PETER PAZMANY SEMMELWEIS UNIVERSITYCATHOLIC UNIVERSITY Development of Complex Curricula for Molecular Bionics and Infobionics Programs within a consortial* framework** Consortium leader PETER PAZMANY CATHOLIC UNIVERSITY Consortium members SEMMELWEIS UNIVERSITY, DIALOG CAMPUS PUBLISHER The Project has been realised with the support of the European Union and has been co-financed by the European Social Fund *** **Molekuláris bionika és Infobionika Szakok tananyagának komplex fejlesztése konzorciumi keretben * * * A p ro j e kt a z Eu ró p a i U n i ó t á m o g a t á sá v a l , a z E u r ó p a i S z o c i á l i s A l a p t á rsf i n a n sz í ro z á sá v a l v a l ó su l m e g . INTRODUCTION TO BIOPHYSICS (Bevezetés a biofizikába) THERMODYNAMICS OF SOLUTIONS (Oldatok termodinamikája) GYÖRFFY DÁNIEL, ZÁVODSZKY PÉTER . Reactions in solutions are described bysomewhat different laws than reactions of gases because molecules of solvent have asignificant effect on the kinetics of reactions . Although the number of collisions is the samein solutions as is gas phase, the rate ofreactions can be very different . Depending on the ratio of the rate oftransformation following a collision to the rateof diffusion we can deviate two types ofsolution phase reactions . For the components of a solution, partial molarquantities can be defined which show how thegiven quantity changes when an infinitesimalamount of a component is added to thesolution . Because of its importance, the partial molarfree energy is called chemical potential . Chemical potential tends to balance by flowingmaterial . In solutions, there is a relationship betweenthe mole fraction of a component and itschemical potential . For ideal (sufficiently diluted) solutions, thisrelation is linear . Raoult's law states a proportionality betweenthe mole fraction and the vapour pressure of acomponent of a solution . For some diluted but not ideal solutions, this relationship is described by Henry's rather than Raoult's law . In a gas, significant interactions occur onlybetween particles taking part in the chemicalreaction . In solutions, molecules of solvent also affect the kinetics of reaction so we have to take into account collisions of the reactant molecules with the solvent molecules . The solvent molecules form a 'cage' around the reactant molecules . For small solvent molecules such as water molecules, the reactant molecule collides about 200 times with the solvent cage before itdiffuses a distance corresponding to its diameter Introduction to biophysics: Thermodynamics of solutions Reactant in a solvent milieu . Eventually, two reactant molecules, A and B,should diffuse together and occupy the samesolvent cage . Then A and B will collide about 200 times in this one 'encounter' before they move awayfrom each other . This is in contrast to the gas phase reactionwhere A and B will collide only once beforemoving apart . Now, let us compare the collisions in the gas and solution phases over time . Each slash in the following figure represents acollision between A and B . In gas phase, randomly uniform collisionsoccur while in solution phase collisions takeplace as 'packets' which are called'encounters' Comparison of the collision distributions of gas and solution phase reactions . Over time, the total number of collisions for the gas and solution phase reactions are aboutthe same . It does however change the pattern of distribution of collisions Introduction to biophysics: Thermodynamics of solutions . Naturally, we are interested in how thepresence of solvent affects the reaction rate . For an improbable reaction – where theactivation energy is large – which occurs onlyonce in about 106 collisions, there is no difference, i.e. the reaction goes at the samerate in the solution phase as in the gas phase . For a reaction with low activation energy, which occurs about once in every 10 collisions, the reaction in the solution phase will occurevery time the molecules encounter eachother . Formally, let us consider a reaction betweenthe reactants A and B A.B . AB . P . The velocity of formation of an encountercomplex, which can be assumed to be a first-order reaction with respect to both A and B v=k1 [ A][ B] . Having formed the complex, the reactants canmove away from each other or the reaction can occur AB . A.Bv=k-1 [ AB] or AB . Pv=k2 [ AB] where P is the product of the reaction . The concentration change rate of the encounter complex is d [ AB ] =k1 [ A][ B]-k-1 [ AB]-k2 [ AB ] dt . Using the steady-state approximation, theconcentration of the encounter complex is obtained as k1 [ A][ B][ AB]= k-1 .k2 . Thus the rate law for the formation of the product is d [ P ] k2 k1 .k2 [ AB]. dt k 2 .k-1 . The case when k-1 << k2, i.e. the rate of formation of the product is far larger than the disintegration of the encounter complex, corresponds to a law activation energy . These reactions are called diffusion controlled reactions . In contrast, the case when k2 >> k-1, i.e. the rate of formation of the product is far smallerthan the disintegration of the encounter complex, corresponds to a high activation energy . These reactions are called energy or activation controlled reactions . In every packet of collisions, which maycontain about 100-200 collisions, there will certainly be one with sufficient energy andcorrect orientation . For example, in an enzymatic reaction, theenzyme and the substrate will find each other and collide many times while rotating withrespect to each other until they have thecorrect orientation and the reaction occurs . This type of reaction occurs very fast . Enzymatic reactions are limited only by how fast the reactants can diffuse towards each other, and collide . To formalize the velocity of diffusion controlledreactions, let us consider a spherically symmetric system where the centre is occupied by a particleA and this particle is surrounded by B particlesaccording to some distribution . Let [B](r) denote the concentration of particleB at the distance r . Since we assume that every collision leads to areaction, the concentration of B at rAB is [ B].r AB .=0 where rAB is the sum of the radii of A and B . At infinite distance, the concentration of B approaches the value of bulk concentration ofB lim [ B ]. r.=[ B ] r .. . Since the concentration of B continuouslydecreases as we approach A, there is a flux ofB towards A . This flux can be described by the Fick's first law: d [ B].r .J =4. r2 DAB dr where J is the flux, i.e. the amount of substances crossing a spherical unit area centred around A inunit time, DAB=DA+DB is the relative diffusion coefficient of A and B, r is the distance between A and B, and d[B](r)/dr is the concentration gradient of B (1829-1901) . To get the concentration of B as a function of r, we need to integrate the equation above from r to . [ B]... . J . d [ B ]. r.=. dr [ B].r . r 4 . r2 DAB . Thus J [ B].r .=[ B ]- 4 . r DAB . Since [ B].r AB .=0 the flux is J =4. r AB DAB[ B ] . Substituting the expression obtained for theflux into the equation for [B](r) we get [ B].r .=[ B ].1- rAB . r which is the distribution of concentration of B as a function of the distance from A . Since the particle B which crosses the spherical surface with radius rAB immediately reacts, we can calculate the velocity of thereaction based on the flux at rAB . The total velocity of the reaction is v=J [ A] . And since v=k [ A][ B ] the rate constant is v J k= ==4 . rAB D AB [ A][ B][ B] particle particles around an A particle Chemical potential . Let us consider two containers containing solutions with different concentrations . If they are mixed, we observe that the concentration becomes uniform . We know that the equilibrium of a systemwhere the temperature, the pressure and theamounts of the components are constant ischaracterized by the minimum of the Gibbsfree energy Mixing of solutions . Based on these observations and knowledge itis obvious that the Gibbs free energy of asolution is a function of the amounts of components . We know that the free energy is G=U -TS . According to the first law of thermodynamicsthe differential of internal energy is dU =. q.. w . The work in our simple case contains a termrelated to the volume and a term related to the amount of material of components . w=. .i dni- pdV i . Substituting the term describing the change ofinternal energy into the equation describingthe change of Gibbs free energy, we get dG=-SdT - pdV ...i dni i . If we consider dG as a total derivative, the expression above has the following form .G . G .G dG=dT .dV .. dni ..T ... V ... ni . p,n j T,n j i p,T,nj .ni . It is obvious based on the equations abovethat .G.i = .. dni . p,T ,n j. ni where µi is the partial molar free energy with respect to the substance i. This is called chemicalpotential, and it expresses how the free energyof a solution changes when we add or remove aninfinitesimal amount of i . For simplicity, let us examine a solution of two components . Let us suppose that we have a solution of ureawith free energy G . If we add a little urea (component 1), we change G . Similarly, if we add a little water (component2), we also change G Introduction to biophysics: Thermodynamics of solutions www.itk.ppke.hu . If the temperature and the pressure are constant then the change of the free energy is .G . G dG=dn1.dn2.. n1 .T ,p,n2 .. n2 .T ,p,n1 . This is the fundamental equation of solution thermodynamics Free energy of an urea solution . This result in a surface in 3D space . If you vary n1, you move along a curve whose projection onto the n1, n2 plane is parallel to the n1 axis . The slope of this curve at any point n1 is the partial derivative . G .. n1 .T,p,n2 . Likewise for n2 . . G . . n2 T ,p,n1 . The equation for the differential of the freeenergy can be written in terms of the chemicalpotentials of the two components dG=.1 dn1..2 dn2 . It is worth noting that µ is a function oftemperature, pressure and composition but not of absolute amount . For example, if two beakers contain differentvolumes of the same solution, i.e. n1.n1 ' and n2.n2 ' but n2 / n1 =n2 ' / n1 ' then they have the same composition and thusthe same chemical potential . Now, let us construct a solution, keeping theabove principle in mind . Let us take an infinitesimal amount of urea and an infinitesimal amount of water and mix them . The free energy for this infinitesimal amountof solution is . G=.1 dn1..2 dn2 . Repeat this 106 times and mix all of the little volumes of solutions . Because chemical potentials have remainedconstant, you can get the total free energy, G, of the solution by adding up all of .G's . This allows us to integrate .G over the whole solution .. G=.1.dn1..2.dn2 . Performing this integration we get G=.1 n1..2 n2 which is called the additivity rule . Now let us differentiate the additivity rule, taking the total derivative: dG=.1 dn1..2 dn2.n1 d .1.n2 d .2 . Since we know that dG=.1 dn1..2 dn2 thus n1 d .1.n2 d .2 =0 which is the Gibbs-Duhem equation . The Gibbs-Duhem equation is valuablebecause if you know one chemical potentialyou can calculate the other n1 d .2=-..d .1 n2 . Integrating it .2 .1 n1 .d .2 =-. d .1 .20 .10 n2 we get .1 .2-.02=-. n1 d .1 .10 n2 the integrated Gibbs-Duhem equation where µ20 is an integration constant (1861-1916) . For an ideal solution, by definition .1=.10. RT ln x1 where µ10 is the standard chemical potential and X1 is the mole fraction of component 1 . This is an empirical relationship that is it was obtained by experiments (1830-1901) . Raoult's law or the Law of Dilute Solutions can also be stated as follows: p1 x1 = 0 for x1.1 p1 i.e. the vapour pressure p1 of the solvent is proportional to the mole fraction of it . All real solutions approach this at high dilution . The “real” equation for µ1 is p1 .1=.10. RT ln0 =.01 .RT ln x1 .. p1 since Raoult's law relates the partial pressure ofthe solvent (p1/p10) to the mole fraction x1 of the solvent . This expression can be used to find µ2 using the integrated form of the Gibbs-Duhem equation d .1 d .10 d ln x1 =. RT dx1 dx1 dx1 . From this we get d .1 =RT 1 dx1 x1 Introduction to biophysics: Thermodynamics of solutions dx1 d .1 =RT x1 . Since x1. x2 =1 we can write that dx1 .dx2 =0 and after rearrangement dx1 =-dx2 . The mole fractions of components can bewritten as n1 x1 = n1.n2 and n2 x2 = n1 .n2 . Dividing the mole fraction of component 1 bythe mole fraction of component 2 we get x1 n1 n2 n1 =/= x2 n1.n2 n1 .n2 n2 Duhem equation .1 n1 .2=.02-. d .1 .10 n2 . Let us substitute the expression, we obtainedfor the relationship between mole fractionsand amounts of material into the Gibbs-Duhem equation 0 -. x1 .2 =.2 d .1x2 . Further rearrangements can be performedbased on the expressions above to get dx1.2 =.02-RT . x1 x2 x1 and then dx1.2=.02-RT . x2 . And making use of the equation describing therelationship between the differentials of the two components we get dx 2 .2 =.02.RT . x2 . Performing the integration, the expression willbe .2=.02.RT ln x2 . We have now expressions for µ1 and µ2 in terms of composition . These hold for real solutions only when x1.1 and x2 .0 diluted solutions . Some rules of thumb for real solutions approaching ideal solutions – Non-ionizable solutes (for example glucose) • Ideality up to 0.01-0.1 M – Ionizable solutes (for example NaCl) • Ideality up to . 0.001 M – Proteins • Ideality up to 10-6-10-5 M . Now let us assign numerical values to µ's . Let us set out from Raoult's law .1=.10. RT ln x1 . Let us plot µ1 as a function of ln x1 µ1 vs. ln x1 . In the figure, µ10 represents the chemical potential of pure solvent, which means the change in the free energy, G, with addition of pure water and xsat represents the mole fraction of water at saturation . The purple curve applies to ideal solutions andthe red curve applies to real ones wheresaturation may occur . The ideal law is a limit law . At low concentrations, the two curves coincide . Deviation occurs at higher concentration andthe “real” curve stops at saturation . To get the real curve we should plot µ1 vs. ln (p1/p10) . Likewise for the solute .2=.02.RT ln x2 . Let us plot µ2 vs. ln x2 µ2 vs. ln x2 . µ20 has no physical meaning, it is impossible to make solution at this concentration . The expression for µ2 is valid over the same region (x2.0, x1.1) as the expression for µ1 since µ2 was derived from µ1 . For any system at equilibrium, the chemical potential of an uncharged substance is the same in all phases between which it can pass . It was recognized by William Henry that for some solutions, Raoult's law does not hold . Although the mole fraction of the solute inthese solutions continues to be proportional tothe vapour pressure, the proportionalityconstant is not the vapour pressure but adifferent constant with pressure dimension . So Henry's law is p2 =x2 K 2 where p2 is the vapour pressure of the solute above the solution, x2 is the mole fraction of the solute and K2 is a constant of pressure dimension William Henry (1775-1836) . Since measuring the mole fractions of thecomponents of a solution is difficult butmeasureing the molar concentration is quitesimple, biochemists prefer expressioncontaining molar concentrations rather thanmole fractions . To be able to work with such expressions, wehave to find the relationship between thesetwo quantities . We know that in a simple two-componentsolution: n2x2 = n1 .n2 where x2 is the mole fraction of the solute and n1 and n2 are the amounts of material of the solvent and the solute, respectively . We can divide the numerator and denominator by the same factor so that the value of thefraction does not change, thus n2 /dm3 c2 x2 = n1 / dm3.n2 / dm3 = c1.c2 . Since biochemists usually work with dilute solutions, the concentration of water will remain approximately that of pure water 1000 g·dm-1 cH2 O..55.56 M 18 g·mol-1 . For dilute solutions c2 « c1 where c2 and c2 are the molar concentrations of solute and solvent, respectively . Thus c2 c2 x2 .= c1 55.56 M . Thus, the chemical potential of the solute willbe c2 .2 =.02.RT ln 55.56 M . We can introduce a new constant µ. instead of µ0 ..=.0-RT ln 55.56 . Thus, the expression for the chemical potentialof the solute is c2 .2 =..2 . RT ln 1 M . Now, let us consider again the plot showing thechemical potential of the solute as a functionof its molar concentration . Let us introduce a new quantity such that itmakes the real solution have a straight line onthe plot . This new quantity is called activity and denoted by a .. real .-...ideal . ln a= RT and after rearrangement ..real .=...ideal .. RT ln a . At great dilution .2 . real.=.2 .ideal . and thus a2 =c2 so lim a2 =c2 c2 . 0 . If we plot µ2 vs. ln a2, the real solution gives a straight line . The ideal curve now deviates from linearity athigher concentration µ2 vs. ln a2 . Now, however, there is an additional problem: we still do not know how µ (real) varies with concentration . To solve this problem, we define the activity coefficient, which is denoted by ., such that a2 =.2 c2 . The advantage of introducing activity and activity coefficient is that the laws for real solutions have the same form as the laws for ideal solutions . The activity has a dimension of concentration . Because of the definition of the activity coefficient, it is unitless . Furthermore, since a.c at low concentrations, for the activity coefficient lim .=1 c .0 . . can be obtained from any plot containing anideal and a real curve . RT·ln . is the vertical distance between the two curves at some concentration Calculating the free energy change of a chemical reaction in solution We can relate µ to the change in free energyfor a reaction occurring in solution For the reaction G .. what is .G when 1 mole of A converts to 2 moles of B? A 2 B . We can use the additivity rule to get the total G of the whole solution Gafter=.H2 OnH 2 O..A .nA -1...B. nB.2. and Gbefore=.H2 OnH 2 O..AnA..BnB . Let us note that since µ is a function of concentration, this subtraction to get .G is not straightforward . As the n's change so do the µ's . To resolve this problem, let us assume that weare working with a huge volume of solution.Then allowing 1 mole of A to convert to 2moles of B will not noticeably change thecomposition . Therefore µ's before and after are essentially the same . Alternatively, if we allow only an infinitesimalchange of the amount of A, the compositionwill effectively remain constant .G=Gafter-Gbefore =2 .B-.A . Substituting .A=.0 A. RT ln aA and .B=.0 B. RT ln aB into the expression above, we get .G=2 .0 B-.0 A.RT ln .a2 B . aA where 2 .0 B..0 A=.G0 is the standard reaction free energy or expressed in terms of molar concentrations 2 [ B]2 .B .G=.G0. RT ln .A [ A]