Chem 4501 Introduction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics. Fall Semester Homework Problem Set Number 10 Solutions

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1 Chem 4501 Introdction to Thermodynamics, 3 Credits Kinetics, and Statistical Mechanics Fall Semester 2017 Homework Problem Set Nmber 10 Soltions 1. McQarrie and Simon, Paraphrase: Apply Eler s theorem to internal energy to derive a fndamental eqation. Given that U depends on 3 extensive variables, i.e., U =U ( S,V, n) we can apply Eler s theorem to obtain ( ) = S U U S,V, n " % # S V,n " +V U % # V S,n " + n U % # n and we know from the natral variables of U that the partial derivative qantities are themselves simple thermodynamic qantities we may rearrange to U ( S,V, n) = S( T ) +V ( P) + n ( µ ) = TS PV + G G =U TS + PV = A + PV = H TS which is the definition of the Gibbs free energy. S,V

2 2. McQarrie and Simon, (ignore the reference to problem 10-14, which is not needed). Paraphrase: From the lever rle compte the relative amonts of liqid and vapor phases if a mixtre has an overall 50:50 composition and the liqid and vapor phases have mole fractions for one component of 0.38 and 0.57, respectively. The lever rle tells s that the total nmber of moles of liqid relative to vapor can be compted from n liq n vap = y 2 x a x a x 2 = = = 0.58 Or, pt differently, there is abot a 37:63 ration of liqid to vapor. 3. McQarrie and Simon, and Paraphrase: Derive the analytic form of the pressre/mole-fraction crves for the liqid and vapor phases of an ideal binary soltion as a fnction of the pre component vapor pressres. Discss the relationship between the vapor pressres of the two pre components and the relative enrichment of the liqid vs. vapor phases in one of them. For an ideal soltion, the vapor pressre is the mole-fraction weighted average of the pre component vapor pressres. Ths, P tot = x 1 P 1 + x2 P 2 If we take advantage of the relationship that x1 = (1 x2), then we have

3 P tot = ( 1 x 2 )P 1 + x2 P 2 ( ) x 2 + P 1 = P 2 P1 which indicates that the total vapor pressre is a linear fnction of x2 with slope P2 P1 and intercept P1. In the gas-phase, Dalton s law says that the total pressre will derive from the contribtions of all individal partial pressres, which are weighted eqally on a per-mole basis, so the mole fraction of component 2 will be y 2 = x 2P 2 P tot or, expressed in terms of x2 x 2 = y 2 P tot P 2 Inserting this expression into the total vapor pressre expression derived above provides P tot = P 2 ( P1 ) y 2P tot P 2 which may be rearranged to give + P 1 P tot = P 1 P2 ( ) P 2 y2 P 2 P1

4 and a crved plot of total pressre will be observed as a fnction of y2. Retrning to the expression for y2 above, and sbstitting for Ptot as derived frther above, we have y 2 = x 2 P 2 P 2 ( P1 ) x 2 + P 1 The goal is to express the relationship of y2 to x2 in terms of the pre component vapor pressres. So, first we can divide both sides by x2 y 2 x 2 = P 2 P 2 ( P1 ) x 2 + P 1 Since we want to nderstand how the ratio on the right-hand side varies as a fnction of the ratio of the pre component vapor pressres, we shold divide nmerator and denominator on the r.h.s. by P1. y 2 x 2 = P1 # ( P1 % ) 1 ( x 2 +1 As we re interested in whether the ratio on the l.h.s. is less than or greater than 1, we shold sbtract 1 from both sides, arriving at

5 y 2 x 2 1 = ( ) 1 ( ) 1 # P P1 2 / P1 % # ( P1 % ) 1 ( x # 2 +1 P1 % ( P1 )( 1 x 2) ( 1 x 2 ) = # ( P1 % ) 1 ( x 2 +1 # ( P1 ) 1 % ( ( 1 x 2) = # ( P1 % ) 1 ( x 2 +1 ( x 2 +1 ( x 2 +1 We are finally in a position to evalate the behavior of the r.h.s. for the alternative cases of P1 > P2 and vice versa. For the former case, the qantity in sqare brackets on the r.h.s. (in both the nmerator and denominator) is negative. In the denominator, since the magnitde of the vale in brackets is less than nity (think abot it) and the vale of x2 is a positive nmber less than nity, the net sign of the denominator is positive, so the net sign of the r.h.s. is negative, and ths y2 < x2. A similar argment establishes the converse. This proves the intitive reslt that the vapor phase above an ideal binary soltion will be enriched in the component with the higher vapor pressre. 4. McQarrie and Simon, Paraphrase: Prove that the partial molar volme of a component in an ideal binary soltion is eqal to the molar volme of the pre component liqid. The partial molar volme is defined from the total volme, generally, as V tot = n 1 V 1 + n 2 V 2

6 In addition, we know from Maxwell relations that # V tot = G % ( P T,n 1,n 2 In the case of an ideal binary soltion, G is eqal to the sm of the chemical potentials of the components, written as G = n 1 µ 1 + n 2 µ 2 ( ) + n 2 ( µ 2 + RT ln x 2 ) = n 1 µ 1 + RT ln x1 So, if we differentiate with respect to P holding the other three variables constant, we have # G # # µ % ( = n 1 P 1 % P ( n 2 µ 2 % T,n 1,n 2 P T,n 1,n 2 = n 1 V 1 + n2 V 2 ( + 0 T,n 1,n 2 Comparing the two approaches to Vtot then makes evident that the partial molar volme in soltion is the same as that for the pre component liqid. 5. McQarrie and Simon, Paraphrase: Show that if the chemical potential of a solte varies according to Henry s law as its mole fraction concentration goes to zero then the chemical potential of the solvent follows Raolt s law as its mole fraction correspondingly goes to nity. We start with the definition of the chemical potential of the solte (component 2)

7 µ 2 = µ 2 + RT ln P 2 P 2 = µ 2 + RT ln P2 RT ln P 2 Under Henry s law conditions, P2 approaches kh,2x2 as x2 à 0, so nder those conditions we may write µ 2 = µ 2 + RT ln kh,2 x 2 RT ln P 2 = µ 2 RT ln P2 + RT ln kh,2 + RT ln x 2 = µ 2 o + RT ln kh,2 + RT ln x 2 where in the final line we ve sed the relationship between the pre liqid standard state () and the 1-bar ideal gas standard state ( o ) to simplify notation. We can now take the differential of each side, noting that the first two terms on the r.h.s. are simply constants, to determine dµ 2 = RT dx 2 x 2 From the Gibbs-Dhem eqation, we have x 1 dµ 1 = x 2 dµ 2 Solving for dµ1 after sbstitting for dµ2 sing or above derivation provides

8 dµ 1 = x 2 x 1 dµ 2 = x 2 x 1 RT dx 2 x 2 = RT dx 1 x 1 where in the last line we ve sed dx2 = dx1 (from mass balance). Using dmmy integration variables in this eqation and integrating from x1 = a (a vale jst slightly less than 1) to x1 = 1 gives µ " 1 ( x 1 =1) µ " 1 ( x 1 =a) dµ " 1 = RT 1 a dx 1 " x 1 " µ 1 µ1 ( x 1 = a) = RT ln1 RT ln a = RT ln a Or, rearranged, and replacing a with x1 µ 1 = µ 1 + RT ln x1 which is the desired reslt, noting that if x2 à 0 then x1 à McQarrie and Simon, Paraphrase: Derive the dependence of the molar qantities of Δ mix H, Δ mix S, and Δ mix G on, x 1, and x 2 when the vapor pressres of the two components in a binary soltion behave as P a = x a P a e x b 2 / RT

9 We can begin with the definition of the free energy of a soltion as G = n 1 µ 1 + n 2 µ 2 " % " P = n 1 µ 1 + RT ln 1 # P + n 2 µ 2 + RT ln P % 2 1 # P 2 " = n 1 µ 1 + RT ln x1 e x 2 2 / RT % " + n # 2 µ 2 + RT ln x2 e x 1 # 2 / RT = n 1 µ 1 + n1 RT ln x 1 + n 1 x n2 µ 2 + n2 RT ln x 2 + n 2 x 1 2 % Dividing both sides by the total nmber of moles n1 + n2 provides G = x 1 µ 1 + x1 RT ln x 1 + x 1 x x2 µ 2 + x2 RT ln x 2 + x 2 x 1 2 ( ) = x 1RT ln x 1 + x 1 x x2 RT ln x 2 + x 2 x 1 2 G x 1 µ 1 + x2 µ 2 ( ) + x 1 x 2 ( x 1 + x 2 ) Δ mix G = RT x 1 ln x 1 + x 2 ln x 2 Noting that x1 + x2 = 1, and dividing both sides by, we have Δ mix G = RT ( x 1 ln x 1 + x 2 ln x 2) + x 1 x 2 Note the correct behavior that if either x1 or x2 is zero, there is no mixing, and there is no free energy of mixing. To determine the molar entropy of mixing, we can differentiate G with respect to T to obtain

10 Δ mix S = T % Δ mix G ( ) ( ) + 0 = R x 1 ln x 1 + x 2 ln x 2 ( ) Δ mix S = R x 1 ln x 1 + x 2 ln x 2 A good sanity check is to ensre that this qantity is always non-negative, and eqal to zero for either x1 or x2 = 0, and one can see that this is satisfied. Finally, the molar enthalpy of mixing may be compted from H = G + TS or Δ mix H = Δ mixg = RT + TΔ mixs ( x 1 ln x 1 + x 2 ln x 2 ) + x 1 x 2 T R x 1 ln x 1 + x 2 ln x 2 % ( ) ) ( = x 1 x 2 which, interestingly, is always positive if is positive and always negative if is negative. 7. McQarrie and Simon, Paraphrase: Plot the behavior of the molar mixing free energy vs. x 1 in the prior problem as a fnction of RT/. Determine conditions where the crvatre of the mixing free energy plots can be negative. For the vales of RT/ given (0.60, 0.50, 0.45, 0.40, and 0.35) it is a straightforward spreadsheet exercise to generate the necessary plots (in the plot, w is sed as a variable name instead of ).

11 To evalate regions of negative crvatre, we need the second derivative of the mixing free energy as a fnction of x1. First derivative first x 1 % Δ mix G ) =, ( x = x 1 = RT RT RT Second derivative next - ( x 1 ln x 1 + x 2 ln x 2 ) + x 1 x 2 /. 4 [ x 1 ln x 1 + ( 1 x 1 ) ln( 1 x 1 )] + x 1 ( 1 x 1 ) 5 6 [ ( ) 1] +1 2x 1 ln x 1 +1 ln 1 x 1

12 2 x 1 2 % Δ mix G ) = ( x 1 = RT = RT +, - [ ( ) 1] +1 2x 1 RT ln x 1 +1 ln 1 x ) 2 % x 1 1 x 1 ( x 1 ( 1 x 1 ) 56 2 and the qestion is, when is the r.h.s. negative? This is eqivalent to asking when is the first term on the r.h.s. less than 2. To answer that qestion, it is helpfl to ask what is the minimm vale that the qantity in brackets can take on in the range 0 < x1 < 1. To address that, we evalate the first derivative 1 x 1 % x 1 1 x 1 ( ) ) () = 1 x x 1 1 x1 1 % ( ) 1 () 2 = x 1 ( 1 x1 ) x 1 ( 1 x1 ) 2 = (1 x 1 ) + x 1 x x1 ( ) 2 = 2x 1 1 ( ) 2 x x1 It is evident that the only critical point is x1 = 1/2 (the nmerator is zero). Simple consideration of the endpoints makes it clear that this critical point is indeed a minimm. And, plgging 1/2 into the original expression we see that its vale is simply 4. So, if RT/ is less than 0.5, it is possible for the prodct of RT/ and the qantity in brackets to be less than 2.. / 0

13 8. McQarrie and Simon, Paraphrase: What are the solvent and solte activities and activity coefficients for a component having 25% mole fraction and obeying the vapor pressre eqation P 1 = 78.8x 1 exp(0.65x x 2 3 )? We may evalate the vapor pressre of pre component 1 by setting x1 = 1 and x2 = 0, which gives Within the solvent standard state (Raolt s law), we have a 1 = P 1 P = 78.8x 1 e ( ) = x 1 e 0.65x x 2 = 0.25e = 0.39 " # 0.65x 2 3 ( x 2 ) 78.8 ( ) ( 0.75) 3 And, as the activity coefficient γ is eqal to the activity divided by the mole fraction, it is 0.39 / 0.25 = 1.6. In the solte standard state (Henry s law), % a 1 = P 1 k H,1 We re not given the Henry s constant, bt it is defined as the slope of P1 as x1 goes to zero. To find the slope, we shold take the derivative of P1 with respect to x1

14 dp k H,1 = lim 1 x 1 0 dx 1 = lim x 1 0 = lim x 1 0, = 78.8 lim + x 1 0 -, d 78.8x dx 1 e 0.65x ( +0.18x 2 ) % d 78.8x dx 1 e x 1 1 = 78.8e % e x 1 ( ) ( 1 x 1 ) 3 ( ) ( 1 x 1 ) 3 [ ] + 0 something finite = 78.8e [ ] =180.7 So, we can now evalate ( () ( ) a 1 = P 1 = 78.8x 1e k H, % e = = 0.17 # % ) d + x1 e x 1 dx x 2 3 ( x 2 ) ( ) ( ) 3 ( ( ) ( 1 x 1 ) 3 and the Henry s Law standard-state activity coefficient is 0.17 / 0.25 = ()., / 0,