Aine Intervall, K (kJ/mol) (J/mol*K) a b*103 c*10-5 CO -110,53 197,55 28,41 4,10 -0,46 298-2500 H2O -241,81 188,72 30,00 10,71 0,33 298-2500 CO2 -393,51 213,66 44,14 9,04 -8,54 298-2500 H2 0 130,52 27,28 3,26 0,5 298-3000 1. G0298 = H0298 - 298S0298 H0298 = Hf(H2) + Hf(CO2) - Hf(H2O) - Hf(CO) = -393,51 + 241,81 + 110,53 = -41,17 kJ/mol = -41170 J/mol S0298 = S0298(H2) + S0298(CO2) - S0298(H2O) - S0298(CO) = 130,52 + 213,66 Â 188,72 Â 197,55 = -42,09 J/mol*K G0298 = -41170 Â 298*(-42,09) = -28627,18 J/mol = -28,62718 kJ/mol -G/RT 2. Kp = e Kp = e28627,18/(8,314 * 298) = 1,0425*105 3. T M0 M1*10-3 M2*105 1400 0,7595 0,4336 0,34835
Aine H0f,298(kJ/mol) S0298(J/K*mol) a b* 103 c *106 c´*10-5 K CH3OH(g -201,00 239,76 15,28 105,20 -31,04 - 298-1000 ) CO -110,53 197,55 28,41 4,10 - -0,46 298-2500 H2 0 130,52 27,28 3,26 - 0,50 298-3000 1) G0298 = H0298  298*S0298 H0298 = H0f,298 (CH3OH(g)) - H0f,298 (CO) = -201,00 +110,53 = -90,47 kJ/mol = -90470 J/mol S0298 = S0298 (CH3OH(g)) - S0298 (CO)  2*S0298 (H2) = = 239,76  197,55  2*130,52 = -218,83 J/K*mol G0298 = -90470  298 * (-218,83) = -25258,66 J/mol = -25 kJ/mol - G 2) Kp = e RT Kp = e 25258,66 ¿/8,314298 = 2,7 * 104 3) Tasakaalu konstandi leidmine T = 400 K juures
a b 103 c´ 10-5 N2 0 191,5 27,88 4,27 - 298-2500 O2 0 205,04 31,46 3,39 -3,77 298-3000 NO 91,26 210,64 29,58 3,85 -0,59 298-2500 1. Arvutan Gibbsi energia muudu standardtemperatuuril G0298 = H0298 + T S0298 Selleks arvutan reaktsiooni tekke-entalpia standardtemperatuuril: H0298 =2 Hf (NO) - Hf (N2)  Hf (O2) = 291,26  0- 0 = 182,52 kJ/mol = 182520 J/mol ning entroopia muudu: S0298 =2 S0 (NO) - S0 (N2)  S0 (O2) = 2210,64  205,04- 191,5 = 24,74 J/mol K Saame Gibbsi energia muudu: G0298 = 182520 J/mol  298 K 24,74 J/mol K = 175147,48 J/mol 2. Arvutame tasakaalukonstandi G0298 = - RT ln Kp R= 8,314 J/molK Kp = e - = e - (175147,48/(8,314 (J/molK) 298K) = e - 70,69
kolb III: 71,5 mL kolb IV: 71,6.mL keskmine: 71,55 mL Lähteandmed teoreetilise tasakaalukonstandi arvutamiseks: CH3COOH + C2H5OH CH3COOC2H5 + H2O 0 -1 H f 298K kJ mol Â487 Â278 Â481 Â286 0 -1 -1 S 298K J mol K 160 161 259 70 G0298=H0298 - 298*S0298 H0298=-481-286+487+278= -2 S0298=70+259-160-161= 8 G0298=-2000-298*8= -4384 J - G 0 --4384 RT K p=e =e 8,314 298 =5,87 Arvutused I ja II kolvi arvutused: 2014 mHCl = VNaOH * CN(NaOH) * EHCl mHCl =0,030450,516036,5=0,573 g Kuna EHCl = MHCl, siis
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