"90470" - 1 õppematerjal

F��sikaline keemia kodut��

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 T M0 M1 *10-3 M2 * 10-6 M-2*105