Chapter 17 Energy and Chemical Change

Thermochemistry

• The study of heat changes in chemical reactions

Law of Conservation of Energy

• Defn – energy can be converted from one to another, but neither created nor

destroyed

• Ex: potential to kinetic

solar to chemical

Energy and Chemical Change

• Energy (defn) – ability to do work or produce heat

• Exists in two formsa) potential energy – stored energy

b) kinetic energy – energy in motion

Chemical Potential Energy

• Defn – energy stored in a substance based on composition

• Ex: methane (CH4) vs. propane (C3H8)

C HH

H

H

C HH

H

H

C

H

H

C

H

H

Propane has more energy b/c it has more bonds

Heat (q)

• Defn – energy that goes from warmer object to colder object

• Energy transfer moves from an area of high energy to an area of low energy

Units of Heat

• calorie (cal) – amount of heat required to raise the temp of 1 gram of water one degree Celsius– Calorie vs calorie

1 Calorie = 1000 calories = 1 kilocalorie (kcal)• Joule – SI unit of heat and energy

1 calorie = 4.18 J

Comparing Specific Heatwater (liquid)

aluminum

iron

water (ice)

4.18

2.03

0.897

0.449

ethanol 2.44

Specific Heat

• Defn – amount of heat required to raise temp of one gram of any substance by one degree Celsius

q = m c ΔT

heat mass specificheat

changeIn temp

Unit:J

g ̊ C

Specific Heat

• If heat is released from the reaction: -q

• If heat is absorbed into the reaction: +q

• Change in temperature = tfinal - tinitial

Sample problem (a)

• The temperature of a 10.0 g sample of iron is changed from 50.4°C to 25.0°C with the release of 114 J of heat. What is the specific heat of iron?

q = mcΔT

114 J = (10.0 g) c (25.4°C)

c = 0.449 J/g·°C

Sample problem (b)

• If the temperature of 34.4 g ethanol increases from 25°C to 78.8°C, how much heat is absorbed by ethanol? (specific heat of ethanol = 2.44 J/g°C)

q = mcΔT

= (34.4 g) ( 2.44 J/g°C) (53.8°C)

= 4515 J

Heat In Chemical Reactions

• Calorimeter –

insulated device used to measure amount of heat absorbed or released during a chemical or physical process

• Thermochemistry – study of heat changes in chemical reactions

3 parts we look at

• 1) system – specific part of universe that contains the reaction

• 2) surroundings – everything in universe other than the system

• 3) universe – system + surroundings

Enthalpy (H)

• Defn – heat content of a system

• Enthalpy (heat) of rxn (ΔHrxn)

a) defn – change in enthalpy for a reaction

b) formula

ΔHrxn = Hproducts – Hreactants

A + B C + D

Enthalpy (H)

• c) endothermic vs exothermic rxns

if +ΔHrxn = endothermic rxn

if –ΔHrxn = exothermic rxn

Reaction Energy Diagrams

• This is an exothermic reaction – the reactants have more energy than the products, so energy has been released

Reaction Energy Diagrams

• This is an endothermic reaction – the reactants have less energy than the products, so energy has been absorbed

Reaction Energy Diagrams

• A: energy held by the activated complex

• B: energy of the reactants

• C: energy of the products

• F: heat of reaction (ΔH)

• I: activation energy

H H

products

reactants products

reactants

EXOTHERMIC ENDOTHERMIC

ΔH ΔH

ΔH < 0 ΔH > 0

Example reactions

• 4 Fe + 3 O2 2 Fe2O3 + 1625 kJ

i) exo- or endo-?

ii) what is ΔHrxn?

Exothermic (heat written on right side of equation)

ΔHrxn = -1625 kJ

Another way to write equation:4 Fe + 3 O2 2 Fe2O3 ΔHrxn = -1625 kJ

2 Fe2O3

4 Fe + 3 O2

ΔH = -1625 kJH

Example reactions

• 27 kJ + NH4NO3 NH4+ + NO3

1-

i) exo- or endo-?

ii) what is ΔHrxn?

Endothermic (heat written on left side of equation)

ΔHrxn = +27 kJ

Another way to write equation:NH4NO3 NH4

+ + NO31- ΔHrxn = +27 kJ

H

NH4+ + NO3

-

NH4NO3

ΔH = +27 kJ

Standard Enthalpy of Formation (ΔHf

°)• Defn – change in enthalpy when one mole

of a compound is formed from its elements in their standard states

• Standard State – normal physical state of substance at room conditions (25°C and 1 atm)

ex: standard state of Hg is liquid

N2 is gas

Standard Enthalpy of Formation (ΔHf

°)• Examples

H2 (g) + S (s) H2S (g) ΔHf° = -21 kJ

S (s) + O2 (g) SO2 (g) ΔHf° = -297 kJ

Hess’s Law

• Defn – overall enthalpy change of reaction is equal to the sum of the enthalpy

changes of individual steps

A D ΔH = ?

A + B C ΔH = x

C D + B ΔH = y

Overall: A D ΔH = x + y

Example problem #1

• Calculate the enthalpy of reaction, ΔH, for the reaction:

2 H2O2 2 H2O + O2

(a) H2 + O2 H2O2 ΔH = -188 kJ

(b) 2 H2 + O2 2 H2O ΔH = -572 kJ

Example problem #1

2 H2O2 2 H2 + 2 O2 ΔH = +376 kJ

2 H2 + O2 2 H2O ΔH = -572 kJ

2 H2O2 + 2 H2 + O2 2 H2O + 2 H2 + 2 O2

2 H2O2 2 H2O + O2

ΔH = -196 kJ

1

Example problem #2

• Calculate the enthalpy of reaction, ΔH, for the reaction:

2 S + 3 O2 2 SO3

(a) S + O2 SO2 ΔH = -297 kJ

(b) 2 SO3 2 SO2 + O2 ΔH = 198 kJ

Example problem #2

2 S + 2 O2 2 SO2 ΔH = -594 kJ

2 SO2 + O2 2 SO3 ΔH = -198 kJ

2 S + 3 O2 + 2 SO2 2 SO2 + 2 SO3

2 S + 3 O2 2 SO3 ΔH = -792 kJ

Heat of Reaction (ΔHrxn)

• Defn – amount of heat lost or gained in a reaction

• Formula

ΔHrxn = Σ ΔHf (products) – Σ ΔHf (reactants)

Example problem

• Find ΔHrxn for the following reaction. Is reaction endo- or exothermic?

H2S + 4 F2 2 HF + SF6

ΔHrxn =

= -1745 kJ

[(2)(-273) + (1)(-1220)] - [(1)(-21) + (0)(4)]

exothermic

-273 kJ -1220 kJ-21 kJ 0 kJ

Entropy (S)

• Defn – measure of disorder or randomness in a system

• Formula

ΔS = Sproducts – Sreactants

• Reaction Tendency– Nature favors a disordered state– The more entropy/disorder, the greater ΔS

Unit: J/K

Entropy (S)

• Predicting ΔS

- go to more disorder + ΔS

- go to less disorder - ΔS• Ex problem: predict ΔSsystem for these

changes

a) H2O (s) H2O (l)+ ΔS; Solid to liquid is more disorder

Entropy (S)

b) 2 SO3 (g) 2 SO2 (g) + O2 (g)

• Keep in mind: the reverse reactions have opposite signs

There are more product particles (3) than reactant particles (2)

+ ΔS

Entropy example problem

• Calculate entropy change for this reaction:

2 PbO2 (s) 2 PbO (s) + O2 (g)

ΔSrxn =

= +209.2 J/K

[(2)(68.7) + 205] - [(2)(66.6)]

68.7 20566.6

Entropy, the Universe, and Free Energy

• The universe entropy

ΔSuniverse > 0

• Two natural universe processes

a) things tend to go towards lower energy (-ΔH, exothermic)

b) things tend to go towards higher disorder (+ΔS)

Spontaneous Process

• Defn – physical or chemical change that occurs with no outside intervention

Free Energy (G)

• Defn – energy available to do work• Gibbs Free Energy Equation

ΔG = ΔH – TΔS

T is in Kelvin• What does ΔG tell you?

- ΔG spontaneous rxn

+ ΔG not spontaneous rxn

Free Energy Problem

• calculate the change in free energy, ΔG, for the reaction at 25°C. Is the reaction spontaneous or nonspontaneous?

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

ΔHrxn = -91800 J, ΔSrxn = -197 J/K

Free Energy Problem

ΔGrxn = ΔHrxn – TΔSrxn

= -91800 J – (298 K)(-197 J/K)

= -33100 J Spontaneous

How does ΔH and ΔS affect spontaneity?

+ΔS

-ΔS

-ΔH +ΔH

Alwaysspontaneous

neverspontaneous

Spontaneity depends on

temp

Spontaneity dependson temp