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Chapter 8 1

Thermochemistry: Chemical EnergyThermochemistry: Chemical EnergyChapter 8

Chapter 8 2

Chemical Thermodynamics• Chemical Thermodynamics is the study of the energetics of a

chemical reaction.– Thermodynamics deals with the absorption or release of energy (generally

as heat) that accompanies chemical reactions.• These dynamics generally give us an idea of whether a reaction

will be spontaneous– Spontaneous describes a process that can occur without outside

intervention (i.e. changing the temperature, pressure or concentration)– Spontaneity does not imply that the process occurs quickly, but rather

describes a capability to proceed.– If a chemical reaction is found to be spontaneous in one direction, then

under the same conditions, the reverse reaction will be non-spontaneous– This does not mean that the reaction cannot occur, but that it will need

some outside help.

Evisdom

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Chapter 8 3

Energy• Energy is the capacity to do work, or

supply heat.

Energy = Work + Heat• Potential energy, PE, is stored

energy; it results from position or composition.

• Kinetic energy, KE, is the energy matter has as a result of motion.

KE = 1/2 mv2

Units for KE: 1 Joule = 1 kg⋅m2/s2 OR 1 calorie = 4.184 J

Chapter 8 4

• The SI unit for energy is the Joule (J)

• The joule is a small amount of energy so scientists generally refer to Kilojoules (kJ)

• The calorie (cal) is another unit of energy.– It is defined as the amount of

energy needed to raise the temperature of 1.00 g of water by one degree Celsius

Units of Energy

1 cal = 4.184 J

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Chapter 8 5

KE, Temperature & State• All substances have kinetic energy no matter

what physical state they are in.

• Solids have the lowest kinetic energy, and gases have the greatest kinetic energy.

• As you increase the temperature of a substance, its kinetic energy increases.

Chapter 8 6

Forms of Energy• There are six forms of energy:

– Heat, Radiant, Electrical, Mechanical, Chemical and Nuclear

– We can convert among these types of energy

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Chapter 8 7

• Thermal Energy is the kinetic energy of molecular motion (translational, rotational, and vibrational).– We measure this energy by finding the temperature of an object

• Thermal energy is proportional to the temperature in degrees Kelvin

Ethermal α T(K)

• Chemical Energy is the PE of a chemical compound where the chemical bonds act as “storage” containers for the energy

Thermal Energy

Chapter 8 8

The System and the Surroundings• The system is the specific part of the universe that is of interest to us

(i.e. the reaction with reactants & products)• The surroundings is the rest of the universe• When performing experiments, we measure changes to the

surroundings then apply those findings to the system.• Energy is transferred between the system and the surroundings

– Energy that is transferred from the system to the surroundings has a negative sign – Energy that is transferred from the surroundings to the system has a positive sign

∆E = Efinal - EInitial

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Chapter 8 9

State Functions• A State Function is a function or property whose value

depends only on the present state (condition) of the system not the path used to arrive at that condition.

– The change in a state function is zero when the system returns to its original condition.

– For non-state functions, the change is not zero if the path returns to the original condition

Chapter 8 10

State Functions• Here’s an example with a chemical reaction:• The two paths below give the same final state:

N2H4(g) + H2(g) → 2 NH3(g) + heat (188 kJ)

N2(g) + 3 H2(g) → 2 NH3(g) + heat (92 kJ)– State properties include for the formation of NH3:

temperature, total energy, pressure, density, and [NH3]

– Non-state properties include for the formation of NH3: heat and reactants used

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Chapter 8 11

• Work (w) is the application of a force (F) through a distance (d)– This work produces an object’s movement– If the system performs work on the surroundings, the sign of w is negative.– If the surroundings perform work on the system, the sign of w is positive.

Work = Force • Distancew = F • d

• A system that contains one or more gases performs a specific type of work called “PV Work” or expansion work.

Work

Units for w:Newton-meters (N-m) or Joules (J)

1 J = 1 N-m

Chapter 8 12

• A system that contains a gas can perform work as the gas expands against an opposing pressure (P) exerted by the surroundings.

• This type of work is called Expansion Work or PV Work

• When a gas expands, w is negative because the system transfers energy to the surroundings

• When a gas contracts, w is positive because the surroundings transfer energy to the system

Expansion Work

w = - P∆V∆V = VFinal - VInitial

1 L • atm = 101 J

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Chapter 8 13

• How much work is done (in kilojoules), and in which direction, as a result of the following reaction?

Work

Chapter 8 14

• Heat (q) is the energy that flows from a warmer object (higher temperature) to a cooler object (lower temperature)– Heat is associated with the

movement of particles

– The faster the particles are moving, the more heat you generate (and vice versa)

– If the system gives heat to the surroundings, the sign of q is negative.

– If the surroundings give heat to the system, the sign of q is positive.

Heat

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Chapter 8 15

The First Law of Thermodynamics• In a reaction, energy is conserved. It is neither created nor destroyed.

• It can be converted into a different type of energy though!

• The energy change in a system equals the work done on the system + the heat added.

Total Energy of Reactants = Total Energy of Products

∆E = Efinal – Einitial = q + w

Chapter 8 16

• Most chemical reactions are performed in containers that are open to the atmosphere.

• The volume of the system is allowed to change but the pressure (atmospheric pressure) remains constant.

• If the work performed is limited to PV work, a state function called Enthalpy (H) can be defined.

• The change in enthalpy (∆H) is equal to the heat absorbed by the system under conditions of constant pressure.

Energy and Enthalpy

H = E + PV

∆H = ∆E + P∆V

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Chapter 8 17

• The following reaction has ∆E = –186 kJ/mol.– Is the sign of P∆V positive or negative?– What is the sign and approximate magnitude of ∆H?

Energy and Enthalpy

Chapter 8 18

Enthalpy Changes: Physical Change• There are two types of Enthalpy changes:

– The Enthalpy of Physical Change.– The Enthalpy of Chemical Change

• The Enthalpy of Physical Change is the amount of heat required to change the state of matter

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Chapter 8 19

• The Enthalpy of Chemical change deals with the energy requirements of chemical reactions.

• The Heat of the Reaction (∆Hrxn) gives the amount of heat absorbed or released due to a chemical reaction at constant pressure.

• If a reaction has more than one step, the ∆Hrxn is the sum of the ∆H for each individual reaction.

Enthalpy Changes: Chemical Change

∆Hrxn = Hproducts – Hreactants

(1) Reactants → Elements Heat Absorbed = ∆H1(2) Elements → Products Heat Absorbed = ∆H2

(Overall) Reactants → Products Heat Absorbed = ∆Hrxn

Chapter 8 20

Enthalpy Changes: Chemical Change• In an Exothermic reaction, the reaction

releases energy in the form of heat to the surroundings (you feel the reaction get hot!)

– The energy of the reactants is greater than that of the products.

C (s) + O2 (g) → CO2 + Energy

• In an Endothermic reaction, the reaction absorbs energy in the form of heat from the surroundings (you feel the reaction get cold!)

– The energy of the products is greater than that of the reactants.

2 H2O (l) + Energy → 2 H2 (g) + O2 (g)

Reactants

Products

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Chapter 8 21

• The reaction between hydrogen and oxygen to yield water vapor has ∆H = –484 kJ. How much PV work is done, and what is the value of ∆E (in kilojoules) for the reaction of 0.50 mol of H2 with 0.25 mol of O2at atmospheric pressure if the volume change is –5.6 L?

• The explosion of 2.00 mol of solid TNT with a volume of approximately 0.274 L produces gases with a volume of 489 L at room temperature. How much PV (in kilojoules) work is done during the explosion? Assume P = 1 atm, T = 25°C.2 C7H5N3O6(s) → 12 CO(g) + 5 H2(g) + 3 N2(g) + 2

C(s)

Enthalpy Changes: Chemical Change

Chapter 8 22

• How much heat (in kilojoules) is evolved or absorbed in each of the following reactions? State whether the reactions are endothermic or exothermic.

– Burning of 15.5 g of propane:C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ∆H = –2219 kJ

– Reaction of 4.88 g of barium hydroxide octahydrate with ammonium chloride:

Ba(OH)2·8 H2O(s) + 2 NH4Cl(s) → BaCl2(aq) + 2 NH3(aq) + 10 H2O(l) ∆H = +80.3 kJ

Stoichiometry and Enthalpy Changes

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Chapter 8 23

• To insure a standardization of enthalpy measurements from different experimental sources, a standard set of conditions has been selected.

• Thermodynamic Standard State:– Most stable form of a substance at 1 atm pressure and 25°C– 1 M concentration for all substances in solution.

• Measurements made under these conditions are indicated by a superscript ° to the symbol of the quantity reported.

• Standard enthalpy change is indicated by the symbol ∆H°.

Standard State

Chapter 8 24

Hess’s Law• Hess’s Law: The overall enthalpy change for a

reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.

3 H2(g) + N2(g) → 2 NH3(g) ∆H° = –92.2 kJ

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Chapter 8 25

• Reactants and products in individual steps can be added and subtracted to determine the overall equation.

Hess’s Law

(a) 2 H2(g) + N2(g) → N2H4(g) ∆H°1 = ?(b) N2H4(g) + H2(g) → 2 NH3(g) ∆H°2 = –187.6 kJ(c) 3 H2(g) + N2(g) → 2 NH3(g) ∆H°3 = –92.2 kJ

∆H°1 = ∆H°3 – ∆H°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ

Chapter 8 26

• Reversing a reaction changes the sign of ∆H for a reaction:

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ∆H = –2219 kJ3 CO2(g) + 4 H2O(l) → C3H8(g) + 5 O2(g) ∆H = +2219 kJ

• Multiplying a reaction increases ∆H by the same factor because ∆H is also stoichiometrically related to the coefficients of the balanced chemical equation:

3(C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ∆H = –2219 kJ)3 C3H8(g) + 15 O2(g) → 9 CO2(g) + 12 H2O(l) ∆H = –6657 kJ

Manipulating Chemical Enthalpy Changes

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Chapter 8 27

• The industrial degreasing solvent methylenechloride (CH2Cl2, dichloromethane) is prepared from methane by reaction with chlorine:

CH4(g) + Cl2(g) → CH2Cl2(g) + HCl(g)

• Use the following data to calculate ∆H° (in kilojoules) for the above reaction:

CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) ∆H° = –98.3 kJ

CH3Cl(g) + Cl2(g) → CH2Cl2(g) + HCl(g) ∆H° = –104 kJ

Hess’s Law

Chapter 8 28

• What is the equation and ∆H° for the net reaction?

• Which arrow on the diagram corresponds to which step and which arrow corresponds to the net reaction?

• The diagram shows three energy levels. The energies of which substances are represented by each?

Hess’s Law

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Chapter 8 29

Standard Heats of Formation• Where do all of these ∆H° values come from that we

have been using? Has someone determined these values for all of these reactions?

• Standard Heats of Formation (∆H°f): The enthalpy change for the formation of 1 mole of substance in its standard state from its constituent elements in their standard states.

• The standard heat of formation for any element in its standard state is defined as being ZERO.

C (s) + 2 H2 (g) → CH4 (g) ∆Hf° = -74.5 kJ

∆H°f = 0 for an element in its standard state

Chapter 8 30

• The standard enthalpy change for any chemical reaction is found by subtracting the sum of the heats of formation of all reactants from the sum of the heats of formation of all products

∆H° = Σ ∆H°f (Products) – Σ ∆H°f (Reactants)• For a balanced equation, each heat of formation must be

multiplied by the stoichiometric coefficient.

aA + bB → cC + dD∆H° = [c∆H°f (C) + d∆H°f (D)] – [a∆H°f (A) + b∆H°f (B)]

Standard Heats of Formation

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Chapter 8 31

Standard Heats of Formation

• Calculate ∆H° (in kilojoules) for the reaction of ammonia with O2 to yield nitric oxide (NO) and H2O(g), a step in the Ostwald process for the commercial production of nitric acid.

• Calculate ∆H° (in kilojoules) for the photosynthesis of glucose from CO2 and liquid water, a reaction carried out by all green plants.

Chapter 8 32

Bond Dissociation Energy• There are over 18 million chemical compounds known at this

time. Who do you think is making all of the ∆H°fdeterminations for those compounds?

• It is possible to get an approximate ∆H°f for a compound based on the bonds present.

• The Bond Dissociation Energy can be used to determine an approximate value for ∆H°f .

∆H°f = D (Bonds Broken) – D (Bonds Formed)∆H°f = D (Reactant Bonds) – D (Product Bonds)

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Chapter 8 33

Bond Dissociation Energy• To work these problems you will have to draw the Lewis

Structures of the reactants and products. Refer to your notes from Chapter 7

• For the reaction between H2 and Cl2 to form HCl:H2 (g) + Cl2 (g) → 2 HCl (g)

∆H°f = D (Bonds Broken) – D (Bonds Formed)∆H°f = (DH-H + DCl-Cl) – 2(DH–Cl)

= (436 kJ/mol + 243 kJ/mol) – 2(432 kJ/mol)= -185 kJ

Chapter 8 34

Bond Dissociation Energy

• Calculate an approximate ∆H° (in kilojoules) for the synthesis of ethyl alcohol from ethylene:

C2H4(g) + H2O(g) → C2H5OH(g)• Calculate an approximate ∆H° (in kilojoules) for the synthesis of hydrazine

from ammonia:

2 NH3(g) + Cl2(g) → N2H4(g) + 2 HCl(g)

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Chapter 8 35

Calorimetry and Heat Capacity• The Bond energies and Heats of Formation give us an

approximate idea of ∆Hf°.• How do we experimentally measure the amount of heat

transferred during a reaction?• Calorimetry is the science of measuring heat changes (q)

for chemical reactions. There are two types of calorimeters:

– Bomb Calorimetry: A bomb calorimeter measures the heat change at constant volume such that q = ∆E.

– Constant Pressure Calorimetry: A constant pressure calorimeter measures the heat change at constant pressure such that q = ∆H.

Chapter 8 36

Constant Pressure Bomb

Calorimetry and Heat Capacity

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Chapter 8 37

• Heat capacity (C) is an extensive property that gives the amount of heat required to raise the temperature of an object or substance a given amount.

– The greater the Heat Capacity, the greater the amount of heat needed to increase the temperature

Calorimetry and Heat Capacity

C = q ∆T

Units: J/K or J/°C

• Specific Heat (SH): The amount of heat required to raise the temperature of 1.00 g of substance by 1.00°C.

• Molar Heat Capacity (CM):The amount of heat required to raise the temperature of 1.00 mole of substance by 1.00°C.

q = SH • m • ∆T q = CM • n • ∆T

Chapter 8 38

• What is the specific heat of lead if it takes 96 J to raise the temperature of a 75 g block by 10.0°C?

• When 25.0 mL of 1.0 M H2SO4 is added to 50.0 mL of 1.0 M NaOH at 25.0°C in a calorimeter, the temperature of the solution increases to 33.9°C. Assume specific heat of solution is 4.184 J/(g·°C), and the density is 1.00 g/mL, calculate ∆H for the reaction.

Calorimetry and Heat Capacity

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Chapter 8 39

Calorimetry and Heat Capacity

A 12.6 g block of Iron has a temperature of 25.3°C. What is the final temperature of the block if 106.2 J of heat are added?

Chapter 8 40

Introduction to Entropy• We have stated that chemical and physical processes occur

spontaneously only if they go “downhill” energetically (give off energy) so that the final state is more stable and lower in energy than the initial state.

• However, we know that reactions that require energy (endothermicreactions) will also occur.

• This is possible because energy is more than just heat. There are other forms of energy present in reactions that allow it to be spontaneous despite absorbing heat.

• Another factor that effects spontaneity of a reaction is the molecular disorder of the reaction.–The amount of molecular disorder (or randomness) in a system is called the system’s Entropy (S)

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Chapter 8 41

• Entropy has units of J/K (Joules per Kelvin).

∆S = Sfinal – SinitialPositive value of ∆S indicates increased disorder.

Negative value of ∆S indicates decreased disorder.

Entropy

Second Law of Thermodynamics:

Reactions proceed in the direction that increases the entropy of the system plus

surroundings.

Chapter 8 42

2 CO(g) + O2(g) → 2 CO2(g)CaCO3(s) → CaO (s) + CO2(g)

2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g)

Entropy

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Chapter 8 43

• To decide whether a process is spontaneous, both enthalpy and entropy changes must be considered:

Spontaneous process: Decrease in enthalpy (–∆H).Increase in entropy (+∆S).

Nonspontaneous process: Increase in enthalpy (+∆H).Decrease in entropy (–∆S).

Entropy and Enthalpy

Chapter 8 44

Introduction to Gibbs Free Energy• Gibbs Free Energy Change (∆G) weighs the relative

contributions of enthalpy and entropy to the overall spontaneity of a process.

∆G = ∆H – T∆S∆G < 0 Spontaneous∆G = 0 Equilibrium∆G > 0 Non-spontaneous

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Chapter 8 45

Gibbs Free Energy

∆G = ∆H – T∆S

Chapter 8 46

Which of the following reactions are spontaneous under standard conditions at 25°C?

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq) ∆G° = –55.7 kJ

2 C(s) + 2 H2(g) → C2H4(g) ∆G° = 68.1 kJ

N2(g) + 3 H2(g) → 2 NH3(g) ∆H° = –92 kJ; ∆S° = –199 J/K

Gibbs Free Energy

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Chapter 8 47

• Equilibrium (∆G° = 0): Estimate the temperature at which the following reaction will be at equilibrium. Is the reaction spontaneous at room temperature?

N2(g) + 3 H2(g) → 2 NH3(g)∆H° = –92.0 kJ ∆S° = –199 J/K

Gibbs Free Energy at Equilibrium

Chapter 8 48

• Benzene, C6H6, has an enthalpy of vaporization, (∆Hvap) equal to 30.8 kJ/mol and boils at 80.1°C. What is the entropy of vaporization, ∆Svap, for benzene?

C6H6 (l) → C6H6 (g) ∆Hvap = 30.8 kJ/mol

Gibbs Free Energy