Chapter 8Chapter 8Chapter 8Chapter 8

Thermochemistry: Chemical Energy

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 kgm2/s2)

(1 calorie = 4.184 J)

• Potential Energy: stored energy

Thermodynamics 02

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

Thermodynamics 03

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

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

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

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 of view 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 = –PV

QuickTime™ and aSorenson Video decompressorare needed to see this picture.

Sign of w08

negative positive

positive negative

w = -PV expansion

w = -PV contraction

Work Units 09

w = -PV (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

E = w + q = q - PV

q = E + PV

When a person does work, energy diminishes

w = negative

E = negative

Heat and Enthalpy 11

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

q = E + PV

At constant volume (V = 0): qv = E

At constant pressure: qp = E + PV = H

enthalpy

State Functions 12

State Function:

• value depends only on the present state of the system

• path independent

• when returned to its original position, overall change is zero

State Functions 13

• State and Nonstate Properties: 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)

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

• nonstate properties include heat and work

Enthalpy 14

• Enthalpy or heat of reaction:

•H = 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) Hº = -92.2kJ

Enthalpy Changes 16

• Most changes in a system involve a gain 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:

QuickTime™ and aSorenson Video decompressorare needed to see this picture.

Chemical Changes 18

• Enthalpies of Chemical Change: Often called heats of reaction (Hreaction).

Endothermic: Heat flows into the system from the

surroundings H is positive

Exothermic: Heat flows out of the system into the

surroundings H is negative

Enthalpy Changes 19

• Reversing a reaction changes the sign of H for a reaction.

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

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

• Multiplying a reaction increases H by the same factor.

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

H =(-2219kJ x 3) = –6657 kJ

Example 20

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

w = -0.25kJ

Expansion, system loses -0.25kJ

Example 21

• The following reaction has E = –186 kJ/mol.• Is the sign of PV positive or negative?

• What is the sign and approximate magnitude of H?

Contraction, PV is negative, w is positive

H = E + PV

H = (-186kJ) + (1atm) (-1mole)

H = negative (slightly more than E)

Example 22

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 (kJ) for the reaction of 0.50 mol of H2 with 0.25 mol of O2 at atmospheric pressure if the volume change is

–5.6 L?

PV = -0.57kJ

Contraction, so w is positive

E = -120.43kJ

Example 23

The explosion of 2.00 mol of solid TNT with a volume of approximately 0.274 L produces gases with a volume of 448 L at room temperature. How much 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)

PV = 45.2kJ

Expansion, so w = -45.2kJ

Example 24

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

1.) Burning of 15.5 g of propane:

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) Hº = –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)

Hº = +80.3 kJ

-780kJ (exothermic)

+1.24kJ (endothermic)

Hess’s Law 25

• 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

Hess’s Law 26

(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

Standard Heats of Formation 27

Where do H° values come from?

• Standard Heats of Formation (H°f): enthalpy

change for the formation of 1 mole of substance in its standard state

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

Standard Heats of Formation 28

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

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

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

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

Standard Heats of Formation 29

• Calculating H° for a reaction:

H° = H°f (products) – H°f (reactants)

Heat of formation must be multiplied by the coefficient of the reaction

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

H° = [2H°f(ethanol) + 2H°f(CO2)] - H°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, Some Heats of Formation, HHff° ° (kJ/mol)(kJ/mol)

Bond Dissociation Energy 31

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

D = Approximate Hº

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

H2 + Cl2 2HCl

H° = (DCl-Cl + DH-H) - (2 D H-Cl)

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

= -185 kJ

Bond Dissociation Energy 32

Calorimetry and Heat Capacity 33

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

• Constant Pressure Calorimetry: measures the

heat change at constant pressure q = H

• Bomb Calorimetry: measures the heat change

at constant volume such that q = E

Calorimetry and Heat Capacity 34

Constant Pressure Bomb

Calorimetry and Heat Capacity 35

• Heat capacity {C}: amount of heat required to raise the temperature 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

T

Calorimetry and Heat Capacity 36

Example 37

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

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

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

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

H° = -98.3 + -104 = -202kJ

Example 38

Calculate H° (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 H° (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

Example 40

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

of ethyl alcohol from ethylene:

C2H4(g) + H2O(g) C2H5OH(g)

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

of hydrazine from ammonia:

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

Introduction to Entropy 42

• Second Law of Thermodynamics: Reactions proceed in the direction that increases the entropy of the system plus surroundings.

• A spontaneous process is one that proceeds on its own without any continuous external influence.

• A nonspontaneous process takes place only in the presence of a continuous external influence.

Introduction to Entropy 43

• The measure of molecular disorder in a system is called the system’s entropy; this is denoted S.

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

S = Sfinal – Sinitial

Positive value of S indicates increased disorder.

Negative value of S indicates decreased disorder.

Introduction to Entropy 44

Introduction to Entropy 45

• 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).

Introduction to Entropy 39

• Predict whether S° is likely to be positive or negative for each of the following reactions. Using tabulated values, calculate S° for each:

a. 2 CO(g) + O2(g) 2 CO2(g)

b. 2 NaHCO3(s) Na2CO3(s) + H2O(l) + CO2(g)

c. C2H4(g) + Br2(g) CH2BrCH2Br(l)

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

Introduction to Free Energy 40

• Gibbs Free Energy Change (G): Weighs the relative contributions of enthalpy and entropy to the overall spontaneity of a process.

G = H – TS

G < 0 Process is spontaneous

G = 0 Process is at equilibrium

G > 0 Process is nonspontaneous

Introduction to Free Energy 41

• Situations leading to G < 0:H is negative and TS is positiveH is very negative and TS is slightly negativeH is slightly positive and TS is very positive

• Situations leading to G = 0:H and TS are equally negativeH and TS are equally positive

• Situations leading to G > 0:H is positive and TS is negative H is slightly negative and TS is very negativeH is very positive and TS is slightly positive

Introduction to Free Energy 42

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

a. AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

G° = –55.7 kJ

b. 2 C(s) + 2 H2(g) C2H4(g)

G° = 68.1 kJ

c. N2(g) + 3 H2(g) 2 NH3(g)

H° = –92 kJ; S° = –199 J/K

Introduction to Free Energy 43

• 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

Equilibrium is the point where G° = H° – TS° = 0

Introduction to Free Energy 44

• 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?