Chapter 14Heat, Work, Energy, Enthalpy

1 The Nature of Energy 2 Enthalpy 3 Thermodynamics of Ideal Gases 4 Calorimetry 5 Hess's Law 6 Standard Enthalpies of Formation 7 Present Sources of Energy (skip)8 New Energy Sources (skip)

James Prescott Joule (1818-1889)

Highlights– English physicist and !!brewer!!– Studied the nature of heat, and discovered its

relationship to mechanical work – Challenged the "caloric theory”l that heat

cannot be created nor destroyed.– Led to the theory of conservation of energy,

which led to the development of the first law of thermodynamics

– Worked with Lord Kelvin to develop the absolute scale of temperature

– Pupil of John Dalton (Atomic Theory) – Joule effect: found the relationship between

the flow of current through a resistance and the heat dissipated, now called Joule's law

Moments in a Life (Death)– Gravestone is inscribed with the number

“772.55” the amount of work, in ft lb, he determined experimentally to be required to raise the temperature of 1 lb of water by 1° Fahrenheit.

“772.55”

Thermite Reaction

Reactants → Products

2 Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (s)

Energy

• Definition of Energy: Capacity to do work.

• The study of energy and its interconversions is called Thermodynamics.

• 1St Law of Thermodynamics: Conservation of energy.

– Energy can be converted from one form to another but cannot be created of destroyed.

– The energy of the universe is constant.– The energy of a closed system is constant.– Heat and work are interconvertable

Forms of Energy

• Chemical Energy• Potential energy (e.g., ΔEpotential= mgΔh)• Kinetic energy (e.g., ΔEKinetic= Δ (1/2mv2)• Electrical Energy• Nuclear Energy•

• Energy transfer– Through heat– Through work

Energy

• Energy is a state property, which means it depends on the initial and final state, not the path between them.

ΔE = Efinal – Einitial

ΔE = Eproducts – Ereactants

What is a “State Property”?

Thermodynamics: Systems and Surroundings

A thermodynamic system is the part of the universe that is under consideration. A boundary separates the system from the rest of the universe, which is called the surroundings. A system can be anything that you wish it to be, for example a piston, and engine, a brick, a solution in a test tube, a cell, an electrical circuit, a planet, etc.

Heat (q) and Work (w)

ΔE = q + w

• q = Heat absorbed by the system If q > 0 , heat is absorbed If q < 0 , heat is given off

• w = Work done on the system

• Increase the energy of a system by heating it (q>0) or by doing work on it (w>0).

Work w = Force x distance

= (Pressure x area) x distance = pressure x ΔV = PΔV

Work is Pressure-Volume work (P-V)

Exothermic reactions release energy to surroundings (by heating the surroundings)

CH4 (g) + 2O2 (g) → CO2 (g) +2 H2O (g) + heat

In an exothermic process, the energy stored in chemical bonds/molecular interactions is converted to thermal energy (random kinetic energy).

Endothermic reactions absorb energyN2 (g) + O2 (g) + Energy (heat) → 2 NO (g)

An endothermic reaction consumes heat and stores energy in the chemical bonds/molecular interactions of the products.

Work

w = F x Δh P=F/AP x A x Δh ΔV = A Δh

W=P ΔV

• Force exerted by heating = P1A– Where P1 is the pressure inside the vessel– Where A is the Area of the piston

Pext = P1 if balanced

w = F x d = F x Δh = P x A x ΔhΔV = A Δh

w = −Pext ΔV– Units are atm·L or L atm– Where 1 L atm = 101.3 Joules

• Expansion (w<0)– The system does work on the surroundings.

• Compression (w>0)– The surroundings do work on the system.

• Energy is the capacity to do work.

• Energy is a state property, which means it depends on the initial and final state, not the path between them.

ΔE = Efinal – Einitial

ΔErxn = Eproducts – Ereactants

Energy

• Heat is a means by which energy is transferred.• For processes carried out at constant pressure, the heat

absorbed equals a change in Enthalpy, ΔH.

qp= ΔH Enthalpy Change

(p denotes constant pressure)

• Enthalpy is a state property, which means it depends on its initial and final state, not the path between them.

ΔH = Hfinal – Hinitial

ΔHrxn = Hproducts – Hreactants

Enthalpy

Enthalpy

• ΔE=q+w = ΔH + w (at constant P)

• If a chemical reaction occurs in solution with no PV work (no change in V), then w=0 and ΔE = ΔH.

• ΔH for a reaction may be either positive (absorbs heat from surroundings) or negative (disperses heat to the surroundings).

• For a gas phase reaction, where ΔV is allowed, then ΔE ≠ ΔH.

Phase ChangesPhase changes (are not chemical reactions) but involve enthalpy

changes

Melt ice: H2O(s) → H2O(l) ΔHfusion= + 6 kJ at 0C

ΔHfusion is positive means endothermic (add heat)

if reversed

H2O(l) → H2O(s) ΔHfreeze= – 6 kJ

ΔHfreeze= – ΔHfusion

ΔHvaporization = – ΔHcondensation

ΔHsublimation = – ΔHdeposition

GAS

SOLID LIQUID

Freezing

Melting/Fusion

Subl

imat

ion

Dep

ositi

on

Condensation

Evaporation

Thermodynamics of Ideal Gases(volume changes)

Energy ("heat") required 3 = R T2

Δ

(KE)avg is the average translational energy of 1 mole of a perfect gas.

The energy (“heat”) required to change the translational energy of 1 mole of an ideal gas by ΔT is

ave3(KE) = RT2

Heat Capacity (C)

• How much heat is required to raise the temp of one mole of something by 1 degree? C is a proportionality constant linking T and q.

CΔT=q

If no work is allowed (ΔV = 0)CvΔT=q

If work is allowed (ΔP=0)CpΔT=q

Cp>Cv

Heating an Ideal Gas

• Molar Heat Capacity: Cv = 3/2 R (constant volume) (for a perfect monatomic gas)

• Molar Heat Capacity: Cp = 5/2 R (constant pressure) (for a perfect monatomic gas)

• polyatomic molecules have larger Cv‘s & Cp’s due to vibrational and rotational degrees of freedom.

Molar Heat Capacity (C) of a substance is defined as the heat (q) required to raise the temperature of 1 mole of a substance by 1K.

Thermodynamic Properties of an Ideal Gas

ΔE = q + w = ΔH + w (const P)

ΔH = ΔE + (PΔV) = qp

ΔH = ΔE + PΔV = ΔE + nRΔT = qp

Example

Consider 2.00 mol of a monoatomic ideal gas that is take from

state A (PA = 2.00 atm, VA = 10.0 L) to

state B (PB = 1.00 atm, VB = 30.0 L)

Calculate q, w, ΔE, and ΔH

Know: PV=nRT, and Cp, Cv, for monatomic gas.

Conversion: L-atm to Joules = 1 L atm = 101.3 Joules

Two pathways from A to BI) Step 1. A to C (constant P)

Step 2. C to B (constant V)

II) Step 3. A to D (constant P)Step 4. D to B (constant V)

State AV=10 LP=2 atm

State BV=30 LP=1 atm

State A State B

Vi = 10L Vf = 30L

Pi = 2 atm Pf = 1atm

Calculate q, w, ΔE, ΔH

p p

p

P(ÄV)q = nC ÄT where ÄT =

nR5 P(ÄV)

q = n R2 nR

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

v v

v

(ÄP)Vq = nC ÄT where ÄT =

nR3 (ÄP)V

q = n R2 nR

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

q

q1, q4: Steps 1 and 4 are constant pressure, use Cp

q2, q3: Steps 2 and 3 are constant volume, use Cv

w = - PΔV

w1= -2(20) = -40 l-atm

w2 = 0 (ΔV = 0)

w3 = 0 (ΔV = 0)

w4= -1(20) = -20 l-atm

State A State B

Vi = 10L Vf = 30L

Pi = 2 atm Pf = 1atm

Calculate q, w, ΔE, ΔH

w

q2, q3: Step 1 and 4 are constant volume, w2=w3=0

State A State B

Vi = 10L Vf = 30L

Pi = 2 atm Pf = 1atm

Calculate q, w, ΔE, ΔH

ΔE

Use ΔE1 = q1+w1

ΔE2 = q2+w2

ΔE3 = q3+w3

ΔE4 = q4+w4

ΔEAB

Use ΔEAB = ΔE1 + ΔE2

or

ΔEAB = ΔE3 + ΔE4

• Heat Capacity, Cp, is the amount of heat required to raise the temperature of a substance by one degree at constant pressure.

Cp is always a positive number q and ΔT can be both negative or positive

If q is negative, then heat is evolved or given off and the temperature decreases If q is positive, then heat is absorbed and the temperature increases

Heat Capacity =Cp =qΔT

=heat absorbed

ΔTwhere Cp denotes Heat Capacity at constant pressure

ΔT= Tf -Ti

Heat Capacity

Heat Capacity (continued)

• Cp units are energy per temp. change

• Cp units are Joule/oK or JK-1

• Units are J/K/mole or JK-1mol-1

• Molar heat capacity

Specific Heat Capacity• The word Specific before the name of a physical

quantity very often means divided by the mass

• Specific Heat Capacity is the amount of heat required to raise the temperature of one gram of material by one degree Kelvin (at constant pressure)

Standard State Enthalpies

• Designated using a superscript ° (pronounced naught)• Is written as ΔH°

• OFB text appendix D lists standard enthalpies of formation of one mole of a variety of chemical species at 1 atm and 25°C.

• Standard State conditions– All concentrations are 1M or 1 mol/L– All partial pressures are 1 atm

Standard Enthalpies of Formation

ΔH°rx is the sum of products minus the sum of the reactants

For a general reaction

a A + b B → c C + d D

ΔHro = cΔHf

o (C) + dΔHfo (D) − aΔHf

o (A) − bΔHfo (B)

CO (g) + ½ O2 (g) → CO2 (g)

Enthalpy of Reaction: ΔHrxn

€

ΔHreactiono = ΣΔHformation

o (products) − ΣΔHformationo (reactants)

ΔHformationo (CO2) = −394kJ /mol

ΔHformationo (CO ) = −111kJ /mol

ΔHformationo (O2) = 0kJ /mol

ΔHreactiono = −394 −111 −

0

2= −283kJ /mol

CO (g) + ½ O2 (g) → CO2 (g)

ΔH°rxn= – 283kJ/mol

An exothermic reaction

If stoichiometric coefficients are doubled (x2), then double ΔH°2CO (g) + 1 O2 (g) → 2 CO2 (g)

ΔH°rxn= (2)(-283kJ/mol)

If the reaction direction is reversed

CO2 (g) → CO (g) + ½ O2 (g)

Change the sign of ΔH°rxn

How much heat heat at constant pressure is needed to decompose 9.74g of HBr (g) (mwt =80.9 g/mol) into its elements?

H2 + Br2 → 2HBr (g)

ΔH°rxn = -72.8 kJmol-1

Strategy:• Decomposition is actually the reverse reaction• ΔH to decompose 2 moles of HBr is +72.8kJ/mol

Hess’s Law

• Sometimes its difficult or even impossible to conduct some experiments in the lab.

• Because Enthalpy (H) is a State Property, it doesn’t matter what path is taken as long as the initial reactants lead to the final product.

• Hess’s Law is about adding or subtracting equations and enthalpies.

• Hess’s Law: If two or more chemical equations are added to give a new equation, then adding the enthalpies of the reactions that they represent gives the enthalpy of the new reaction.

Principle of Hess's law

N2 (g) + 2 O2 (g) → 2 NO2 ΔH1 = +68kJ

N2 (g) + O2 (g) → 2 NO ΔH2 = +180 kJ

2 NO (g) + O2 (g) → 2 NO2 ΔH3 = –112 kJ

Suppose hydrazine and oxygen react to give dinitrogen pentaoxide and water vapor:

2 N2H4 (l) + 7 O2 (g) → 2 N2O5 (s) + 4 H2O (g)

Calculate the ΔH of this reaction, given that the reaction:

N2 (g) + 5/2 O2 (g) → N2O5 (s)

has a ΔH of -43 kJ.

2 N2H4 (l) + 7 O2 (g) → 2 N2O5 (s) + 4 H2O (g)

ΔHr° = 2 x ΔHf°N2O5 (s) + 4 x ΔHf°H2O (g)

– 2 x ΔHf° N2H4 (l) – 7 x ΔHf°O2 (g)

What values to use?ΔHf°N2O5 (s) given ΔH° = ─ 43 kJmol-1

ΔHf°H2O (g) look up in App. 4

ΔHf° N2H4 (l) look up in App. 4

ΔHf°O2 (g) look up in App. 4

=

ΔHr° =