Fundamental equations of Thermodynamics, Study notes of Thermodynamics

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Fundamental equations of Thermodynamics Fundamental equations of Thermodynamics (1) The combined first and second law dWdqdU +=From the first law: From the second law: T dq dS ≥ Where, for irreversible system T dq dS > and, for reversible system dq dS = T For a closed system in which only reversible pV work is involved pdVdW −= and T dq dS = Fundamental equationpdVTdSdU −=∴ The internal energy is a function of S and V Where U, T, S, P, and V are state functions 2 U and H provide criteria for whether a process can occur spontaneously in a system when the corresponding natural variables are held constant. dVpdW, T dq dS ext−=≥From: At infinitesimal change (rev.) with constant S and V “A change in a process can occur spontaneously if the internal energy decreases dWdqdU +=And substitute in : dVpTdSdU ext−≤∴We obtain: 0)dU( V,S ≤∴ when the change occurs at constant entropy and volume” The meaning: dU = zero equilibrium dU < zero spontaneous dU > zero non-spontaneous At constant S and p dH = zero equilibrium dH < zero spontaneous dH > zero non-spontaneous 0)dH( p,S ≤∴ 5 Helmholtz Energy (A) TSUA −=It is defined by: T and V are the natural variables of A SdTTdSdUdA −−=By differentiating: pdVTdSdU −=∴But pdVSdTdA −−=∴ If an infinitesimal change takes place in a system of constant T and V, thus: For irreversible process, A decrease. 6 0)dA( V,t ≤ For reversible process, A is constant. It is more practical to use the criterion 0)dA( V,t ≤ The Helmholtz energy can be used to show that the pressures of two phases must be equal at equilibrium For two-phase system in a container of fixed volume surrounded by a heat reservoir When the system at equilibrium Suppose that the volume of phase α is increased by dV and the volume of phase ß is decreased by dV. So, the total volume is constant pdVSdTdA −−=∴ pß p α dA = 0 = dAα + dAß 7 -pαdV +pß dV=0 pα= pß Thus for a spontaneous process at constant T and p the Gibbs energy must decrease. If the process is at equilibrium then dG = 0. This criteria is the most important developed so far because it leads to all future analyses of spontaneous and equilibrium processes such as equilibrium constants, electrode potentials and the Nernst equation, gas, liquid and solid equilibria, solution processes, etc. etc. While there are now two separate conditions for the two different paths we will focus almost exclusively on the Gibbs Energy since the constant pressure path is so much more convenient to establish experimentally. We only need to do it in the open under constant atmospheric pressure. Thus we will have tables of values for ∆G rather than ∆A. Recall that this is exactly the same reason why we focus on H and not U and why there are tables of ∆H and not ∆U. Thus to determine if a constant T, p process will be spontaneous we only need to find whether the change in one state function dG is negative i.e. that the state function G decreases. How does G behave overall as the state of a system changes? Since the new criteria for spontaneous change is that G must decrease, we can sketch how it must behave as a system changes spontaneously. From the plot we can see that the system naturally tends to roll down a Gibbs energy hill until it reaches the lowest point. The system is then at equilibrium. We can see that a system tends to stay in a state G of equilibrium because it must climb a Gibbs energy hill to get out. Thus the reason for change, the lowering of the Gibbs free energy can be viewed as a “driving force” or a tendency for change. The steeper the hill the greater the tendency for the system to change. The Gibbs (free) energy driving force is relatively simple in that it is made up of two recognizable factors, the enthalpy and entropy. It is useful in analyzing how each of these factors separately behaves when a change occurs i.e. the Gibbs energy changes. In general, any process in which the enthalpy (or energy) decreases is favorable to a decrease in G and any process in which the entropy goes up is also favorable to a decrease in G. In other words, systems, like most people, seek a position of minimum energy and maximum disorder. Thus the Gibbs energy, enthalpy (energy) and entropy are the three main properties to keep uppermost in mind when thinking about a change of state. This leads to minimum Gibbs energy and a state from which a system is reluctant to move. ANALYSIS OF ∆G IN TERMS OF ∆H AND ∆S For useful work to be done, w, ∆G and (almost always) ∆H must be negative. If the entropy term T∆S is positive [ then - T∆S will be negative] it makes ∆G more negative than just ∆H and so increases the work done for us. In this case T∆S = qrev > 0 i.e. heat is transferred from the surrounding to the system to help fuel the work. In the reverse case where the system entropy decreases T∆S < 0 [ -T∆S is positive and ∆G is not as negative as ∆H] heat must flow from the system to the surroundings and so all of the ∆H is not available to do work.. The heat flowing to the surroundings increases the entropy of the surroundings sufficiently so that the process remains spontaneous. Thus part of the energy of the system has to be sacrificed in order to maintain spontaneity. 15 0) For adds to. AH subtracts from AH --| 1 | AH TAS AH TAS AG AG = AH-TAS = AH-TAS AS=0 maximum net work AG equal to |.AH | AS > more than 1.AH | AS <0 less than |.AH | 16 Example: Check if this process is spontaneous or not? NH4Cl(s) + H2O → aqueous solution at 25 oC ∆Hθ(solution) = + 34.7 kJ mol-1 (endothermic) ∆Sθ(solution) = + 167.1 J K-1 mol-1 ∆Gθ(solution) =∆Hθ (solution) - T∆Sθ (solution) unfavorable to spontaneity favorable to spontaneity = 34.7 kJ mol-1 – 298 K (.1671 kJ K-1 mol-1) = -15.1 kJ mol-1 In this case a favorable entropy change overcomes an unfavorable energy change. In fact, this process is spontaneous at any temperature above: 17 dev =—Wyq = ORT In| Pe! py! =1molx 8314] K' mot'x 300K Inf 0/1) = 5.74 k) as —fe 574 8 og gr Kt T 300K AG=AH-TAS =0- 300Kx.01I9kI K' mol! = -3.74 kJ 20 STANDARD MOLAR GIBBS ENERGY In the same way that standard heats of formation were defined for compounds we define standard molar Gibbs free energy of formation. This will allow us to calculate free energies of reactions at 298 K and hence whether a reaction will be spontaneous IF the reaction is carried out at constant temperature and pressure. ∆fG θ ≡ Standard Gibbs (free energy) of formation of a compound [formed from constituent elements in their standard state]. Values in tables are given for T = 298 K ∆ Gθ = 0 for elements in their standard state i.e. for O (g), I (s), C(s; graphite)f 2 2 The standard free energy of the reaction is obtained in the same manner as the heat of reaction and the entropy of reaction. The standard Gibbs energy of formation of a compound and the standard Gibbs energy of a reaction at constant temperature can also be calculated from heats of reaction and entropies of reaction using: ∆Hθ is almost independent of the temperature ∆Sθ is moderately dependent on the temperature ∆Gθis strongly dependent on the temperature 21 Example: oxidation of α-D glucose C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) If the ∆Gθ for CO2(g) = -394.4 kJ mol-1, ∆Gθ for H2O(g)= -237.2 kJ mol-1 , ∆Gθ for C6H12O6(s) = -910.9 kJ mol-1 Calculate ∆Gθ = ∆GθProduct - ∆GθReactants ∆Gθ =[ 6 ∆Gθ CO2(g) + 6 ∆Gθ H2O(l))]- [∆Gθ C6H12O6(s) + 6 ∆Gθ O2(g)] ∆Gθ = 6(-394.4) + 6(-237.2) - 1(-910.9) - 0 = -2879 kJ mol-1 If this reaction is carried out at constant temperature and pressure it would be spontaneous. Is this reaction is spontaneous or not? Example: Iodine sublimes at 25 °C as I2(s) → I2(g) where the heat and entropy of sublimation are ∆H = 39.37 kJ mol-1 and ∆S = 86.19 J K-1 mol-1. What is equilibrium sublimation temperature if ∆H and ∆S are assumed to be independent of temperature. We use the fact that at equilibrium ∆G = ∆H – T ∆S = 0 22 Using the fundamental equation we can obtain new equations for dH, dA and dG and use the same procedures as above to obtain three more equations like [I] as well as three other Maxwell Relations. These turn out to be particularly useful in manipulating partials as will be shown shortly. G = U – TS + pV ∴dG = dU – TdS –SdT + pdV + Vdp H = U + pV and G = H – TS For a constant T and p process dT = 0 and dp = 0 and thus Substitute the fundamental equation dU = TdS – pdV into the equation for dG to get: dG = Vdp – SdT This equation for dG suggests that we take p and T as the variables for the Gibbs energy. pT G G dG dp dT p T  ∂ ∂  = +    ∂ ∂   Comparing coefficients of dp and dT for the two equations gives S T G andV p G pT −=      ∂ ∂ =      ∂ ∂ Applying the state function condition for G p T V S T p  ∂ ∂  = −   ∂ ∂    This is a Maxwell Relation The four Maxwell relations are ∂ ∂    p T V S T p  ∂ ∂  = −    ∂ ∂   V T p S T V ∂ ∂    =    ∂ ∂     ∂ ∂ S V T p V S =−    ∂ ∂   PS T V p S   =    ∂ ∂   Maxwell relations allow us to develop different equations pdVTdSdU −=∴ The differential of U dV dV dU dS dS dU dU SV       +      =∴ Thus, we can calculate T and p as dU T     =∴ dU p     =−and Equation I VdS  SdV  S and V are natural variables of U represented as U(S,V) 30 Also since U is a state function we can apply the state function condition This is a Maxwell Relation V SS V U U V S S V    ∂ ∂ ∂ ∂    =       ∂ ∂ ∂ ∂       S V T p V S ∂ ∂    = −    ∂ ∂    Using the fundamental equation we can obtain new equations for dH, dA and dG and use the same procedures as above to obtain three more equations like [I] as well as three other Maxwell Relations. These turn out to be particularly useful in manipulating partials as will be shown shortly. 31 G = U – TS + pV ∴dG = dU – TdS –SdT + pdV + Vdp H = U + pV and G = H – TS For a constant T and p process dT = 0 and dp = 0 and thus Substitute the fundamental equation dU = TdS – pdV into the equation for dG to get: dG = Vdp – SdT This equation for dG suggests that we take p and T as the variables for the Gibbs energy. pT G G dG dp dT p T  ∂ ∂  = +    ∂ ∂   Comparing coefficients of dp and dT for the two equations gives S T G andV p G pT −=      ∂ ∂ =      ∂ ∂ Applying the state function condition for G p T V S T p  ∂ ∂  = −   ∂ ∂    This is a Maxwell Relation 32 THE CHANGE of GIBBS ENERGY WITH T AND p It is easy to determine the spontaneity of a reaction at 1 atm and 298 K since we can get the free energy of a reaction from the free energy of formation of compounds given in tables. But, there are many reactions that are carried out at very different temperature and pressure conditions. It is important to know how the free energy changes with temperature and pressure if we want to determine spontaneity at any temperature and pressure. This is especially true for temperature changes since G is strongly dependent on temperature for solids, liquids, gases and solutions. Pressure effects on G are substantial only for gases. 35 constant pressure \ i \ constant temperature liquid ee TEMPERATURE PRESSURE 36 Change of G with temperature The change of G with T was given by S T G p −=      ∂ ∂ This is the slope of the plots of G vs T The fact that the slope is just the negative of the entropy makes the interpretation of the plots very simple. The entropy of a gas is much larger than that of a liquid which in turn is larger than that of the solid. Thus, it is easy to rationalize the relative steepness of the G vs T lines for gas, liquid and solid. 37 Change of G with pressure: It was proved before that V p G T =      ∂ ∂ The slope of the plot of G vs p is just the volume of the system. Since V is always positive, the free energy must always increase with pressure. The slopes of the plot for the gas, liquid and solid should decrease in that order since the molar volumes of these phases decreases as we go from gas to liquid to solid. For a finite change in ∆G when p changes : V p G T ∆=      ∂ ∆∂ This equation can be applied to the water ↔ steam equilibrium at 100 °C, 1 atm. What will happen to the equilibrium if the pressure is decreased to 0.5 atm? T G G V G ( V) p p p  ∂∆ δ∆ ≈ = ∆ ∴δ∆ = ∆ δ  ∂ δ  40 BUT ∆V = V(steam) – V(water) >> 0 and δp = 0.5 – 1.0 = - 0.5 < 0 and as a result δ∆G < 0 Water turning into steam (water vapor) will occur spontaneously Deriving equation for the change of Gibbs energy with pressures at constant temperature: From the above partial derivative, at constant temperature: T T T T G dG V and dG | Vdp | p dp  ∂ = = =  ∂  Thus the understanding that for this integration, T is constant dG Vdp=∫ ∫ f i p f i p G(p ) G(p ) V(p,T)dp− = ∫ Always true for any isothermal process 41 )pp(VGG 1212 −+=∴ a) For solids and liquids ( V is independent of p) f i p f i p G(p ) G(p ) V(p,T)dp− = ∫ )pp(VGG oo −+= Where, Go and po are the standard value Since for most solids and liquid, the molar volume is small the value of ∆G is also very small unless the pressure change is huge. Thus as a first approximation we can say that G is independent of pressure for solids and liquids. This is seen in the plot of G vs p where the lines for liquid and solid are almost flat. 42 ∆fG O (Standard Gibbs energy of formation) for liquid CH3OH at 298K is -166.27 kJmol-1, and that for gaseous CH3OH is -161.96 kJmol-1 . The density of the liquid methanol at 298K is 0.7914 g.cm-3. Calculate ∆fG (CH3OH, g) and ∆fG (CH3OH, liq ) at 10 bar at 298 K and ∆fG (CH3OH, g) ∆fG = ∆fG o +nRT ln(p/po) = -161.96 kJmol-1+(1 mol x 8.314J K-1 mol-1x 298K ln(10/1) = -156.25 kJmol-1 ∆fG (CH3OH, liq) 1 bar = 100 kPa ∆fG = ∆fG o + Vm(p-po) and Vm= molar mass/ density = 32 g mol-1 / 0.7914 g.cm-3= 40.49 cm3 mol-1 Vm= 40.49 x 10-6 m3 mol-1 Vm(p-po) = 40.49 x 10-6 m3 mol-1 (10 x 105- 1 x 105)Pa x (1 x 10-3 kJJ-1) ∆fG = -166.27 kJmol-1+ 40.49 x 10-6 m3 mol-1 (10 x 105- 1 x 105)Pa x (1 x 10-3 kJJ-1) = -166.23 kJmol-1 45 An ideal gas at 27oC expands isothermally and reversibly from 10 to 1 bar against a pressure that is gradually reduced. Calculate q, W, ∆U, ∆H, ∆G, ∆A, and ∆S. Analysis: Isothermally and reversibly, ∆U, ∆H, equal zero Gas expand, W is negative, q = -W, ∆A is negative, ∆S is positive and ∆G is negative Calculations: Wmax = ∆A = -RTln(p1/p2) = -(8.314 JK-1 mol-1)(300 K) ln (10/1) = -5746 J mol-1 46 q = - Wmax = 5746 J mol-1 ∆U = ∆H = 0 ∆S = (qrev/T )= (5746 J mol-1 )/(300 K) = 19.14 J K-1 mol-1 ∆G = ∆H - T ∆S= 0 – (300 K) (19.14 J K-1 mol-1 ) = 5746 Jmol-1 OR ∆G = RTln(p2/p1) = (8.314 JK-1 mol-1)(300 K) ln (1/10) = 5746 Jmol-1 INTRODUCTION OF THE CHEMICAL POTENTIAL The change in number of moles of any substance must be considered. Starting with pure substances and determine how the Gibbs function will change if infinitesimal amount of the same substance is added at constant temperature and pressure. This partial derivative is defined as the Chemical Potential : pTn G ,       ∂ ∂ ≡µ For a pure substance , µ is just the molar free energy Gm. 47 G = n Gm pT m m pT m n G nG n nG ,,       ∂ ∂ +=      ∂ ∂ ≡µ But for a pure substance Gm is constant with respect to the n and thus µ = Gm. This is true of any molar quantity. For example, the molar volume Vm of water is 0.018 L mol-1 and is independent of whether we are talking about 1 mole or 23 moles. Molar quantities are “intensive” like density - they don’t depend on the amount. If fugacity is an “effective pressure” ie the pressure that gives the right value for the chemical potential of a real gas. The only way we can get a value for fugacity and hence for µ is from the gas pressure. Thus we must find the relation between the effective pressure f and the measured pressure p. let f = φ p φ is defined as the fugacity coefficient. φ is the “fudge factor” that modifies the actual measured pressure to give the true chemical potential of the real gas.     +µ=µ θ θ f lnRT)T()T,p( 50 By introducing φ we have just put off finding f directly. Thus, now we have to find φ. Substituting for φ in the above equation gives: p (p,T) (T) RT ln RT ln (ideal gas) RT ln p θ θ   µ = µ + + φ = µ + φ    φ=µ−µ lnRT)gasideal()T,p( This equation shows that the difference in chemical potential between the real and ideal gas lies in the term RT ln φ. This is the term due to molecular interaction effects. p The equation relating f or φ to the measured pressure p: Note that as p → 0, the real gas → ideal gas, so that f→p and φ→ 1 The chemical potential for an ideal gas and a real gas at two pressures p and p′ is p ideal,m ideal ideal ideal p p V dp d (p,T) (p ,T) RT ln p′   ′= µ = µ − µ =  ′  ∫ ∫ p m p f V dp d (p,T) (p ,T) RT ln f′   ′= µ = µ − µ =  ′  ∫ ∫ Subtracting the first equation from the second gives 51 ( ) p m ideal,m p f p V V dp RT ln RT ln f p′    − = −   ′ ′    ∫ or ( ) p m ideal,m p f / p 1 ln V V dp f / p RT ′   = − ′ ′  ∫ Let p′ → 0, then in the initial state the real gas → the ideal gas. Thus f′ → p′ ( ) p m ideal,m 0 f 1 ln V V dp p RT   = −    ∫ ideal,m RT V p = m RT V Z p   =     Since for an ideal gas and for a real gas where Z is the compressibility factor p p 0 0 f 1 RT RT Z(p,T) 1 ln Z dp dp ln p RT p p p         − = − = = φ                  ∫ ∫ p Z(p,T) 1 f p exp dp p   −  =        ∫ 52 0 The fugacity coefficient φ = f/p is given by             − =φ ∫ p 0 dp p 1)T,p(Z exp Thus the fugacity of a gas is readily calculated at same pressure p if Z is known as a function of pressure up to that particular pressure.