703 g/mL.


b. What is the volume of the gases at 475 K (the temperature of a car engine)?

c. Use the equation, w = – P∆V, to calculate the work (in kJ) done by the system (the reaction) during the combustion of 1.000 gal of octane. (Use 1 Latm = 101.325 J).


d. If 1 kJ 1 kWs (kilowatt·second), then calculate the energy (in kJ) in a kilowatt·hour (kWh).

e. Consider a new kind of vehicle that could be powered by an electrical current similar to that used in our homes. Calculate the cost of electricity needed to produce the same amount of energy (determined in part c) as the combustion of 1.00 gal of gasoline if Seattle City light charges 9.55 cents/kWh.


f. The combustion of 1.000 gal of octane produces about 1.2105 kJ of heat. Compare the change in internal energy due to work calculated in part c with the heat of the reaction. How much of the internal energy change is due to the work done by the system due to gas expansion? (Hint: Compare the absolute value for heat versus work.)

For the remainder of the chapter, we will focus mainly on the heat of reaction and assume the change in energy due to gas expansion work is negligible for the reactions considered.

enthalpy and enthalpy changes
Enthalpy and Enthalpy Changes

Enthalpy(H): the sum of a system’s internal energy and the product of its pressure and volume

H = E + PV

Thus, the enthalpy change (∆H) is: ∆H = ∆E + ∆(PV)

or at constant atmospheric pressure is: ∆H = ∆E + P∆V

Since the previous definition of internal energy, ∆E = q – P∆V, the equation can be rewritten as∆H = q – P∆V + P∆V = qp

(the subscript p means “at constant pressure”)

enthalpy and enthalpy changes1
Enthalpy and Enthalpy Changes

Thus, the enthalpy change (∆H) refers to heat flow into and out of a system under constant pressure (usually the case since reactions occur under atmospheric pressure),

qreaction= ∆H = Hproducts – Hreactants

For an endothermic reaction: ∆H = positive

For an exothermic reaction: ∆H = negative

heating curves and heating capacity
Heating Curves and Heating Capacity

Consider the changes that H2O undergoes when a block of ice is taken from a freezer and heated in a pan until it is completed converted into steam.

  • A heating-cooling curve shows the changes in physical state with temperature and heat added to or removed from any system.

100 °C


0 °C

Heat Added

specific heat
Specific Heat

specific heat (cs):amount of heat necessary to raise the temperature of 1 gram of any substance by 1°C; has units of J/g°C

  • Water has a relatively high specific heat (4.184 J/g·°C) compared to the specific heats of rocks and other solids (1.3 J/g·°C for dry Earth, 0.9 J/g·°C for concrete, 0.46 J/g·°C for iron)
  • Because water covers most of the Earth, water can absorb a lot more energy before its temperature starts to rise.
    • Water helps to regulate temperatures on Earth within a comfortable range for humans.
    • Why coastal regions have less extreme temperatures compared to desert regions.
heat capacity
Heat Capacity

molar heat capacity (cp): heat capacity per mole of a substance (in J/mol·°C)

heat capacity: amount of heat necessary to raise the temperature of a given amount of any substance by 1°C; in units of J/°C

Use the following equations to solve calorimetry problems:

q = n cp∆T or q = csm ∆T

where ∆T=change in temperature, n=# of moles, and m=mass.


a. If 279.9 J is required to raise the temperature from 23.0°C to 99.5°C for a 15.5-g sample of silver, what is the specific heat of silver?

b. A beaker with 100.0 g of water is heated from 25.0°C to its boiling point. If the specific heat of water is 4.184 J/g·°C, how much heat is required to heat the water?


c. Determine the final temperature for a 100.0 g sample of iron at 25.0°C heated with the amount of heat calculated in a. given the molar heat capacity of iron is 25.1 J/mol·°C.

units of energy1
Units of Energy

calorie (cal): unit of energy used most often in the US. Is equal to the amount of energy required to raise the temperature of 1 g of water by 1˚C

1 cal 4.184 J (Note: This is EXACT!)

But a nutritional calorie (abbreviated Cal) is 1000 cal:

1 Cal = 1 kcal = 4.184 kJ


When consuming an ice-cold drink, one must raise the temperature of the beverage to 37.0°C (normal body temperature). Can one lose weight by drinking ice-cold beverages if the body uses up about 1 calorie per gram of water per degree Celsius (i.e. the specific heat of water = 1.00 cal/g·°C) to consume the drink?

a. Calculate the energy expended (in Cal) to consume a 12-oz beer (about 355 mL) if the beer is initially at 4.0°C. Assume the drink is mostly water and its density is 1.01 g/mL.


b. If the label indicates 103 Cal, what is the net calorie gain or loss when a person consumes this beer? Is this a viable weight loss alternative?

c. Calculate the amount of heat (in kJ) required to heat 1.00 kg (~1 L) of water at 25°C to its boiling point.

units of energy2
Units of Energy

Another energy unit is the British thermal unit (abbreviated Btu). A Btu is the energy required to raise the temperature of 1 pound of water by 1°F when water is most dense (at 39°C).

The heating power of many gas cooktops is often given in Btu’s. Calculate the time (in minutes) required to heat 1.00 kg of water at 25°C to boiling using a 12,000 Btu per hour burner. (Assume complete energy transfer from the burner to the water.) Use 1 kWh = 3412 Btu and 1 kJ = 1 kWs.