chapter 14 heat, work, energy, enthalpy 1 the nature of energy 2 enthalpy 3 thermodynamics of ideal...
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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”
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
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.
• 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.