Unit 11 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K) Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Unit 11 (Chp 5,8,19): Thermodynamics (∆H, ∆S, ∆G, K) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

Chapters 5,8: Energy (E), Heat (q), Work (w), and Enthalpy (∆H) Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapters 5,8: Energy (E), Heat (q), Work (w), and Enthalpy (∆H) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K) Free Energy (G) (kJ)

Energy (E) What is it? 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) 1 Cal = 1000 cal 1 cal = 4.18 J 2000 Cal ≈ 8,000,000 J ≈ 8 MJ!!!

System and Surroundings molecules to be studied (reactants & products) Surroundings: everything else (container, thermometer,…)

1st Law of Thermodynamics Energy is neither created nor destroyed. total energy of an isolated system is constant (universe) (no transfer matter/energy) (conserved) Internal Energy (E): E = KE + PE (motions) (Thermal Energy) (attractions) (calculating E is too complex a problem) E = Efinal − Einitial released or absorbed

Changes in Internal Energy Energy is transferred between the system and surroundings, as either heat (q) or work (w). E = q + w E = ? E = (–) + (+) Surroundings E = + System q in (+) q out (–) E = q + w w on (+) w by (–)

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. PV = −w ΔV “–” b/c work done BY system ON surroundings Zn + H+ Zn2+ + H2(g)

Enthalpy Enthalpy (H) is: H = E + PV H = E + PV E = q+w PV = −w internal work done energy work done by system heat/work energy in or out of system H = E + PV E = q+w PV = −w (change in) H or DH (at constant P) : H = E + PV H = (q+w) + (−w) H = q (change in) enthalpy IS heat absorbed/released H = heat

CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) Enthalpy of Reaction CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) enthalpy is… …the heat transfer in/out of a system (at constant P) H = E + PV H = q H = Hproducts − Hreactants exergonic exothermic endergonic endothermic

Endothermic & Exothermic Endothermic: H > 0 (+) H(+) = Hfinal − Hinitial products reactants Exothermic: H < 0 (–) H(–) = Hfinal − Hinitial reactants products H(–) is thermodynamically favorable

Enthalpy of Reaction 2 H2(g) + O2(g)  2 H2O(g) Hrxn, is the enthalpy of reaction, or “heat” of reaction. units: kJ molrxn kJ/molrxn kJ∙molrxn Demo –242 kJ per 1 mol O2 (OR) H = –242 kJ/molrxn –242 kJ per 2 mol H2 Demo – Zn in narrow mouth Erlenmeyer flask. Add 40-50 mL of 6 M H2SO4. Stretch out large balloon well and place over mouth. Blow some (not much) O2 into balloon and tie. (OR) –1 –121 kJ per 1 mol H2

CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) Enthalpy HW p. 207 #34,35,38,45 H depends on amount (moles, coefficients) Hreverse rxn = –Hforward rxn Hrxn depends on the state (s, l, g) of products & reactants H1 = –802 kJ if 2 H2O(g) because… 2 H2O(l)  2 H2O(g) H = +88 kJ/molrxn CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)

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 E internuclear distance When bonds/attractions form, energy is _________. released – + + – + + What repulsive forces? What attractive forces? E Where is energy being released? internuclear distance

Potential Energy of Bonds High PE chemical bond Low PE (energy released when bonds form) High PE (energy absorbed when bonds break) + +

Bond Enthalpy (BE) p.330 BE: ∆H for the breaking of a bond (all +) aka… bond dissociation energy

Enthalpy of Reaction (∆H) BE: ∆H for the breaking of a bond (all +) To determine H for a reaction: compare the BE of bonds broken (reactants) to the BE of bonds formed (products). Hrxn = (BEreactants)  (BEproducts) (bonds broken) (bonds formed) (released) (stronger) H(+) = BEreac − BEprod (NOT on equation sheet) H(–) = BEreac − BEprod (stronger)

Enthalpy of Reaction (∆H) CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g) Hrxn = [4(C—H) + (Cl—Cl)]   [3(C—H) + (C—Cl) + (H—Cl)] = (4x413 + 242)  (3x413 + 328 + 431) Hrxn = 104 kJ/molrxn HW p. 339 #66, 68

Chp. 5,8: Calculate ∆H (4 Ways) 1) Bond Energies Hrxn = (BEreactants)  (BEproducts) 2) Hess’s Law 3) Standard Heats of Formation (Hf ) 4) Calorimetry (lab) (NOT given) (+ broken) (– formed) 

(NOT on equation sheet) Hess’s Law H = Hfinal − Hinitial prod. react. Hrxn is independent of path taken Hrxn = sum of H of all steps (NOT on equation sheet) Hoverall = H1 + H2 + H3 …

Calculation of H by Hess’s Law C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) ∆Hcomb = ?  Given: + 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 3( ) 3( ) 4( ) 4( ) Used: 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 kJ ∆Hcomb = –2222 kJ C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)

Standard Enthalpy of Formation (Hf ) heat released or absorbed by the formation of a compound from its pure elements in their natural states. o (25oC , 1 atm) o Hf = 0 for all elements in natural state 3 C(gr.) + 4 H2(g)  C3H8(g) ∆Hf = –104 kJ o o o Hf = 0 Hf = 0 Hf = –104 kJ H = Hfinal − Hinitial Recall… …therefore ---> prod. react.

Calculation of H by Hf’s …we can use Hess’s law in this way: H = nHf(products) – nHf(reactants) n (mol) is the stoichiometric coefficient. “sum”    (on equation sheet)

Calculation of H by Hf’s C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) ∆Hcomb = ?   H = nHf(products) – nHf(reactants) Appendix C (p. 1123 )     H = [3(Hf CO2) + 4(Hf H2O)] – [1(Hf C3H8) + 5(Hf O2)] H = (3 ∙ –393.5 + 4 ∙ –285.83) – (–103.85 + 5 ∙ 0) H = (–2323.7) – (–103.85) H = –2220.0 kJ HW p. 209 #60,63,66, 72,73

Chp. 5,8: Calculate ∆H (4 Ways) 1) Bond Energies Hrxn = (BEreactants)  (BEproducts) 2) Hess’s Law Hoverall = Hrxn1 + Hrxn2 + Hrxn3 … 3) Standard Heats of Formation (Hf ) H = nHf(products) – nHf(reactants) 4) Calorimetry (lab) (NOT given) (+ broken) (– formed) (NOT given)     (given)

Calorimeter nearly isolated Calorimetry We can’t know the exact enthalpy of reactants and products, so we calculate H by calorimetry, the measurement of heat flow. By reacting (in solution) in a calorimeter, we indirectly determine H of system by measuring ∆T & calculating q of the surroundings (calorimeter). Calorimeter nearly isolated (on equation sheet) heat (J) q = mcT mass (g) [of sol’n] Tf – Ti (oC) [of surroundings]

Specific Heat Capacity (c) (or specific heat) energy required to raise temp of 1 g by 1C. (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 = mcT – q = Hrxn in J of surroundings (thermometer) HW p. 208 #49, 52, 54 in J of surroundings q = mcT (thermometer) – q = Hrxn (in kJ/mol) of system 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). q = (4.50 + 200)(4.18)(28.3–22.4) qsurr = 5040 J H = –5.04 kJ 0.1125 mol Hsys = –5.04 kJ 4.50 g NaOH x 1 mol = 0.1125 mol NaOH 40.00 g = –44.8 kJ mol

Chp. 5,8: Calculate ∆H (4 Ways) 1) Bond Energies Hrxn = (BEreactants)  (BEproducts) 2) Hess’s Law Hoverall = Hrxn1 + Hrxn2 + Hrxn3 … 3) Standard Heats of Formation (Hf ) H = nHf(products) – nHf(reactants) 4) Calorimetry (lab) q = mc∆T (surroundings or thermometer) –q = ∆H ∆H/mol = kJ/mol (molar enthalpy) (NOT given) (+ broken) (– formed) (NOT given)     (given) (given)

Chapter 19: Thermodynamics (∆H, ∆S, ∆G, K) Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 19: Thermodynamics (∆H, ∆S, ∆G, K) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall, Inc.

Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K) Free Energy (G) (kJ) ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w) ΔE = q + w PΔV = –w (at constant P) ΔH = q (heat) ΔH = ? ΔS = ? ΔG = ?

Big Idea #5: Thermodynamics Bonds break and form to lower free energy (∆G). Chemical and physical processes are driven by: a decrease in enthalpy (–∆H), or an increase in entropy (+∆S), or both.

1st Law of Thermodynamics Energy cannot be created nor destroyed (is conserved) or…total energy of the universe is constant. Hsystem = –Hsurroundings OR Huniv = Hsystem + Hsurroundings = 0 if (+) then (–) = 0 if (–) then (+) = 0

Thermodynamically Favorable Thermodynamically Favorable (spontaneous) processes occur with no outside intervention. If Favorable in one direction, then UNfavorable in reverse.

Thermodynamically Favorable Processes that are favorable (spontaneous) at one temperature… …may not be at other temperatures. HW p. 837 #7, 11 melting freezing

Entropy (S) S = Sfinal  Sinitial (okay but oversimplified) disorder/randomness (more correct) dispersal of matter & energy among various motions of particles in space at a temperature in J/K. “The energy of the universe is constant.” “The entropy of the universe tends toward a maximum.” (ratio of heat to temp) S = ∆H T S = Sfinal  Sinitial S = + therm fav S = – therm UNfav (more dispersal) (less dispersal) (structure/organization)

Entropy (S) S = ∆H T change in entropy (S) depends on… height AND weight (a part) S = ∆H T (the rest) change in entropy (S) depends on… …heat (∆H)…AND…temp. (T) same ∆H diff. ∆S System A (100 K) 50 J System B (25 K) 50 J Surroundings (100 K) Visualize a 1 ft tall dog. Now, that dog is a 100 lb. dachshund! Surroundings (25 K) ∆S = ____J/K +0.5 ∆S = ____J/K +2.0 (loud restaurant, less disturbed) (quiet library, more disturbed) Cough!

Entropy Example: melting 1 mol of ice at 0oC. Shand = ? Hhand = –6000 J 6000 J Hice = +6000 J Sice = ?

Entropy S = ∆H T +Hice = –Hhand Huniv = 0 Suniv = + +Sice > –Shand The melting of 1 mol of ice at 0oC. Hfusion T (1 mol)(6000 J/mol) 273 K Sice = = = +22.0 J/K (gained by ice) Assume the ice melted in your hand at 37oC. Hfusion T (1 mol)(–6000 J/mol) 310 K Shand = = = –19.4 J/K (lost by hand) Suniv = Ssystem + Ssurroundings Suniv = (22.0 J/K) + (–19.4 J/K) = (ice) (hand) Suniv +2.6 J/K

Universe (isolated system) 1st Law: 6000 J = 6000 J Huniv = 0 Suniv = + 2nd Law: +22 J/K > –19 J/K 5290 J (usable E) (usable E) 6000 J + (dispersed E) 710 J (in hand) (in ice) +2.6 J/K x 273 K = (∆Suniv) (T) Universe (isolated system) Initial Energy “dispersed” energy (unusable) Free energy (useful for work) Final Energy

2nd Law of Thermodynamics Suniv = Ssystem + Ssurroundings For thermodynamically favorable (spontaneous) processes… … +∆S gained always greater than –∆S lost, so… Suniv = Ssystem + Ssurroundings > 0 2nd Law of Thermodynamics (formally stated): All favorable processes increase the entropy of the universe (Suniv > 0) HW p. 837 #20, 21

Entropy (Molecular Scale) Ludwig Boltzmann described entropy with molecular motion. Motion: Translational , Vibrational, Rotational  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) S = k lnW Boltzmann constant 1.38  1023 J/K microstates (max number possible) Entropy increases (+∆S) with the number of microstates in the system. < < <

Entropy (Molecular Scale) S : dispersal of matter & energy at T 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 (T)

Entropy (Molecular Scale) S : dispersal of matter & energy at T Entropy increases with the freedom of motion. S(s) < S(l) < S(g) S(s) < S(l) < S(aq) < S(g) gas solid T more microstates (s) + (l)  (aq) V H2O(g) H2O(g)

Standard Entropy (So) Standard entropies tend to increase with increasing molecular size. larger molecules have more microstates

Entropy Changes (S) In general, entropy increases when liquids or solutions form from solids moles of gas increases total moles increase Predict the sign of S in these reactions: 1. Pb(s) + 2 HI(aq)  PbI2(s) + H2(g) 2. NH3(g) + H2O(l)  NH4OH(aq) S = + S = –

3rd Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero is 0. (not possible) S = k lnW S = k ln(1) S = 0 increase Temp. only 1 microstate 0 K S = 0 > 0 K S > 0

Standard Entropy Changes (∆So) HW p. 838 #29, 31, 40, 42, 48 Standard entropies, S. (Appendix C) So = nSo(products) – nSo(reactants) (on equation sheet) n (mol) is the stoichiometric coefficient.

dispersal of matter & energy at T Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K) Free Energy (G) (kJ) ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w) (disorder) microstates dispersal of matter & energy at T ΔE = q + w PΔV = –w (at constant P) ∆Suniv = + ΔH = q (heat) ΔS = ΔH T ΔS = ? ΔG = ?

Big Idea #5: Thermodynamics Bonds break and form to 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 Chemical and physical processes are driven by: decrease in enthalpy (–∆Hsys) increase in entropy (+∆Ssys) causes (+∆Ssurr) (+) (+) Suniv = Ssystem + Ssurroundings > 0 Thermodynamically Favorable: (defined as) increasing entropy of the universe (∆Suniv > 0) ∆Suniv > 0 (+Entropy Change of the Universe)

(∆Suniv) ↔ (∆Gsys) Hsystem For all thermodynamically favorable reactions: Suniverse = Ssystem + Ssurroundings > 0 (Boltzmann) Hsystem T Suniverse = Ssystem + (Clausius) multiplying each term by T: –TSuniverse = –TSsystem + Hsystem rearrange terms: –TSuniverse = Hsystem – TSsystem Gsystem = Hsystem – TSsystem (Gibbs free energy equation)

(∆Suniv) & (∆Gsys) –TSuniv = Hsys – TSsys Gsys = Hsys – TSsys (Gibbs free energy equation) Gibbs defined TDSuniv as the change in free energy of a system (Gsys) or G. Free Energy (G) is more useful than Suniv b/c all terms focus on the system. If –Gsys , then +Suniverse . Therefore… –G is thermodynamically favorable. “Bonds break & form to lower free energy (∆G).”

(not react to completion) Gibbs Free Energy (∆G) ∆G : free energy transfer of system as work –∆G : work done by system (–w) favorably +∆G : work done on system (+w) to cause rxn (not react to completion) +DG –DG

Q & ∆G (not ∆Go) [P] R  P Q = [R] Q < K Q > K +DG –DG –DG Q = K –DG (release), therm. fav. +G (absorb), not therm. fav. DG = 0, system at equilibrium. (Q = K) (not react to completion) can cause with electricity/light Q > K +DG –DG –DG Q = K Gmin  0 DG = 0 DGo (1 M, 1 atm) Q = 1 = K (rare)

Standard Free Energy (∆Go) and Temperature (T) (on equation sheet) (consists of 2 terms) DG = DH – TS free energy (kJ/mol) enthalpy term (kJ/mol) entropy term (J/mol∙K) units convert to kJ!!! max energy used for work energy transferred as heat energy dispersed as disorder The temperature dependence of free energy comes from the entropy term (–TS).

Standard Free Energy (∆Go) and Temperature (T) DG = DH  TS Thermodynamic Favorability ∆Go = (∆Ho) ∆So – T( ) ( ) –T( ) (high T) – (low T) + (fav. at high T) (unfav. at low T) + + = ( ) – T ( ) + + – (unfav. at ALL T) + = ( ) – T( ) + (fav. at ALL T) – = ( ) – T( ) – + (high T) + (low T) – ( ) –T( ) (unfav. at high T) (fav. at low T) – – = ( ) – T ( ) – –

+ = Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K) Free Energy (G) (kJ) ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w) (disorder) microstates –T∆Suniv as: ΔHsys & ΔSsys at T dispersal of matter & energy at T max work done by favorable rxn ΔE = q + w PΔV = –w (at constant P) ∆Suniv = + ΔH = q (heat) ΔS = ΔH T ΔG = ΔH – TΔS ΔG = ? sys sys –T∆Suniv

Calculating ∆Go (4 ways) Standard free energies of formation, Gf : Gibbs Free Energy equation: From K value (next few slides) From voltage, Eo (next Unit) DG = SnG(products) – SnG(reactants) f (given equation) HW p. 840 #52, 54, 60 DG = DH – TS (given equation) (may need to calc. ∆Ho & ∆So first) (given equation) (given equation)

Free Energy (∆G) & Equilibrium (K) Under any conditions, standard or nonstandard, the free energy change can be found by: G = G + RT lnQ Q = [P] [R] RT is “thermal energy” RT = (0.008314 kJ)(298) = 2.5 kJ at 25oC At equilibrium: Q = K G = 0 therefore: 0 = G + RT lnK rearrange: G = –RT lnK

Free Energy (∆G) & Equilibrium (K) G = –RT ln K (on equation sheet) If G in kJ, then R in kJ……… R = 8.314 J∙mol–1∙K–1 = 0.008314 kJ∙mol–1∙K–1 –∆Go RT = ln K –∆Go RT Solved for K : (NOT on equation sheet) K = e^

Free Energy (∆G) & Equilibrium (K) G = –RT ln K ∆Go = –RT(ln K) K @ Equilibrium – + = –RT ( ) > 1 product favored (favorable forward) + – = –RT ( ) < 1 reactant favored (unfavorable forward)

+ = Energy (E) Enthalpy (H) (kJ) Entropy (S) (J/K) Free Energy (G) (kJ) ΔH = ΔE + PΔV internal work by energy system (KE + PE) (–w) (disorder) microstates –T∆Suniv as: ΔHsys & ΔSsys at a T dispersal of matter & energy at T max work done by favorable rxn ΔE = q + w PΔV = –w (at constant P) K > 1 means –∆Gsys & +∆Suniv ∆Suniv = + ΔH = q (heat) ΔS = ΔH T ΔG = ΔH – TΔS sys sys –T∆Suniv

p. 837 #6 What does x quantify? ΔGo What is significant at this point? G of Reactants What does x quantify? ΔGo What is significant at this point? G of Products ΔG = 0 (at equilibrium)

p. 837 #4 In what T range is this favorable? What happens at 300 K? ΔG = ΔH – TΔS T > 300 K ΔH = TΔS so… ΔG = ΔH – TΔS ΔG = 0 (at equilibrium) HW p. 841 #62, 63, 76

∆Go & Rxn Coupling Rxn Coupling: Unfav. rxns (+∆Go) combine with Fav. rxns (–∆Go) to make a Fav. overall (–∆Gooverall ). goes up if coupled (zinc ore)  (zinc metal) (NOT therm.fav.) ZnS(s)  Zn(s) + S(s) ∆Go = +198 kJ/mol S(s) + O2(g)  SO2(g) ∆Go = –300 kJ/mol ZnS(s) + O2(g)  Zn(s) + SO2(g) ∆Go = –102 kJ/mol (therm.fav.)

∆Go & Biochemical Rxn Coupling (weak bond broken, stronger bonds formed) 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

∆Go & Biochemical Rxn Coupling 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: 1st of 10 steps of Glycolysis is phosphorylation of glucose at C-6 (ends with ATP production). ∆Govr ∆Govr = ∆G1 + ∆G2

∆Go & Biochemical Rxn Coupling Glucose (C6H12O6) ATP Proteins + O2 (oxidation) Free Energy (G) –∆G (fav) +∆G (not fav) –∆G (fav) +∆G (not fav) ADP CO2 + H2O Amino Acids

Thermodynamic vs Kinetic Control 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 . A  B ∆Go = +10 Ea = +20 (kinetic product) (initially pure reactant A) path 1 (low Ea , Temp , time) B A A  C ∆Go = –50 Ea = +50 (thermodynamic product) Free Energy (G)  +10 kJ path 2 –50 kJ C (–∆Go, Temp, Q<<K, time)

Thermodynamic vs Kinetic Control Thermodynamic Product: ___ E Kinetic Product: ___ D Rxn A  E will be under ______________ control at low temp and Q > K . kinetic (high Ea) Pain 