The vapor pressure of a liquid in the temperature range 200 K to 260 K was found

Question

The vapor pressure of a liquid in the temperature range 200 K to 260 K was found to fit the expression In(p/Torr)- 18.361 3036.8/(7). 6) i) What is the standard enthalpy of vaporization of the liquid? (10 pts) ii) What is the boiling point of liquid at normal pressure (1 atm) (10 pts) Hint: Plug the temperature into the formula to get corresponding pressure and apply Clausius-Clapeyron equation to get normal boiling point ii) Estimate the standrad entropy of vaporization of the liquid. (10 pts) iv) What is the free-energy change at 30°C? (10 pts)

Explanation / Answer

6

i)

std. enthalpy of vaporization

Recall that in equilibrium; especially in vapor-liquid equilibriums, we can use Clasius Clapyeron combination equation in order to relate two points in the same equilibrium line.

The equation is given as:

ln(P2/P1) = -dHvap/R*(1/T2-1/T1)

Where

P2,P1 = vapor pressure at point 1 and 2

dH = Enthalpy of vaporization, typically reported in kJ/mol, but we need to use J/mol

R = 8.314 J/mol K

T1,T2 = Saturation temperature at point 1 and 2

if we linearize

lnP = -H/R*1/T + dS/R

then

slope = -H/R

H = -R*slope = -8.314*-3036.8

H = 25247.95 J/mol = 24.24 kJ/mol

ii)

at P = 1 atm, 760 torr, find T

ln(760) = 18.361 - 3036.8/(T)

T = -3036.8 / (ln(760) -18.361 )

T = 258.94 K = 258.94-273.15 = -14°C

iii)

find entropy given

dSrxn/R = y.-intercept

dSRxn/8.314* = 18.361

Srxn = 18.361*8.314 = 152.65 J/molK

iv)

dG = dH - T*dS

dG = 25247.95 - (30+273) * 152.65

dG = -21005 J/mol

dG = -21.005 kJ/mol

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