Effect of adding an ideal inert gas, M

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Derick Hudson
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1 Effect of adding an ideal inert gas, M Add gas M If there is no change in volume, then the partial pressures of each of the ideal gas components remains unchanged by the addition of M. If the reaction were taking place in an isolated system, M would almost surely also affect the final temperature G = RT ln K P aa + bb yy + zz K K = y ( P ) ( P ) ( P ) ( P ) y z m ( P ) ( P ) ( P ) o Y Z M = P a b m ( P ) ( P ) ( P ) A B M y ( P ) ( P ) Y a Z ( P ) ( P ) A z o Y Z = P a b mm+ aa + bb yy + zz+ mm A B B z b No change in any partial pressures. K is unchanged! However, M might well affect the rate at which the process occurs.

2 Change the volume Changing the volume can have an effect if the number of moles of gas on the left and right sides of the equation are different. N 2 O 4 2 NO 2 Let α be the fraction of the dimer that has dissociated. N(1-α) 2Nα N in moles Total number of moles of gas = N(1 - α) + 2Nα = N(1 + α) We now write the mole fractions χ of each component: χ N O ( ) ( ) ( ) N χno ( ) N ( ) ( ) N 1 α 1 α 2α 2α = = = = N 1+ α 1+ α 1+ α 1+ α 2 4 2

3 Change the volume (2) We use the mole fractions to obtain the partial pressures of each component: 1 α 2α P = P P = P N2O 4 NO2 1+ α 1+ α Use these to write the pressure equilibrium constant Solve for α: α = K P P 2 P 2 NO 4α 2 = = P 2 P α NO KP K + 4P The degree of dissociation decreases when the pressure increases. Another manifestation of the Le Chatelier Principle

4 Change quantity of one component 2 NO 2 (g) N 2 O 4 (g) K P P 2 NO = 2 P NO The system will shift in a direction away from the component that was increased. This is a method that is often used to determine whether or not equilibrium has actually been attained. (Rather than simply having an equilibration time >> observation time) 2 4

5 Effect of Temperature on Equilibrium We know G = RT ln K P We couple this ΔG relationship with the Gibbs-Helmholtz equation derived in Chapter 3, 0 0 G H = (3.56) 2 T T T P to obtain an expression that gives the temperature dependence of K ( ) P d 0 ln K 0 P = H the van't Hoff equation 2 dt RT Since ( 0 ) 1 dt d ln K P H d =, we can rewrite, giving = 2 T T d T R ( 1/ ) This last form shows that a plot of ln K P vs 1/T has slope - H /R 0

6 The van t Hoff Equation G = RT lnk P lnk lnk dlnk d(1/t) P 0 P = G RT 0 0 P H = 0 H = + RT 0 0 R S R Very useful!

Note that the positive slope means that H is negative, an exothermic reaction.

7 Effect of Temperature on Equilibrium This last form shows that a plot of ln K P vs 1/T has slope - H /R, if H does not depend strongly on temperature. Note that the positive slope means that H is negative, an exothermic reaction. Here H has a large T dependence. Can the slope be negative? What would a negative slope imply? The Y intercept of this line is also interesting!

8 Effect of Temperature on Equilibrium The Y intercept is also interesting! We know that So ln G = RT ln K P ( K ) P G H T S = = + RT R T RT The y-axis is located at 1 T So the value of the intercept is S /R = 0 These plots can be a very useful way to determine basic thermochemical properties of molecules and reactions. S/R

9 Mixed Phase Equilibria Consider CaCO 3 (s) CaO (s) +CO 2 (g) As always, CaO CO 2 = K ' C CaCO3 However, the concentrations of solids or liquids are constant, and it is conventional to incorporate them into the equilibrium constant. Thus CaCO ' 3 = = CO K, where K K 2 C C C CaO A similar thing is done in a saturated solution Consider KCl (s) K + (aq)+ Cl - (aq) + K Cl - ' = K, but we combine to get + K - Cl = K C KCl SP

10 Partial Molar Quantities Recall that G of a pure substance is an extensive property. Thus if G m is the molar free energy of the substance, then the Gibbs energy of k moles of the substance is k G m. The same is true for U, H and A. We encountered the concept of partial molar properties a couple of lectures ago, discussing the entropy of mixing gases. This was done when there was no possibility of reaction. Let s revisit this process when reactions are possible. Consider a three component mixture of ideal gases at pressures P A, P B, and P C. P = xp, where x is the mole fraction of component c c c c and P is the total pressure nc x = with n = n + n + n + c A B C n Thus P + P + P + = x + x + x + P = P ( ) A B C A B C

11 Partial Molar Quantities That last relation is true for both ideal and real gases. When the gases are perfect, the partial pressures as defined are also the pressures that each component would have if it were the only gas in the container. This is Dalton s law from earlier in the term. For real gases, the interactions between molecules change the behavior away from ideal. We will use these ideas to introduce partial molar volumes

12 Partial Molar Volume Consider a large volume of water at 25 C. When I add one additional mole of water at 25 C to the pot, the volume increases by 18 cm 3. We have the result that 18 cm 3 /mol is the molar volume V molar of pure water at 25 C. When we note that the volume V of n moles of water is n H2O V molar, we see a mathematical way to generalize the result: V molar ( HO) 2 V = nho 2 PT, While it makes no difference here, this construct will be very valuable when dealing with multicomponent systems

13 Partial Molar Volume for an ethanol-water mixture 25 C H 2 O C 2 H 5 OH

14 Partial Molar Volume for an ethanol-water mixture 25 C H 2 O C 2 H 5 OH

15 Partial Molar Volume for an ethanol-water mixture V molar ( A) V = n A PT, V molar can be positive, negative or zero!

16 Two Component (A and B) P,T, all other n Molar Volume Definition: Molar volume, V A, of substance A V V = A n A So V V dv = d n + d n = V d n + V d n n n A B A Integrating, A B A A B B PT,, n B PT,, n n A n B n n nv nv V Vd Vd = + = + A A B B A A B B 0 0 Correct, IF V A and V B do not depend on n A and n B, respectively. We now apply the same concepts to the Gibbs Free Energy

17 Partial Molar Gibbs Energies We can extend this concept to any extensive state function! For a substance in a mixture, the partial molar Gibbs Energy has a well-known name, the chemical potential, μ. µ J G G = n' all n J n n J J P,T,all other n P,T, n' In analogy to partial molar volumes in a binary mixture, G = n µ + n µ A A B B where μ A and μ B are the chemical potentials of A and B at the composition of the mixture.

18 Partial Molar Gibbs Energies µ J G n J p, T, n'

19 Partial Molar Gibbs Energies Remember from last time that dg = VdP SdT for constant composition! But for a multicomponent system, G depends on composition, G = n µ + n µ + n µ + (,,,, ) G = G pt n n n and A A B B C C A B C G G G G dg = dp + dt + dn + dn + B p T n n A T,n' p,n' A p,t,n' B p,t,n'

20 Chemical Potential (7) So dg = Vdp SdT + μ A dn A + μ B dn B + μ C dn C +... the fundamental equation of chemical thermodynamics Suppose we carry out a process at constant temperature and pressure dg = μ A dn A + μ B dn B + μ C dn C +... We see that dg under these conditions gives the maximum amount of non-expansion work available by simply changing the composition of the system. electrochemical cells

21 Chemical Potential (8) The chemical potential appears in many other contexts. Thus, since G = H -TS = U + pv TS, we have U = pv + TS + G, and du = -pdv Vdp +TdS +SdT +dg = -pdv Vdp +TdS +SdT +Vdp -SdT +μ A dn A +μ B dn B +... = -pdv +TdS +μ A dn A +μ B dn B +... This is the composition-dependent generalization of eq 3.46 from earlier Now, at constant V and S, du= μ A dn A +μ B dn B +..., which means that U = etc. eq 5.10, text B n B S,V,n' Thus the chemical potential also shows how internal energy changes with composition. µ

22 Chemical Potential (9) This result can be derived for all extensive properties to see how they all change with composition. µ µ µ B B B U = n B A = n B H = n B S,V,n' T,V,n' S,P,n' etc. etc. etc. eq 5.10, text eq 5.11, text eq 5.11, text Thus we see that the chemical potential µ tells us how all of the extensive thermodynamic properties, U, A, H and G, depend upon composition. This is really why this concept is so central!.

23 General Chemistry revisited A(g) + B(g) C(g) + D(g) We said that G = H T S where, eg, H = f H(C) + f H(D) - f H(A) - f H(B) G < 0 implied spontaneous to right G > 0 implied spontaneous to left (ignoring mixing may or may not have been mentioned) In a very unconnected way, we discussed the equilibrium with equilibrium constant A(g) + B(g) C(g) + D(g) K eq = PP C D PP We talked about the reaction proceeding in both directions to establish equilibrium. ie, sometimes proceeding when G > 0 A B What s going on??

24 General Chemistry revisited A(g) + B(g) C(g) + D(g) We said that G = H T S where, eg, H = f H(C) + f H(D) - f H(A) - f H(B) G < 0 implied spontaneous to right G > 0 implied spontaneous to left The description above basically treats A,B, C and D as pure components that never mix. That is to say, as though the composition is constant. This has to be made more sophisticated if we are to couple this concept with that of equilibrium. We now know about chemical potentials and molar Gibbs energies, so that we can follow G as the composition changes, gaining a much more sophisticated viewpoint. We now do just that!

25 No Mixing The importance of free energy of mixing G(A) A B A B G = (1-n A )G B n A G A < 0 With Mixing G(B) P A, P B o P B G = RT lnk = RT ln r P A A B

26 Gibbs Energy Minimum The composition that makes the chemical potentials equal is the equilibrium composition. ξ

27 Gibbs Duhem equation Consider a binary mixture of A and B G = μ A n A +μ B n B So we expect dg = μ A dn A + dμ A n A +μ B dn B + dμ B n B BUT, we already know that dg = μ A dn A + μ B dn B So it must be that 0 = n A dμ A + n B dμ B A special case of the general result ndµ J J = J 0 Gibbs Duhem shows that a change of the chemical potential of one component must change at least one of the other chemical potentials

General Chemistry revisited

General Chemistry revisited A(g) + B(g) C(g) + D(g) We said that G = H TS where, eg, H = f H(C) + f H(D) - f H(A) - f H(B) G < 0 implied spontaneous to right G > 0 implied spontaneous to left In a very