coligativeproperties.pdf compound amount, concentration, ions and colligative...
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Compound amount, concentration, ionsand colligative – osmolar osmosis
Water circulation attractors:evaporation<=>condensation, osmosis through membrane aquaporin channels
Latin colligatus - its working as bound or working together,Greek Osmos - water squeeze through (membrane)
Oxidation, reduction,
http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
Oxidation, reduction,Nernst’s and membrane potentialOxidation-reduction Nernst’s half reactions electron balancing and
on phases inerface of metals or of cell membranes formed potential
Latin oxidation meaning add oxygen,Latin reduction meaning oxygen reduce,
Latin potencia meaning might and force is electrochemistry power in volts
http://aris.gusc.lv/BioThermodynamics/ColigatConcOsmosOxRed.pdf
Water circulation attractor evaporation-condensation Roul's I law
Relative vapor depression Δp/p ̊ is equal to non electrolyte solute x
mol fraction N concentration
evapo ration
conde nsation
inter face
H2Oliquid H2Ogas
(a) p ̊ > pH2O (b)
=
p°- H O2
p
p°
p
p°
3rd page :http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
solute x water solublecompound nx
number of moles
mol fraction Nx concentration
H2O waternH2O
number of moles
= Nx=n x
H O2
n + n x
p
p°
Relative vapor depression Δp/p ̊ is equal to non electrolyte solute x mol fraction Nx
C6H12O6 1 M water solution of glucose mol fraction is calculated from expression
= NxC6H12O6=
One liter contains glucose nC6H12O6= 1 mol.
Molar mass is MC6H12O6=6C+12H+6O=6*12+12*1+6*16=180 g•mol-1.
nC6H12
H O2
n+
O6
nC6H12 O6
p
p°
3rd page:http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
Water circulation attractor evaporation-condensation Roul's I law
If density ρ = 1.0 g•cm-3 mass of water mH2O=1000g – 180g=820g .
Molar mass MH2O=2H+O=2*1+16=18 g•mol-1.
Water Number of moles nH2O= = =45.5(5) mol .
Solute x glucose mol fraction NxC6H12O6 concentration is
NxC6H12O6= = =0.0215=nC6H12
H O2
n+
O6
nC6H12 O6
H O2
M
H O2
m
1molg18
g820
mol1mol5.45
mol1
p
p°
ISOTONIC COEFFICIENT
Isotonic coefficient i (or Vant Hoff’s coefficient) is the proportionality factorbetween the total concentration in to water dissolved solute molecules and
concentration of water dissolved particles.
Cparticles= i*Ctotal
Swante Arrenius, Wilhelms Ostwalds in Riga during year 1886.recovering acid, base and salt propereties in
dissociation degree α principlesfor strong (α =>1) and weak (α =>0) electrolytes:
α = ; C = α * C .Cdiss
1st page :http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
α = ; Cdiss= α * Ctotal .where α is dissociation degree ,
Ctotal is total concentration of molecules andCdiss is dissociated molecule concentration .
So isotonic coefficient calculates as : i = 1 + α (m–1) ,where m is the number of formed ions:
electrolyte dissociationFeCl3 => Fe3+ + Cl–+ Cl–+ Cl– (1 + 1+ 1+ 1) = 4 = m ions
Ctotaldiss
Fe3+
ClCl
Cl
Fe3+
Cl Cl Cldissociation
+ + +formed number of ions in products
Solubility products as strong and weak electrolytes THERMODYNAMICS
O:O
O
O
Na+
:O
:OH
H
H
H H
H
H
H
HH
H
H
hexagonal
H
H
OOH
H
O HH
OHH
OH
H
Cl
OH
H
hexagonal
6th page : http://aris.gusc.lv/BioThermodynamics/CO2O2Thermodynamic15A.pdfCl
ClCl
Cl
Cl
Cl
Cl
ClCl
Cl
ClCl
Cl
Na+
Na+
Na+
Na+
Na+
Na+
Na+ Na
+
Na+
Na+
Na+
Na+
Na+
Na+
1) crystalline Na+Cl- =>Na+aqua+Cl- aqua
Solubility 36 g/100g H2O, demsity 1,23 g/mL, CNaCl= 5,571039 mol/L ; w%=26,5%;
2) hydration with six water molecules .
Hess solubility free energy change: ΔGHess = ΔHHess–TΔSHess negative, exoergic, favored;ΔGHess = 3,82•1000 – 298,15•43,5 = -9150 J/mol = -9,15 kJ/mol;
Dissociation degree α=1,as Na+Cl- dispersed in separate ionos.
The ionic crystalline solubility and dissociation of electrolyte solution (4.1)
Heat accumulates in productes endothermic
cooling ΔHHess= -240.1-167.2+411.12=+3.82 kJ/mol.
Solid compound mol fraction is one [Na+Cl-]solid=1 and solubility constant Keq = e-ΔGeq/RT. (4.3)is ions solubility product K = K =[Na+ ]*[ Cl-- ]=5,43516* 5,43516=31,0365=101,492; (4.2)
Strong electrolytes ΔGr<0 negative exoergic, favored and Keq>1are water soluble salt, bases and strong acids.
Weak electrolytes ΔGr>0 positiveendoergic, unfavoredare water insoluble salts, basesbut weak acids are water soluble.
Conclusions: Na+Cl- salt solubility exoergic, favored and Ksp=Keq=31,036 >1
unfavored
Note: human bodyconstitute onlyweak acids whichare water soluble.
is ions solubility product Ksp= Keq=[Na+aqua]*[ Cl--aqua]=5,43516* 5,43516=31,0365=101,492; (4.2)
ΔGsp= -R•T•ln(Ksp)=-8,3144•298,15•ln(31,0365)= -8,51556 kJ/mol,
0% 50% 100%
A 50% B+CNa+Cl-ciets izejvielaproduktiNa+
aq+Cl-aq
favored
Reaching solubility constant Ksp= Keq=[Na+aqua]*[ Cl--aqua]=31,036 for mixture free energy change
minimised 8,516 kJ/mol =ΔGeq2<ΔGHess2=9,15 kJ/mol . Absolute value decreases.
Keq<1
Phase diagram water evaporation condensation3rd page: http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
water
0 ° 100 °
water
ice
vapor
Freezing point depression II Roul’s law states: Δtfreezing= i KcrCm
Kcr =1.86 is the
Molal non-electrolyte (i = 1)
solution concentration Cm shows
mole number of solute
present in 1000 grams of solution.
4th page:http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
0 °
cryoscopy constant of the water(Greek kryos is freezing)
Cryoscopy constant of water 1.86 shows the freezing point depression in a 1 molalnon-electrolyte solution (where i = 1)
which freezes at temperature –1,86°C less zero 0 ̊ C.
180 grams of glucose dissoluted in water 820 grams has freezing point –1,86 ̊ C
Boiling point raise II Roul’s law states: Δtboiling= i Keb Cm
Keb =0.52 is the
non-electrolyte (where i = 1)
solution concentration Cm shows
mole number of solutepresent in 1000 grams of solution.
4th page:http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
100 °
ebullioscopy constant of the water(Greek ebulios means boiling)
Ebullioscopy constant of water 0.52 shows the boiling point raise in a 1 molal nonelectrolyte solution (where i = 1)
which boils at temperature 100,52 ̊ C over 100 ̊ C.
180 grams of glucose dissoluted in water 820 grams has boiling point 100,52 ̊ C
Prigogine attractor in osmosis of aquaporins is concentration gradient
Osmosis organise through an membrane aquaporins pressure by flux of H2O andO2aqua against osmolar concentration gradient ΔCosm across membrane
Pressure π=ΔCosmRT (kPa=J/L) shows free energy amount the Joules in 1 liter volumeof cells, where R=8,3144 J/(mol•K) universal gas constant,
T temperature in Kelvin’s degree T=t ̊+273.15=37 ̊+273.15=310.15 K.
Osmosis is water flow right side against concentration gradient 0<ΔCosm. , as Na+Cl-
ions close flow of water to left side and make osmo molar concentration gradient
Na+Cl- =Na++Cl- electrolyte dissociation α =1 double pressure on cell membrane, asosmolarity ΔCosm=2CM ; i=1+α(m–1)=1+1(2-1)=2; π= iCM RT = 2CM RT = ΔCosm RT .
6th page: http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
O
H
H
O
H
H
OOOO
Cl
Na+
aquaporines
membrane
membrane
+
gradient
Cright- Cleft =CNa++CCl- - CNaCl = iCM - 0 = 2CM- 0 = ΔCosm :
left side zero
CNaCl=0
osm
ions close flow of water to left side and make osmo molar concentration gradient
<= pressure π= ΔCosm RT , (kPa)<= pressure on cell membrane withright side ions Na+, Cl- summaryconcentration make gradient ΔCosm
CNa+ + CCl- = Cosm = Cright = ΔCosm
right side of membraneCright = Cosm= iCM :
Note:Water rate 3•109 sec-1 through membrane aquaporins in erythrocytes both directionstransfer 3000 oxygen molecules per second.Blood Cosm =0,305 M, beet Cosm =0,1 M, alveola epithelia Cosm from 0,25 M to 0,20 M
Na+Cl- measured values of α are smaller than 1 - they often are around α = 0.8-0.9.
For medical application of 0.305 M isotonic solution osmo molar concentrationis necessary to keep constant 0.305 M.
Total ionic strength I = μ = α 1/2 Σ Cizi2 of salts in to solution should be evaluated
by real α=0.8-0.9 for maintenance constant osmo molar concentration 0.305 M.
For 0.01 M solution of Na2SO4 evaluated ionic strength I = μ = α 1/2 Σ Cizi2 calculated
from electrolyte dissociation stoichiometry Na SO => 2 Na+ + SO 2- are m=3 ions
Strong electrolyte ions concentration sum as ionic forceis total electrolyte ions stoichiometry sum half for concentration Ci times
ion charge exponent zi2 times dissociation degree α
I=μ= α 1/2Σ Cizi2
2nd page :http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
from electrolyte dissociation stoichiometry Na2SO4 => 2 Na+ + SO42- are m=3 ions
Stoichiometry, molarity of total ions concentration if α =1 :
[Na+]= 2•0.01 M= 0.02 M, [SO42-] = 0.01 M .
Electrolyte Na2SO4 ionic strength is sum: 2•0.01 M + 0.01 M=0.03 M as
μ = 1 ½(12•0.02+22•0.01)= ½(1•0.02+4•0.01)= ½ (0.02+0.04) = ½ (0.06) = 0.03 M
total ions stoichiometry molarity concentration sum 0.03 M (2+1=3; 2Na++1SO42- ).
Note: (CRC Handbook of Chemistry and Physics, 2010, biochemistry conditions),constant water concentration [H2O]=55,3 M
Ionic force cytosolic I = μ from 0,20 M to 0,25 M, in plants from 0,05 M to 0,10 M .
Human blood osmo molar concentration sum of all solutes:Cosm = i1•C1 + i2•C2 + i3•C3 + .... = Σ ik•Ck = 0,305 M,Glucose, salts, amino acids, proteins, bicarbonate etc.
Hypertonic, isotonic, hypotonic H2O, O2aqua osmosis - movementagainst osmo molar concentration gradient across cell membranes
CHyperton >= 0,4 M ;
Hypertonic solution Isotonic medium
Cblood= 0.305 M
8th page :http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
Hypotonic medium distilled water 0 Mas well concentration CHypoton<=0,2 Mis hypotonic to osmo molar 0.305 M.
Hypertonic salt solutions to apply forpurulent wounds, because pumps outwater with toxic compounds andstimulates blood circulation.
Hypotonic medium the flow of water isgreater towards the cell (as theconcentration of solutes in the cell is higherthan outside), the cell puffs up until itsmembrane is broken.Note:
Transfer water molecule through membrane aquaporin tunnel in erythrocyteswith rate 3•109 sec-1 in both directions transfer 3000 oxygen molecules in second.
8th and 9th page: http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
Human blood osmo molar concentration sum of all solutes:Cosm = i1•C1 + i2•C2 + i3•C3 + .... = Σ ik•Ck = 0,305 M,Glucose, salts, amino acids, proteins, bicarbonate etc.
H2O, O2aqua osmosis through cell membrane aquaporinsdrive oxygen transport in living organisms
OH
H
O
H
H
OO OO
H
CO OH
+O
H
H
H+
OH
H H
aquaporines
channels
membrane
+
proton
+ bicarbonate
gas
HypotonC = 0.2 MosmC = 0.305 M
<=alveolar epithelial surface∆Cosm=0.105 M
∆Cosm=0.305–0.2=0.105 M
as CHypoton=0,2 M;Alveolar cells not broken, because collagenelastic frame the cells against π=270 kPa
like as plant cells cellulose frame.
Erythrocyteis broken
http://aris.gusc.lv/ChemFiles/Aquaporins/WCPsAQPsIUBMBlife09/AQP0-11.pdf
Osmosis is H2O and O2 flow against gradient of concentration ΔCosm=0.105 M ,with energy π=(Cosm-CHypoton)RT= ΔCosm RT=0,105*8,3144*310=270 J/L.
O
H
H
OC
OOH
H
H O 3
OC
OO
HC
O
H
HO
H
H
++ +
bicarbonatechannels gas
membrane
like as plant cells cellulose frame.Plant root cells osmolar concentration
Cosm= 0,1 M resist pressure π=255 kPa.
Venous deoxy HbT shuttle adsorbs four oxygen 4O2 molecules releasing 4H+, 4HCO3-
during one blood circulation cycle amounts 459*6•10–5 M=0,0275 M=[HCO3-]=[H+]
O2Solutions.pdf, which shifts equilibrium to right H+ +HCO3-+ Q↔H2O +CO2gas via membrane
channels. Physiologic pH=7,36 stabilises as removed CO2 gas right sidethrough respiration forming Chypoton=0.2 M .
8th and 9th page: http://aris.gusc.lv/BioThermodynamics/ColigativeProperties.pdf
Cosm = i1•C1 + i2•C2 + i3•C3 + .... = Σ ik•Ck,
Cosm = = =0,305 M
Isotonic osmo molar blood concentration
Biological Human liquid measurement(consider blood, sweat, saliva, tear, urine,
etc.).
cr
frezing
K
t
86.1
567.0
Osmosis drive H2O, O2aqua across cell membrane aquaporin channels
Glycolysis and Krebs cycle oxidative phosphorilation :increases cell and mitochondria osmolar concentrationfrom 1 glucose molecule to 12, forming difference 11 molecules = ΔCosm , drivingosmosis of , shift oxidative phosphorilation to produce increase ΔCosm.
By H2O, O2aqua osmosis driven exoergic and exothermic process into cellorganism maintain the body temperature with supplied heat amount Q.
Green plants Photosynthesis reaction alone:thermodynamic forbidden but joint in tandem withPhotosynthesis enzyme complexes synthesises products , decreasesconcentration from 12 molecules to 1 in chlorophyl tylakoids, forms difference11 = ΔC , driving osmosis out . At the same time creates
10th page :http://aris.gusc.lv/BioThermodynamics/ColigativePropertiesL.pdf
ΔGreact= -2570,4 kJ/mol ; ΔHreact= -2805,27 kJ/mol
exoergic exothermic
C6H12O6+ 6O2aqua+6H2O=>6HCO3-+6H3O
++ΔGreact +Q
6O2aqua+6H2O
ΔGreakc= +2570,4 kJ/mol ; ΔHreakc= +2805,27 kJ/mol
endoergic endothermic
6HCO3-+6H3O
++ΔGreakc +Q => C6H12O6+ 6O2aqua+6H2O
C6H12O6+ 6O2aqua+6H2O
6HCO3-+6H3O
+
11 = ΔCosm , driving osmosis out for synthesised products. At the same time createsconcentration gradient increasing influx in tylakoid from environmentaccording Le Chatelier principle shift equilibrium to products .Photosynthesis drive H2O, O2aqua osmosis process endoergic and endothermic coolingserrouding in summer time but in winter stop the process.
O
H
H
O
H
H
OOOO
O
H+
H O 3OC
OOH
H+
OH
H
C
aquaporins
membrane
membrane
|-20Å-| gradientsC H6 12 6
+
+
membrane
channels
+
gradients
Krebscycle
Glycolysis-
6O2aqua+6H2O
6HCO3-+6H3O
+
C6H12O6+ 6O2aqua+6H2O
O
H
H
O
H
H
O O O O
H+
OC
OO H
H+O
H
H
O
H OC 3
aquaporins
membrane
membrane
|-20Å-|gradients
+
+
membrane
channels
+
gradientsplantsenvironment
Photosynthesis
C H6 12 6
Type of electrodes:
free electrons transfer Type I,
free electrons transfer Type II
free electrons transfer Red-Ox electrode
Ionic transfer through channels in Membrane electrode
Product energy G2 minus initial metal freeenergy G1 is free Gibbs energy change
ΔG ̊ = G2 - G1
Nernst’s metallic electrode potential expression formationNernst’s awarded Nobel Prize 1920
free electrons in metal n e-
nFE = RT ln Keq
E= ln KeqnF
RT
Keq = = ٠[e-]n]d[Re
]e[]Ox[ n
]d[Re
]Ox[
E ̊ = •log [e-]n
Electric charge work equal to electrochemicalMen+ transfer work from metal to solution
half reaction Red Oxn+ + n e- ;metal free electrons n e- surface molfraction [e-]n is constant and standard
potential calculates E ̊ = log [e-]n constant
2nd page: http://aris.gusc.lv/BioThermodynamics/ElektrodsAM.pdf
nF
RT
W = qE = nFE = - ΔG ̊ = Wwork = nFE = RTlnKeq,where F=96485 Coulombs Faraday constant
E = E ̊ + •log n
0591.0
]d[Re
]Ox[
]d[Re ]d[Re
n
0591.0
E = log [e-]n + log
]d[Re
]Ox[n
0591.0
n
0591.0
Half reaction metallic hydrogen (Pt)H, water and hydronium ion H3O+ :
Nernst’s equation
Metal interface / to its cation solution classic zero hydrogen electrode
(Pt)H + H2O H3O+ + e- ; equilibrium constant K = [H3O
+]/[H2O]
EH=E ̊H+0,0591•log([H3O+]/[H2O])= 0,103 V+0,0591•log([H3O
+]/[H2O])
saturated platinum (Pt)H sheet represents metallic hydrogenelectrode immersed in hydronium ions H3O
+ acid solutions.
EoH =E ̊H+0,0591•log(1/[H2O])= 0 is classic hydrogen scale.
In hydrogen potential scale reference point zero incorporatewater constant concentration [H O]=55,3 M,
3rd page: http://aris.gusc.lv/BioThermodynamics/ElectrodsAM.pdf
water constant concentration [H2O]=55,3 M,which converts thermodynamic E ̊H= 0,103 V to classic Eo=0 V.
Standard potentials given half reactions values in volts refers to thermodynamic
hydrogen potential including water real concentration [H2O]=55,3 M application and in
brackets like 0 V classic values after. These absolute
values from brackets 0 V are unusable in thermodynamic equilibrium calculations.
OH
H O
E,V(0.00 V)
(E = -0.0591*pH)
0.103 V
H/H+ E =E° + 0.0591*log( )(Pt)H/H+
[ ]2
[ ]+3
H
Oxidation – Reduction half reaction (RED-OX system Table)
Half reaction shows two states of compound at equilibriumby gaining electrons for oxidized changes to reduced form and
loosing electrons for reduced changing to oxidized form.Half reactions are present in standard potential tables for all known
studied and complete redox systems.Five columns refers to: 1. Chemical element symbol of responsible atom;
2. oxidized form;3. number of loosing electrons reduced form and gaining electrons to oxidized form;
4. reduced form;5. standard potential E ̊ in volts.
Number of Standard potential
3rd page: http://aris.gusc.lv/BioThermodynamics/OxRedBiologicalW.pdf
Element Oxidized formNumber ofelectrons e- Reduced form
Standard potential E ̊H2O , V(classik Eo standard potential)
HH3O
+ 1 (Pt)H + H2O +0,103 (0,00 Nernsta skala)
O O2(g) + 4 H3O
+ 4 6H2O +1,383 (1,229) SuhotinaH2O2+ 2 H3O
+ 2 4H2O +1,982 (1,776) Suhotina
O2aq + 2 H3O+ 2 H2O2 + 2 H3O +0,7975 (0,6945) Alberta University
Mn (H+) MnO4– + 8 H3O
+ 5 Mn2+ + 12H2O +1,76 (1,51)(H2O) MnO4
– + 2H2O 3 MnO2↓+ 4OH– +0,531 (0,60)(OH-) MnO4
– 1 MnO42– +0,56
Fe Fe3+ 1 Fe2+ +0,77
Silver /silver chloride/chloride ion II-type electrode consists
of silver metal, AgCl precipitate insoluble salt and
K+Cl- solution, containing the counter-ions Cl- of
AgCl insoluble salt half reaction is:
AgCl +e- Ag++ Cl-
Metal/insoluble salt/ion II-type electrode
Nernst’s equation
Eag/AgCl = E ̊AgCl - 0.0591•log [Cl-]
The main application of II-type electrodes is their use as reference electrodes,
because potential value depends only on chloride ion concentration.
Chloride concentration is precise controlled technology for instruments use.
4th page: http://aris.gusc.lv/BioThermodynamics/ElectrodsAM.pdf
Electric potential in volt measurement by couple of electrodesElectric Motion Force
http://aris.gusc.lv/BioThermodynamics/ElektrodsAM.pdf
+
e -
+
+
++++
+ + + + +
VV
- +
EMF
J=0
Voltmeter with minus "-" andplus "+" clamps measuresdifference of potentials
EMF = EI - EII , called
Electric Motion Force
Between two MeI (Indicator) and MeII (Standard)on electric circuit linked electrodes
can be expressed MeI Indicator EI as sum : EI = EMF + EII
Indicator electrode having EI –has reactivity with solution - electrode of investigations,
Standard reference electrode having EII =constant -and has no reactivity with environment into solution.
+
e - MeIMeIIElectric Motion Force
EMF .
Nernst’s potential studies two half reaction balanced electrons sum
Ox oxidising reagent half reaction: MnO4- + 8H3O
+ + 5e- Mn2++12H2O ; E˚MnO4-=-1,76 V
Red reducing reagent half reaction: 5(Pt)H + 5H2O 5H3O+ + 5e- ; E˚H =0,103 V
Balanced electrons Red-Ox reaction: 5(Pt)H + MnO4- + 3H3O
+ Mn2++8H2O ;
5
0591.0
OH
Mn
Mn
H OO[ ]-
[ ].[ ]22+
.[ ]+34
8
12 mol/g_18
L/g_996EMnO4=1.76 V + •lg ; [H2O]=55,3 M=
OH
H O
[ ]2
[ ]+3EH=E ̊ +0,0591•lg = 0,103 V +0,0591•lg
OH
H O
[ ]2
[ ]+3 Viela ΔH˚H,kJ/mol ΔS˚H,J/mol/K ΔG˚H,kJ/mol
H2O -285,85 69,9565 -237,191H2O -286,65 -453,188 -151,549H3O
+ -285,81 -3,854 -213,2746H2(aq) 23,4 -130 99,13MnO4
- -541,4 -191,2 -447,2
Exothermic and exoergic MnO4- reduction by 5(Pt)H
Hess free energy change negativeΔGHess=ΔGOxRed=-1286 kJ/mol , but minimises reachingΔGmin= ΔGeq= -799,38 kJ/mol equilibrium mixture
5th page: http://aris.gusc.lv/BioThermodynamics/ElektrodsAM.pdf
MnO4- -541,4 -191,2 -447,2
Mn2+ –220,8 -73,6 -228,1
GHess=G˚Mn2+8G˚H2O-3G˚H3O-G˚MnO4-5G˚(Pt)H= -1286 kJ/mol
ΔGeq=(E˚H-E˚MnO4-)•F•1•5=(0,103-1,76)*96485*5=-1,863*96485*5=-799378,225 J/mol=-799,38 kJ/mol
Keq=exp(-ΔGeq/R/T)=exp(799378,225/8,3144/298,15)= 1,112*10140;
ΔGmin= ΔGeq= -799,38 kJ/mol equilibrium mixture
1,112*10140 =Keq=OH
Mn
Mn
H OO[ ]-
[ ].[ ]22+
.[ ]+34
3
8
[(Pt)H] .5
Prigogine attractor free energy change minimum ΔGmin .Free energy change minimum reaching establishes equilibrium.
ΔGmin=ΔGeq = -799,4 kJ/mol < GHess= -1286 kJ/mol;5A+B+3C 50% D+8E5(Pt)H+MnO4
-+3H3O+
produkti Mn2++8H2O
Ox oxidising reagent half reaction: O2aqua + 4H3O+ + 4e- <=> 6 H2O ;
Red reducing reagent half reaction:: 4(Pt)H + 4H2O 4 H3O+ + 4e- ;
Balanced electrons Red-Ox reaction : 4(Pt)H + O2aqua- 2 H2O ;
OH
H O
[ ]2
[ ]+3EH = E ̊ +0,0591•lg = 0,103 V +0,0591•lg
OH
H O
[ ]2
[ ]+3
Substanc ΔH˚Hess,kJ/mol
ΔS˚Hess,J/mol/K
ΔG˚Hess,kJ/mol
H2O -285,85 69,9565 -237,191H2O -286,65 -453,188 -151,549H3O
+ -285,81 -3,854 -213,2746H2(aq) 23,4 -130 99,13
Exothermic and exoergic O2aqu reduction by 4(Pt)H Hess freeenergy change negative ΔGHess=ΔGOxRed= -689 kJ/mol , butminimises ΔGmin=ΔGeq= -494 kJ/mol reaching equilibrium
5
0591.0OH
H OO2[ ]
[ ]2
.[ ]+3
4
6aqua
E=E ̊O2 + •lg5
0591.0OH
H OO2[ ]
[ ]2
.[ ]+3
4
6aqua
=1.383 V + •lg
Nernst’s potential O2aqu /(Pt)H studies two half reaction balansed electrons sum
OH[ ]2
5th page: http://aris.gusc.lv/BioThermodynamics/ElektrodsAM.pdf
H2(aq) 23,4 -130 99,13O2aqua -11.70 -94,2 16,4O2aqua -11.715 110.876 16,4
mixture ; 3,518*1086 =Keq= ;
Prigogine attractor free energy change minimum ΔGmin .Free energy change minimum reaching establishes equilibrium.
4A+B 50% 2D4(Pt)H+ O2aqu
products 2H2O
ΔGmin=ΔGeq = -494 kJ/mol < GHess= -689 kJ/mol;
ΔGeq=(E˚H-E˚O2)•F•1•4=(0,103-1,383)*96485*4=-1,28*96485*4=-494003,2 J/mol=-494 kJ/mol
Keq=exp(-ΔGeq/R/T)=exp(799378,225/8,3144/298,15)= ; 3,518*1086;
OH
O2
[ ]22
[(Pt)H] .4 aqua[ ]
GHess= 2G˚H2O-4G˚(Pt)HG˚O2= -689 kJ/mol
O2aqua/ H2O red-ox system biochemic mechanism of acidosis andoxidative stress (forced oxidising agent power by potential E increase)
ΔE>0 as oxidative stres acidify and increase O2aqua concentration:1) as ΔE= +0.236 V if increases hydrogen ion concentration [H3O
+] 10 times acidosis;2) ΔE>0 increase about ΔE= +0.01 V if increases oxygen concentration 5 times ;
3) water concentration [H2O] = 55.3 M diminishing oxidative stress risk potential aboutΔE= - 0.154 V from thermodynamic E ̊=1,383 V to classic Eo=1,229 V.
6th page: http://aris.gusc.lv/BioThermodynamics/ElektrodsAM.pdf
Note: Oxidative stress causes chaos of non enzymatic oxidation in multiple radical-chain reactions and parallel products!
O2aqua + 4 H3O+ + 4 e- 6 H2O ; E ̊ = 1,383 V
oxidised form free electrons reduced form
Oxygen solubility Prigogine attractor free energy change minimum as Hess law isendoergic change positive unfavored :endoergic change positive unfavored :
but minimises reaching equilibrium mixture ΔGmin=ΔGeq=26,58 kJ/mol
GHess=G ̊H2O+G ̊O2aqua-G ̊H2O-G ̊O2gas=16,4 -(-61,166)= 77,57 kJ/mol ,
Keq =O
O
OH[ ]2 air
[ ]2 aqua
[ ]2
. =2,205*10-5= 10-4,66;
ΔGeq=-R•T•ln(Keq)=-8,3144*298,15*ln(2,2*10-5)=26,58 kJ/molA+B 50% C+Dreactants O2↑air+H2Oproducts O2aqua-Asinis
Free energy change minimisation ΔGmin is Prigogine attractor.
ΔGmin=ΔGeq =26,58 kJ/mol < GHess= 77,57 kJ/mol ;
O2air + H2O <=> O2aqua ;
[O2air]=0,2095oxygen mol fraction
Physiologic equilibrium constant is KO2blood=[O2aqua]/[O2air]= 9,768·10-5/0,2095=4,663*10-4=10-3,3314.Arterial [O2aqua]=6·10-5 M and venous [O2aqua]=1,85·10-5 M concentration determines pKO2blood=3,33.
Nernst’s potential studies reducing with vitamin B3 ethanal H3CCH=Oand oxidising H3CCH2OH ethanol
8th page: http://aris.gusc.lv/BioThermodynamics/ElektrodsAM.pdf
ΔE˚ =E˚2H2O-E˚1=0,2415-(-0,113)= 0.3545 V, ΔGeq=ΔE˚•F•n=0.3545•2•96485= 68,408 kJ/mol
Aerobic H3C-CH2-OH+NAD++H2O+ΔG+Q => H3C-CH=O+NADH+H3O+
Ox NAD+ + H-(2e-) <=> NADH ; E˚1 = -0,113 VRed CH3CH2OH+2H2O<=>CH3CHO+2H3O
++H(2e-); E˚2H2O=0,2415 V=0,190+0,0591/2*log([H2O]2)
OHOH
H O
CH3CH
2
OHCH3C
.[NAD ]
[NADH].[ ]2
+ [ ]
.[ ]+3[ ].
TR
Geq
e
15.298314.8
68408
e ΔGeq= -R•T•ln(Keq); Keq= = = =1,036•10-12=10-11,985
GHess=G˚H3O+G˚CH3CHO+G˚NADH-G˚CH3CH2OH-G˚H2O-G˚NAD+=159,1 kJ/mol;
anaerobic
aerobic
Prigogine attractor free energy change absoluteminimum ΔGmin reachable at inverse equilibria points:
anaerobic
A+B+C 50% D+E+FNAD++H3CCH2OH+H2ONADH+H3CCHO+H3O
+
Anaerobic H3C-CH=O+NADH+H3O++ΔG+Q =>H3C-CH2-OH+NAD++H2O;
GHess=G˚CH3CH2OH+G˚H2O+G˚NAD+-G˚H3O-G˚CH3CHO-G˚NADH= -159,1 kJ/mol;
Red NADH <=> NAD+ + H-(2e-); E˚1 = -0,113 VOx CH3CHO+2H3O
++H(2e-)<=>CH3CH2OH+2H2O; E˚2H2O=0,2415 V
ΔE˚=E˚1-E˚2H2O=-0,113 -0,2415=-0.3545 V, ΔGeq=ΔE˚•F•n=0.3545 V•2 mol•96485 C/mol= -68,408 kJ/mol
OHOH
H O
CH3CH2
OHCH3C
.[NAD ][NADH]
. [ ]2+ [ ]
.[ ]+3[ ]. TR
Geq
e
15.298314.8
68408
e
ΔGeq= -R•T•ln(Keq); Keq= = = =9,65•1011=1011,985
anaerobic
D+E+F 50% A+B+CNADH+H3CCHO+H3O
+
NAD++H3CCH2OH+H2O
minimum ΔGmin reachable at inverse equilibria points:
ΔGmin=68,4....... kJ/mol=ΔGeq<ΔGHess= 159........kJ/mol.
10-11,985 = Keq_aerobi; Keq_anaerobi=1011,985;Keq_aerobi=
anaerobieqK _
1
H+
H+
H+
H+
H+
H+
H+
H+
H+
membrane
channels
membranepH=5
pH=7.36
mitochondria extramitochondria
space
Membrane potentials for hydrogen ions concentration gradient as H+ protons movementin to channel based on mitochondria pH=7.36 and extra mitochondria space pH=5
Wwork=qE=nFE=ΔGr=RTlnKeq;chondriaextra_mito]H[
http://aris.gusc.lv/BioThermodynamics/MembraneElektrodsAM.pdf
Wwork=qE=nFE=ΔGr=RTlnKeq;n charge of ion; Keq =
nF
RT
iamitochondr
chondriaextra_mito
]H[
]H[Emembrane = *ln = *log
EH+ =Plg = 0.06154V*lg = 0.06154V*log(102.36) = 0,14523V
F
T·R)·10ln(C96485
K15.310)·K/mol/J(3144.8·3,2 P= = =0.06154 V
iamitochondr
chondriaextra_mito
]H[
]H[
iamitochondr
chondriaextra_mito
]H[
]H[
pH
pH
10
10
iamitochondr
iamitochondrextra
36.7
5
10
10
1
06154,0
Membrane potentials for ions H3O+ concentration gradient in mitochondria
[H3O+
Mitochon]=10-7,36 M and extra inter membrane space [H3O+
extraMi]= 10-5 M
H+
H+
O
H
H
H+
O
H
H
H+
H+
O
H
H
H+
O
H
H
membrane
channels
membrane
MitochondriaextraMitochondrialmembranespace pH=5
pH=7,36
[ ]-
Emembrane = 0.06154*log( )
EH3O+Mitochon=-Plg( )= -0,06154V*log = 0,14523 V
F
T·R)·10ln(P= = =0.06154 V
[ ]O3-
extraMi
[ ]O3-
mitochondria
H
H
C96485
K15.310)·Kmol
J(3144.8)·10ln(
3rd page: http://aris.gusc.lv/BioThermodynamics/MembraneElektrodsAM.pdf
[ ]O3-
extraMi
[ ]O3-
mitochondria
H
H
36,7
5
10
10
Membrane potentials for ions HCO3- concentration gradient
in mitochondria [HCO3-Mitochon]=0.0338919 M and cytosol [HCO3
-cytosol]= 0.0154 M
H C O
H
OC
OO
H
OC
OO
H C O
HO
C OO
HO
C O
O
3
membrane
channels
membrane
3Mitochondria
cytosol
[ ]=0,0154 MHCO3-
[ ]=0,03389 MO3-HC
[ ]-Emembrane = 0.06154/(-1)*log( )
EHCO3-Mitochon,=-Plg( )= -0,06154V*log( ) = 0,0210821 V
F
T·R)·10ln(P= = =0.06154 V
[ ]HCO3-
cytosol
[ ]O3-
mitochondriaHC
[ ]HCO3-
cytosol
[ ]O3-
mitochondriaHC 0338919.0
0154.0
C96485
K15.310)·Kmol
J(3144.8)·10ln(
3rd page: http://aris.gusc.lv/BioThermodynamics/MembraneElektrodsAM.pdf
Hydrogen and bicarbonate total membrane potential sum is :EH3O+Mitochon,+EHCO3-Mitochon=0,14523V+0.0210821V = Emembr=0,1663V
Electric free energy change for H+ :ΔG=-Emembr•F•n=-0,1663*96485*+1=-16,045 kJ/mol
Free energy change for concentration gradient driven through proton H+ channelscrossing lipid bilayer membranes:
3rd page: http://aris.gusc.lv/BioThermodynamics/MembraneElektrodsLat.pdf
The proton H+ concentration gradient sum with electrochemical free energy change:ΔG= ΔGmembr+Gkanāls = -16,045 kJ/mol+ -14,013 kJ/mol = -30,058 kJ/mol
drive ATPase nano engine to synthesizing ATP molecules.
GH+=RTln([H3O+]extraMit/[H3O
+]Mitohon) =8,3144*310,15*ln(10-5/10-7,36) = -14,013 kJ/mol
Both free energy negative changes sum, consuming four protons 4 H+,4*-30,058 kJ/mol = -120,232 kJ/mol
drive ATPase nano engine rotation to synthesizing one ATP mole 503 grams.
drive ATPase nano engine to synthesizing ATP molecules.
Macro ergic ATP phosphate anhydride bond in human erythrocyte hydrolyze releasesΔG = -55,16 kJ/mol free energy. ATP accumulated chemical free energy efficiency 45,9 %of theoretically 100% (-120.2 kJ/mol) . Oxidative phosphorylation at least 54,1 % of used
four proton transfer energy consumes the friction of ATPase rotor to heat production andATP movement in cytosol water medium forming the concentration gradients across lipid
bilayer membranes as transportation free energy source to drive ATP molecules.
http://aris.gusc.lv/BioThermodynamics/BioThermodynamics.pdf. (19th. lpage)
Kin= Glass membrane electrode Kout= [H+out]
↓↓↓↓↓↓↓↓↓
Kmembrane=Kin*Kout=
Innerconcentration
is constant[H+ ]=const
Emembrane=0.0591/(+1)*log ;
Emembrane= 0.0591*log +0.0591*log([H+out]) ;
[H+in]=const ; Econst= 0.0591*log ;
]H[
]H[
in
out
]H[
]H[
in
out
]H[
1
in
]H[
1
]H[
1
in
H+in + SiO−
3−SiO2 HSiO3−SiO2 /////SiO2//// SiO2−SiO3H SiO2−SiO−3+ H+
out
[H+in]=const
[H in]=const ; Econst= 0.0591*log ;
Emembrane= Econst +0.0591*log([H+out]) ;
as pH= -log([H+out])
Eglass= Econst -0.0591*pH .
Glass electrode potential is
proportional to pH of solution.
]H[ in
7th, 8th pages: http://aris.gusc.lv/BioThermodynamics/ElektrodsAM.pdf