Question

The vapor pressure of dichloromethane, CH2Cl2, at 0 ∘C is 134 mmHg. The normal boiling point of dichloromethane is 40. ∘C. How would you calculate its molar heat of vaporization?

Answer 1

Answer:

The molar heat of vaporization of dichloromethane is 30.8kJ/mole

Explanation:

Using Clausius Clapeyron equation

ln (P1/P2) = (ΔHvap/R) (1/T2-1/T1)

At initial temperature of Ooc , the vapour pressure is 134mmHg

Therefore T1 = 0+273 =273K

And P1 = 134mmHg

At normal boiling point of 40oC , the vapour pressure is 760mmhg

T2 = 40 +273 = 313K

P2 = 760mmHg

ln (134/760) = ΔHvap/(8.3145 J/molK)

( 1/313K - 1/273K)

ΔHvap = 30800J/mol

= 30.8kJ/mol

Therefore, the molar heat of vaporization can be calculated using

Clausius Clapeyron equation using the above steps

Answer 2

Answer: The molar heat of vaporization is 30.81 kJ/mol

Explanation:

To calculate the molar heat of vaporization, we use the Clausius-Clayperon equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = initial pressure which is the pressure at normal boiling point = 1 atm  = 760 mmHg    (Conversion factor used:  1 atm = 760 mmHg)

$P_2$ = final pressure = 134 mmHg

$\Delta H_{vap}$ = Molar heat of vaporization

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $40^oC=[40+273]K=313K$

$T_2$ = final temperature = $0^oC=[0+273]=273K$

Putting values in above equation, we get:

$\ln(\frac{134}{760})=\frac{\Delta H_{vap}}{8.314J/mol.K}[\frac{1}{313}-\frac{1}{273}]\\\\\Delta H_{vap}=30814.6J/mol=30.81kJ/mol$

Hence, the molar heat of vaporization is 30.81 kJ/mol