Thermodynamics. Lecture 9. The Third Law Free Energy Functions. NC State University

Transcription

1 Thermodynamics Lecture 9 The Third Law Free Energy Functions NC State University

2 Thermodynamics The Third Law NC State University

3 The Third Law of Thermodynamics The third law of thermodynamics states that every substance has a positive entropy, but at zero Kelvin the entropy is zero for a perfectly crystalline substance. The third law introduces a numerical scale for the entropy. Stated succinctly: S(0) = 0 for all perfectly ordered crystalline materials. It is commonly said that all motion ceases at absolute zero. Certainly all translation and rotation have ceased at absolute zero. However, the nuclei still vibrate about their equilibrium positions in so-called zero point motion. However, all molecules are in their lowest vibrational state and entropy is zero in any case provided there are no imperfections in the crystal.

4 Residual Entropy The first statistical picture of entropy was that of Boltzmann. The function W that represents the number of ways we can distribute N particles into a number of states. Using the function W the entropy can be expressed as: S = k B ln W At zero Kelvin the system is in its lowest energy state. For a perfect crystal there is only one way to distribute the energy and W = 1, therefore S = 0. However, the entropy not equal to zero at T = 0 K if the substance is not a perfect crystal. Although the residual entropy in such cases is a small correction to the entropy calculated for chemical reactions it still leads to an important concept.

5 CO: an Imperfect Crystal The molecule CO has a very small dipole moment and there is a finite chance that CO will crystallize as CO:CO:CO instead of CO:OC:CO. For each CO molecule there are two possible orientations of the molecule, therefore there are two ways each CO can exist in the lattice. The number of ways per molecule is w = 2 for each CO. If we have N CO molecules there are w N ways or 2 N ways that all of the CO can be distributed. Therefore, the entropy at zero Kelvin is S = k ln W = k ln(w N ) = Nk ln w = nr ln 2. The entropy at zero Kelvin is known as residual entropy. There are number of substances that show similar statistical variations in orientation that lead to a residual entropy.

6 Residual entropy of ice Because of the multiple hydrogen bond partners possible In ice, there is a residual entropy of S = R ln(3/2). The Bernal-Fowler rule allows the rotation of water molecules within the ice lattice by hydrogen atoms jumping sites. A single hydrogen atom lies on a line between each oxygen atom. The angle between oxygen atoms in ice is109, thus only a small variation of the H-O-H angle in lattice. There are six possible configurations of H atoms around O atoms.

7 The Temperature Dependence of Entropy We have seen calculations for the entropy change for processes. However, it is also possible to calculate the absolute entropy. We can begin with the definition ds = dq rev /T. The heat transferred during a process at constant volume is dq v,rev = C v dt. Thus, the entropy change at constant volume is: S = S(T) S(0) = 0 T C V (T)dT T We have kept the equation general by showing C v (T) as a function of temperature. This calculation of the entropy is valid only at constant V. At constant P we find an analogous expression. Starting with the heat transferred, dq p,rev = C p dt. T We have S = S(T) S(0) = 0 C P (T)dT T

8 The temperature dependence of the heat capacity The third law of thermodynamics shows that C p 0 as T 0 K. It is necessary to treat C p (T) as a function of temperature for this reason alone. The Einstein and Debye theories of heat capacity can be used to determine the functional form of the heat capacity of at these low temperatures. The Debye law states that the heat capacity depends on T 3 near T = 0 K. At temperatures close to T = 0, C V,m = at 3 The constant a is an empirical constant. For practical calculation of the entropy experimental values can be used and the integrals are evaluated numerically.

9 Absolute Entropy The absolute entropy can be calculated from 0 to any temperature T using the integral of the function C P (T)/T. If there is a phase transition between 0 and temperature T we can also calculate the contribution to the entropy from the transition. S(T) = 0 T fus C p (T)dT T + fush T fus + T vap T fus C p (T)dT T + vaph T vap + T T vap C p (T)dT T The heat capacity of each phase is different and the heat capacities are also a function of temperature. Remember that the heat capacity approaches zero and the temperature goes to zero so that the C p is a strong function of T at very low temperature.

10 Graphical representation Gas Solid Liquid S(T) = 0 T fus C p (T)dT T + fush T fus + T vap T fus C p (T)dT T + vaph T vap + T T vap C p (T)dT T

11 Graphical representation Gas Solid Liquid S(T) = 0 T fus C p (T)dT T + fush T fus + T vap T fus C p (T)dT T + vaph T vap + T T vap C p (T)dT T

12 Absolute Entropies can be used to calculate Reaction Entropies Entropies are tabulated in order to facilitate the calculation of the entropy change of chemical reactions. For the general reaction aa + bb yy + zz the standard entropy change is given by r S o = ys o [Y] + zs o [Z] - as o [A] - bs o [B] where the absolute entropies S o are molar quantities. Note the significant difference compared to enthalpy. There is no such thing as an absolute enthalpy. Instead, we used a reference of the elements in their standard states. In that case the enthalpy of formation was set arbitrarily to zero. Since entropies are zero and T = 0 K we can use S o as an absolute quantity.

13 The standard reaction entropy The standard reaction entropy, r S, is the difference between the standard molar entropies of the reactants and products, with each term weighted by the stoichiometric coefficient. r S = S m (products) S m (reactants) The standard state is for reactants and products at 1 bar of pressure. The unit of energy used is J/mol-K. IMPORTANT: Do not confuse entropy and enthalpy. One common mistake is to set the entropies of elements equal to zero as one does for enthalpies of formation. Elements have an entropy that is not zero (unless the temperature is T = 0 K).

14 An example: formation of H 2 O We apply the absolute entropies of H 2, O 2 and H 2 O to the calculation of the entropy of reaction for: 2H 2 (g) + O 2 (g) 2 H 2 O(l) The entropy change is: S = 2 S (H 2 O, l) - S(O 2, g) - 2S(H 2, g) = 2(70) - 2(131) J/mol-K = J/mol-K This result is not surprising when you consider that 3 moles of gas are being consumed. A gas has a greater number of translational degrees of freedom than a liquid. On the other hand, it is difficult (at first) to understand how a spontaneous reaction can have such a large negative entropy.

15 The spontaneity of chemical reactions The spontaneity of chemical reactions must be understood by consideration of both the system and the surroundings. This is one of the most subtle points of thermodynamics. The heat dissipated by the negative enthalpy change in the reaction to make water results in a large positive entropy change in the surroundings. In fact, the reaction to make water proceeds with explosive force. The heat dissipated in the surroundings r H is also an entropy term. The entropy change in the surroundings is: r S surr = rh T The entropy change of the surroundings is: r S = -(-572 kj/mol)/298 K = x 10 3 J/mol-K which is much larger than the negative entropy change of the system (-327 J/mol-K).

16 System and surroundings both play in role in the entropy In an isolated system the criterion ds > 0 indicates that a process is spontaneous. In general, we must consider ds sys for the system and ds surr for surroundings. Since we can think of the entire universe as an isolated system ds total > 0. The entropy tends to increase for the universe as a whole. If we decompose ds total into the entropy change for the system and that for the surroundings we have a criterion for spontaneity for the system that also requires consideration of the entropy change in the surroundings. The free energy functions will allow us to eliminate consideration of the surroundings and to express a criterion for spontaneity solely in terms of parameters that depend on the system.

17 Thermodynamics Free Energy NC State University

18 Free Energy at Constant T and V Starting with the First Law du = dw + dq At constant temperature and volume we have dw = 0 and du = dq Recall that ds dq/t so we have du TdS which leads to du - TdS 0 Since T and V are constant we can write this as d(u - TS) 0 The quantity in parentheses is a measure of the spontaneity of the system that depends on known state functions.

19 A Definition of Helmholtz Free Energy We define a new state function: A = U -TS such that da 0. We call A the Helmholtz free energy. At constant T and V the Helmholtz free energy will decrease until all possible spontaneous processes have occurred. At that point the system will be in equilibrium. The condition for equilibrium is da = 0. time

20 Definition of Helmholtz Free Energy Expressing the change in the Helmholtz free energy we have A = U T S for an isothermal change from one state to another. The condition for spontaneous change is that A is less than zero and the condition for equilibrium is that A = 0. We write A = U T S 0 (at constant T and V) If A is greater than zero a process is not spontaneous. It can occur if work is done on the system, however. The Helmholtz free energy has an important physical interpretation. Noting the q rev = T S we have A = U q rev According to the first law U q rev = w rev so A = w rev (reversible, isothermal) A represents the maximum amount of reversible work that can be extracted from the system.

21 Definition of Gibbs Free Energy Most reactions occur at constant pressure rather than constant volume. Using the facts that dq rev TdS and dw rev = -PdV we have: du TdS PdV which can be written du - TdS + PdV 0. The = sign applies to an equilibrium condition and the < sign means that the process is spontaneous. Therefore: d(u - TS + PV) 0 (at constant T and P) We define a state function G = U + PV TS = H TS. Thus, dg 0 (at constant T and P) The quantity G is called the Gibb's free energy. In a system at constant T and P, the Gibb's energy will decrease as the result of spontaneous processes until the system reaches equilibrium, where dg = 0.

22 Comparing Gibbs and Helmholtz The quantity G is called the Gibb's free energy. In a system at constant T and P, the Gibb's energy will decrease as the result of spontaneous processes until the system reaches equilibrium, where dg = 0. Comparing the Helmholtz and Gibb's free energies we see that A(V,T) and G(P,T) are completely analogous except that A is valid at constant V and G is valid at constant P. We can see that G = A + PV which is exactly analogous to H = U + PV the relationship between enthalpy and internal energy. For chemical processes we see that G = H T S 0 (at constant T and P) A = U T S 0 (at constant T and V)

23 Conditions for Spontaneity We will not use the Helmholtz free energy to describe chemical processes. It is an important concept in the derivation of the Gibbs energy. However, from this point we will consider the implications of the Gibbs energy for physical and chemical processes. There are four possible combinations of the sign of H and S in the Gibbs free energy change: H S Description of process >0 >0 Endothermic, spontaneous for T > H/ S <0 <0 Exothermic, spontaneous for T < H/ S <0 >0 Exothermic, spontaneous for all T >0 >0 <0 Never spontaneous

24 Gibbs energy for a phase change For a phase transition the two phases are in equilibrium. Therefore, G = 0 for a phase transition. For example, for water liquid and vapor are in equilibrium at K (at 1 atm of pressure). We can write vap G m = G m H 2 O(g) G m H 2 O(l) where we have expressed G as a molar free energy. From the definition of free energy we have vap G m = vap H m T vap S m The magnitude of the molar enthalpy of vaporization is 40.7 kj/mol and that of the entropy is J/mol-K. Thus, vap G = kj mol K J K 1 mol 1 = 0

25 Gibbs energy for a phase change However, if we were to calculate the free energy of vaporization at K we would find that it is +1.1 kj/mol so vaporization is not spontaneous at that temperature. If we consider the free energy of vaporization at K it is kj/mol and so the process is spontaneous ( G < 0).

26 State Function Summary At this point we summarize the state functions that we have developed: U (internal energy) H = U + PV (enthalpy) S (entropy) A = U - TS (Helmholtz free energy) G = U + PV - TS = H - TS (Gibbs free energy) Please note that we can express each of these in a differential form. This simply refers to the possible changes in each function expressed in terms of its dependent variables. dh = du + PdV + VdP da = du - TdS - SdT dg = dh - TdS - SdT

27 The internal energy expressed in terms of its natural variables We can use the combination of the first and second laws to derive an expression for the internal energy in terms of its natural variables. If we consider a reversible process: du = dq + dw dw = -PdV (definition of work) dq = TdS (second law rearranged) Therefore, du = TdS - PdV This expression expresses the fact that the internal energy U has a T when the entropy changes and a slope -P when the volume changes. We will use this expression to derive the P and T dependence of the free energy functions.

28 The Gibbs energy expressed in terms of its natural variables To find the natural variables for the Gibbs energy we begin with the internal energy: du = TdS - PdV and substitute into: dh = du + PdV + VdP to find: dh = TdS + VdP (S and P are natural variables of enthalpy) and using: dg = dh - TdS - SdT we find: dg = -SdT + VdP (T and P are natural variables of G) Once again we see why G is so useful. Its natural variables are ones that we commonly experience: T and P.

29 The variation of the Gibbs energy with pressure We have shown that dg = VdP - SdT. This differential can be used to determine both the pressure and temperature dependence of the free energy. At constant temperature: SdT = 0 and dg = VdP. The integrated form of this equation is: G = P 2 P 1 VdP For one mole of an ideal gas we have: G m = RT P 2 dp = RT ln P 2 P P 1 P 1 Note that we have expressed G as a molar quantity G m = G/n.

30 The variation of the Gibbs energy with pressure We can use the above expression to indicate the free energy at some pressure P relative to the pressure of the standard state P = 1 bar. G m T = G 0 T + RT ln P 1 bar G 0 (T) is the standard molar Gibb s free energy for a gas. As discussed above the standard molar Gibb s free energy is the free energy of one mole of the gas at 1 bar of pressure. The Gibb s free energy increases logarithmically with pressure. This is entirely an entropic effect. Note that the 1 bar can be omitted since we can write: RT ln P 1 bar = RT ln P RT ln 1 = RT ln P

31 The pressure dependence of G for liquids and solids If we are dealing with a liquid or a solid the molar volume is more or less a constant as a function of pressure. Actually, it depends on the isothermal compressibility, k = -1/V( V/ P) T, but k is very small. It is a number of the order 10-4 atm -1 for liquids and 10-6 atm -1 for solids. We have discussed the fact that the density of liquids is not strongly affected by pressure. The small value of k is another way of saying the same thing. For our purposes we can treat the volume as a constant and we obtain G m T = G 0 T + V m P 1

32 Systems with more than one component Up to this point we have derived state functions for pure systems. (The one exception is the entropy of mixing). However, in order for a chemical change to occur we must have more than one component present. We need generalize the methods to account for the presence of more than one type of molecule. In the introduction we stated that we would do this using a quantity called the chemical potential. The chemical potential is nothing more than the molar Gibbs free energy of a particular component. Formally we write it this way: i = G n i T,P,j i Rate of change of G as number of moles of i changes with all other variables held constant.

33 Example: a gas phase reaction Let s consider a gas phase reaction as an example. We will use a textbook example: N 2 O 4 (g) = 2 NO 2 (g) We know how to write the equilibrium constant for this reaction. 2 K = P NO 2 P N2 O 4 At constant T and P we will write the total Gibbs energy as: dg = NO2 dn NO2 + N2 O 4 dn N2 O 4 dg = 2 NO2 dn N2 O 4 dn We use the reaction stoichiometry to obtain the factor 2 for NO 2.

34 Definition of the Gibbs free energy change for chemical reaction We now define rxn G: rxn G = dg dn T,P This is rxn G but it is not rxn G o! Note that we will use rxn G and G interchangably. If we now apply the pressure dependence for one component, to a multicomponent system: G m T = G 0 T + RT ln i T = i0 T + RT ln P 1 bar P i 1 bar These two expressions are essentially identical. The chemical potential, i, is nothing more than a molar free energy.

35 Application of definitions to the chemical reaction We can write the Gibbs energy as: G = 2 NO2 N2 O 4 and use the chemical potentials: 0 NO2 = NO2 + RT ln P NO2 0 N2 O 4 = N2 O 4 + RT ln P N2 O 4 to obtain the following: 0 0 G = 2 NO2 N2 O 4 + 2RT ln P NO2 RT ln P N2 O 4 G = G 0 + RT ln P NO 2 2 P N2 O 4, G = 2 NO2 N2 O 4

36 Note the significance of G and G o The change G is the change in the Gibbs energy function. It has three possible ranges of value: G < 0 (process is spontaneous) G = 0 (system is at equilibrium) G > 0 (reverse process is spontaneous) On the other hand G o is the standard molar Gibbs energy change for the reaction. It is a constant for a given chemical reaction. We will develop these ideas for a general reaction later in the course. For now, let s consider the system at equilibrium. Equilibrium means G = 0 so: 0 = G 0 + RT ln K, G 0 = RT ln K, K = P NO 2 2 P N2 O 4

37 Temperature dependence of G o The van t Hoff equation We use two facts that we have derived to determine the temperature dependence of the free energy. G 0 = RT ln(k) G 0 = H 0 T S 0 ln(k) = H0 RT + S0 R ln(k 2 ) ln(k 1 ) = H0 R 1 T 1 1 T 2 If we plot ln(k) vs 1/T the slope is - H o /R. This is a useful expression for determining the standard enthalpy change.

38 Van t Hoff plots Slope = - H o /R Note: H o > 0 The standard method for obtaining the reaction enthalpy is a plot of ln K vs. 1/T

39 van t Hoff plot for exothermic process Slope = - H o /R Note: H o < 0 In this example, the slope is positive because the enthalpy of binding is negative (i.e. binding is exothermic).

40 Pressure dependence We can see from the gas phase form of the equilibrium constant that the equilibrium concentrations of species depend on pressure. This dependence is inside the equilibrium constant. The equilibrium constant is not changed by the species are affected for a given value of K. The pressure dependence of free energy contained in the the term dg = VdP is a different dependence, which can result in a shift in the equilibrium constant itself.

41 Pressure dependence of species We can see from the gas phase form of the equilibrium constant that pressure of species depend on pressure. For the general gas phase reaction, we can write the equilibrium constant as And the free energy is From Dalton s law

42 Pressure dependence of species If we substitute these mole fractions and total pressure into the equilibrium constant we have Which depends on the total pressure unless z c d = 0. This expression shows that, in general, the free energy depends on the total pressure. This means that for the fixed pressure may affect the proportion of products to reactants.

43 Pressure dependence N 2 O 4 (g) 2 NO 2 (g) In the last lecture we treated this problem using If there is a constraint on the total pressure then We must use Dalton s law to calculate the mole Fractions.

44 Pressure dependence The equilibrium constant is N 2 O 4 (g) 2 NO 2 (g) For a given initial pressure of N 2 O 4 we have N 2 O 4 NO 2 Total Initial 0 Delta -x 2x +x Final

45 To solve a gas phase problem you need to know whether the pressure can change or not. If it is fixed, then P tot has a value that does not change. The mole fractions are, For example, if we assume that the initial pressure is maintained, then the above values should be substituted into the equilibrium constant

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47 Using the same values as yesterday (K = 0.11 and an initial pressure of 1 atm), the calculated value for x = Therefore, the pressures are: N 2 O 4 The percent dissociation is 28%. If the total pressure is reduced to 0.1 atm, the value of x = and the pressure of NO 2 is This Corresponds to a dissociation of 63%. Can this trend be explained in simple terms?

48 Shift in equilibrium due to pressure: the formation of diamond Graphite and diamond are two forms of carbon. Given that the free energy of formation of diamond is: C(s, graphite) C(s, diamond) r G o = kj/mol and the densities are: r(graphite) = 2.26 and r(diamond) = 3.51 calculate the pressure required to transform carbon into diamond. Solution: Graphite will be in equilibrium with diamond when: 0 = G 0 + V m P 1 P = 1 G0 V m = 1 G0 M r d M r gr

49 Example: the formation of diamond Plugging the values we find: 0 = G 0 + V m P 1 P = 1 G0 = 1 G0 V m M r M d r gr = J/mol kg/mol = 1.5 x 10 9 Pa = bar