H γ0 - the standard enthalpy of g phase. H β γ - the transition enthalpy of b to g

Transkrypt

1 H α0 - the standard enthalpy of a phase H β0 - the standard enthalpy of b phase H γ0 - the standard enthalpy of g phase H α β - the transition enthalpy of a to b phase at T t1 H β γ - the transition enthalpy of b to g phase at T t2

2 S 0 0 a = S X(a) - the standard entropy of a phase S 0 0 β = S X(β) - the standard entropy of b phase S 0 0 γ = S X(γ) - the standard entropy of g phase S α β = S (α β) - the transition entropy of a to b phase at T t1 S β γ = S (β γ) - the transition entropy of b to g phase at T t2

3 Exercise 1 Calculate the change in the enthalpy and entropy of solid and liquid zinc in the temperature range from T 0 = K to T = 1000 K, step =100 K based on following data: MPT= K BPT= 1180K s E-3.199E H s 0 = 0. S s 0 = 9.95 l * H s l = g * H l g =

4 0 Calculation of the standard enthalpy and entropy of the liquid phase H lt 0 0 and S lt 0 Step I. Establishing the equations for Cp and enthalpis and entropies of solid, liquid and gas Zn C p(s) = a + b T + c T 2 + d T 2 C p(s) = E 3 T +.199E+5 T T 2 C p(s) = T T 2 C p(l) = 7.5 C p(g) = 4.968, (b=0, c=0, d=0) H 0 (s) =0 H s l =1750 H l g = S 0 s = 9.95 S s l = H s l T t1 = = S l g= H l g T t2 = =

5 Step 2. Calculation of the standard enthalpy of liquid and gas phase for Zn H 0 (l) = H 0 (s) + T m C p(s) dt + H (s l ) T 0 T 0 C p(l) dt T m 0 H (g) = H 0 (l) + T b Cp(l) dt + H (l g) T 0 C p(g) dt T m T b H 0 (l) = 0 + T m ( T T 2 ) dt T 0 T dT T m 0 H (g) = H 0 (l) + T b 7. 5dT + H(l g) T dT T m T b

6 Step 2. Calculation of the standard entropy of liquid and gas phase for Zn S 0 (l) = S 0 (s) + T m C p(s) T 0 T dt + S (s l ) T m T 0 C p(l) T dt 0 S (g) = S 0 (l) + T b C p(l) T m T C p(l)dt + S (l g ) T 0 C p(g) T b T dt S 0 (l) = T m ( T T 2 ) dt + T T T T m T dt 0 S (g) = S 0 (l) + T b 7. 5 T m T dt + S (l g) T T dt T b

7 Calculated the standard entropy and enthalpy of liquid and gas Zn H 0 (l) = [cal mol 1 ] 0 H (g) = [cal mol 1 ] S 0 (l) 0 S (g) = cal K 1 mol 1 = cal K 1 mol 1

8 Step III. Calculation of the enthalpy and entropy of solid, liquid and gas Zn 0 H Zn(l) = H Zn(l) + a (l) T T 0 = (T ) 0 H Zn(g) = H Zn(g) + a (g) T T 0 = (T ) 0 S Zn(l) = S Zn(l) + a (l) ln T T 0 = ln T S Zn(g) = S Zn(g) + a (g) ln T T 0 = ln T

9 Step III. Calculation of the enthalpy and entropy of solid, liquid and gas Zn 0 H Zn(s) = H Zn(s) + a (s) T T 0 + b Zn s T2 T c Zn s T T0 T T 0 H Zn(s) = T T T T T 0 S Zn(s) = a (s) ln T T 0 + b s b T T 0 + c (s) T 2 T 02 2 T 2 T 02 S Zn(s) = ln T T T T 2 T 02 2 T 2 T 02

10 T(K) H (s) H (l) H (g) T(K) S (s) S (l) S (g)

11 H 0 0 (s) = H Zn(s) H 0 0 (l) = H Zn(l) 0 H (g) 0 = H Zn(g) H s T = H Zn s H l T = H Zn l H g T = H Zn g H s l Tm = H Zn s l H l g Tb = H Zn l g

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14 The change of the Gibbs Energy of the chemical reaction A + B = C w T [K] = const G R = G p - G sub G R = G C - ( G A + G B ) G R = G C - G A - G B G 2 G 1 = 1 2 C p dt T 1 2 C p T dt C p = a + bt + c T 2 + dt2 +

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16 Binary solutions G A = G A 0 + RTlnX A + G A G B = G B 0 + RTlnX B + G B G AB = X A G A + X B G B G AB = X A (G A 0 + RTlnX A + G A ) + X B (G B 0 + RTlnX B + G B ) X A = 1 X B X B = 1 X A

17 Binary solutions - continuation 0 G AB = X A G A 0 + X B G B 0 G AB = X A (RTlnX A + G A ) + X B (RTlnX B + G B ) Roztwory doskonałe G id AB = G 0 AB + X A RTlnX A + X B RTlnX B G id AB = X A RTlnX A + X B RTlnX B

18 Excess Gibbs free energy G AB = X A G A + X B G B G AB = X A TRlnγ A + X B RTlnγ B G AB = X A (RTlnX A + G A ) + X B (RTlnX B + G B ) G AB = X A RTlnX A γ A + X B RTlnX B γ B a A = X A γ A a B = X B γ B G AB = X A RTlna A + X B RTlna B

19 Partial Gibbs free energy of component G AB = X A RTlna A + X B RTlna B G A = RTlna A G B = RTlna B

20 Component activity

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24 Gibbs-Duhem equation X 1 dg 1 + X 2 dg 2 = 0, T=const G 1 = G 1 (T, X 2 ) G 2 = G 2 (T, X 2 ) dg 1 = dg 2 = dg 1 dg 2 T T

25 Interphase equilibria Thermodynamic analysis

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28 Applying the Gibbs-Duhem equation a) Partial function of component 2 is known X 1 dg 1 + X 2 dg 2 = 0, G 2 = V(X 2 ), T=const G 1 = G 1 (X 2 ) G 2 = G 2 (X 2 ) dg 1 = dg 1 T dg 2 = dg 2 dg 1 dg 2 X 1 = X 2 dg 1 = X 2 dg 2 1 X 2 T

29 Applying the Gibbs-Duhem equation dg 1 = X 2 dg 2 1 X 2 dg 1 X 2 dg 2 dx = dx 2 1 X 2 2 dg 1 = X 2 dv(x 2 ) dx 1 X 2 2 G 1 = W X 2 + C

30 G 1 = W X 2 + C To calculate C constant the G 1 value for any X 2 (X 1 ) value must be known. From the figur on the left side one can notice that: for X 2 = 0 (X 1 = 1), G 1 = 0 and it gives C = W(X 2 = 0)

31 Example: The partial cess Gibbs energy of component A in binary liquid alloys A-B is given by following equation: G B = 2500(1 X B ) 2 Calculate the cess Gibbs Energy the component B. dg B dx B = 5000(1 X B ) From the Gibbs-Duhem equation it is known that: dg A X B dg B = dx B 1 X B dx B T = const.

32 dg A dx B dx B = 5000 G A = 5000 X B 2 X B 1 X B (1 X B ) dx B 2 = 2500X B 2 G B = 2500(1 X B ) 2 = 2500X A 2 G AB = X A G A + X B G B G AB = 2500X A X B

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35 G = X 1 G 1 + X 2 G 2 = 1 X 2 G 1 + X 2 G 2 dg dg = dg 1 dg Applying the Gibbs-Duhem equation b) Molar Gibbs energy of solution (1-2) is known = d G 1 + X 2 G 1 G dg X 2 + X 2 G 2 + G dg X 2 = dg 1 G dg 1 dx 1 X 2 + G dg X 2 2

36 dg = dg 1 G dg 1 1 X 2 + G dg 2 + X 2 2 dg = dg 1 G 1 1 X 1 dg dg 1 = dg 1 G 1 + G 2 1 X 1 dg + G dg 2 + X 2 2 dg 1 dg 2 + X 2 = dg 1 + G G 1 dg 1 dg 1 dg 2 + X 2 dx 1 + X 2 2 dg = G 2 G 1 / X 2

37 X 2 dg = X 2 G 2 X 2 G 1 X 2 dg = X 2 G 2 1 X 1 G 1 X 2 dg G 2 = G + X 1 dg = X 2 G 2 G 1 + X 1 G 1 G 1 = G X 2 dg dx 1 = G + (1 X 1 ) dg

38 Measurement of the thermodynamic properties G AB = X A (G A 0 + RTlnX A + G A ) + X B (G B 0 + RTlnX B + G B ) G AB = X A G A 0 + X B G B 0 + X A RTlnX A + X B RTlnX B + X A G A + X B G B G id AB = [X A G 0 A + X B G 0 B + X A RTlnX A + X B RTlnX B ] G AB = X A G A + X B G B G = H TS G i = H i TS i

39 Electromotive force measurements of activity (Au-Li) Li (l) (KCl-LiCl) eut Au-Li (s, l) G Li = nfe = RTln(X Li ) a Li = p n F E R T G E Li = G Li RTln X Li = nfe RTln(X Li ) S Li = d G Li dt p = nf de dt p = nfb

40 H Li = G Li + T S Li = nf E nf de dt p = nfa S E Li = S Li + Rln(X Li where: F Faraday constant, R universal gas constant, T absolute temperature [K], E electromotive force as a function T, X Li - concentration of lithium in the alloy, a Li activity of Li G Li change in the partial free enthalpy of lithium, S Li change in the partial entropy of lithium, H Li change in the partial enthalpy of lithium, S E Li partial cess entropy of lithium.

41 The calorimetric study of the mixing enthalpy H DISS X = H Signal K H X T R T M n X H mix = H DISS X n Au +n Li +n Sn where H Signal is the heat effect measured after each drop of metal samples (Au, Li or Sn) into the metallic bath (Au, Li, Au-Li or Au-Sn), K is the calibration constant, T R and T M are the drop temperature (room temperature) and the calorimeter temperature of the respective measurement in Kelvin. H X T R T M is the enthalpy change of the pure metals (Au, Li or Sn) from the room temperature (T R ) to the measurement temperature (T M ), n Au, n Li and n Sn are the numbers of moles of gold, lithium and tin, respectively. H DISS-X is the enthalpy of dissolution of pure gold, lithium or tin.

42 Direct reaction method X A A (T R + X B B (T R = A XA B XB T + Q where: T R is the temperature of the components before the introduction into the reaction zone (crucible), T is the temperature at which the reaction goes on (crucible), and Q is the heat effect measured by the calorimeter. In such a case, the formation enthalpy f H at temperature T is calculated according to the following equation: f H = Q X A H A + X B H B Where: X A and X B are the mole fractions of the reacted components (A, B) and ΔH A and ΔH B are the enthalpy changes of the solid metals (Au or Li) between temperatures T R and T.

43 Task to do: Korzystając z definicji funkcji mola roztworu funkcji namiarowej oraz równania Gibbsa Duhema wyprowadzić równanie dla nadmiarowej energii swobodnej Gibbsa składnika 2 pisemnie.

44 Al (s) (LiCl - LiF) eut + AlF 3 Al-Ni-Ti (s)

45 Entalpię tworzenia faz w układach trójskładnikowych określa się z wykorzystaniem zależności: 1. Tygiel alundowy. 2. Stos termoparowy NiAl-NiCr. 3. Mieszadełko. 4. Doprowadzenie próbek do kąpieli z temperatury pokojowej. 5. Termopara mierząca temperaturę w tyglu. 6. Korpus bloku kalorymetrycznego. 7. Pojemnik ze stali żaroodpornej. 8. Piec kalorymetru. 9. Pojemnik pośredni. 10. Doprowadzenie próbek do pojemnika pośredniego. 11. Termopara mierząca temperaturę w pojemniku pośrednim. Schemat bloku kalorymetru typu rozpuszczania f H X A H ef A X B H ef B X C H ef C H ef A XA B XB C XC

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