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 is 30.81 kJ/mol

Step-by-step explanation:

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

`\ln((P_2)/(P_1))=(\Delta H_(vap))/(R)[(1)/(T_1)-(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((134)/(760))=(\Delta H_(vap))/(8.314J/mol.K)[(1)/(313)-(1)/(273)]\\\\\Delta H_(vap)=30814.6J/mol=30.81kJ/mol`

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

Answer 2

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

Step-by-step 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