 G  r (T) = -RTlnK  G  r (T) =  H  r (T) -  S  r (T) How do we evaluate  H  r (T) ? How do we evaluate  S  r (T) ?

Since both ethylene and cyclobutane are gases at room temperature and above, all we need is the heat capacity of the gas phase. How do we measure the Cp(g)(298) of benzoic acid (solid), or of methyl salicylate (liquid)?

Ethylene at 298 K  S f (298) = S ethylene - 2 S f H 2 -2 S fC S ethylene (298) = J mol -1 K -1 2 S f H 2 = 2(4.184)(31.2) = J mol -1 K -1 2 S fC = 2(4.184)(1.361) = 11.4 J mol -1 K -1  S f (298) (C 2 H 4 ) = = J mol -1 K -1

Cyclobutane at 298 K  S f (298) = S cyclobutane - 4 S f H S fC S cyclobutane (298) = J mol -1 K -1 4 S f H 2 = 4(4.184)(31.2) = J mol -1 K -1 2 S fC = 2(4.184)(1.361) = 22.8 J mol -1 K -1  S f (298) (C 4 H 8 ) = = J mol -1 K -1

 S f (298) (C 4 H 8 ) = J mol -1 K -1  S f (298) (C 2 H 4 ) = J mol -1 K -1 Cyclobutane  2 ethylene  S r (298) = 2(-53.1) - (-279.8) = J mol -1 K -1  S r (298) = 2 S ethylene - S cyclobutane  S r (298) = 2(219.5) = 173.6

 G f (298) =  H f (298) - T  S f (298) Ethylene:  H f (298) = 52300;  S f (298) =  G f (298) = (298)(-53.1) = J mol -1 Cyclobutane:  H f (298) = 28400;  S f (298) =  G f (298) = (298)(-279.8) = J mol -1 Cyclobutane  2 ethylene  G r (298) = 2(68.13) = 24.5 kJ mol -1  G r (298) = (298)(173.6) = 24.5 kJ mol -1

S ethylene (773) S ethylene (298) = J mol -1 K -1 S ethylene (773) = S ethylene (298) + S ethylene (773) = S ethylene (298) + Cp(g) ln(773/298) S ethylene (773) = = J mol -1 K -1 Cp(g) = 67 J mol -1 K -1 Ethylene

S cyclobutane (298) = J mol -1 K -1 S cyclobutane (773) = S cyclobutane (298) + Cp(g)ln(773/298) S cyclobutane (773) = = 394 J mol -1 K -1 Cyclobutane Cp(g) = 135 J mol -1 K -1

 G f (773) =  H f (773) - T  S (773) Ethylene:  H f (773) = 84125; S (773) = Cyclobutane:  H f (773) = 92525; S (773) = -394  G f (773) =  H f (773) - T  S (773)  G f (773) = 2(84125) T[2(-283.4)-(-394)]  G f (773) = (773)(-172.8) = J mol -1