Hour 15-26 Thermodynamics PDF

Hour 15 Maximum work and Free energy 15.1 The Clausius inequality We now test to see if entropy is really a measure of spontaneous change. Consider a system in thermal and mechanical contact with the surroundings, but not necessarily in equilibrium. 𝒅𝑺𝒔𝒚𝒔 + 𝒅𝑺𝒔𝒖𝒓𝒓 ≥ 𝟎 for any spontaneo...

Hour 15 Maximum work and Free energy 15.1 The Clausius inequality We now test to see if entropy is really a measure of spontaneous change. Consider a system in thermal and mechanical contact with the surroundings, but not necessarily in equilibrium. 𝒅𝑺𝒔𝒚𝒔 + 𝒅𝑺𝒔𝒖𝒓𝒓 ≥ 𝟎 for any spontaneous change. This implies that: 𝒅𝑺𝒔𝒚𝒔 ≥ −𝒅𝑺𝒔𝒖𝒓𝒓 𝑑𝑞 But recall that 𝑑𝑆𝑠𝑢𝑟𝑟 = −. This is because the heat supplied to the surroundings (which 𝑇 increases its entropy) is the heat that escaped the system so dqsurr = - dqsys. Hence: 𝒅𝒒𝒔𝒚𝒔 𝒅𝑺𝒔𝒚𝒔 ≥ 𝑻 This is known as the Clausius inequality. However, if the system is isolated from the surroundings, then dq = 0 and hence this becomes 𝑑𝑆𝑠𝑦𝑠 ≥ 0. This means that the entropy of an isolated system cannot decrease in the course of a spontaneous change. 15.2 Concentrating on the system Entropy was found to be a measure of spontaneity; however, it is difficult to use because we have to consider both system and surroundings. We shall try to devise criteria for spontaneity which rely on system parameters only. Consider a system in thermal equilibrium with its surroundings at a temperature T. When a change occurs with transfer of heat between system and surroundings, the Clausius inequality gives: 𝑑𝑞 𝑑𝑆 − ≥0 𝑇 (1) When heat is transferred at constant volume, dU = dw + dq, but dw = 0 since there is no change in volume. Hence dU = dq in the absence of non-expansion work. Substituting for dq we get: 𝑑𝑈 𝑑𝑆 − ≥0 𝑇 Which gives: 𝑻𝒅𝑺 ≥ 𝒅𝑼 1  Note that only system parameters are used to describe spontaneity. At constant U, 𝑑𝑆𝑈,𝑉 ≥ 0. This means that, for an isolated system, entropy increases with spontaneous change (Second Law). At constant entropy, 𝑑𝑈𝑆,𝑉 ≤ 0. This means that the internal energy of a system decreases as heat is given to the surroundings for a change to be spontaneous at constant entropy. (2) When heat transfer occurs at constant pressure, we know dqP = dH, therefore, 𝑑𝐻 𝑑𝑆 − ≥ 0. This gives 𝑇𝑑𝑆 ≥ 𝑑𝐻. 𝑇 At constant H, dS ≥ 0. This means that since no heat can be given to the surroundings, the entropy of the system must increase during a spontaneous change. At constant entropy, dH ≤ 0; this means that enthalpy must decrease as heat is given to the surroundings for a change to be spontaneous. We therefore have found ways of describing spontaneous changes without considering the surroundings directly. The two relationships are: 𝒅𝑼 − 𝑻𝒅𝑺 ≤ 𝟎 and 𝒅𝑯 − 𝑻𝒅𝑺 ≤ 𝟎 To trace these relationships, two thermodynamic quantities are introduced: (a) Helmholtz energy, A, defined as A = U – TS such that dA = dU – TdS (b) Gibbs energy, G, defined as G = H – TS such that dG = dH – TdS These two definitions imply that the criteria for spontaneity can be summed up as: 𝒅𝑨𝑽,𝑻 ≤ 𝟎 and 𝒅𝑮𝑷,𝑻 ≤ 𝟎 For the remainder of the course we will focus on G since, for convenience most observations are made at constant pressure (atmospheric pressure) rather than at constant volume. 15.3 Maximum non-expansion work and Gibbs energy Since most chemical reactions occur at constant pressure, Gibbs energy is more important to our study of chemical thermodynamics. At constant T and P, spontaneous reactions are those with decreasing Gibbs energy. dG = dH - TdS explains why some endothermic processes are spontaneous. For instance, endothermic reactions have dH > 0, however dG < 0 which means that the negative term “TdS” is large enough to overcome and outweigh dH; in this case, dS is large so that the process is entropy- driven. 2 Conversely, some spontaneous changes are enthalpy-driven whereby a very exothermic process can compensate for a decrease in entropy which makes the term “TdS” into a positive term. It can be shown that ΔG is the maximum non-expansion work that the system can do at constant T and P. That is: 𝜟𝑮 = 𝒘𝒂𝒅𝒅,𝒎𝒂𝒙 3 15.4 Combining First and Second Laws The Fundamental equation: Since dU = dq + dw and dqrev = TdS and dwrev = -PdV, we can substitute for dq and dw to get the Fundamental equation: 𝒅𝑼 = 𝑻𝒅𝑺 − 𝑷𝒅𝑽 This equation applies to any change of a closed system of constant composition that does no (non- expansion) work. We have defined it reversibly but dU is a state function which means the path is not important and so the fundamental equation applies to any change (whether reversible or irreversible) of a closed system. Since U is a function of S and V, then: 𝜕𝑈 𝜕𝑈 𝑑𝑈 = ( ) 𝑑𝑆 + ( ) 𝑑𝑉 𝜕𝑆 𝑉 𝜕𝑉 𝑆 But we have just seen that 𝑑𝑈 = 𝑇 𝑑𝑆 − 𝑃 𝑑𝑉 which means that: 𝜕𝑈 𝜕𝑈 𝑇 = ( ) 𝑎𝑛𝑑 − 𝑃 = ( ) 𝜕𝑆 𝑉 𝜕𝑉 𝑆 If 𝑑𝑓 = 𝑔𝑑𝑥 + ℎ𝑑𝑦, for df to be an exact differential then: 𝜕𝑔 𝜕ℎ ( ) =( ) 𝜕𝑦 𝑥 𝜕𝑥 𝑦 Since dU is an exact differential then: 𝜕𝑇 𝜕𝑃 ( ) = −( ) 𝜕𝑉 𝑆 𝜕𝑆 𝑉 This is a Maxwell Relation. Similar relations can be derived from other exact differentials. Try finding Maxwell relations using the differentials for thermodynamic quantities H, G and A. Maxwell relations allow us to derive unusual and useful relationships between quantities which may not seem related. They give us the flexibility to pursue changes in the system along paths that are most convenient to us. 4 Hour 16 – Temperature and pressure dependence of the Gibbs energy 16.1 General relationships In this section, we will explore how Gibbs free energy (G) varies with pressure and temperature since these are the variables we typically can control in a chemical reaction. Recall that by definition, G = H – TS. When the system undergoes a change of state, G may change because H, T and S change. For infinitesimal changes in each property: 𝑑𝐺 = 𝑑𝐻 − 𝑇𝑑𝑆 − 𝑆𝑑𝑇 Because H = U+PV we can see that: 𝑑𝐻 = 𝑑𝑈 + 𝑃𝑑𝑉 + 𝑉𝑑𝑃 For a closed system doing no additional work, we can replace dU by the fundamental equation: 𝑑𝑈 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉 The result of these two steps is: 𝑑𝐺 = 𝑇𝑑𝑆 − 𝑃𝑑𝑉 + 𝑃𝑑𝑉 + 𝑉𝑑𝑃 − 𝑇𝑑𝑆 − 𝑆𝑑𝑇 This simplifies to: 𝑑𝐺 = 𝑉𝑑𝑃 − 𝑆𝑑𝑇 This expression shows that G changes with P or temperature. This is particularly important in Chemistry because not only does G change with the variables under our control, but also G is the product of, and therefore carries the consequences of, the first and second laws. So, understanding how G changes is critical to understanding and predicting how the system will behave. G is a state function and has an exact differential dG. Partial derivatives for G are: 𝜕𝐺 (𝑎) ( ) = −𝑆 𝜕𝑇 𝑃 And 𝜕𝐺 (𝑏) ( ) =𝑉 𝜕𝑃 𝑇 5 According to (a), G decreases when temperature is raised at constant pressure and composition. Moreover, the decrease is sharpest for systems with a high entropy (greater for gases and liquids than for solids). Shown graphically: According to (b), G always increases when the pressure of a system is increased at constant temperature and composition. This increase is sharpest for systems with high molar volumes (gases again). Shown graphically: 6 16.2 Variation of Gibbs energy with temperature As we introduced before: 𝜕𝐺 ( ) = −𝑆 𝜕𝑇 𝑃 But since G = H – TS we can easily see that: 𝐺−𝐻 −𝑆 = 𝑇 Which when we substitute into (a) we get: 𝜕𝐺 𝐺−𝐻 ( ) = 𝜕𝑇 𝑃 𝑇 It can be shown that : 𝜕 𝐺 𝐻 ( ) =− 2 𝜕𝑇 𝑇 𝑃 𝑇 This can alternatively be written: 𝜕 𝛥𝐺 𝛥𝐻 ( ) =− 2 𝜕𝑇 𝑇 𝑃 𝑇 This expression is called the Gibbs-Helmholtz equation. It shows that if we know the enthalpy of the system, then we know how G/T varies with temperature. A more useful form of the equation is; 𝛥𝐺 𝛩 (𝑇2 ) 𝛥𝐺 𝛩 (𝑇1 ) 1 1 − = 𝛥𝐻 𝛩 ( − ) 𝑇2 𝑇1 𝑇2 𝑇1 16.3 Variation of Gibbs energy with pressure 𝜕𝐺 Since (𝜕𝑃) = 𝑉, to find the Gibbs energy at one pressure in terms of its value at another pressure 𝑇 (keeping temperature constant): 𝑃𝑓 𝑃𝑓 ∫ 𝑑𝐺 = ∫ 𝑉𝑑𝑃 𝑃𝑖 𝑃𝑖 7 This will yield: 𝑃𝑓 𝐺(𝑃𝑓 ) = 𝐺(𝑃𝑖 ) + ∫ 𝑉𝑑𝑃 𝑃𝑖 For liquids and solids The volume changes only slightly as the pressure changes since these phases are not very compressible. So V is more or less constant. This makes the pressure dependence relationship look like: 𝐺(𝑝𝑓 ) = 𝐺(𝑝𝑖 ) + 𝑉𝑚 (𝑝𝑓 − 𝑝𝑖 ) 𝐺(𝑝𝑓 ) = 𝐺(𝑝𝑖 ) + 𝑉𝑚 𝛥𝑝 The term VmΔP is very small and under normal laboratory conditions may be neglected without great consequence. Take home message: G is independent of pressure for solids and liquids. Gases Recall that since the molar volume of gases is large, G changes markedly with pressure for gases. Also, V changes with pressure and cannot be taken as a constant. 𝑝𝑓 𝐺(𝑝𝑓 ) = 𝐺(𝑝𝑖 ) + ∫ 𝑉𝑑𝑝 𝑝𝑖 For a perfect gas, pV = nRT so V = nRT/p: 𝑝𝑓 𝑑𝑝 𝐺(𝑝𝑓 ) = 𝐺(𝑝𝑖 ) + 𝑛𝑅𝑇 ∫ 𝑝𝑖 𝑝 𝑝𝑓 𝐺(𝑝𝑓 ) = 𝐺(𝑝𝑖 ) + 𝑛𝑅𝑇𝑙𝑛 ( ) 𝑝𝑖 Also, if we set pi = pϴ (standard pressure of 1 bar), then the Gibbs energy of a gas at a pressure P is related to standard Gibbs energy by: 𝑝 𝐺(𝑝) = 𝐺 𝛳 + 𝑛𝑅𝑇𝑙𝑛 ( 𝛳 ) 𝑝 8 Hour 17 Chemical potential & the fundamental equation of chemical thermodynamics Fugacities The Chemical potential (μ) The standard state of a perfect gas is established at 1 bar; (hence standard pressure of 1 bar is denoted pθ) 𝑝 We saw previously that 𝐺(𝑝𝑓 ) = 𝐺(𝑝𝑖 ) + 𝑛𝑅𝑇𝑙𝑛 ( 𝑝𝑓 ) 𝑖 It therefore follows that if we set pi = pθ, then the standard Gibbs function is Gθ and: G (p) = Gθ + nRT ln (p/ pθ) The molar Gibbs function Gm = G/n therefore gives; Gm (p) = Gθm + RT ln (p/ pθ) The chemical potential µ is defined as the Molar Gibbs function when dealing with pure substances, that is: μ = μθ + RT ln (p/pθ) Since systems spontaneously strive for lower G then it is fair to say that spontaneous systems strive for lower μ. Looking at real gases Recall; Gm (p) = Gθm + RT ln (p/ pθ) This equation is obeyed when gases show perfect behaviour (ideality). However, real gases show departures from this behaviour. In an attempt to preserve the form of the original equation it is necessary to replace the true pressure p by an effective pressure called the fugacity (f). Fugacity is derived from the latin word for “fleetness” or “escaping tendency”. The equation therefore becomes; Gm (p) = Gθm + RT ln (ƒ / pθ) or μ = μθ + RT ln (ƒ / pθ) The fugacity is defined so that the equation is exactly true for all gases. For instance, consider the reaction; H2 (g) + Br2 (g) ⇋ 2 HBr(g) the equilibrium constant is given by; 2 𝑝𝐻𝐵𝑟 𝐾=𝑝 𝐻2 𝑝𝐵𝑟2 where pJ is the partial pressure of a substance J. 9 For real gases this is only an approximation; the thermodynamically exact expression is: 2 𝑓𝐻𝐵𝑟 𝐾= 𝑓𝐻2 𝑓𝐵𝑟2 where ƒJ is the fugacity of substance J. We therefore need to be able to see a relationship between ƒ and p. If fugacity is written as ƒ = ø p , where ø is the dimensionless fugacity coefficient; then it can be 𝒁−𝟏 shown that: 𝒍𝒏∅ = ∫ 𝒅𝒑 𝒑 10 Hour 18 – Phase Diagrams of Pure Substances Phase – a form of matter, uniform throughout in chemical composition and physical state: gas, liquid, solid or allotropes. Phase Diagram – a map of the pressure and temperature at which each phase of a substance is most stable. Change of phase without change of chemical composition – examples include vapourisation, melting or packing changes (ie. lattice changes). Phase Transition – the spontaneous conversion of one phase to another, this occurs at a characteristic temperature for a given pressure or vice versa. Recall that a spontaneous change is one which results in the lowering of Gibbs free energy and hence the lowering of chemical potential (μ). μ = ∂G ∂n NB. For a particular temperature and Pressure, the phase with the lowest chemical potential is the most stable. Eg. At 1 atm, ice is the stable phase of H2O below 0 0C but above 0 0C liquid H2O is more stable, that is, the chemical potential of ice is lower than that of liquid water below 0 0C; the opposite being true above. Transition Temperature, Ttrs – the temperature at which two chemical potentials are equal and two phases are in equilibrium at the prevailing pressure. NB. Thermodynamics ≠ Kinetics; even though a transition is predicted as spontaneous thermodynamically, kinetics may be slow as to be indiscernible. Eg. Diamond and graphite, in this case, thermodynamic instability is frozen-in in the solids (different from liquids and gases). Metastable phases – thermodynamically unstable phases that persist due to kinetic hindrances, eg. diamond is the metastable phase of carbon under normal conditions. Phase Boundaries – lines separating the regions on a phase diagram; they show values of T and P at which two phases co-exist in equilibrium. Boiling point – a heated liquid in an open vessel reaches a point where vapour pressure equals external pressure and vapourisation can occur throughout the liquid. Free vapourisation throughout the liquid is called boiling. The temperature at which vapour pressure equals external pressure is the boiling temperature. At 1 atm, this is the normal boiling temperature, denoted Tb. At 1 bar, this is the standard boiling temperature. (1 bar = 0.987 atm) Therefore, std. boiling temp. < Tb Eg. If Tb = 100 0C then std. boiling temp. = 99.6 0C A liquid heated in a closed vessel does not boil. The vapour pressure and hence the vapour density keep increasing with increasing temperature. The density of the liquid decreases because it expands. Eventually density of vapour = density of remaining liquid and the surface between the two phases disappears resulting in one phase. The temperature at which one phase results is called the critical temperature, Tc. Vapour pressure at Tc is called the critical pressure, Pc. At and above Tc the one phase remaining is called a supercritical fluid. Melting point – the temperature at which solid and liquid phases coexist in equilibrium for a given pressure (if P = 1 atm; normal melting/freezing point, Tf). Triple Point – the point where three different phases (usually gas, liquid and solid) are in equilibrium and coexist. The triple point, T3 occurs at a single definite pressure and temperature characteristic of the substance. 11 Phase Stability & Conditions (T,p) Overview: At equilibrium, μ of a substance is the same throughout a sample, regardless of how many phases are present. For instance, when solid and liquid are in equilibrium, μ is the same in all parts of solid and liquid and the same in both phases. Temperature & Pressure Dependence At low temperatures, the solid phase has the lowest μ and if pressure is not too low, it will be the most stable phase. μ of phases change with T in different ways; as T is raised, μ of another phase may fall below that of solid (ie. if kinetics does not hinder) – a phase transition occurs. Temperature Dependence of Stability Temperature dependence of Gibbs energy The schematic temperature dependence of the chemical potential of the solid, liquid, and gas phases of a substance (in practice, the lines are curved). The phase with the lowest chemical potential at a specified temperature is the most stable one at that temperature. Chemical potentials of the two phases are equal at the transition temperatures Since Sm (g) > Sm (l); the slope of μ vs T is steeper (more negative) for gas than for liquids. -Sm (g) > -Sm (l) Similarly, since Sm (l) > Sm (s) {almost always}; therefore the slope will be steeper for liquids than for solids. 12 Pressure Dependence of Phase Stability can be viewed in two parts: 1. Effect of pressure on melting (solid to liquid) 2. Effect of pressure on vapour pressure (liquid to gas) NB. Solids are virtually incompressible, so we don’t expect a response in μ(s) with ∆p. 1. Response of melting to applied pressure Most substances (solid phase) require a higher temperature to melt when subjected to pressure; the pressure behaves as if it prevents the formation of the disordered, less dense liquid phase. Exception: H2O has a denser liquid phase so it behaves unusually (higher pressure makes a solid melt (or liquid freeze) at a lower temperature). Since Vm (l) > Vm (s) for most substances, an increase in pressure causes the μ of a liquid to increase faster than that of a solid therefore at high pressures μ(s) would be less than μ(l) (the solid phase is more stable at higher pressures). (a) In this case the molar volume of the solid is smaller than that of the liquid and μ(s) increases less than μ(l). As a result, the freezing temperature rises. (b) Here the molar volume is greater for the solid than the liquid (as for water), μ(s) increases more strongly than μ(l), and the freezing temperature is lowered. 13 2. Effect of applied pressure on vapour pressure When an external pressure is applied to a condensed phase, its vapour pressure increases. It behaves as if the molecules are squeezed out of the liquid phase into the gas phase. It can be shown that; p = p*eVm∆p/RT 14 Hour 19 – Phase boundaries and the Clapeyron equation Focus question: What are phase diagrams and why are they important? Recall that if two phases are in equilibrium, then μ is the same everywhere in the two phases: 𝜇𝛼 (𝑃, 𝑇) = 𝜇𝛽 (𝑃, 𝑇) To find the phase boundary, solve this equation in terms of P and T. Let P and T change infinitesimally so that α and β remain in equilibrium: 𝑑𝜇𝛼 = 𝑑𝜇𝛽 Recall that dμ = -Sm dT + VmdP so substituting for μ on both sides of this equation we get: −𝑆𝛼,𝑚 𝑑𝑇 + 𝑉𝛼,𝑚 𝑑𝑃 = −𝑆𝛽,𝑚 𝑑𝑇 + 𝑉𝛽,𝑚 𝑑𝑃 Taking all the like terms together on the same side we get: 𝑉𝛽,𝑚 𝑑𝑃 − 𝑉𝛼,𝑚 𝑑𝑃 = 𝑆𝛽,𝑚 𝑑𝑇 − 𝑆𝛼,𝑚 𝑑𝑇 (𝑉𝛽,𝑚 − 𝑉𝛼,𝑚 )𝑑𝑃 = (𝑆𝛽,𝑚 − 𝑆𝛼,𝑚 )𝑑𝑇 𝑑𝑃 (𝑆𝛽,𝑚 − 𝑆𝛼,𝑚 ) 𝛥𝑆𝑡𝑟𝑠 = = 𝑑𝑇 (𝑉𝛽,𝑚 − 𝑉𝛼,𝑚 ) 𝛥𝑉𝑚,𝑡𝑟𝑠 𝒅𝑷 𝜟𝑺𝒕𝒓𝒔 = 𝒅𝑻 𝜟𝑽𝒎,𝒕𝒓𝒔 We have derived the general form of the Clapeyron equation which is the relationship used to find the slope of any phase boundary of any pure substance. 15 19.1 The solid-liquid boundary Melting (fusion) is accompanied by ΔHfus and occurs at a specific temperature for a specific pure substance. So 𝛥𝐻𝑓𝑢𝑠 𝛥𝑆𝑡𝑟𝑠 = 𝑇 Hence 𝑑𝑃 𝛥𝐻𝑓𝑢𝑠 = 𝑑𝑇 𝑇𝛥𝑉𝑓𝑢𝑠 Usually, ΔHfus is positive, and ΔVfus is positive and small. So for most substances, the slope dP/dT for the solid-liquid is usually positive and large (it slopes steeply upward). 𝑃 𝛥𝐻𝑓𝑢𝑠 𝑇 𝑑𝑇 ∫ 𝑑𝑃 = ∫ 𝑃∗ 𝛥𝑉𝑓𝑢𝑠 𝑇 ∗ 𝑇 𝛥𝐻𝑓𝑢𝑠 𝑇 𝑃 − 𝑃∗ = 𝑙𝑛 𝛥𝑉𝑓𝑢𝑠 𝑇 ∗ However, when x > Vm (l), ΔVvap ≈ Vm (g). If perfect behaviour is shown, then: 𝑅𝑇 = 𝑉𝑚 𝑃 16  𝑑𝑃 𝛥𝐻𝑣𝑎𝑝 𝛥𝐻𝑣𝑎𝑝 = = 𝑑𝑇 𝑇(𝑅𝑇) 𝑅𝑇 2 𝑃 𝑃 With rearrangement this becomes the liquid-vapour form of the Clausius-Clapeyron equation: 𝑑𝑃 1 𝛥𝐻𝑣𝑎𝑝 ( ) = 𝑃 𝑑𝑇 𝑅𝑇 2 But since dx/x = d lnx, we get: 𝑃 𝑇 𝛥𝐻 𝑣𝑎𝑝 ∫ 𝑑𝑙𝑛𝑃 = ∫ 2 𝑑𝑇 𝑃∗ 𝑇 ∗ 𝑅𝑇 𝑃 𝛥𝐻𝑣𝑎𝑝 1 1 𝑙𝑛 ( ∗ ) = [− + ∗ ] 𝑃 𝑅 𝑇 𝑇 Taking ln-1 of both sides we get: 𝜟𝑯𝒗𝒂𝒑 𝟏 𝟏 𝑷 = 𝑷∗ 𝒆−𝝌 𝑤ℎ𝑒𝑟𝑒 𝝌 = [ − ∗] 𝑹 𝑻 𝑻 19.3 The solid-vapour boundary In this case, we need the enthalpy of sublimation, ΔHsub. 𝛥𝐻𝑠𝑢𝑏 𝛥𝑆 = 𝑇 Everything else follows as for the liquid-vapour boundary giving: 𝜟𝑯𝒔𝒖𝒃 𝟏 𝟏 𝑷 = 𝑷∗ 𝒆−𝝌 𝑤ℎ𝑒𝑟𝑒 𝝌 = [ − ∗] 𝑹 𝑻 𝑻 17 Hour 20 – Partial molar quantities and the thermodynamics of mixing. Colligative properties. So far we have been dealing with pure substances in one-component systems. We now wish to describe simple mixtures and their thermodynamic properties. For chemical applications, we need to predict the behaviours of mixtures, particularly those with reacting components. As a lead up, we will examine the properties of mixtures with non-reacting components. 20.1 Partial molar quantities (i) Partial molar volume This is the contribution that a component of a mixture makes to the total volume of the sample. By definition: The partial molar volume of a substance A in a mixture is the change in volume per mole of A added to a large volume of the mixture. This is expressed mathematically as: 𝜕𝑉 𝑉𝐽 = ( ) 𝜕𝑛𝐽 𝑃,𝑇,𝑛′ Where VJ is the partial molar volume of J and n’ indicates that the amounts of all other substances in the mixture are constant. NB The partial molar volume of the components change with composition because the environment of each type of molecule changes. From the mathematical definition, we see that VJ is the slope of the plot of V (total volume) as the amount of J is changed (provided that P, T and nothers are all constant). So for a two-component system, the total change in volume dV can be expressed as: 𝜕𝑉𝐴 𝜕𝑉𝐵 𝑑𝑉 = ( ) 𝑑𝑛𝐴 + ( ) 𝑑𝑛 𝜕𝑛𝐴 𝑃,𝑇,𝑛𝐵 𝜕𝑛𝐵 𝑃,𝑇,𝑛𝐴 𝐵 This is equivalent to: 𝑑𝑉 = 𝑉𝐴 𝑑𝑛𝐴 + 𝑉𝐵 𝑑𝑛𝐵 18 If this were integrated, we would get: 𝑉 = 𝑉𝐴 𝑛𝐴 + 𝑉𝐵 𝑛𝐵 (ii) Partial Molar Gibbs Energies The concept of partial molar volume can be extended to any extensive state function. Recall that μ = Gm. For a substance in a mixture, chemical potential μJ is defined as the partial molar Gibbs energy: 𝜕𝐺 𝜇𝐽 = ( ) 𝜕𝑛𝐽 𝑃,𝑇,𝑛′ μJ is the slope of the plot is G vs nJ. For a two-component mixture: 𝐺 = 𝑛𝐴 𝜇𝐴 + 𝑛𝐵 𝜇𝐵 μA is the contribution of A to GTOTAL in mixture. Recall the fundamental equation of thermodynamics (this is for constant composition): 𝑑𝐺 = 𝑉𝑑𝑃 − 𝑆𝑑𝑇 For a two component system, this becomes: 𝒅𝑮 = 𝑽𝒅𝑷 − 𝑺𝒅𝑻 + 𝝁𝑨 𝒅𝒏𝑨 + 𝝁𝑩 𝒅𝒏𝑩 At constant temperature T and pressure P, the first two terms fall out and we get: 𝑑𝐺 = 𝜇𝐴 𝑑𝑛𝐴 + 𝜇𝐵 𝑑𝑛𝐵 = 𝑑𝑤𝑎𝑑𝑑,𝑚𝑎𝑥 So μ shows how G varies with composition. 19 Now recall: G = H – TS which gives G = U + PV – TS Rearranging for U we get: 𝑈 = −𝑃𝑉 + 𝑇𝑆 + 𝐺 Hence: 𝑑𝑈 = −𝑃𝑑𝑉 − 𝑉𝑑𝑃 + 𝑇𝑑𝑆 + 𝑆𝑑𝑇 + 𝑑𝐺 Therefore: 𝑑𝑈 = −𝑃𝑑𝑉 − 𝑉𝑑𝑃 + 𝑇𝑑𝑆 + 𝑆𝑑𝑇 + (𝑉𝑑𝑃 − 𝑆𝑑𝑇 + 𝜇𝐴 𝑑𝑛𝐴 + 𝜇𝐵 𝑑𝑛𝐵 ) This reduces to: 𝑑𝑈 = −𝑃𝑑𝑉 + 𝑇𝑑𝑆 + 𝜇𝐴 𝑑𝑛𝐴 + 𝜇𝐵 𝑑𝑛𝐵 However, at constant V and S, the first two terms drop out and we get: 𝑑𝑈 = 𝜇𝐴 𝑑𝑛𝐴 + 𝜇𝐵 𝑑𝑛𝐵 𝝏𝑼 𝝁𝑱 = (𝝏𝒏 ) 𝑱 𝑺,𝑽,𝒏′ So μ also shows how internal energy varies with composition (at constant S, V and nothers). With similar derivations, it can be shown that: 𝜕𝐻 𝜇𝐽 = ( ) 𝜕𝑛𝐽 𝑆,𝑃,𝑛′ And 𝜕𝐴 𝜇𝐽 = ( ) 𝜕𝑛𝐽 𝑉,𝑇,𝑛′ All extensive state functions can be expressed this way and that is why μ is a central parameter to chemistry. 20 20.2 Thermodynamics of mixing We now focus on spontaneous changes of composition (spontaneous mixing). Eg two gases introduced to the same container will spontaneously mix. (a) Gibbs energy of mixing 𝛥𝐺𝑚𝑖𝑥 = 𝑛𝑅𝑇(𝑥𝐴 𝑙𝑛𝑥𝐴 + 𝑥𝐵 𝑙𝑛𝑥𝐵 ) Since all mole fractions are less than one, the ln term will be negative so ΔGmix is negative and hence this is a spontaneous change. So we can conclude that gases mix spontaneously in all proportions; ΔGmix is proportional to T but independent of the total pressure. (b) Entropy and enthalpy change of mixing Since: 𝜕𝐺 ( ) = −𝑆 𝜕𝑇 𝑃,𝑛 Then: 𝜕 𝛥𝑆𝑚𝑖𝑥 = − [ 𝛥𝐺 ] 𝜕𝑇 𝑚𝑖𝑥 𝑃,𝑛𝐴,𝑛𝐵 Therefore: 𝛥𝑆𝑚𝑖𝑥 = −𝑛𝑅(𝑥𝐴 𝑙𝑛𝑥𝐴 + 𝑥𝐵 𝑙𝑛𝑥𝐵 ) Since the ln terms are always negative, then ΔSmix is always positive (ΔSmix > 0 ). Question: Show that ΔHmix = 0. 21 20.3 Chemical potential of liquids To discuss the equilibrium properties of liquid mixtures we need to know how the Gibbs energy of a liquid varies with composition. To calculate its value, we use the fact that, at equilibrium, the chemical potential of a substance present as a vapour must be equal to its chemical potential in the liquid. (a) Ideal solutions (see derivation on pages 143-144 of Atkins Physical Chemistry) μA = μA* + RT ln χA Some solutions depart significantly from Raoult’s law. Nevertheless, even in these cases the law is obeyed increasingly closely for the component in excess (the solvent) as it approaches purity. The law is therefore a good approximation for the properties of the solvent if the solution is dilute. (b) Ideal-dilute solutions For ideal solutions, the solvent and solute obey Raoult’s law. However, for real solutions, when they are dilute (that is the solute is in low concentration), the solute obeys Henry’s law: 𝑃𝐵 = 𝑥𝐵 𝐾𝐵 Where PB is the vapour pressure of the solute, xB is the mole fraction of the solute and KB is the Henry’s law constant (which has dimensions of pressure). Mixtures for which the solute obeys Henry’s law and the solvent obeys Raoult’s law are called “ideal-dilute” solutions. 20..3 Properties of solutions Now we consider the thermodynamics of mixing liquids; first considering the simple case of forming ideal solutions, then on to more complex mixtures. (i) Liquid mixtures (both components volatile), ΔGmix is calculated in the same way as for two gases: (ii) Colligative properties These are properties that depend only on the NUMBER of solute particles and NOT their identity. In order to simplify calculations, two assumptions are used: a. That the solute is not volatile; it does not contribute to the vapour; (this is a reasonable assumption) and b. That the solute does not dissolve in the solid solvent (this is not always true). 22 Boiling point elevation The strategy is to use the phase diagram to look for the temperature (at 1 atm) at which the chemical potential of the pure solvent vapour is equal to that of the solvent in the solution. This is the new equilibrium temperature for the transition and is therefore, the new boiling temperature. 𝑅𝑇 ∗2 𝛥𝑇 = 𝐾𝜒𝐵 where 𝐾 = 𝛥𝐻𝑣𝑎𝑝 23 Justification Consider a boiling solution; equilibrium is established between the solvent vapour (denoted A in this case) and the solvent in solution. Hence: 𝜇𝐴∗ (𝑔) = 𝜇𝐴∗ (𝑙) + 𝑅𝑇𝑙𝑛𝑥𝐴 Rearranging we get: 𝜇𝐴∗ (𝑔) − 𝜇𝐴∗ (𝑙) = 𝑅𝑇𝑙𝑛(1 − 𝑥𝐵 ) Now since ΔG = Gf – Gi = μf – μi 𝜇𝐴∗ (𝑔) − 𝜇𝐴∗ (𝑙) = 𝛥𝐺𝑣𝑎𝑝 𝛥𝐺𝑣𝑎𝑝 ln(1 − 𝑥𝐵 ) = 𝑅𝑇 24 Because the equation 𝛥𝑇 = 𝐾𝜒𝐵 makes no reference to the identity of the solute, only to its mole fraction, we conclude that the elevation of boiling point is a colligative property. The value of ΔT does depend on the properties of the solvent, and the biggest changes occur for solvents with high boiling points. For practical applications of the equation 𝛥𝑇 = 𝐾𝜒𝐵 , we note that the mole fraction of B is proportional to its molality, b, in the solution, and write ΔT = Kbb where Kb is the empirical boiling-point constant of the solvent Freezing point depression The heterogeneous equilibrium now of interest is between pure solid solvent A and the solution with solute present at a mole fraction xB. At the freezing point, the chemical potentials of A in the two phases are equal: 𝜇𝐴∗ (𝑠) = 𝜇𝐴∗ (𝑙) + 𝑅𝑇𝑙𝑛𝑥𝐴 The only difference between this calculation and the last is the appearance of the solid’s chemical potential in place of the vapour’s. Therefore, we can write the result directly from the previous equation: 𝑅𝑇 ∗2 𝛥𝑇 = 𝐾 ′ 𝜒𝐵 where 𝐾′ = 𝛥𝐻𝑓𝑢𝑠 where ΔT is the freezing point depression, T* − T, and ΔHfus is the enthalpy of fusion of the solvent. Larger depressions are observed in solvents with low enthalpies of fusion and high melting points. When the solution is dilute, the mole fraction is proportional to the molality of the solute, b, and it is common to write the last equation as ΔT = Kfb where Kf is the empirical freezing-point constant. Once the freezing point constant of a solvent is known, the depression of freezing point may be used to measure the molar mass of a solute in the method known as cryoscopy; however, the technique is of little more than historical interest. Other colligative properties Refer to pages 153 – 156 in the Atkins textbook for compulsory reading on solubility and osmosis. 25 Hour 21 - Activities of solvents We now have tools to adjust the expressions for ideal behaviour to cater for real behaviour. Recall that fugacity (f) accounts for the imperfections of gases and still allowed us to use a form of the thermodynamic equations that resembled the equations used for ideal systems. Now we will use some the of the same tools to show how the expressions encountered in the treatment of ideal solutions can also be preserved almost without changes by introducing the concept of “activity” (a). 21.1 The solvent activity Recall that the chemical potential of a real or ideal solvent is given by: 𝑃𝐴 𝜇𝐴 = 𝜇𝐴∗ + 𝑅𝑇𝑙𝑛 ( ∗ ) 𝑃𝐴 Where PA* is the vapour pressure of pure A and PA is the vapour pressure of A when it is a component of a solution. For an ideal solution, as we have seen, the solvent obeys Raoult’s law at all concentrations and we can express this relation as 𝜇𝐴 = 𝜇𝐴∗ + 𝑅𝑇𝑙𝑛𝑥𝐴. The form of this relation can be preserved when the solution does not obey Raoult’s law by writing: 𝜇𝐴 = 𝜇𝐴∗ + 𝑅𝑇𝑙𝑛𝑎𝐴 The quantity aA is the activity of A, a kind of effective mole fraction, just as the fugacity is an effective pressure. Because the latter equation is true for both real and ideal solutions, we need to understand how activity is related to mole fraction. For this we need to introduce the activity coefficient γ, where 𝒂 = 𝜸𝒙. This coefficient is defined so that as 𝑥 → 1, 𝛾 → 1. This means that as the solvent approaches purity (the state of ideality), γ will approach 1 which will give 𝑎 = 𝑥. For practical purposes, we can calculate the activity of a solvent by measuring the vapour pressure 𝑃𝐴 and then using the equation: 𝑎𝐴 = 𝑃𝐴∗ 26 Hour 22 - Activities of solutes The difference with defining activity coefficients and standard states for solutes is that they approach ideal-dilute (Henry’s law) behaviour as xB→0, not as xB→1 (corresponding to pure solute). We shall show how to set up the definitions for a solute that obeys Henry’s law exactly, and then show how to allow for deviations. 22.1 Ideal-dilute solutions A solute B that satisfies Henry’s law has a vapour pressure given by: 𝑃𝐵 = 𝐾𝐵 𝑥𝐵 where KB is an empirical constant. In this case, the chemical potential of B is: 𝑃𝐵 𝐾𝐵 𝜇𝐵 = 𝜇𝐵∗ + 𝑅𝑇𝑙𝑛 ∗ ∗ = 𝜇𝐵 + 𝑅𝑇𝑙𝑛 ∗ + 𝑅𝑇𝑙𝑛𝑥𝐵 𝑃𝐵 𝑃𝐵 Both KB and PB* are characteristics of the solute, so the second term may be combined with the first to give a new standard chemical potential: 𝐾𝐵 𝜇𝐵𝛩 = 𝜇𝐵∗ + 𝑅𝑇𝑙𝑛 𝑃𝐵∗ It then follows that the chemical potential of a solute in an ideal-dilute solution is related to its mole fraction by: 𝜇𝐵 = 𝜇𝐵𝛩 + 𝑅𝑇𝑙𝑛𝑥𝐵 If the solution is ideal, KB = PB* and we simply get 𝜇𝐵𝛩 = 𝜇𝐵∗ as we should expect. 22.2 Real solutes Now we will look at deviations from ideal-dilute, Henry’s law behaviour. For the solute, we introduce aB in place of xB in the chemical potential equation to get: 𝜇𝐵 = 𝜇𝐵𝛩 + 𝑅𝑇𝑙𝑛𝑎𝐵 The standard state remains unchanged in this last stage, and all the deviations from ideality are captured in the activity aB. The value of the activity at any concentration can be obtained in the 𝑃 same way as for the solvent: 𝑎𝐵 = 𝐾𝐵 𝐵 We will need to use an activity coefficient for the solvent: 𝑎𝐵 = 𝛾𝐵 𝑥𝐵. Now all the deviations from ideality are captured in the activity coefficient γB. Because the solute obeys Henry’s law as 27 its concentration goes to zero, it follows that: aB→xB and γB→1 as xB→0 at all temperatures and pressures. Deviations of the solute from ideality disappear as zero concentration is approached. Component Basis Standard state Activity Limits Solid or liquid Pure a=1 Solvent Raoult Pure solvent 𝑎 = P/P*, γ→1 as x→1 𝑎 = 𝛾𝑥 (pure solvent) Solute Henry (1) A hypothetical (1) a= p/K, γ→1 as x→0 state of the pure a = γx solute (2) A hypothetical (2) a = γb/ bϴ γ→1 as b→1 state of the solute at molality bϴ Comment: The activities of ions in solution Since the choice to standard state is arbitrary, we can choose one that best suits our purpose depending on the system. Often chemical compositions are reported as molalities instead of mole fractions. Such close agreement between reality and the model is obtained by expressing composition as molalities, that they are often used to simplify chemical potential equations. However they are not always a good approximation to activities. Interactions between ions are so strong that the approximation of replacing activities by molalities is valid only in very dilute solutions (less than 10−3 mol kg−1 in total ion concentration) and in precise work, activities themselves must be used. We need, therefore, to pay special attention to the activities of ions in solution, especially in preparation for the discussion of electrochemical phenomena. 28 Hours 24-25 Extent of reaction (Equilibrium); Homogenous and Heterogenous Equilibria; Effect of Temperature and Pressure on Equilibrium The aim of chemical reactions is towards dynamic equilibrium i.e. both reactants and products are present but have no tendency to undergo further change – (forward & backward reactions are occurring at the same rate). “Complete reactions”: [products] >> [reactants] in the equilibrium mixture. We can use thermodynamics to predict equilibrium composition. We know that for spontaneous chemical reactions: change is towards lowering of Gibbs energy. Gibbs Energy Minimum The equilibrium position of a reaction mixture is the one corresponding to the MINIMUM value of Gibbs energy. The equilibrium composition is known by calculating the Gibbs energy and identifying the composition which corresponds to minimum G. Finding ∆Grxn (Gibbs energy and reaction) Definition: Slope of graph of Gibbs energy, G vs extent of reaction, ξ (epsilon) Suppose for a reaction: A ↔ B we find a small amount dξ of A converts to B, then change in amount of A is: dnA = -dξ (negative sign since A is being used up) And change in amount of B is: dnB = +dξ So ∆Grxn = ∂G NB. ∆ in this case is a derivative (i.e. gradient of G vs ξ) ∂ξ p,T As the reaction advances (represented by motion from left to right along the horizontal axis) the slope of the Gibbs energy changes. Equilibrium corresponds to zero slope, at the foot of the valley. So, ∆Grxn is the difference between the chemical potentials of the reactants and products at the composition of the reaction mixture. NB. Composition varies with extent of reaction and μ varies with composition so, ∆Grxn will vary with extent of reaction. Note also that the reaction is spontaneous in the direction that lowers G; since ∆Grxn = μB – μA then the reaction A→ B is spontaneous if μA > μB whereas the reverse reaction is spontaneous if μB > μA. When the slope is zero, the reaction is spontaneous in neither direction: we have ∆Grxn = 0 (i.e. μB – μA = 0; μB = μA) Hence the system is at equilibrium. If ∆Grxn < 0 : forward reaction is spontaneous (forward rxn is exergonic; work producing) If ∆Grxn > 0 reverse reaction is spontaneous (forward rxn is endergonic; work consuming) If ∆Grxn = 0 reaction is at equilibrium (forward rxn is neither exergonic nor endergonic) 29 Perfect gas equilibria (Homogenous equilibria) Q, the reaction quotient. It ranges from 0 (pure A and no B; that is pB = 0) to ∞ (no A and pure B; that is pA = 0) ∆GӨrxn is the Standard reaction Gibbs energy and is defined as the difference in standard molar Gibbs energies of reactants and products: [∆GӨrxn = GӨm,B - GӨm,A] Recall that; ∆GӨrxn = ∆GӨƒ,B - ∆GӨƒ,A Hence we can calculate ∆GӨrxn Now at equilibrium ∆GӨrxn = 0; Let K represent Q at the equilibrium: 0 = ∆GӨrxn + RT ln K Therefore; RT ln K = -∆GӨrxn where K = PB/PA at equilibrium NB: This equation is MOST important since it relates to the thermodynamic parameter ∆GӨrxn with the chemically important equilibrium constant K. Hence ∆GӨrxn may be used to predict the equilibrium composition. From the equation it can been seen that when ∆GӨrxn < 0; then ln K is positive and K > 1 (Products are favoured at equilibrium) Conversely when ∆GӨrxn > 0; then ln K is negative and K < 1 (reactants are favoured at equilibrium. 30 General reaction equilibria (Heterogenous equilibria) We can easily extend the argument that led to the equation RT ln K = -∆GӨrxn to a general reaction. First, we need to generalize the concept of extent of reaction. Consider the reaction 2 A + B→3 C + D. A more sophisticated way of expressing the chemical equation is to write it in the symbolic form 0=3C+D−2A−B by subtracting the reactants from both sides (and replacing the arrow by an equal sign). This equation has the form where J denotes the substances and the νJ are the corresponding stoichiometric numbers in the chemical equation. In our example, these numbers have the values νA= −2, νB= −1, νC = +3, and νD = +1. A stoichiometric number is positive for products and negative for reactants. Then we define ξ so that, if it changes by Δξ, then the change in the amount of any species J is νJΔξ. The reaction Gibbs energy, ΔrG, is defined in the same way as before. It can be shown that the Gibbs energy of reaction can always be written as ΔrG = ΔrGΘ + RT ln Q with the standard reaction Gibbs energy calculated from or, more formally The reaction quotient, Q, has the form Q = activities of products activities of reactants 31 32 EFFECT OF TEMPERATURE & PRESSURE ON EQUILIBRIUM /VAN’T HOFF EQUATION. Effect of catalyst: K is unaffected by presence of a catalyst (or enzyme). A catalyst affects the rate of attainment of equilibrium but not its position. Response to pressure The value of K is not affected by pressure as ∆GӨrxn is defined at a single standard pressure. However, the equilibrium composition may be affected by pressure particularly when reactants and products are gaseous. There are two ways of increasing pressure: addition of an inert gas to reaction vessel – (if gases are perfect, there is no charge in partial pressure) compression of gases by modifying volume of reaction vessel – (concentrations or partial pressures change hence the value of K can only remain the same if changes in equilibrium position occur). The changes on equilibrium position are summarized in Le Chatelier’s principle: A system at equilibrium, when subjected to a disturbance, responds in a way that tends to minimize the effect of disturbance. E.g. If gases at the equilibrium are compressed, the system will respond in such a way as to minimize the increase in pressure. Consider A ↔ 2B K = P2B/PA.PӨ For K to remain constant, an increase in PA must be accompanied by an increase in the square of PB. In the case of compression and increase in pressure the system will shift in the direction 2B → A so as to reduce the number of gas particles. Quantitatively 33 34 Response to temperature Le Chatelier’s principle predicts that a system at equilibrium will tend to shift in the endothermic direction if the temperature is raised, for then energy is absorbed as heat and the rise in temperature is opposed. Conversely, an equilibrium can be expected to shift in the exothermic direction if the temperature is lowered, for then energy is released and the reduction in temperature is opposed. Summary For exothermic reactions; increased temperature favours reactants For endothermic reactions; increased temperature favours products The van ’t Hoff equation, which is derived in the Justification below, is an expression for the slope of a plot of the equilibrium constant (specifically, ln K) as a function of temperature. It may be expressed in either of two ways: 𝑑𝑙𝑛 𝐾 𝛥𝑟 𝐻 𝛩 = 𝑑𝑇 𝑅𝑇 2 𝑑𝑙𝑛 𝐾 𝛥𝑟 𝐻 𝛩 =− ……… equ 7.23b 𝑑(1/𝑇) 𝑅 Justification 35 Rationale: ∆Grxn = ∆Hrxn - T∆Srxn Dividing throughout by –T gives; -∆Grxn = -∆Hrxn + ∆Srxn T T change of S of surr. change of S of system When ∆Hrxn < 0 (exothermic) this corresponds to a positive change in entropy of surroundings so products are favoured. When T is raised, the contribution of a negative ∆H is less important to entropy of the surroundings so the products are less favoured and equilibrium lies less to the right. When ∆Hrxn > 0 (endothermic) the main push factor is to increase the entropy of the reaction system. Heat taken from the surroundings decreases the entropy of the surroundings (unfavourable) so raising the temperature reduces this negative effect. Value of K at different temperatures Suppose we wish to find the value of the equilibrium constant, K2 at a different temperature T2, once we know K1 and T1, we integrate Van’t Hoff equation (b). 36 37 HR 26 BIOCHEMICAL APPLICATION OF THERMODYNAMICS Biological Activity: Thermodynamics of ATP Adenosine triphosphate (ATP) is an important molecule which stores energy from the metabolism of food then supplies it to drive biological processes. It provides energy through hydrolysis to ADP (Adenosine diphosphate) ATP(aq) + H2O(l) → ADP(aq) + Pi (aq) + H3O+(aq) (exergonic) Inorganic phosphate eg. H2PO4 The reaction provides energy to drive an endergonic reaction if suitable enzymes are present. To describe the thermodynamics of a biological system we have to define biological standard states. Conventional standard state is aH3O+ = 1 (pH = 0). This condition is not suitable for a biological system so we adopt biological std. state pH = 7 (aH3O+ = 10-7); a neutral soln. The ΔGΘ for ATP hydrolysis is -30kJ/mol. Hence the maximum work that can be done is 30 kJ/mol. The action occurs both ways: It can donate phosphate (release energy) to an “acceptor” but it can also accept phosphate (and so become recharged) from a more powerful phosphate donor. Anaerobic and Aerobic Metabolism The efficiency of some biological processes can be gauged in terms of the value of Anaerobic (metabolism where the oxygen inhaled plays no part). Energy source: glycolysis - partial oxidation of glucose to lactic acid The reaction is couple to a reaction where 2 ADP are converted into 2 ATP: Glucose + 2 ADP + 2 Pi → 2 Lactate- + 2 ATP + 2H2O ATP is now “recharged”. Compare with Aerobic Respiration (combustion of glucose) = -2880 kJ/mol Aerobic respiration manages to capture as much of the energy released as possible. So aerobic respiration is a much more efficient use of glucose than anaerobic respiration. 38 ATP molecules are generated for each glucose molecule consumed; 1 mol glucose supplies 2880 kJ of energy 1 mol ATP extracts 30 kJ for storage (30 kJ/mol × 38 mol) = 1140 kJ stored for later use. Endergonic reactions can then be driven by stored energy in ATP. Eg. Protein synthesis is endergonic owing largely to a huge decrease in entropy occurring when amino acids are linked in a precise order. Eg. A peptide link formation requires 3 ATP molecules For myoglobin synthesis (150 peptide links, require 450 ATP molecules) therefore 12 mol glucose per mol protein. 38