Question

solution Un can be ignoles S. A mixture at 25°C of methanol (MeOH) and cyclohexane (CyHx) has am (XMeOH) of 0.40. The B-parameter is +3 kJ/mol. clohexane (CyHx) has a mole fraction of MeOH a. Why is B positive? (+3) ß is positive because on - 2 - - ng = x o4. moles Etol If XeoH = .4 you told us is just this once Total mole . b. Calculate AG mixing under the assumption that the solution is ideal. (+3) - Gmix- NRT ((Xalla XA) + (XA) IngA) / 8XA= "A No 8x I also AG=AGO ryn + RT in Q (If Q = mpregner ano partaal AG mix = N(108205)(293)(-4) 11(04) +04 Inga) there A Grix = -8.81 N + 4 ln til molerentur - mrealtants - .4 noles produe C. Calculate AG mixing under the assumption that the solution is regular (approximation for a real solution). (+4)

Ignore the answer written

Answer 1

a) '$\beta$' is positive because the mixture of methanol and cyclohexane shows positive deviation from Raoult's law.

b) $\Delta$G_mixing = RT (n_Aln$\chi$A + n_Bln$\chi$B), Where A = methanol, and B = cyclohexane, n_A = 4, n_B = 6, $\chi$A = 0.4, $\chi$B = 0.6

= 8.314 J.mol^-1.K^-1 * (25+273) K * {4*ln(0.4) + 6*ln(0.6)}

= -16674.35 J or -16.674 kJ