what is the molar heat of vaporization.?

Question

The vapor pressure of dichloromethane, CH_2Cl_2, at 0 degrees C is 134 mmHg. The normal boiling point of dichloromethane is 40 degrees C.

Answer 1

the amount of heat necessary to boil (or condense) 1.00 mole of a substance at its boiling point
Note the two important factors:
1) It's 1.00 mole of a substance
2) there is no temperature change
Keep in mind the fact that this is a very specific value. It is only for one mole of substance boiling. The molar heat of vaporization is an important part of energy calculations since it tells you how much energy is needed to boil each mole of substance on hand. (Or, if we were cooling off a substance, how much energy per mole to remove from a substance as it condenses.)

Every substance has its own molar heat of vaporization.

The units are usually kilojoules per mole (kJ / mol). Sometimes the unit J/g is used. The first unit is technically the most correct unit to use.

The molar heat of vaporization for water is 40.7 kJ/mol. Remember the value!!!

Molar heat values can be looked up in reference books. They are determined by experiment. In the ChemTeam classroom, you will not need to memorize any molar heat values EXCEPT those for water.

The molar heat of vaporization equation looks like this:

q = DHvap (mass / molar mass)

The meanings are as follows:

1) q is the total amount of heat involved
2) DHvap is the symbol for the molar heat of vaporization. This value is a constant for a given substance.
3) (mass / molar mass) is the calculation to get the number of moles of substance

Answer 2

Molar Heat Of Vaporization

Answer 3

You take the grams of Nitrogen in your problem and multiply it by 201J then you divide that by the number of moles To find the number of moles just take the grams of Nitrogen given and divide it by 28 (N2 Molar Mass because I'm sure its liquid Nitrogen) Im guessing this is an Owl problem if you still dont get it just email me the problem before 5am

Answer 4

The vapour pressure at 273.15K is 134 torr
The vapour pressure at 313.15K is 760 torr

Depending on what you have been taught, you can either use an Arrhenius-type equation, or use the concepts of free energy, reaction enthalpy, and reaction entropy, to find your answer. (Hint: think of the vapour pressure as a kind of equilibrium constant, which it is)

Vaporization is classified as a physical process, but the thermodynamic arguments apply to any process.