1 atm for the gaseous methanol Putting all of this stuff together…Thermodynamics Let’s consider the process CH3OH(g) → CH3OH(l) Write a sentence describing what is happening in this process: Calculate DH° by using Hess’s Law “the long way”: Calculate DH° by using “the short cut”: Calculate DS°: Calculate DG° in two different ways: In reference to this reaction what does “standard state” mean? Is this reaction spontaneous under standard state conditions? Under standard state conditions at what temp. would there be no tendency for either the forward of reverse reaction to occur? At 25.0°C DH°= -38 kJ S°= 240 J/K S°= 127 J/K The condensation of methanol C(s) +2H2(g) + ½O2(g) → CH3OH(g) DH°f= -201 kJ (Flipped) becomes C(s) +2H2(g) + ½O2(g) → CH3OH(l) DH°f= -239 kJ Use these two CH3OH(g) → C(s) +2H2(g) + ½O2(g) DH°= 201 kJ CH3OH(g) → CH3OH(l) DH°= SnDH°f(prod.) - SnDH°f(react.) = -239kJ – (-201kJ) = -38 kJ DS°= SnS°(prod.) - SnS°(react.) = 127 J/K – (240 J/K) = -113 J/K = -0.113 kJ/K DG°= DH° - TDS° = -38kJ – (298 K)(-0.113 kJ/K) = -4 kJ DG°= SnDG°f(prod.) - SnDG°f(react.) = -166kJ – (-163kJ) = -3 kJ 1 atm for the gaseous methanol Yes….DG° = negative value DG°= 0 DH° = TDS° -38kJ = T (-0.113 kJ/K) T= 336 K (63° C) (The normal boiling point)

You already know the V.P. of methanol at another temp… If the initial conditions of the system were such that the vapor of the methanol above the liquid were at a pressure of 0.10 atm, what would the change in free energy, DG, be at that point for this process? Is the reaction spontaneous under the initial conditions? What is the equilibrium constant for the reaction at 25.0°C? What is the V.P. of methanol at 25.0°C? What is the heat of vaporization, DHvap of methanol? Q = 10 DG= DG° + RTlnQ = -4 kJ + (8.31J/K)(298 K) ln (1/0.1) = 2 kJ No….DG = positive Q>K…. DG° = -RTlnK Reverse Reaction favored -4000 J = -(8.31J/K)(298 K) ln K K = 5 K = 1/P(g) ←mass action expression V.P. = 0.2 atm at 25.0°C You already know the V.P. of methanol at another temp… V.P. = 1.00 atm at 63°C Use C-C equation ln V.P.1/V.P.2= (DHvap/R) (1/T2-1/T1) DHvap = 35 kJ/mol

1. DG is + (spontaneous in the reverse direction) 2. But…free energy changes as reaction proceeds… hence DG for the forward reaction can become negative at some point in the process (spontaneous in the forward direction) G R (Free energy of reactants and products are equal) The system is at its minimum free energy → R + P (DG = 0) (Equilibrium) 3. There must be another “target” for the reaction 4. This is a situation where both processes are favored equally Time →