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Chapter 8

Thermochemistry:Chemical Energy

Thermodynamics 01

• Thermodynamics: study of energy and it’s transformations

• Energy: capacity to do work, or supply heat

Energy = Work + Heat

• Kinetic Energy: energy of motion

EK = 1/2 mv2 (1 Joule = 1 kg⋅m2/s2)

(1 calorie = 4.184 J)

• Potential Energy: stored energy

Thermodynamics 02

• Conservation of energy law: Energy cannot becreated or destroyed; it can only be converted fromone form into another.

Thermodynamics 03

• Thermal Energy: kinetic energy of molecularmotion (translational, rotational, and vibrational)

• Heat: the amount of thermal energy transferredbetween two objects at different temperatures

• Chemical Energy: potential energy stored inchemical bonds released in the form of heat or light

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Thermodynamics 04

• First Law of Thermodynamics: energy of an

isolated system must be kept constant

Thermodynamics 05

• System ⇒ reactants + products• Surroundings ⇒ everything else

• Energy changes are measured from the point ofview of the system!

• ∆E is negative ⇒ energy flows out of the system• ∆E is positive ⇒ energy flows into the system

Thermodynamics 06 Work 07

w = –PΔV

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Sign of w 08

negative positive

positive negative

w = -PΔV ⇒ expansion

w = -PΔV ⇒ contraction

Work Units 09

w = -PΔV (J or kJ)

1L x 1000mL x 1cm3 x 1m3

1L 1mL (100cm)3

1000

101 x 103 kg

ms2

= 101 kgm2 = 101J

s2

m2w = L x atm =

Energy and Heat 10

Energy = Work + Heat

ΔE = w + q = q - PΔV

q = ΔE + PΔV

When a person does work, energy diminishes

w = negative

ΔE = negative

Heat and Enthalpy 11

• The amount of heat exchanged between thesystem and the surroundings is given the symbol q.

q = ΔE + PΔV

At constant volume (ΔV = 0): qv = ΔE

At constant pressure: qp = ΔE + PΔV = ΔH

enthalpy

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State Functions 12

State Function:

• value depends only on the present state of thesystem

• path independent

• when returned to its original position, overallchange is zero

State Functions 13

• State and Nonstate Properties: The two pathsbelow 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)

• temperature, total energy, pressure, density,volume, and enthalpy (∆H) ⇒ state properties

• nonstate properties include heat and work

Enthalpy 14

• Enthalpy or heat of reaction:

•ΔH = H(products) - H(reactants)

• States of the reactants and products are important!⇒ (g, l, s, aq)

• Thermodynamic standard state: P = 1atm, [ ] = 1M,

T = 298.15K (25ºC)

Standard Enthalpy of Reaction 15

Thermodynamic standard state: P = 1atm, [ ] = 1M,

T = 298.15K (25ºC)

Standard enthalpy of reaction (Hº)

N2(g) + 3H2(g) → 2NH3(g) ΔHº = -92.2kJ

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Enthalpy Changes 16

• Most changes in a system involve again or loss in enthalpy

• Physical (melting of ice in a cooler)

• Chemical (burning of gas in your car)

Physical Changes 17

• Enthalpies of Physical Change:

Chemical Changes 18

• Enthalpies of Chemical Change: Often calledheats of reaction (ΔHreaction).

Endothermic: Heat flows into the system from the

surroundings ⇒ ΔH is positive

Exothermic: Heat flows out of the system into the

surroundings ⇒ ΔH is negative

Enthalpy Changes 19

• Reversing a reaction changes the sign of ΔH for areaction.

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

3 CO2(g) + 4 H2O(l) → C3H8(g) + 5 O2(g) ΔH = +2219 kJ

• Multiplying a reaction increases ΔH by the same factor.

3 x [C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ]

ΔH =(-2219kJ x 3) = –6657 kJ

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Example 20

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

w = -0.25kJ

Expansion, systemloses -0.25kJ

Example 21

• The following reaction has ΔE = –186 kJ/mol.

• Is the sign of PΔV positive or negative?

• What is the sign and approximate magnitude of ΔH?

Contraction, PΔV is negative,w is positive

ΔH = ΔE + PΔV

ΔH = (-186kJ) + (1atm) (-1mole)

ΔH = negative (slightly morethan ΔE)

Example 22

The reaction between hydrogen and oxygen to yieldwater vapor has ΔH° = –484 kJ. How much PVwork is done, and what is the value of ΔE (kJ) forthe reaction of 0.50 mol of H2 with 0.25 mol of O2 atatmospheric pressure if the volume change is

–5.6 L?

PΔV = -0.57kJ

Contraction, so w is positive

ΔE = -120.43kJ

Example 23

The explosion of 2.00 mol of solid TNT with avolume of approximately 0.274 L produces gaseswith a volume of 448 L at room temperature. Howmuch PV (kJ) 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)

PΔV = 45.2kJ

Expansion, so w = -45.2kJ

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Example 24

How much heat (kJ) is evolved or absorbed in eachof the following reactions?

1.) Burning of 15.5 g of propane:

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

2.) 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)

ΔHº = +80.3 kJ

-780kJ (exothermic)

+1.24kJ (endothermic)

Hess’s Law 25

• Hess’s Law: The overall enthalpy change for areaction is equal to the sum of the enthalpychanges for the individual steps in the reaction.

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

Hess’s Law 26

(a) 2 H2(g) + N2(g) N2H4(g) ΔH°1 = ?

(b) N2H4(g) + H2(g) 2 NH3(g) ΔH°2 = –187.6 kJ

(c) 3 H2(g) + N2(g) 2 NH3(g) ΔH°3 = – 92.2 kJ

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

Standard Heats of Formation 27

Where do ΔH° values come from?

• Standard Heats of Formation (ΔH°f): enthalpychange for the formation of 1 mole of substance inits standard state

• ΔH°f = 0 for an element in its standard state!

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Standard Heats of Formation 28

H2(g) + 1/2 O2(g) → H2O(l) ΔH°f = –286 kJ/mol

3/2 H2(g) + 1/2 N2(g) → NH3(g) ΔH°f = –46 kJ/mol

2 C(s) + H2(g) → C2H2(g) ΔH°f = +227 kJ/mol

2 C(s) + 3 H2(g) + 1/2 O2(g) → C2H5OH(g) ΔH°f = –235 kJ/mol

Standard Heats of Formation 29

• Calculating ΔH° for a reaction:

ΔH° = ΔH°f (products) – ΔH°f (reactants)

Heat of formation must be multiplied by the coefficient of thereaction

C6H12O6 (s) 2C2H5OH (l) + 2CO2 (g)

ΔH° = [2ΔH°f(ethanol) + 2ΔH°f(CO2)] - ΔH°f (glucose)

Standard Heats of Formation 30

-1131Na2CO3(s)49C6H6(l)-92HCl(g)

-127AgCl(s)-235C2H5OH(g)95.4N2H4(g)

-167Cl-(aq)-201CH3OH(g)-46NH3(g)

-207NO3-(aq)-85C2H6(g)-286H2O(l)

-240Na+(aq)52C2H4(g)-394CO2(g)

106Ag+(aq)227C2H2(g)-111CO(g)

Some Heats of Formation, ΔHf° (kJ/mol)

Bond Dissociation Energy 31

• Bond Dissociation Energy (D): Amount of energy needed tobreak a chemical bond in gaseous state

D = Approximate ΔHº

ΔH° = D(reactant bonds broken) – D(product bonds formed)

H2 + Cl2 2HCl

ΔH° = (DCl-Cl + DH-H) - (2 D H-Cl)

= [(1 mol)(243 kJ/mol) + (1)(436 kJ/mol] - (2)(432 kJ/mol)

= -185 kJ

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Bond Dissociation Energy 32 Calorimetry and Heat Capacity 33

• Calorimetry: measurement of heat changes (q)for chemical reactions

• Constant Pressure Calorimetry: measures theheat change at constant pressure q = ΔH

• Bomb Calorimetry: measures the heat changeat constant volume such that q = ΔE

Calorimetry and Heat Capacity 34

Constant Pressure Bomb

Calorimetry and Heat Capacity 35

• Heat capacity {C}: amount of heat required to raise thetemperature of an object or substance a given amount

Specific Heat: amount of heat required to raise the

temperature of 1.00 g of substance by 1.00°C

Molar Heat: amount of heat required to raise the

temperature of 1.00 mole of substance by 1.00°C

C =

q

Δ T

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Calorimetry and Heat Capacity 36 Example 37

The industrial degreasing solvent methylene chloride(CH2Cl2, dichloromethane) is prepared from methane byreaction with chlorine:

CH4(g) + 2 Cl2(g) CH2Cl2(g) + 2 HCl(g)

Calculate ΔH° (kJ)

CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) ΔH° = –98.3 kJ

CH3Cl(g) + Cl2(g) CH2Cl2(g) + HCl(g) ΔH° = –104 kJ

ΔH° = -98.3 + -104 = -202kJ

Example 38

Calculate ΔH° (kJ) 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.

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

= [(4)(90.2kJ/mol) + (6)(-241.8)] - [(4)(-46.1) + (5)(0)]

= -905.6kJ

Example 39

Calculate ΔH° (kJ) for the photosynthesis of glucose from

CO2 and liquid water, a reaction carried out by all green

plants.

6CO2(g) + 6H2O(l) C6H12O6(s) + 6O2(g)

= [(1 mole)(-1260kJ/mol) + (6)(0)] - [(6)(-393.5) + (6)(-285.8)]

= 2816kJ

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Example 40

• Calculate an approximate ΔH° (kJ) for the synthesis

of ethyl alcohol from ethylene:

C2H4(g) + H2O(g) → C2H5OH(g)

• Calculate an approximate ΔH° (kJ) for the synthesis

of hydrazine from ammonia:

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

Example 41

• What is the specific heat of lead if it takes 96 J toraise 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 mLof 1.0 M NaOH at 25.0°C in a calorimeter, thetemperature of the solution increases to 33.9°C.Assume specific heat of solution is 4.184 J/(g–1·°C–1),and the density is 1.00 g/mL–1, calculate ΔH for the

reaction.