Question

1.

NAME (CAPS): Bruce Washington CHEM441 Homework • Shwe will dietails of the work in the area wit of rural and www.t leonly find i with explicity councing or the c ost . Ara que The vapor pressure of a liquid in the temperature range 200 K to 260 K was found to fit the expression Init/Tom) 18.361 - 30368/C/K). A) What is the enthalpy of vaporization of the liquid? (2 pts) B) What is the entropy of vaporization of the liquid? (2 pts) What is the free-energy change at 30°C? (2 pts)

Answer 1

Clausius-clapyron equation is

lnP = $\frac{\Delta H^{0}}{RT}$ +   $\frac{\Delta S^{0}}{R}$ (1)

Given equation is

ln(P/torr) = 18.361 -

3036.8 T

a)

So, comparing with eq.1

$\frac{\Delta H^{0}}{R}$  = 3036.8

Or, $\Delta$ H = 3036.8×R

= 3036.8×8.314

= 24294 J/mole = 24.294 KJ/mole

Hence, enthalpy of vaporisation = 24.294 KJ/mole

b

$\frac{\Delta S^{0}}{R}$ = 18.361

Or, $\Delta$ S⁰ = 18.361× R = 18.361 × 8.314 = 152.65 J/K.mol

Entropy of vaporisation 152.65 J/K.mol

c)

$\Delta$G^o = $\Delta$ H^o - T$\Delta$S^o

T = 30 +273 = 303 K.

$\Delta$G^o = 24294 - (303 × 152.65 )

Or, $\Delta$ G^o =( 24294 - 46253)

= - 21959 J/mol

= - 21.959 KJ/mol

Hence, free energy change at 30^oc = - 21.959 KJ/mole