QBio102 Biomolecules Lecture 8 Thermodynamics PDF

B.Sc. Quantitative Biology Biomolecules Lecture 8 – Thermodynamics Dr. James Birrell University of Essex, UK hhu.de  What is thermodynamics?  In order to exist, a living species has to:  collect energy from the surroundings  convert it to a us...

B.Sc. Quantitative Biology Biomolecules Lecture 8 – Thermodynamics Dr. James Birrell University of Essex, UK hhu.de  What is thermodynamics?  In order to exist, a living species has to:  collect energy from the surroundings  convert it to a useful form (ion gradients, ATP, etc)  use it to grow, change, replicate, etc.  Living systems can only do something if the energy needed is balanced by the energy that is available.  Thermodynamics is the study of energy converting from one form to another  Thermo = heat  Dynamics = power  Dealing with bulk properties  No assumptions made about the system  Tells us which processes will happen and which will not  Empirical – we use it because it works 2 hhu.de  Equilibrium  At equilibrium - no macroscopic changes  Where k1 and k-1 are forward and reverse rate constants  Rate = k1[A] - k-1[B]  At equilibrium Rate = 0, k1[A]eq = k-1[B]eq  Where [A]eq and [B]eq are the equilibrium concentrations of A and B  In static equilibrium the system is stationary and there are also no microscopic changes  k1[A] = k-1[B] = 0  In dynamic equilibrium there are microscopic changes but no macroscopic changes  k1[A] = k-1[B] > 0  We define the equilibrium position with an equilibrium constant Keq  Keq = [B]eq/[A]eq and Keq = k1/k-1 3 hhu.de  Standard states and conditions  It is useful to have a common reference point  Standard states  Pure solid  Pure liquid  Solute of 1 M (1 mole per litre) concentration  Pure gas at 1 atmosphere of pressure  Usually defined at 25 oC (298 K) and atmospheric pressure  For reactions involving H+  Standard state is 1 M H+ or pH 0  In biochemistry it is common to refer to a biochemical standard state with 10-7 M H+ or pH 7 4 hhu.de  Types of system  Isolated  No matter transfer  No energy transfer  Simplest case - not realistic  Except entire universe...to our knowledge  Closed  No matter transfer ideal gases have closed systems  Energy transfer  Still simple - good approximation to a test tube  May be applicable to some biological situations  Open  Matter transfer  Energy transfer  Complex – not easy to model – but represents most realistic situations 5 hhu.de  State and path functions  A state function depends only on the state of the system not on the path taken to reach that state  Intensive state function  Independent of the size of the system  E.g. Temperature, pressure  Extensive state function  Dependent on the size of the system  E.g. Mass, volume, energy  Most thermodynamic quantities are extensive state functions  A path function depends on the path taken  E.g. Heat, work 6 hhu.de  Thermodynamic laws  0th law:  If two systems are at equilibrium with a third system then they are also at equilibrium with each other  If A ⇌ B and B ⇌ C then A ⇌ C  1st law:  Energy cannot be created or destroyed but can be converted from one form to another  2nd law:  Everything tends towards maximum disorder  3rd law:  Disorder increases with temperature 7 hhu.de  Ideal gas law  To start with, we will consider the simplest system "an ideal gas"  An ideal gas is one in which:  The particles interact negligibly  The particles are small compared with the spaces between them  pV = nRT  Where p = pressure in pascals (N/m2)  V = volume in m3  n = number of moles of gas  R = the molar gas constant (8.31 m2 kg s-2 K-1 mol-1)  T = temperature in K 8 hhu.de  1st law of thermodynamics  The Law of Conservation of Energy  Energy cannot be created or destroyed – simply changed from one form to another  Alternatively, all energy changes in an isolated system sum to zero  We can define the internal energy of a system as U  The sum of the kinetic and potential energies  And the change in internal energy (ΔU) as  ΔU = Ufinal – Uinitial  According to the first law, for an isolated system ΔU = 0  For a closed system:  ΔUsystem + ΔUsurroundings = 0  ΔUsystem = -ΔUsurroundings  The energy gained by the system comes from the surroundings and the energy lost by the system goes to the surroundings 9 hhu.de  Work and heat  We distinguish between two types of energy exchange: Work (w) and Heat (q)  q is the heat exchanged between the system and the surroundings  Random motion in the system – contributes to temperature  Positive = heat from surroundings to system  Negative = heat from system to surroundings  w is the work done on the system by the surroundings  Orderly motion of the system  Positive = work done on system by surroundings  Negative = work done by system on surroundings  ΔUsystem = -ΔUsurroundings = q + w  w consists of many different kinds of work e.g. Electrical, mechanical, expansion etc 10 hhu.de  Work and heat  For a process where the volume of the system does not change (isochoric)  No work is done by the system on the surroundings or by the surroundings on the system  The energy can only be echanged as heat  ΔUv = qv  At constant pressure (isobaric):  Work done by the system on the surroundings – expansion  Or by the surroundings on the system - compression  ΔUp = qp + wp  wp = -pΔV (where p is pressure and ΔV is the change in volume of the system)  Negative because an increase in volume (+ve ΔV) does work on the surroundings (-ve wp) 11 hhu.de  Work and heat  Now we can remove the “expansive” work from the change in internal energy:  ΔUp – wp(expansion) = ΔUp + pΔVp  = (Ufinal – Uinitial) + (pVfinal – pVinitial)  = (Ufinal + pVfinal) – (Uinitial + pVinitial)  For convenience we define the enthalpy (H): H = U + pV  (Ufinal + pVfinal) – (Uinitial + pVinitial) = Hfinal – Hinitial = ΔH = ΔUp + pΔVp  Such that the change in the internal energy due to non-expansive work (pV work) is the change in enthalpy or ΔH  If no other work is done except expansion then ΔH is simply the heat transferred between the system and the surroundings  The question is, how much of ΔH can be captured to do useful work 12 hhu.de  Enthalpy as a state function  Enthalpy can often be measured directly as long as ΔH = qp  Calorimetry  If the reaction is difficult to do then can do in a series of steps and add  ΔHtotal = ΔH1 + ΔH2 + etc  Hess’s law of constant heat summation  Enthalpy is an extensive state function – affected by size of the system  The first law of thermodynamics only deals with the sum of energies  It does not tell us whether a reaction will occur  Reactions with a positive ΔH = endothermic → heat in → feels cold  Reactions with a negative ΔH = exothermic → heat out → feels hot  In order for a reaction to happen some of the energy must be available as work energy 13 hhu.de  Boltzmann distribution  For bulk systems  Made up of large numbers of molecules  Thermal collisions and Brownian motion allows energy exchange between molecules  Not all will have the same energy  Energy distributed randomly over all energy levels  Boltzmann distribution: 𝑛𝑛𝑖𝑖 − 𝐸𝐸𝑖𝑖 − 𝐸𝐸𝑗𝑗 = 𝑒𝑒𝑒𝑒𝑒𝑒 𝑛𝑛𝑗𝑗 𝑘𝑘𝐵𝐵 𝑇𝑇  The ratio of the number of molecules with energies Ei and Ej  Depends on temperature and the energy difference  ni always smaller than nj when Ei is bigger than Ej  Because lower energy levels always more populated  Applicable for rotational, vibrational, electronic energy levels 14 hhu.de  Maxwell-Boltzmann distribution  For translational levels  Particles can move in 3D  Need to consider the Maxwell-Bolzmann distribution  Temperature increases the average speed of molecules 15 hhu.de  The second law of thermodynamics  A process will happen as long as the final state is more disordered than the initial state  E.g. consider two boxes – one with red particles, one with blue  We open a door between them  The particles will move until they have a random macroscopic distribution  Why don’t they return to their original distribution with all red on one side and all blue on the other side?  Of course this is possible but is very unlikely to happen  Particles can also occupy energy levels  Different vibrational, rotational, translational levels  If there is more thermal energy in the system then more particles will be able to access the higher energy levels  Therefore, the disorder increases with temperature 16 hhu.de  Carnot cycle  A Carnot engine or Carnot cycle is an example of a cyclic reversible process  The work done is related to the heat transfer  w = qH – qC  The efficiency is Eff = w/qH  Because some of the heat energy in goes to work  The rest goes out as qC  Not possible to have qC = 0  Always less efficient than 100%  Eff = (qH – qC)/qH = 1 - qC/qH  Because qH is proportional to TH  And qC is proportional to TC  Eff = 1 – TC/TH  This means that qC/qH = TC/TH  Which rearranges to qH/TH = qC/TC  This ratio of q/T is defined as the entropy change ΔS 17 hhu.de  Entropy  Disorder in a system is expressed as the entropy (S)  Arrangement of internal energy into microscopic states in a system:  S = kblnΩ where kb is a constant and Ω is the number of microscopic states  If a process releases heat then this will increase the entropy  If we try and do work then this will decrease the entropy of the system  Thus we need to release some heat to compensate for the work  The total available energy is the enthalpy H  The heat energy is TS  The remaining energy available to do work is H - TS  Likewise, total energy change in available energy is ΔH  The heat energy change is TΔS  And the energy released in order to do work is ΔH - TΔS 18 hhu.de  Gibbs free energy  We define the available energy for doing work (which allows a reaction to happen) as the Gibbs free energy G and the change in that energy that happens during the reaction as the ΔG: G = H – TS ΔG = ΔH – TΔS  Negative ΔG = process is spontaneous and can do useful work  Positive ΔG = process is NOT spontaneous – need to do work on the system  ΔG = 0 → process is at equilibrium and the value of G is a minimum  Recall that H = U + pV (slide 12) and G = H – TS  Therefore G = U + pV – TS  Using this and the ideal gas law pV = nRT (slide 8) 𝑝𝑝  We get 𝐺𝐺 = 𝐺𝐺 𝑜𝑜 + 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑝𝑝𝑜𝑜 Note: the derivation is very complex  Where Go is the Gibbs energy under standard conditions and po is the standard pressure of 1 atm  When p is equal to 1 atm then ln(p/po) will be 0 and G = Go 19 hhu.de  Gibbs free energy 1) Not at equilibrium  Reactants will be converted into products  Lowering G (ΔG is negative)  Until G is a minimum value (ΔG is zero)  Equilibrium  Rate NOT determined by thermodynamics  Very negative ΔG does not mean a high rate 2) At equilibrium  No bulk conversion of reactants to products  Rates of forward/backward reaction are the same  Adding more reactants gives negative ΔG  More reactants converted to products  Adding more products gives positive ΔG  Products converted back to reactants 20 hhu.de  Chemical potential  G is an extensive state function  Bigger system = bigger G  It is useful to define an intensive state function called the chemical potential (μ): 𝐺𝐺 𝜇𝜇 = or 𝐺𝐺 = 𝑛𝑛𝜇𝜇 𝑛𝑛  Where n is the number of moles  For a system with multiple components each can have its own chemical potential (μx): n1μ1 + n2μ2 … = G  Alternatively, Δμ can be expressed as the change in G as the amount of n changes: ∆𝐺𝐺 ∆𝜇𝜇 = ∆𝑛𝑛  This expression allows us to deal with an open system where transfer of matter as well as energy occur 21 hhu.de  Spontaneous reactions  Let’s consider a simple reaction of n1 moles of A converting to n moles of B: nA → nB  ΔG = nΔμ = nμB – nμA  If μB is less than μA then ΔG is negative and the reaction will be spontaneous  At equilibrium ΔG = 0 so μA = μB  Now consider: aA + bB → cC + dD  ΔG = aμA + bμB – cμC – dμD  Now for a spontaneous reaction aμA + bμB > cμC + dμD  For equilibrium μAa + μBb = μCc + μDd  Generally:  For a spontaneous reaction the weighted sum of the chemical potentials must decrease  For equilibrium the weighted sum of chemical potentials of the products and reactants are equal 22 hhu.de  Chemical potential  For a pure ideal gas: 𝑝𝑝  𝐺𝐺 = 𝐺𝐺 𝑜𝑜 + 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑝𝑝𝑜𝑜 𝐺𝐺 𝐺𝐺 𝑜𝑜  𝜇𝜇 = and 𝜇𝜇𝑜𝑜 = 𝑛𝑛 𝑛𝑛 𝑝𝑝  𝜇𝜇 = 𝜇𝜇𝑜𝑜 + 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑝𝑝𝑜𝑜  Can be applied for a component in a mixture of pure gases 𝑝𝑝𝑥𝑥  𝜇𝜇 = 𝜇𝜇𝑜𝑜 + 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑝𝑝𝑥𝑥𝑜𝑜  Where px is the partial pressure of x  Since at constant pressure px tells us the molar fraction of x: 𝑝𝑝𝑥𝑥 𝑛𝑛 [𝑥𝑥]  can be substituted for 𝑛𝑛𝑥𝑥𝑜𝑜 , which in turn is the same as [𝑥𝑥]𝑜𝑜Therefore, more generally: 𝑝𝑝𝑥𝑥𝑜𝑜 𝑥𝑥 [𝒙𝒙]  𝝁𝝁 = 𝝁𝝁𝒐𝒐 + 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 [𝒙𝒙]𝒐𝒐  Where [x] is the concentration of x and [x]o is the concentration in the standard state  But only for ideal solutions! 23 hhu.de  ΔG for a reaction  So now we know that: [𝑥𝑥] ΔG = μx1Δnx1 + μx2Δnx2 … and 𝜇𝜇 = 𝜇𝜇 𝑜𝑜 + 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 [𝑥𝑥]𝑜𝑜  We can combine these to get an equation for the ΔG in terms of [products] and [reactants] Π[𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑]𝒏𝒏 Π[𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑]𝒐𝒐,𝒏𝒏  ∆𝑮𝑮 = ∆𝑮𝑮𝒐𝒐 + 𝑹𝑹𝑹𝑹𝑹𝑹𝑹𝑹 Π[𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓]𝒏𝒏 Π[𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓]𝒐𝒐,𝒏𝒏  Where Π indicates a multiplication of all the values of [products] or [reactants]  Often Π[reactants]o and Π[products]o are left out because they equal 1  As an example:  aA + bB → cC + dD 𝑜𝑜 [𝐶𝐶]𝑐𝑐 ×[𝐷𝐷]𝑑𝑑  ∆𝐺𝐺 = ∆𝐺𝐺 + 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 [𝐴𝐴]𝑎𝑎 ×[𝐵𝐵]𝑏𝑏  Note that [A]o, [B]o, [C]o and [D]o have been left out because they are all equal to 1 M 24 hhu.de  Mass action ratio and equilibrium constants Π[𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑𝒑]𝒏𝒏  We define the ratio of as the mass action ratio Γ Π[𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓]𝒏𝒏  ΔG = ΔGo + RTlnΓ  At equilibrium ΔG = 0  Therefore: 0 = ΔGo + RTlnΓ  ΔGo = -RTlnΓ  And we define Γ at equilibrium as the equilibrium constant Keq  ΔGo = -RTlnKeq  For any other mixture of reactants and products:  ΔG = -RTlnKeq + RTlnΓ  ΔG = RTln(Γ/Keq)  So for ΔG < 0 (spontaneous reaction), Γ < Keq  If Γ > Keq then the reaction will not be spontaneous and will require energy to make it happen 25 hhu.de  Ideal vs non-ideal behaviour  Ideal gases:  No interactions between the gas molecules and container apart from collisions  pV = nRT for a pure gas  pxV = nxRT for a gas mixture where px is the partial pressure and nx is the molar fraction of the gas x  No enthalpy change on mixing gasses ΔHmix = 0  Ideal solutions  We cannot ignore the interactions but  Interactions between molecules same as in the standard state  Thus, we can simply use concentrations to describe the behaviour of the system  For dilute solutions this is a good approximation 26 hhu.de  Ideal vs non-ideal behaviour  For a non-ideal system:  There will be significant interactions between molecules that will give the gas a less random distribution  To account for this we talk about the effective concentration or activity:  ax = γx[x]  Where γx is the activity coefficient of x under those conditions  Then μx = μxo + RTln(ax/axo)  The activity coefficient will depend on a variety of factors including the kind of molecule involved but also its concentration, the temperature 27 hhu.de  Summary  Thermodynamics is the study of energy changes  We define the internal energy of a system U as the sum of kinetic and potential energies  The 1st law states that the total energy never changes i.e. ΔUsystem = -ΔUsurroundings  We define the energy without expansive work as the enthalpy (ΔH)  ΔH is negative = exothermic (heat out)  ΔH is positive = endothermic (heat in)  We also define the concept of entropy (S) as  Disorder in the system  Related to the number of energy levels and amount of energy S = kBlnΩ  ΔS = q/T or q = TΔS  TΔS is the amount of energy released from the system that cannot be used to do work  The available energy to do work is called the Gibbs free energy (G)  The amount released during a reaction is the change in G or ΔG  ΔG = ΔH - TΔS 28 hhu.de  Summary  ΔG = ΔH – TΔS  If ΔG is negative then a process is spontaneous (exergonic)  If ΔG is positive then a process is not spontaneous and requires an input of work (endergonic)  If ΔG is zero then a process is at equilibrium and the system has the minimum value of G  Using the ideal gas law we could show that  ΔG = ΔGo + nRTln(p/po)  Where ΔGo is the under standard conditions, p is the pressure, po is the standard pressure  For ideal gas mixtures we defined the chemical potential (μ):  μx = μxo + nRTln(px/pxo)  Where ΔG = μ1 + μ2 …  And for ideal solutions:  μx = μxo + nRTln([x]/[x]o)  Where ΔG = μ1 + μ2 … = ΔGo + RTln(Π[products]n/Π[reactants]n) = ΔGo + RTlnΓ  Where Γ is the mass action ratio and Γ = Keq when ΔG = 0 29 hhu.de  Next time...  We will use these equations to look at some types of reactions:  Association/dissociation  Acid/base equilibria  Redox reactions 30 hhu.de  Some suggested reading 31 hhu.de