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
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
Exercise 1 Calculate the change in the enthalpy and entropy of solid and liquid zinc in the temperature range from T 0 = 298.16 K to T = 1000 K, step =100 K based on following data: MPT=692.655K BPT= 1180K s - 4.956 2.99E-3.199E+5 0. 298.14 H s 0 = 0. S s 0 = 9.95 l - 7.5 3*0. 692.655 H s l = 1750. g - 4.968 3*0. 1180. H l g = 27569.
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) = 4.956 + 2.99E 3 T +.199E+5 T 2 + 0 T 2 C p(s) = 4.956 + 2.99 10 3 T + 0.199 10 5 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 = 27569 S 0 s = 9.95 S s l = H s l T t1 = 1750 692.655 = 2. 5265 S l g= H l g T t2 = 27569 1180 = 23. 3636
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 (4.956 + 2.99 10 3 T + 0.199 10 5 T 2 ) dt + 1750 T 0 T 0 7. 5dT T m 0 H (g) = H 0 (l) + T b 7. 5dT + H(l g) T 0 4. 968dT T m T b
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) = 9. 95 + T m (4.956 + 2.99 10 3 T + 0.199 10 5 T 2 ) dt + T T 0 1750 T 0 692. 655 7. 5 T m T dt 0 S (g) = S 0 (l) + T b 7. 5 T m T dt + S (l g) T 0 4. 968 T dt T b
Calculated the standard entropy and enthalpy of liquid and gas Zn H 0 (l) = 1368. 8 [cal mol 1 ] 0 H (g) = 31170. 65 [cal mol 1 ] S 0 (l) 0 S (g) = 11. 603 cal K 1 mol 1 = 38. 450 cal K 1 mol 1
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 = 1368. 8 + 7. 5 (T 298. 16) 0 H Zn(g) = H Zn(g) + a (g) T T 0 = 31170. 65 + 4. 968 (T 298. 16) 0 S Zn(l) = S Zn(l) + a (l) ln T T 0 = 11. 603 + 7. 5 ln T 298. 16 0 S Zn(g) = S Zn(g) + a (g) ln T T 0 = 38. 450 + 4. 968 ln T 298. 16
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 02 2 + c Zn s T T0 T T 0 H Zn(s) = 0 + 4.956 T 298. 16 + 0. 00299 T2 298. 16 2 2 + 19900 T 298. 16 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) = 4. 956 ln T 298. 16 + 0. 00299 T T0 + 19900 T 2 T 02 2 T 2 T 02
T(K) H (s) H (l) H (g) T(K) S (s) S (l) S (g) 298.2 0 5727 130418 298.2 41.632 48.548 160.875 400 2628 8923 132535 400 49.208 57.769 166.982 450 3954 10492 133574 450 52.33 61.465 169.431 500 5306 12061 134614 500 55.179 64.771 171.621 550 6687 13630 135653 550 57.81 67.762 173.602 600 8096 15199 136692 600 60.262 70.492 175.410 650 9534 16768 137732 650 62.564 73.004 177.074 692.7 10785 18107 138618 692.7 64.428 74.998 178.395 700 11002 18337 138771 700 64.740 75.329 178.615 750 12500 19906 139810 750 66.807 77.494 180.049 800 14029 21475 140850 800 68.780 79.520 181.39 900 17177 24613 142928 900 72.487 83.216 183.838 1000 20449 27751 145007 1000 75.932 86.522 186.028 1100 23844 30889 147085 1100 79.167 89.513 188.01 1180 26648 33400 148748 1180 81.628 91.716 189.469 1200 27362 34027 149164 1200 82.228 92.243 189.818 1300 31005 37165 151243 1300 85.143 94.755 191.482 1400 34772 40303 153321 1400 87.934 97.08 193.022
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
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 +
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
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
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
Partial Gibbs free energy of component G AB = X A RTlna A + X B RTlna B G A = RTlna A G B = RTlna B
Component activity
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
Interphase equilibria Thermodynamic analysis
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
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
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)
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.
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
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 1 1 + X 2 + X 2 G 2 + G dg 2 2 + X 2 = dg 1 G dg 1 dx 1 X 2 + G dg 2 2 + X 2 2
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
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
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
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
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.
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.
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.
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.
Al (s) (LiCl - LiF) eut + AlF 3 Al-Ni-Ti (s)
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