THERMODYNAMICS Exercise 1 6CO 2 gas photosynthesis to 6O 2aqua and C 6 H 12 O 6

TERMDYNAMIS Exercise 1 6 2 gas photosynthesis to 6 2aqua and 6 12 6 alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? 2 gas assimilation reaction in green plants with blue and red photons E=h v absorption in photosynthetic reaction center PR+h v for production 6 2aqua and 6 12 6 using the data table! Mention whether the reaction will be exoergic or endoergic! Initial compounds => products glucose + oxygen Substance Δ r, kj / mol ΔS r, J / mol/k 6 2 gas + 6 2 +Q+ G PR+hv > 6 12 6 + 6 2aqua 6 12 6 (aq) -1263.78 269.45 <=biooxidation (Krebs cycle, Glycolyse) 2aqua -11.715 110.876 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 2-285.85 69.9565 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 2 gas -393.509 213.74 3. ΔG reaction = Δ reaction -T ΔS reaction 1. r = 612 + 6 2-6 2-6 2 =...J/K/mol....=-1263.78-6*11.715-(6*-285.85+6*-393.509) =-1334.07+4076154= +2742.084 kj / mol endothermic... 2. ΔS dispersed = - Δ r / T = -2742.084 /298.15= -9196,995...J/K/mol... S r = S 612 +6 S 2-6 S 2-6 S 2 =.. J / mol/k......=269.45+6*110.876-(6*69.9565+6*213.74)=934.706-1702.179= -767.473 J / mol/k... 3. ΔS total = ΔS r + ΔS dispersed = -767.473-9196,9948 = -9964,4678.J/K/mol... G r = r = +2742.084-298.15*-767.473 = +3942.084+228.822 = +2970.906 endoergic.....j/k/mol... T ΔS total =-9964,4678*298,15=-2970.9...kJ/mol... bound TΔSn accumulate energy ΔG reversereaction.q= -2742.084 kj/mol not spontaneous ΔG reaction = +2970,906 kj/mol... 1

TERMDYNAMIS Exercise II 2 gas assimilation in water 2aqua reaction alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? From 2 gas assimilation in water 2aqua reaction for green plants +stomata enclosed surface of cell membranes on photosynthetic reaction center PR+h v using the data table! Mention whether the reaction will be exoergic or endoergic!.. Initial compounds => products Substance Δ r, kj / mol ΔS r, J / mol/k ΔG r, kj / mol 2 gas + 2 + G Stomata_ell_Membranes > 2 + 2aqua +Q 2aqua -413,7076 117,57-385,98 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 2-285,85 69,9565-237,191 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 2 gas -393,509 213,74-394,359 3. ΔG reaction = Δ reaction -T ΔS reaction 1. r = 2 + 2aqua - 2-2gas =...kj/mol......=-413,7076-285,85-(-393,509+-285,85) =-413,7076-(-393,509) = -20,1986 kj / mol exothermic... ΔS dispersed = - Δ r / T =20,1986/298.15= 67,746436.J/K/mol... 2. S r = S 2 + S 2aqua - S 2 - S 2gas =... J / mol/k......=117,57+69,9565-(213,74+69,9565)= 117,57-(-213,74)= -96,17 J / mol/k... ΔS total = ΔS r + ΔS dispersed =-96,17 +67,746436 = -28,424...J/K/mol... 3. G r = r = -20,1986-298.15*-96,17= -20,1986+28.673 = +8,474......kJ/mol......endoergic T ΔS total =-28,424*298,15K=--8,47...kJ/mol... bound TΔSn accumulate energy ΔG reversereaction.q= -3942.084 kj/mol..not spontaneous ΔGreaction = +8,474kJ/mol... 2

TERMDYNAMIS Exercise III Bicarbonate 6-3 photosynthesis to 6 2 and 6 12 6 alculate Δ r, ΔS r, ΔG r. Reaction is exothermic, athermic, endothermic? From water, Bicarbonate assimilation reaction in green plants with blue and red photon E=h v absorption photosynthesis in photosynthetic reaction center PR+h v for 6 2aqua and 6 12 6 at standard conditions 298.15 K using the data table! Mention whether the reaction will be exoergic or endoergic! Initial compounds => products glucose + oxygen +water Substance Δ r, kj / mol ΔS r, J / mol/k 6-3 +6 3 + +Q+ G PR+hv > 6 12 6 +6 2aqua +6 2 6 12 6 (aq) -1263,78 269,45 <biooxidation (Krebs cycle, Glycolyse) 2aqua -11,715 110,876 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 2-285,85 69,9565 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 3 + -285,81-3,854 3. ΔG reaction = Δ reaction -T ΔS reaction - 3-689,93 98,324 1. r = 6126 +6 2 +6 2-6 3 +6 3 =+2805,27... kj / mol... =6*-689,93+6*-285,81-(-1263,78-6*11,715-6*285,85)=-3049,17+5854,44 = +2805,27 kj / mol endothermic... 2. ΔS dispersed = - Δ r / T=-2805,27*1000/298.15= -9408,9217...J/K/mol... S r = S 6126 +6 S 2 +6 S 2-6 S 3 +6 S 3 =... J / mol/k......=6*98,324+6*-3,854-(269,45+6*110,876+6*69,9565)=566,82-1354,45= -787,625 J / mol/k.. ΔS total = ΔS r + ΔS dispersed =-787,625-9408,9217=-10196,55...J/K/mol... 3. G r = r T* S r = +2805,27-298,15*-0,787625= +2805,27+247,0669=3040,1.kJ/mol...... endoergic... T ΔS total = -10196.55/K/ mol 298,15K= -3040,1...kJ/mol... bound TΔS total accumulate energy ΔG reverse_reaction... Q= -2805,27 kj/mol not spontaneous ΔG reaction = +3040,1 kj/mol p.3:http://aris.gusc.lv/biothermodynamics/biohemicalpprocese.pdf. accumulate energy TΔS total = -3040,1 kj/mol is bound in products: 6 12 6 +6 2aqua +6 2... 3

TERMDYNAMIS Exercise IV Vitamin B3 etanal 3 -= reducing to ethanol 3-2 - product alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? Vitamin B3 for anaerobic ethanal reduction to ethanol product with alcohol dehydrogenase using the data table! Mention whether the reaction will be exoergic or endoergic! Initial compounds => products+ G+Q Substance Δ r, kj / mol ΔS r, J / mol/k 3 -=+NAD+ 3 + => 3-2 -+NAD + + 2 3 -= aq -212,23-281,84 <=aerobic fermentation by alcohol dehydrogenase B3 vitamin NAD (aq) -1036,66-140,50 3 + (aq) -285.81-3,854 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 3-2 - aq -288,3-357,74 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds NAD + (aq) -1007,48-183 3. ΔG reaction = Δ reaction -T ΔS reaction 2 (aq) -285.85 69,96 1. r = 32 + 2 + NAD+ - 3-3 - NAD =...kj/mol.....= -288,3-1007,48-285,85-(-212,23-1036,66-285,81)= -1581,63+1534,7= -46,93- kj / mol exothermic... 2. ΔS dispersed = - Δ r / T = - -46,93/298.15= +157.4..J/K/mol... S r = S 32 + S 2 + S NAD+ - S 3 - S 3 - S NAD =...J/K/mol.....= -357,7394-183+69,956-(-281,838-140,50-3,854)= -470,78+426,192= -44,588 J / mol/k... 3. ΔS total = ΔS r + ΔS dispersed = 157.4-44,588= +112.812..J/K/mol... G r = r = -46,93-298,15*(-44,588)/1000= -33,636...kJ/mol...... exoergic... T ΔS total = +112.812 3 J /K/ mol 298,15 K= +33.635...kJ/mol... bound TΔSn lost energy ΔG reversereaction...q= +46,93kJ/mol spontaneous ΔG reaction = -33,636 kj/mol... 4

TERMDYNAMIS Exercise V For crystalline salt Na + l - hydration reaction alculate Δ r, ΔS r, ΔG at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? For crystalline salt Na + l - hydration reaction with water using the data table! Mention whether the reaction will be exoergic or endoergic! Initial compounds => products Substance Δ r, kj / mol ΔS r, J / mol/k crystalline Na + l - +12 2 free +Q =>Na + aqua+6 2 +l - aqua+6 2 crystalline Na + l - -411.12 72.00 1. Δ reaction = ΣΔ products - ΣΔ initial compounds Na + aqua -240.10 59.00 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds l - aqua -167.2 56.50 3. ΔG reaction = Δ reaction -T ΔS reaction 1. r = Na+ + l- - Na+l- =...kj/mol.... - = -240.1-167.2-(-411.12) =-407.3+411.12= +3.82 kj / mol endothermic... 2. ΔS dispersed = - Δ r / T = -3.82 /298.15= -12.812...J/K/mol... S r = S Na+ + S l- - S Na+l- = 59.00+56.50-(72.00)= 115.5-72.00= +43.5. J / mol/k... 3. ΔS total = ΔS r + ΔS dispersed = -12.812+43.5= +30.688...J/K/mol... G r = r = +3.82-298.15*0.0435 = -9.15 kj/mol......exoergic... T ΔS total =+30.688 J /K/ mol 298,15 K=+9.15.kJ/mol... bound TΔSn lost free energy ΔG reversereaction...q= -3.82 kj/mol..spontaneous ΔG r =-9.15 kj/mol 5

TERMDYNAMIS Exercise VI arbonic Anhydrase made acid/base equilibrium 2 /A/ 2 / 3 + - + 3 alculate Δ r, ΔS r, ΔG at standard conditions (298.15 K). Reaction is exothermic, athermic, endothermic? For the Enzyme arbonic anhydrase (A) made acid/base equilibrium 2 /A/ 2 / 3 + + - 3 using A driven reaction of aqua 2 with water molecules 2 2 using data table! Will be exoergic or endoergic! Initial compounds => products Substance Δ r, kj / mol ΔS r, J / mol/k 2aqua +2 2 + G+Q A => 3 + + - 3 3 + -285.81-3.854 1. Δ reaction = Δ products - Δ initial compounds - 3-689.93 98.324 2. ΔS reaction = ΔS products - ΔS initial compounds 2-285.85 69.9565 3.ΔG reaction =Δ reaction - T ΔS reaction 2aqua -413.7976 117.5704 1. r = 3 + 3-2 2-2 =... =-285.81-689.93-(2*-285.85-413.7976) =-975.74+985.3276= +9.7576 endothermic... kj / mol... ΔS dispersed = - Δ r / T = -9.7576 /298.15= -32.727...J/K/mol...... 2. S r = S 3 + S 3-2 S 2 - S 2 =... J / mol/k......=-3.854+98.324-(2*69.9565+117.5704)= 94.47-257.482= -163.0134 J / mol/k... ΔS total = ΔS r + ΔS dispersed = -32.727-163.0134 = -195.169J/K/mol...... 3. G r = r T* S r =+9.7576+298.15*0.1630134=+58.19 endoergic...kj/mol......... T ΔS total =-195.7404 * 298,15 K=-58.19...kJ/mol... bound TΔSn accumulated free energy ΔGreverse reaction.q=-9.7576 kj/mol endoergic ΔG r eaction=+58.19kj/mol... 6

TERMDYNAMIS Exercise VII 2 + 2 using ionic channels drive reaction 3 + + - 3 to gas 2 alculate Δ r, ΔS r, ΔG at standard conditions (298.15 K). Acid/base equilibrium 2 + 2 using proton and ionic channels drive reaction in lungs + + - 3 to breath out gas 2 with water 2 2 (exothermic, athermic or endothermic?).to use data table! Will be exoergic or endoergic! products Substance Δ r, kj / mol ΔS r, J / mol/k Initial compounds> 3 + + - 3 +Q => 2gas +2 2 + G 3 + -285.81-3.854 <=assimilation for photosynthesis in plant stomata cells - 3-689.93 98.324 1. Δ reaction = Δ products - Δ initial compounds 2-285.85 69.9565 2. ΔS reaction = ΔS products - - ΔS initial compounds 2 gas -393,509 213,74 3.ΔG reaction =Δ reaction - T ΔS reaction 1. r =2 2 + 2-3 - 3 =...kj/mol......=2*-285,85-393,509-(-285,81-689,93)=-965,209+975,74= +10,531 kj / mol endothermic... ΔS dispersed = - Δ r / T =-10,531/298,15= -35,3211..J/K/mol...... 2. S r =2 S 2 + S 2 - S 3 - S 3 =. J / mol/k......=2*69,9565+213,74-(-3,854+98,324)= 353,653-94.47= 259,183 ΔS total = ΔS r + ΔS dispersed =259,183-35,3211= +223,8619..J/K/mol... 3. G r = r T* S r =10,531-298,15*0,259183 =-66,744411exoergic...kJ/mol...... exoergic T ΔS total =223,8619 * 298,15 K= +66.7444kJ/mol... bound TΔSn lost free energy ΔG reversereaction..q=-10,531 kj/mol exoergic spontaneous ΔG reaction =-66,7444 kj/mol... 7

TERMDYNAMIS Exercise VIII glycine + glycine glycylglycine dipeptide synthesis alculate Δ r, ΔS r, ΔG r at standard conditions (298.15 K). Reaction is exothermic, athermic, endothermic? For peptide synthesis polycondensation Enzyme ribosome governed reaction with amino acids glycine Gly (G) using the data table! Mention whether the reaction will be exoergic or endoergic! J.Phys.hem.Ref.Data, Vol. 19, No. 4, 1990; hem. Phys. R, 2010-2005, p.876,882,1220,1223 glycine+ glycine glycylglycine dipeptide synthesis Viela Δ r, kj / mol ΔS r, J / mol/k Q + Gly aqua + Gly aqua + G Ribosome => Gly-Gly aqua + 2... Gly aqua 554.56 76.45 1. Δ reaction = ΣΔ products - ΣΔ initial compounds Gly-Gly aqua -790.99-1 2. ΔS reaction = ΣΔS p r oducts- ΣΔS initial compounds 2-285.83 69.9565 3. ΔG kj reaction, / mol = Δ reaction - T ΔS reaction I=0 M I=0,1 M I=0.25 M ΔG formation kj/mol ΔG 2 = -237.19 kj / mol -180.13-177.07. -176.08 Gly aqua ΔG r = ΣΔG product- ΣΔG initial_compounds -200.55-195.65-194.07 Gly-Gly aqua ; G r =-200,5-213,275-(2*-180,13)= -53.515-213.275-213.275 213.275 2 ; G r =-195.65-213,275-(2*-177.07)= -54,785 kj / mol endoergic... G r =-194,07-213,275 (2*-176,08)= -55,185 kj / mol endoergic... 1. r = Gly-Gly + 2-2 Gly =-790.99-285.83-(2*-554.56)=-1076.82+1109.12=+32.3 kj / mol endothermic...kj / mol 2.ΔS dispersed =-Δ r /T=-32.3/298.15= -108.335...J/K/mol S r = S Gly-Gly + S 2-2 S Gly =... J / mol/k.=-1+69.9565-(2*76.45)=68,957-152.9= - 83,944 J / mol/k.. 3. ΔS total = ΔS r + ΔS dispersed =-83,944-108,335= -192,279...J/K/mol Page 3: http://aris.gusc.lv/biothermodynamics/08thglygly2r10sol.pdf G r = r =32,3-298,15*-0,083944=32,3+25,0279=57,328 endoergic...kj/mol ionic force at G r =57 kj/mol is 1 mol /L (1 M) T ΔS total =-192,279 J /K/ mol 298,15 K= -57,328.kJ/mol bound TΔSn lost free energy ΔG reversereaction.q= -32.3.. kj / mol non spontaneous ΔG reaction =57,328. kj / mol 8

TERMDYNAMIS Exercise IX From water bicarbonate solution for 2 gas evaporation alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Is exothermic, athermic or endothermic? Along concentration gradients through proton + - channels [ 3 ] right /[ 3 ] left and through bicarbonate 3 channels [ 3 ] right /[ 3 ] left evaporation in lungs of water and 2 gas. To use the data table! Mention whether the reaction will be exoergic or endoergic! Initial compounds => products Substance Δ r, kj / mol ΔS r, J / mol/k Q+ 3 + + - 3 =>2 2 + 2 gas + G... 3 + -285,81-3,854 1. Δ reaction = ΣΔ products - ΣΔ initial compounds - 3-689,93 98,324 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 2-285,85 69,9565 3. ΔG reaction = Δ reaction -T ΔS reaction 2 gas -241,8352 188,7402 channels + : G =RTln([ 3 ] right /[ 3 ] left ) 2 gas -393,509 213,74 channels - 3 : G 3 =RTln([ 3 ] right /[ 3 ] left ) 1. r == 2 + 2 gas + 2-3 - 3 = kj / mol... = -285,85-241,8352-393,509-(-285,81-689,93) =-921.19+975.74= +54.546 kj / mol endothermic... kj / mol... 2. ΔS dispersed = -Δ r /T=-54,546/298,15= -182,9475...J/K/mol... S r === S 2 + S 2 gas + S 2 - S 3 - S 3 =..J/K/mol......= 69,956+188,74+213,74-(-3,854+98,324)= 353,652-94.47= +377,966 J / mol/k S =-Rln(10-5,5 /0,02754)=75,42909 J / mol / K... S 3 =-Rln(0,0154/0,0338919)=6,55847 J / mol / K. ΔS r =377,966+75,42909+6,55847 = 459,954 J / mol / K... 3. ΔS total = ΔS r + ΔS dispersed = -182,95+377,966 =195,016 J/K/mol... ΔS total = ΔS r + ΔS dispersed = -182,95+459,954=277,004 J/K/mol.. G r = r = +54,546-298,15*0, 377966 = -58,144... kj / mol... G =RTln(10-5,5 /0,02754)=-22,48918 kj / mol. G 3 =RTln(0,0154/0,0338919)=-1,9554 kj / mol... G r = r = +54,546-298,15*0,459954 = -82,589... kj / mol......exoergic. T ΔS total= 195,016 J /K/ mol 298,15 K=+58.144kJ/mol... T ΔS total= 277,004 J /K/ mol 298,15 K=-58,144-22,48918-1,9554 = +82.589 kj / mol bound TΔSn lost free energy ΔGreverse reaction.q= -54.546 kj/mol.spontaneous ΔGreaction =-58,14 kj/mol... bound TΔSn lost free energy ΔGreverse reaction.q= -54.546 kj/mol.spontaneous ΔGreaction =-82,589 kj/mol... 9

TERMDYNAMIS Exercise X. bicarbonate and proton on membrane lung epithelial cell surface 2 3 alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? For transfer bicarbonate and proton through channels on membrane from cell solution of lung to epithelial cell surface forming 2 3 using the data table! Mention whether the reaction will be exoergic or anenergic or endoergic! Initial compounds products Substance Δ r, kj / mol ΔS r, J / mol/k 3 + - Membrane + 3 2 + 2 3 + G+Q. 3 + -285.81-3.854 1. Δ reaction = ΣΔ products - ΣΔ initial compounds - 3-689.93 98.324 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 2-285.85 69.9565 3. ΔG reaction = Δ reaction -T ΔS reaction 2 3-699,65 187.00 1. r = 2 + 23-3 - 3 =...kj/mol... =-285,85-699,65-(-285,81-689,93) =-985,5+975,74= -9.76 kj / mol exothermic... ΔS dispersed = - Δ r / T = +9.76 /298.15= +32.735..J/K/mol... 2. S r = S 2 + S 23 - S 3 - S 3 =... J / mol/k....=69,956+187-(-3,854+98,324)= 256,956-94,47= +162,486 J / mol/k ΔS total = ΔS r + ΔS dispersed = +32.735 +162,486 =195,221...J/K/mol... 3. G r = r =-9,76-298,15*0,162486= -58.2052...kJ/mol......exoergic.. T ΔS total =195,221 J /K/ mol 298,15 K=+58,2052...kJ/mol... bound TΔSn lost free energy ΔG reversereaction.q= -9.76 kj/mol...spontaneous ΔG reaction =-58.2 kj/mol... 10

TERMDYNAMIS Exercise XI. 2aqua two water molecules 2 2 at lung epithelial cell surface 2 3 alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? For carbon dioxide water solution driven reaction of aqua 2 with water molecules 2 2 at lung epithelial cell surface forming 2 3! Mention reaction will be exoergic or anenergic or endoergic! Substance Δ r, kj / mol ΔS r, J / mol/k Initial compounds products 2aqua -413.7976 117.5704 2 2 + 2aqua Membrane 2 3 + 2 +Q+ G 2-285.85 69.9565 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 2 3-699,65 187.00 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds...3. ΔG reaction = Δ reaction -T ΔS reaction... 1.. r = 2 + 23-2 2-2aqua =... kj / mol... =-285.85-699,65-(2*-285.81 413,7976) =-985.5+985.418= -0,0824 kj / mol athermic... ΔS dispersed = - Δ r / T = +0,0824 /298.15= +0,27637..J/K/mol... 2. S r = S 2 + S 23-2 S 2 - S 2aqua =...J/K/mol....=69.956+187-(117,5704+2*69.956)=256,956-257,4824= -0,5264 J /mol/ K. ΔS total = ΔS r + ΔS dispersed = -0,5264+0,27637 = -0,25...J/K/mol... G r = r =-0,0824-298.15*-0,0005264= 0,07455...kJ/mol.... anenergic... T ΔS total = -0,25 J /K/ mol 298,15 K=-0,0745...kJ/mol... bound TΔSn accumulate free energy ΔG reversereaction.q= +0,0824 kj/mol...anenergic ΔG reaction =+0,0745kJ/mol... 11

TERMDYNAMIS Exercise XII. 2 3 decomposition 2 gas 2 at lung epithelial cell surface alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? From water carbonic acid 2 3 decomposition reaction to carbon dioxide 2 gas and water 2 at lung epithelial cell surface from 2 3 using the data table! Mention whether the reaction will be exoergic or endoergic! Initial compounds => products Substance Δ r, kj / mol ΔS r, J / mol/k Q + 2 3 => 2 + 2 gas + G. endothermic 2 gas -393,509 213,74 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 2-285,85 69,9565 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 2 3-699,65 187,00 3. ΔG reaction = Δ reaction -T ΔS reaction 1. r = 2 + 2-23 =..kj/mol......=-285,85-393,509-(-699,65) =-679,359+699,65= +20,291 kj / mol endothermic.. ΔS dispersed = -Δ r /T =-20,291/298,15= -68,056...J/K/mol... 2. S r = S 2 + S 2 - S 23 =..J/K/mol....=69,956+213,74-(187)= 257,482-94,47= +96,696 J / mol/k ΔS total = ΔS r + ΔS dispersed = -68.056 +96,696=28,64...J/K/mol... 3. G r = r =+20,291-298,15*0, 096696 = -8,538912.kJ/mol.....exoergic. T ΔS total =28,64 J /K/ mol 298,15 K=+8.539...kJ/mol... bound TΔSn lost free energy ΔG reversereaction.q= -20,291 kj/mol spontaneous ΔG reaction =-8,539 kj/mol... 12

TERMDYNAMIS Exercise XIII. 2 gas respiration through membrane aquaporins to form 2aqua-Blood alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K using the data table! Reaction is exothermic, athermic, endothermic? AIR 2 gas assimilation reaction for human body respiration through membrane aquaporins to form 2aqu-Blooda! Mention whether the reaction will be exoergic or endoergic! Initial compounds => products Substance Δ r, kj / mol ΔS r, J / mol/k G+ 2 gas AIR + 2 Aquporin => 2 Blood + 2aqua-Blood +Q 2aqua -11,715 110,876 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 2-285,85 69,9565 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 2 gas 0,0 205,04 3. ΔG reaction = Δ reaction -T ΔS reaction. r = 2 + 2aqua-Blood - 2-2gas-AIR =.kj/mol......=(-285,85+-11,715) (-285,85+0,0)= -285,85-11,715+285,85= -11,715 kj / mol exothermic... ΔS dispersed = - Δ r / T = 11,715/298.15= +39,292..J/K/mol... =110,876+69,9565-(69,9565+205,04)= 110,876-205,04= -94,164 J / mol/k... 2. S r = S 2 + S 2aqua-Blood - S 2 - S 2gas-AIR =...J/K/mol. = 69,9565+110,876-(205,04+69,9565)= 180,83-274,997= -94,164 J / mol/k. ΔS total = ΔS r + ΔS dispersed = -94.164 +39,292 = -54.872...J/K/mol... 3. G r = r T* S r =-11,715 +(298.15*0,094164) = +11,715-19.1305= +16.360...kJ/mol....endoergic... T ΔS total = -0,054872 kj /K/ mol 298,15K= -16.36...kJ/mol... bound TΔSn accumulate energy ΔG reversereaction.q= 11,715 kj/mol non-spontaneous endoergic ΔGreaction =+16,36kJ/mol... 13

TERMDYNAMIS Exercise XIV. 2aqu through green plants membrane aquaporins to form AIR 2 gas alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K using the data table. Reaction is exothermic, athermic, endothermic? From water 2aqu evaporation reaction through green plants membrane aquaporins to form AIR 2 gas for life bodies breathing out to AIR! Mention whether the reaction will be exoergic or endoergic! Initial compounds => products Substance Δ r, kj / mol ΔS r, J / mol/k 2 Blood + 2aqua +Q=> 2 gas AIR + 2 gas AIR + G 2aqua -11,715 110,876 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 2-285,85 69,9565 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 2 gas 0,0 205,04 3. ΔG reaction = Δ reaction -T ΔS reaction 2 gas AIR -241,835 188,74 1. r = 2 gas-air + 2gas-AIR - 2-2aqua =...kj/mol......=-241,835+0,0-(-285,85-11,715) =-241,835+297,565= +55,73 kj / mol endothermic....,,,,kj/mol... ΔS dispersed = - Δ r / T = -55,73/298.15= -186,9193...,J/K/mol......J/K/mol... 2. S r = S 2 gas-air + S 2gas-AIR - S 2 - S 2aqua =...J/K/mol......=205,04+188,74-(69,9565+110,876)= 393,78-180,8325= +212,9475 J / mol/k.....,j/k/mol... ΔS total = ΔS r + ΔS dispersed =212,9475-186,9193= +26,0282...,J/K/mol... 3. G r = r =+55,73-(298.15*0,2129475) = +55,73-63,4903= -7,76 kj/mol......exoergic.. T ΔS total = +0,0260282 kj /K/ mol 298,15K= +7,76...kJ/mol... bound TΔSn accumulate energy ΔG reversereaction.q= -55,73 kj/mol. non-spontaneous exoergic ΔGreaction = -7,76kJ/mol... 14

TERMDYNAMIS Exercise XV. vitamin B3 3 + in to 3 + out inter membrane space alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K using the data table! Reaction is exothermic, athermic, endothermic? For vitamin B3 reduced form NAD and oxidized form NADP + translocate hydrogen ions as protons through membrane from inner 3 + in to out side mitochondria 3 + out inter membrane space Reaction will be exoergic or endoergic!. Initial compounds => products Substance Δ r, kj / mol ΔS r, J / mol/k NADP + +NAD+ 3 + Membrane in =>NADP+NAD + + 3 + out+δg NADP + -1007,48 577,897 1. Δ reaction = ΣΔ products - ΣΔ initial compounds NAD (aq) -1036,66-140,50 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 3 + out -285,81-3,854 3. ΔG reaction = Δ reaction -T ΔS reaction Proton translocating transhydrogenase (E1.6.1.1) ENZYME NADP -1036,66 763,005 found in bacteria and animal mitochondria NAD + (aq) -1007,48-183 that couples the transfer of reducing equivalents between 3 + in -285,81-3,854 NAD() and NADP() to the translocation of protons across the membrane 1. r = 3ou + NADP + NAD+ - NADP+ - 3in - NAD =. kj / mol......= -1036,66-1007,48-285,81-(-1007,48-1036,66-285,81)= -2329,95+2329,95= 0.0 kj / mol athermic neutral....δs dispersed = - Δ r / T = - 0/298.15= +0,0 J/K/mol... 2. S r = S 3ou + S NADP+ + S NAD - S NADP - S 3in - S NAD+ =... J / mol/k......= 763,005-183-3,854-(577,897-140,50-3,854)= 576,2-433,543= +142,6 J / mol/k ΔS total = ΔS r + ΔS dispersed = 0,0 +142,6= = +142,6 J / mol/k... 3. G r = r =0,0-298,15*(+142,6)/1000= -42,516...kJ/mol.....exoergic.. T ΔS total =-142,6 J /K/ mol 298,15 K=+42.516...kJ/mol... bound TΔSn lost energy for proton gradient + out Q= +0,0kJ/mol...spontaneous exoergic ΔG reaction =-42,516 kj/mol... 15

TERMDYNAMIS Exercise XVI. NAD+ 3 + + 2aqua FMN(B2vitamin) NAD + + 2 + 2 2aqua +Q alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K Reaction is exothermic, athermic, endothermic is endoergic or exoergic? For vitamin B3 reduced form NAD or NADP flavin B2 vitamin FMN enzyme using oxygen 2aqua as electron acceptor for the oxidation of NAD with the production of hydrogen peroxide using the data table! Initial compounds => products Substance Δ r, kj / mol ΔS r, J / mol/k NAD+ 3 + + FMN(B2vitamin) 2aqua => NAD + + 2 + 2 2aqua +Q 2aqua -11,715 110,876 1. Δ reaction = ΣΔ products - ΣΔ initial compounds NAD (aq) -1036,66-140,50 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 3 + -285,81-3,854 3. ΔG reaction = Δ reaction -T ΔS reaction 2-285,85 69,9565 NAD + (aq) -1007,48-183 NAD and NADP oxidase (E 1.6.99.1) ENZYME 2 2aqua -191,17 143,9 1 r = 22aqua + 2 + NAD+ - 3+ - 2aqua - NAD =...... r = -1007,48-191,17-285,85-(-1036,66-11,715-285,81)= -1484,5+1334,185= -150,315 kj / mol exothermic...kj/mol... ΔS dispersed = - Δ r / T = +150,315/298.15= +504,0416...J/K/mol... 2. S r = S 22aqua + S 2 + S NAD+ - S 3+ - S 2aqua - S NAD =... J / mol/k......= -183+143,9+69,9565-(110,876-140,50-3,854)= -39,1-29,624=+64,335 J / mol/k... 3. ΔS total = ΔS r + ΔS dispersed = +64,335+504,0416= +568,3766. J / mol/k... G r = r =-150,315-298,15*(+64,335)/1000= -150,315 19,1815 = -169,5..kJ/mol......exoergic... T ΔS total = +568,3766 J /K/ mol 298,15 K=+169.5...kJ/mol... bound TΔSn lost energy NAD + 2aqua + 3 +.Q= +150,315kJ/mol spontaneous exoergic ΔG reaction =-169,5 kj/mol... 16

TERMDYNAMIS Exercise XVII Peroxide 2 2 2(aq) 2 convertion to 2aqua + 2 2 + Q alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? Peroxide 2:- :-:-: -: convertion to 2aqua +2 2 +Q at human temperature (37 ) 310.15 K, using the data table! Mention whether the reaction will be exoergic or endoergic! Initial compounds peroxide => products oxygen+water Substance Δ r, kj / mol ΔS r, J / mol/k 2 2 2(aq) => 2aqua + 2 2 + Q + ΔG 2 2 (aq) -191,17 143,9 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 2aqua -11,715 110,876 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 2-285,85 69,9565 3. ΔG reaction = Δ reaction -T ΔS reaction 1. r =2 2 + 2-2 22 =.. kj / mol...=-11,715-2*285,85-(2*-191,17) =-.133407583,4+383,415= -201,08 kj / mol exothermic... 2. ΔS dispersed = - Δ r / T = -(-201,08) /310,15= 648,33...J/K/mol... S r =2 S 2 + S 2-2 S 22 = J / mol/k...=110,876+2*69,9565-(2*143,9)=250,789-287,8= -37,011 J / mol/k... 3. ΔS total = ΔS r + ΔS dispersed = -37,011+648,33 = +611,319.. J / mol/k G r = r = -201,08-310,15*-0,037011 = -201,08+11,478962 = -189,601. kj / mol.. kj / mol......exoergic... T ΔS total =0,611319*310,15= +189,6 kj / mol bound TΔSn accumulate energy ΔG reversereaction <=.Q= -201,08 kj/mol spontaneous ΔG reaction = -189,6 kj / mol... 17

TERMDYNAMIS Exercise XVIII Glyoxal 2 () convertion to 3 - + 3 + +Q alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? Glyoxal 2 () convertion to acetate 3 = - + 3 + +Q (25 ) 298.15 K, using the data table! Mention whether the reaction will be exoergic or endoergic! Initial compounds Glyoxal => products acetate + 3 + +Q Substance Δ r, kj / mol ΔS r, J / mol/k 2 () + 2 => 3 - + 3 + + Q+ G 2 () -212 272,5 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 3 - -486 85.3 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 2 (aq) -285.85 69,96 3. ΔG reaction = Δ reaction -T ΔS reaction 3 + (aq) -285.81-3,854 Glyoxal + 2 => + 3 + + G + Q 1. r = 3 - + 3-2 + 2 () = kj / mol... kj / mo l..... =-486-285,81-(-212-285,85)= -770,11+497,85= -273,96- kj / mol exothermic... 2. ΔS dispersed = - Δ r / T = 273.96/298.15= +918,966.. J / K/mol... 2. S r = S 3 + S 3 - S 2 - S 2 () =...J / mol/k.....=85,3-3,854-(69,96+272,5)= 81,446+-342,46= -261,014 J / mol/k... 3. ΔS total = ΔS r + ΔS dispersed =918,966-261,014 = 657,952 J / mol/k... J / mol/k 4 G r = r =-273,96+298,15*0,261014= -196,139. kj / mol...exoergic... T ΔS total = +657,952 J /K/ mol 298,15 K= +196,168...kJ/mol... bound TΔSn lost energy ΔG reversereaction...q= +273,96 kj / mol spontaneous ΔG reaction = -196,14 kj / mol... 18

TERMDYNAMIS Exercise XIX Glycolic acid 2 - convertion to 3 - +Q alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? Glycolic acid 2 - convertion to acetate 3 - +Q (25 ) 298.15 K, using the data table! Mention whether the reaction will be exoergic or endoergic! Initial compounds Glycolic acid => products acetate + 3 + +Q Substance Δ r, kj / mol ΔS r, J / mol/k 2 -+NAD+ 3 + => 3 -+NAD + +2 2 +Q 2 - - 651 318,6 B3 vitamin => reduction NAD (aq) -1036,66-140,5 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 3 + (aq) -285.81-3,854 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 3 - -480.6 85.3 3. ΔG reaction = Δ reaction -T ΔS reaction NAD + (aq) -1007,48-183 2 (aq) -285.85 69,96 Glycolic acid + NAD + 3 + => Acetic acid + 2 2 + NAD + 1. r = 3 + 2 2 + NAD+ - 3-2- - NAD = kj / mol.. kj / mol... = -480,6-1007,48-2*285,85-(-651-1036,66-285,81)= -2059,78+1973,47= -86,31- kj / mol exothermic... 2. ΔS dispersed = - Δ r / T = -86,31/298.15= +289,485.. J / K/mol... S r = S 3 + 2 S 2 + S NAD+ - S 3 - S 2- - S NAD =... J / K/mol......=85,3-183+2*69,956-(318,6-140,50-3,854)= 42,212+-174,246= -132 J / mol/k... 3. ΔS total = ΔS r + ΔS dispersed = 289,485-132= +157,485 J / mol/k. J / K/mol..... 4. G r = r = -86,31+298,15*0,132= -46,9542... kj / mol......exoergic.. T ΔS total = +157,485 J /K/ mol 298,15 K= +46,9542...kJ/mol... bound TΔSn lost energy ΔG reversereaction Q= +86,31 kj / mol spontaneous ΔG reaction = -46,9542 kj / mol... 19

TERMDYNAMIS Exercise XX Pyruvate 3 = - - decarboxylation 3 + 3 alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? Pyruvate 3 = - convertion to acetaldehyde 3 + 3 - (25 ) 298.15 K, using the data table! Mention whether the reaction will be exoergic or endoergic! Initial compounds Pyruvic acid+q => products bicarbonate + 3 + Substance Δ r, kj / mol ΔS r, J / mol/k 3 =+2 2 + G+Q => 3 + 3 + + 3-3 = (aq ) -607,82 179,91 3 = - - + 2 + G + Q => 3 + 3 3 = - (aq) -603,7-433,54 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 3 (aq) -212,23 160,2 MassachusettsTinstitute 3 l -192,2 160,2 R 2 (aq) -285,85 69,96 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds 3 + (aq) -285,81-3,854 3. ΔG reaction = Δ reaction -T ΔS reaction - 3-689,93 98,324 3. ΔG reaction = Δ reaction -T ΔS reaction Pyruvic acid +2 2 + G+Q => + 3 + + 3 - ; Pyruvat + 2 - + G +Q => + 3 1. r = 3 + 3 + 3-2 2-3= = -8,45 kj / mol... MassachusettsTinstitute =-212,23-285,81-689,93-(-2*285,85-607,82)= -1187,97+1179,52= -8,45 kj / mol exothermic... 1. r = 3 + 3 + 3-2 2-3= = +11,58 kj / mol... =-192,2-285,81-689,93-(-2*285,85-607,82)= -1187,97+1179,52=11,58 kj / mol endothermic... 1. r = 3 + 3-2 - 3= - = +7,42 kj / mol... =-192,2-689,93-(-285,85-603,7)= -882,13+889,55= +7,42 kj / mol endothermic... 2. ΔS dispersed = - Δ r / T = 8,45/298,.15= +28,35... J / K/mol... MassachusettsTinstitute 2. ΔS dispersed = - Δ r / T = -11,58/298,.15= -38,8395.. J / K/mol... 2. ΔS dispersed = - Δ r / T = -7,42/298,.15= -24,8868.. J / K/mol... 2. S r = S 3 + S 3 + S 3-2 S 2 - S 3= =.. J / mol/k... MassachusettsTinstitute...= 160,2-3,854+98,324-(2*69,96+179,91)= -187,368-319,83= -65,16 J / mol/k... 2. S r = S 3 + S 3 + S 3-2 S 2 - S 3= =... J / mol/k......=160,2+98,324-(69,96+179,91)= 254,67-319,83=-65,16 J / mol/k... 2. S r = S 3 + S 3 - S 2 - S 3= - =... J / mol/k......=160,2-3,854+98,324-(69,96-433,54)= 258,524-249,87=8,654 J / mol/k... 3. ΔS total = ΔS r + ΔS dispersed =-65,16-28,35 = -36,81 J / mol/k.. MassachusettsTinstitute J / mol/k 3. ΔS total = ΔS r + ΔS dispersed =-65,16-38,8395 = -104 J / mol/k... J / mol/k 3. ΔS total = ΔS r + ΔS dispersed =8,654-24,8868 = -16,233 J / mol/k... J / mol/k 4. G r = r =-8,45-298,15*-0,06516= +10,98... kj / mol endoergic MassachusettsTinstitute 4. G r = r = 11,58-298,15*-0,06516= +31,01. kj / mol endoergic 4. G r = r = 7,42-298,15*0,008654= +4,84. kj / mol endoergic T ΔS total = -36,81 J /K/ mol 298,15 K= -10,97.. kj / mol... MassachusettsTinstitute bound TΔSn accumulate energy ΔG reversereaction Q= -8,45 kj / mol not spontaneous ΔG reaction = 10,98 kj / mol... T ΔS total = -104 J /K/ mol 298,15 K= -31,01.... kj / mol... bound TΔSn accumulate energy ΔG reversereaction Q=11,58 kj / mol not spontaneous ΔG reaction = 31,01 kj / mol... T ΔS total = -16,233 J /K/ mol 298,15 K= -4,84.... kj / mol... 20

bound TΔSn accumulate energy ΔG reversereaction Q=7,42 kj / mol not spontaneous ΔG reaction = 4,84 kj / mol... 21

TERMDYNAMIS Exercise XXI Lactate 3 () - - aq decarboxylation 3 2 + 3 alculate Δ r, ΔS r, ΔG r at standard conditions 298.15 K. Reaction is exothermic, athermic, endothermic? Lactate 3 = - convertion to ethanol 3 + - 3 (25 ) 298.15 K, using the data table! Mention whether the reaction will be exoergic or endoergic! Initial compounds Lactate+Q => products ethanol bicarbonate Substance Δ r, kj / mol ΔS r, J / mol/k 3 () - aq+ 2 +Q => 3 2 l + 3 - + G 3 () - aq -686,2-557,71 1. Δ reaction = ΣΔ products - ΣΔ initial compounds 3 2 l -277,6 160,7 R 2 (aq) -285,85 69,96 2. ΔS reaction = ΣΔS products - ΣΔS initial compounds - 3-689,93 98,324 3. ΔG reaction = Δ reaction -T ΔS reaction Lactate + 2 + G +Q =>Ethanol 1. r = 32 + 3 - - 2-32- = 4,52 kj / mol... = -277,6-689,93-(-285,85-686,2)= -967,53+972,06= +4,52 kj / mol endothermic......... 2. ΔS dispersed = - Δ r / T = -4,52/298,15= -15,16... J / K/mol......... 2. S r = S 32 + S 3 - - S 2 - S 32- =.. J / mol/k......= 160,7+98,324-(69,96-557,71)= 259,02 + 487,75= 746,77 J / mol/k......... 3. ΔS total = ΔS r + ΔS dispersed =-15,16 +746,77 = 731,61 J / mol/k.. J / mol/k...... 4. G r = r =4,52-298,15*0,74677= -218,13... kj / mol exoergic...... T ΔS total = 731,61 J /K/ mol 298,15 K= 218,13.. kj / mol... bound TΔSn accumulate energy ΔG reversereaction Q= -4,52 kj / mol spontaneous ΔG reaction = - 218,13 kj / mol... + 3-22

TERMDYNAMIS Exercise XXII. smolar concentration gradient 11= osm in green plants alculate Δ r, ΔS r, ΔG r and the amount of heat exothermic, athermic or endothermic reaction at standard conditions 298.15 K. PR Photo Synthetic Reaction enter ENZYME complex drive green plants products 6 12 6 and oxygen 6 2aqua. by photon E=h v absorption Aquaporin substrates oxygen 6 2 and water 6 2 increase osmotic pressure to outside cell 11 times as concentration in cell decreases from initial times 12= osm =6+6 to one glucose 6 12 6 molecule osm =1. So total flow out of plant organisms through aquaporins increases 11 times: 6-3 +6 3 + +Q PR+hv => 6 12 6 +6 2aqua +6 2 against osmolar concentration gradient 12/1. xygen 6 2 and water 6 2 pushed out 6 2 +6 Aquaporins 2aqua =>6 2aqua +6 2 through Aquaporins in the athermic channel = 0 kj / mol manners (as no heat waste) and used energy gained from PR Photo Synthetic Reaction enter ENZYME complexes G PR =3040,1 kj / mol (Exercise III ) via absorbtion red and blue Photon energy E=h v and through heat supply Q p.3:http://aris.gusc.lv/biothermodynamics/biohemicalpprocese.pdf E=h PR light red Bicarbonate + hydrogen ions + heat supply blue photo synthesis glucose + oxygen water 6-3 +6 3 + + Q + G reaction reverse reaction 6 12 6 + 6 2 +6 2 Mention whether the reaction will be exoergic or endoergic! Universal gas constant R=8,3144 J/mol/K G channel =-RTln( osm [6 2aqua +6 2 ] right / osm [6 2aqua +6 2 ] left )=-12RTln(12/1)= -36.96 kj / mol... Substance before after 2aqua [ 2 ] = 6 10 5 M [ 2 ] = 6 10 5 M 2 55,3 M 55,3 M G 2=-RTln([ 2 ] right /[ 2 ] left )= -RTln(K equilibrium )= -6.1599 = -8,3144*298,15*ln(12/1)= -8,3144*298,15*-2.4849= -6.1599. kj / mol For 6 2aqua G 6 2= -6.1599*6= -36.9596.. kj / mol G 6 2 =-6RTln([ 2 ] right /[ 2 ] left )=6*8,3144*310,15*ln(1/12)= -36.9596.. kj / mol... For 6 2 G 6 2 = -6.1599*6= -36.9596... kj / mol.exoergic S 6 2=-6Rln([ 2 ] right /[ 2 ] left )= -8,3144*ln(1/12)=20,66*6=123.96... J / mol / K... S 6 2 =-6Rln([ 2 ] right /[ 2 ] left )= -8,3144*ln(1/12)=123.96... J / mol / K... channel =.0.. kj / mol...no heat waste... T ΔS 62 = -0, 12396 298,15=36.9596...kJ/mol used energy of PR bound TΔSn=3040,1 kj / mol out off 2... T ΔS 62 = -0,12396 298,15=36.9596..kJ/mol used energy of PR bound TΔSn=3040,1 kj / mol out off 2 For 6 2aqua +6 2 T ΔS 62+62 =36.9596+36.9596=73.919.kJ/mol of PR bound TΔSn=3040,1 kj / mol athermic Δ r eaction =+0 kj/mol; Q= -0 kj/mol...spontaneous ΔG reaction = -73.919 kj/mol.. The Photosynthesis dilutes of osmolar concentration by 6 2aqua +6 2 consuming bicarbonate end hydrogen ions 6-3 +6 3 + drive spontaneous flow of 6 2aqua +6 2 through aquaporins out of PR cells against membrane concentration gradient 12/1 with standard free energy ΔG 62+62 = -73.919 kj/mol per one glucose mol 6 12 6! 23