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Unit 11 (Chp 5,8,19):Thermodynamics
(∆H, ∆S, ∆G, K)
John D. BookstaverSt. Charles Community College
St. Peters, MO 2006, Prentice Hall, Inc.
Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapters 5,8:Energy (E), Heat (q),
Work (w), and Enthalpy (∆H)
John D. BookstaverSt. Charles Community College
St. Peters, MO 2006, Prentice Hall, Inc.
Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Energy (E)• ability to do work OR transfer heat Work (w): transfer of energy by applying a
force over a distance. (moving an object) Heat (q): transfer of energy by DT (high to low)
• unit of energy: joule (J)
• an older unit still in widespread use is… calorie (cal)
What is it?
1 Cal = 1000 cal
2000 Cal ≈ 8,000,000 J ≈ 8 MJ!!!1 cal = 4.18 J
System and Surroundings
• System:molecules to be studied (reactants & products)
• Surroundings:everything else(container, thermometer,…)
• Energy is neither created nor destroyed.
• total energy of an isolated system is constant
Internal Energy (E):E = KE + PE (motions)(Thermal Energy)
(calculating E is too complex a problem)
DE = Efinal − Einitialreleased or absorbed
(attractions)
1st Law of Thermodynamics
(no transfer matter/energy)(universe) (conserved)
Changes in Internal Energy• Energy is transferred between
the system and surroundings, as either heat (q) or work (w).
Surroundings
Systemq out (–)
w by (–)
DE = q + w
DE = q + wDE = ?DE = (–) + (+)DE = +
q in (+)
w on (+)
Changes in Internal Energy
Efinal > Einitial
absorbed energy(endergonic)
Efinal < Einitial
released energy(exergonic)
Work (w)The only work done by a gas at constant P is change in V by pushing on surroundings.
PDV = −w ΔV
Zn + H+ Zn2+ + H2(g)
“–” b/c work done BY system ON surroundings
Enthalpy
(change in) H or DH (at constant P) :DH = DE + PDVDH = (q+w) + (−w)DH = q
• (change in) enthalpy IS heat absorbed/released
DH = DE + PDV
DE = q+w PDV = −w
Enthalpy (H) is: H = E + PV internal work done energy
work done by systemheat/work energy
in or out of system
DH = heat
Enthalpy of Reactionenthalpy is…
…the heat transfer in/out of a system
(at constant P)
DH = DE + PDV
DH = q
endergonicexergonic
endothermicexothermic
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
DH = Hproducts − Hreactants
Endothermic & Exothermic
• Endothermic: DH > 0 (+)
• Exothermic: DH < 0 (–)
DH(+) = Hfinal − Hinitial
DH(–) is thermodynamically favorable
products reactants
DH(–) = Hfinal − Hinitialproducts reactants
Enthalpy of Reaction
DHrxn, is the enthalpy of reaction, or “heat” of reaction.
units: kJmolrxn
kJ/molrxn
kJ∙molrxn
–1
2 H2(g) + O2(g) 2 H2O(g)
DH =–242 kJ/molrxn
–242 kJper 1 mol O2
–242 kJper 2 mol H2
–121 kJper 1 mol H2
(OR)
(OR)
Enthalpy1. DH depends on amount (moles, coefficients)2. DHreverse rxn = –DHforward rxn
3. DHrxn dependson the state (s, l, g) ofproducts & reactants
DH1 = –802 kJ if 2 H2O(g) because… 2 H2O(l) 2 H2O(g)
DH = +88 kJ/molrxn CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
HW p. 207 #34,35,38,45
Chp. 5,8: Calculate ∆H (4 Ways)1) Bond Energies
2) Hess’s Law
3) Standard Heats of Formation (Hf )
4) Calorimetry (lab)
Overlap and Bonding
What attractive forces?
+ +––
What repulsive forces?
Where is energy being released?
++
releasedWhen bonds/attractions form, energy is _________.
internuclear distance
E
(energy absorbed when bonds break)
Potential Energy of BondsHigh PE
Low PE(energy released
when bonds form)
chemical bond
High PE+ +
DHrxn = (BEreactants) (BEproducts)
Enthalpy of Reaction (∆H)
To determine DH for a reaction:• compare the BE of bonds broken (reactants)
to the BE of bonds formed (products).
(bonds broken) (bonds formed)(released)
BE: ∆H for the breaking of a bond (all +)
DH(+) = BEreac − BEprod
DH(–) = BEreac − BEprod
(stronger)
(stronger)
(NOT on equation sheet)
Enthalpy of Reaction (∆H)
= (4x413 + 242) (3x413 + 328 + 431)
DHrxn = 104 kJ/molrxn
CH4(g) + Cl2(g) CH3Cl(g) + HCl(g)
DHrxn =
HW p. 339 #66, 68
[4(C—H) + (Cl—Cl)] [3(C—H) + (C—Cl) + (H—Cl)]
Chp. 5,8: Calculate ∆H (4 Ways)1) Bond Energies
DHrxn = (BEreactants) (BEproducts)
2) Hess’s Law
3) Standard Heats of Formation (Hf )
4) Calorimetry (lab)
(+ broken) (– formed)(NOT given)
Hess’s Law
DHrxn is independent of
path taken
DHoverall = DH1 + DH2 + DH3 …
DHrxn = sum of DH of all steps
DH = Hfinal − Hinitialprod. react.
(NOT on equation sheet)
C3H8(g) 3 C(gr.) + 4 H2(g) ∆H1= +104 kJ
3 C(gr.) + 3 O2(g) 3 CO2(g) ∆H2= –1182 kJ
4 H2(g) + 2 O2(g) 4 H2O(l) ∆H3= –1144 kJC3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
Calculation of DH by Hess’s LawC3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
3 C(gr.) + 4 H2(g) C3H8(g) ∆H1= –104 kJ
C(gr.) + O2(g) CO2(g) ∆H2= –394 kJ
H2(g) + ½ O2(g) H2O(l) ∆H3= –286 kJ
∆Hcomb =–2222 kJ
∆Hcomb = ?
3( ) 3( )
+
4( ) 4( )
Given:
Used:
Standard Enthalpy of Formation (DHf )
o
Standard Enthalpy of Formation (DHf ):heat released or absorbed by the formation of a compound from its pure elements in their natural states.
(25oC , 1 atm)
o
3 C(gr.) + 4 H2(g) C3H8(g)∆Hf = –104 kJ
DHf = 0 for all elements in natural stateo
standard
DHf = 0 DHf = 0 DHf = –104 kJ
DH = Hfinal − Hinitialprod. react.
o o o
…therefore --->Recall…
Calculation of DH by DHf’s
…we can use Hess’s law in this way:
DH = nHf(products) – nHf(reactants)
n (mol) is the stoichiometric coefficient.
“sum”
(on equation sheet)
o
Calculation of DH by DHf’s
DH = [3(DHf CO2) + 4(DHf H2O)] – [1(DHf C3H8) + 5(DHf O2)]
DH = (3 ∙ –393.5 + 4 ∙ –285.83) – (–103.85 + 5 ∙ 0)
DH = (–2323.7) – (–103.85)
DH = –2220.0 kJ
Appendix C (p. 1123 )
HW p. 209 #60,63,66,
72,73
∆Hcomb = ?C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l)
o
DH = nHf(products) – nHf(reactants)
Chp. 5,8: Calculate ∆H (4 Ways)1) Bond Energies
DHrxn = (BEreactants) (BEproducts)
2) Hess’s Law DHoverall = DHrxn1 + DHrxn2 + DHrxn3 …
3) Standard Heats of Formation (Hf )
DH = nHf(products) – nHf(reactants)
4) Calorimetry (lab)
(+ broken) (– formed)(NOT given)
(given)
(NOT given)
By reacting (in solution) in a calorimeter, we indirectly determine DH of system by measuring ∆T & calculating q of the surroundings (calorimeter).
We can’t know the exact enthalpy of reactants and products, so we calculate DH by calorimetry, the measurement of heat flow.
q = mcDT (on equation sheet)
Calorimeter nearly
isolated heat (J)
mass (g)[of sol’n]
Tf – Ti (oC)[of surroundings]
Calorimetry
Specific Heat Capacity (c) • specific heat capacity,(c):
(or specific heat)energy required to raise temp of 1 g by 1C.
(for water)c = 4.18 J/goC + 4.18 J
of heat
Metals have much lower c’s b/c they transfer heat and change temp easily.
Calorimetry
– q = DHrxn
q = mcDT
HW p. 208 #49, 52, 54
When 4.50 g NaOH(s) is dissolved 200. g of water in a calorimeter, the temp. changes from 22.4oC to 28.3oC. Calculate the molar heat of solution, ∆Hsoln (in kJ/mol NaOH).
in J of surroundings
(in kJ/mol) of system
q = (4.50 + 200)(4.18)(28.3–22.4)qsurr = 5040 J
DHsys = –5.04 kJ4.50 g NaOH x 1 mol = 0.1125 mol NaOH 40.00 g
DH = –5.04 kJ 0.1125 mol
= –44.8 kJ mol
(thermometer)
Chp. 5,8: Calculate ∆H (4 Ways)1) Bond Energies
DHrxn = (BEreactants) (BEproducts)
2) Hess’s Law DHoverall = DHrxn1 + DHrxn2 + DHrxn3 …
3) Standard Heats of Formation (Hf )
DH = nHf(products) – nHf(reactants)
4) Calorimetry (lab) q = mc∆T (surroundings or thermometer)–q = ∆H ∆H/mol = kJ/mol (molar enthalpy)
(+ broken) (– formed)(NOT given)
(given)
(NOT given)
(given)
Chapter 19:Thermodynamics
(∆H, ∆S, ∆G, K)
Chemistry, The Central Science, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
John D. BookstaverSt. Charles Community College
St. Peters, MO 2006, Prentice Hall, Inc.
ΔH = ?ΔH = q(heat)
ΔG = ?
ΔE = q + wPΔV = –w (at constant P)
ΔS = ?
Enthalpy (H)(kJ)
Entropy (S)(J/K)
Free Energy (G) (kJ)
Energy (E)
ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)
Big Idea #5: Thermodynamics
Chemical and physical processes are driven by:• a decrease in enthalpy (–∆H), or• an increase in entropy (+∆S), or• both.
Bonds break and formto lower free energy (∆G).
1st Law of Thermodynamics
• Energy cannot be created nor destroyed (is conserved)
• or…total energy of the universe is constant.
DHuniv = DHsystem + DHsurroundings = 0if (+) then (–) = 0if (–) then (+) = 0
DHsystem = –DHsurroundings
OR
Thermodynamically Favorable• Thermodynamically
Favorable (spontaneous) processes occur with no outside intervention.
• If Favorable in one direction, then UNfavorable in reverse.
Thermodynamically Favorable
melting
• Processes that are favorable (spontaneous) at one temperature……may not be at other temperatures.
HW p. 837 #7, 11
freezing
•(okay but oversimplified) disorder/randomness•(more correct)dispersal of matter & energy among various motions of particles in space at a temperature in J/K.
DS = Sfinal Sinitial
DS = + therm favDS = – therm UNfav
(more dispersal)
(less dispersal)(structure/organization)
DS = ∆HT
“The energy of the universe is constant.”
“The entropy of the universe tends
toward a maximum.”
Entropy (S)
(ratio of heat to temp)
DS = ∆HT
(a part)(the rest)
System A(100 K)
∆S = ____J/K+0.5
same ∆Hdiff. ∆S
Entropy (S)
(quiet library, more disturbed)
(loud restaurant, less disturbed) Cough!
50 J
Surroundings (100 K)
System B(25 K)
∆S = ____J/K+2.0
50 J
Surroundings (25 K)
• change in entropy (DS) depends on……heat (∆H)…AND…temp. (T)
heightAND
weight
Entropy
6000 J
DHhand = –6000 J
DSice = ?
• Example: melting 1 mol of ice at 0oC.
DShand = ?
DHice = +6000 J
Entropy
• The melting of 1 mol of ice at 0oC.DHfusion
T=
(1 mol)(6000 J/mol) 273 K = +22.0 J/K
• Assume the ice melted in your hand at 37oC.DHfusion
T=
(1 mol)(–6000 J/mol) 310 K = –19.4 J/K
DSuniv = DSsystem + DSsurroundings
DSuniv = (22.0 J/K) + (–19.4 J/K) =
(gained by ice)
(lost by hand)
DSice =
DShand =
(ice) (hand) DSuniv+2.6 J/K
+DHice = –DHhand
+DSice > –DShand
DS =∆HT
DHuniv = 0DSuniv = +
710 J
5290 J+
(in hand) (in ice)+2.6 J/K x 273 K = (∆Suniv) (T)
Initial Energy
Final Energy
Universe (isolated system)
“dispersed” energy(unusable)
(usable E)6000 J
(usable E)(dispersed E)
1st Law: 6000 J = 6000 J
Free energy(useful for work)
2nd Law: +22 J/K > –19 J/KDHuniv = 0DSuniv = +
2nd Law of Thermodynamics
For thermodynamically favorable (spontaneous) processes…
All favorable processesincrease the entropy of the universe
(DSuniv > 0) HW p. 837 #20, 21
… +∆S gained always greater than –∆S lost,so…
DSuniv = DSsystem + DSsurroundings
DSuniv = DSsystem + DSsurroundings > 0
2nd Law of Thermodynamics (formally stated):
Entropy (Molecular Scale)
Motion: Translational , Vibrational, Rotational
• Ludwig Boltzmann described entropy with molecular motion.
• He envisioned the molecular motions of a sample of matter at a single instant in time (like a snapshot) called a microstate.
Entropy (Molecular Scale)
Boltzmann constant1.38 1023
J/K
microstates (max number
possible)
Entropy increases (+∆S) with the number of
microstates in the system.
<<<
S = k lnW
• The number of microstates and, therefore, the entropy tends to increase with…
↑Temperature (motion as KEavg)
↑Volume (motion in space)
↑Particle number (motion as KEtotal)
↑Particle Size (motion of bond vibrations)↑Particle Type (mixing)
Entropy (Molecular Scale)S : dispersal of matter & energy at T
Maxwell-Boltzmann distribution curve:∆S > 0 by adding heat as…
…distribution of KEavg increases
S : dispersal of matter & energy at TEntropy (Molecular Scale)
(T)
Entropy increases with the freedom of motion.
(s) + (l) (aq)
solid gas
V
more microstates
H2O(g) H2O(g)
T
S(s) < S(l) < S(g) S(s) < S(l) < S(aq) < S(g)
Entropy (Molecular Scale)S : dispersal of matter & energy at T
Standard Entropy (So)• Standard entropies tend to
increase with increasing molecular size. larger
molecules have more microstates
Entropy Changes (DS)
• In general, entropy increases whenliquids or solutions form from solidsmoles of gas increasestotal moles increase
Predict the sign of DS in these reactions:1. Pb(s) + 2 HI(aq) PbI2(s) + H2(g)
2. NH3(g) + H2O(l) NH4OH(aq)
DS =
DS =
+
–
3rd Law of ThermodynamicsThe entropy of a pure crystalline substance at absolute zero is 0. (not possible)
increase
Temp.
0 KS = 0
> 0 KS > 0
S = k lnWS = k ln(1)S = 0
only 1 microstate
Standard Entropy Changes (∆So)
n (mol) is the stoichiometric coefficient.
DSo = nSo(products) – nSo(reactants)
Standard entropies, S.
(on equation sheet)
HW p. 838#29, 31, 40, 42, 48
(Appendix C)
ΔS = ?ΔS = ΔH T
ΔH = q(heat)
(disorder)microstates
dispersal of matter &
energy at T ΔE = q + wPΔV = –w (at constant P)
Enthalpy (H)(kJ)
Entropy (S)(J/K)
Free Energy (G) (kJ)
Energy (E)
ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)
∆Suniv = +
ΔG = ?
Big Idea #5: Thermodynamics
Bonds break and formto lower free energy (∆G).
Chemical and physical processes are driven by:• a decrease in enthalpy (–∆H), or• an increase in entropy (+∆S), or• both.
• Thermodynamically Favorable: (defined as)increasing entropy of the universe (∆Suniv > 0)
Thermodynamically Favorable
∆Suniv > 0 (+Entropy Change of the Universe)
DSuniv = DSsystem + DSsurroundings > 0(+) (+)
Chemical and physical processes are driven by:• decrease in enthalpy (–∆Hsys)
• increase in entropy (+∆Ssys)
causes (+∆Ssurr)
DSuniverse = DSsystem +
(∆Suniv) ↔ (∆Gsys)
DSuniverse = DSsystem + DSsurroundings > 0For all thermodynamically favorable reactions:
multiplying each term by T:
DHsystem
T
–TDSuniverse = –TDSsystem + DHsystem
rearrange terms:–TDSuniverse = DHsystem – TDSsystem
(Boltzmann)(Clausius)
DGsystem = DHsystem – TDSsystem(Gibbs free energy equation)
–DG is thermodynamically favorable.
• Gibbs defined TDSuniv as the change infree energy of a system (DGsys) or DG.
• Free Energy (DG) is more useful thanDSuniv b/c all terms focus on the system.
• If –DGsys , then +DSuniverse . Therefore…
(∆Suniv) & (∆Gsys)–TDSuniv = DHsys – TDSsys
DGsys = DHsys – TDSsys
(Gibbs free energy equation)
“Bonds break & form to lower free energy (∆G).”
∆G : free energy transfer of system as workGibbs Free Energy (∆G)
(not react to completion)
–∆G : work done by system (–w) favorably+∆G : work done on system (+w) to cause rxn
–DG+DG
R P –DG (release), therm. fav.
+DG (absorb), not therm. fav.
DG = 0, system at
equilibrium.(Q = K)
–DG
DG = 0
Q > K
Q < K
–DG
Q & ∆G (not ∆Go)
+DG
Gmin 0
Q =[P][R]
DGo (1 M, 1 atm)Q = 1 = K (rare)
(not react to completion)
Q = K
can cause with electricity/light
Standard Free Energy (∆Go) and Temperature (T)
The temperature dependence of free energy comes from the entropy term (–TDS).
DG = DH – TDS(on
equation sheet)
enthalpy term
(kJ/mol)
entropy term
(J/mol∙K)
free energy (kJ/mol)
energy transferred
as heat
energy dispersed as disorder
max energy used for
work
(consists of 2 terms)
units convert to kJ!!!
∆Go = (∆Ho) ∆So
( ) –T( )( ) – T ( )
( ) –T( )( ) – T ( )
( ) – T( )( ) – T( )
DG = DH TDS
(high T) –(low T) +
+ –
(high T) +(low T) –
+ –– +
+ +
– –
(unfav. at ALL T)
(fav. at ALL T)
(fav. at high T)(unfav. at low T)
(unfav. at high T)(fav. at low T)
– T( )
=
=
=
=
+ +
– –
Standard Free Energy (∆Go) and Temperature (T)
Thermodynamic Favorability
ΔS = ΔH T
ΔH = q(heat)
ΔG = ?
(disorder)microstates
dispersal of matter &
energy at T ΔE = q + wPΔV = –w (at constant P)
+ =Enthalpy (H)(kJ)
Entropy (S)(J/K)
Free Energy (G) (kJ)
Energy (E)
–T∆Suniv as: ΔHsys & ΔSsys at T
max work done by favorable rxn
ΔG = ΔH – TΔS
ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)
∆Suniv = +
–T∆Sunivsys sys
Calculating ∆Go (4 ways)1) Standard free energies of formation, Gf :
2) Gibbs Free Energy equation:
3) From K value (next few slides)4) From voltage, Eo (next Unit)
DG = nG(products) – nG(reactants)f f
(given equation)
DG = DH – TDS(given equation)
(may need to calc. ∆Ho & ∆So first)
HW p. 840 #52, 54, 60
(given equation)(given equation)
Under any conditions, standard or nonstandard, the free energy change can be found by:
DG = DG + RT lnQ
Free Energy (∆G) & Equilibrium (K)
Q =[P][R]
At equilibrium: Q = K DG = 0
0 = DG + RT lnK
rearrange: DG = –RT lnK
therefore:
RT is “thermal energy”RT = (0.008314 kJ)(298) = 2.5 kJ at 25oC
Free Energy (∆G) & Equilibrium (K)
DG = –RT ln K
K = e^–∆Go
RT
(on equation sheet)
(NOT on equation sheet)
R = 8.314 J∙mol–1∙K–1
= 0.008314 kJ∙mol–1∙K–1If DG in kJ,then R in kJ………
Solved for K :
–∆Go
RT= ln K
∆Go = –RT(ln K) K @ Equilibrium–RT ( )
–RT ( )
> 1
< 1
– + product favored
+ – reactant favored
(favorable forward)
(unfavorable forward)
=
=
DG = –RT ln K
Free Energy (∆G) & Equilibrium (K)
ΔS = ΔH T
ΔH = q(heat)
(disorder)microstates
dispersal of matter &
energy at T ΔE = q + wPΔV = –w (at constant P)
+ =Enthalpy (H)(kJ)
Entropy (S)(J/K)
Free Energy (G) (kJ)
Energy (E)
–T∆Suniv as: ΔHsys & ΔSsys at a Tmax work done by favorable rxn
ΔG = ΔH – TΔS
K > 1 means –∆Gsys & +∆Suniv
ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w)
–T∆Sunivsys sys
∆Suniv = +
What is significant at this point?
p. 837 #6
ΔG = 0 (at equilibrium)
What does x quantify?
G of Reactants
G of Products
ΔGo
What happens at 300 K?
ΔH = TΔSso…ΔG = ΔH – TΔSΔG = 0 (at equilibrium)
In what T range is this favorable? ΔG = –ΔG = ΔH – TΔS
T > 300 K
p. 837 #4
HW p. 841 #62, 63, 76
Rxn Coupling:Unfav. rxns (+∆Go) combine with Fav. rxns (–∆Go) to make a Fav. overall (–∆Go
overall ).
ZnS(s) Zn(s) + S(s)(zinc ore) (zinc metal)
∆Go = +198 kJ/mol
S(s) + O2(g) SO2(g)
(NOT therm.fav.)
∆Go = –300 kJ/mol
ZnS(s) + O2(g) Zn(s) + SO2(g) ∆Go = –102 kJ/mol(therm.fav.)
∆Go & Rxn Coupling
goes up if
coupled
∆Go & Biochemical Rxn Coupling
ATP ADP
ATP + H2O ADP + H3PO4 ∆Go = –31 kJ/mol
Alanine + Glycine Alanylglycine ∆Go = +29 kJ/mol(amino acids) (peptide/proteins)
ATP + H2O + Ala + Gly ADP + H3PO4 + Alanylglycine ∆Go = –2 kJ/mol
(weak bond broken, stronger bonds formed)
∆Go & Biochemical Rxn CouplingRxn 1:Rxn 2:
Overall Rxn:
Glu + Pi Glu-6-P ATP ADP + Pi
Glu + ATP Glu-6-P + ADP
+14 (not fav)–31 (fav)–17 (fav)
Overall Reaction:
∆Govr
∆Govr = ∆G1 + ∆G2
∆Go & Biochemical Rxn Coupling
Glucose(C6H12O6)
CO2 + H2O
Proteins
Amino Acids
ATP
ADP
Free
Ene
rgy
(G)
–∆G(fav)
+∆G(not fav)
–∆G(fav)
+∆G(not fav)
+ O2
(oxidation)
Thermodynamic vs Kinetic ControlFr
ee E
nerg
y (G
)
B
C
A
A B ∆Go = +10 Ea = +20
(kinetic product)
A C ∆Go = –50 Ea = +50
(thermodynamic product)–50 kJ
+10 kJ
(initially pure reactant A)
(–∆Go, Temp, Q<<K, time)
(low Ea , Temp , time)
path 1
path 2
Kinetic Control: (path 2: A C )A thermodynamically favored process (–ΔGo) with no measurable product or rate while not at equilibrium, must have a very high Ea .