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Here is the definition of the molar heat of vaporization:

the amount of heat necessary to boil (or condense) 1.00 mole of a substance at its boiling pointNote 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 for the __molar__ heat of vaporization are kilojoules per mole (kJ/mol). Sometimes the unit J/g is used. In that case, it is referred to as the heat of vaporization, the term 'molar' being eliminated.

The molar heat of vaporization for water is 40.7 kJ/mol. To get the heat of vaporization, you simply divide the molar heat by 18.015 g/mol. See Example #3 below.

Molar heat values can be looked up in reference books.

The molar heat of vaporization equation looks like this:

q = (ΔH_{vap}) (mass/molar mass)

The meanings are as follows:

1) q is the total amount of heat involved

2) ΔH_{vap}is the symbol for the molar heat of vaporization. This value is a constant for a given substance.

3) (mass/molar mass) is the division to get the number of moles of substance

**Example #1** 49.5 g of H_{2}O is being boiled at its boiling point of 100 °C. How many kJ is required?

**Solution:**

plug the appropriate values into the molar heat equation shown aboveq = (40.7 kJ / mol) (49.5 g / 18.0 g/mol)

**Example #2:** 80.1 g of H_{2}O exists as a gas at 100 °C. How many kJ must be removed to turn the water into liquid at 100 °C

**Solution:**

note that the water is being condensed. The molar heat of vaporization value is used at the solid-liquid phase change, REGARDLESS of the direction (boiling or condensing).q = (40.7 kJ/mol) (80.1 g / 18.0 g/mol)

**Example #3:** Calculate the heat of vaporization for water in J/g

**Solution:**

divide the molar heat of vaporization (expressed in Joules) by the mass of one mole of water.(40700 J/mol) / (18.015 g/mol) = 2259 J/g

You might see a value of 2257 J/g used. This results from using 40.66 kJ/mol rather than 40.7 kJ/mol. The value used by an author is often the one they used as a student. Just be aware that none of the values are wrong, they arise from different choices of values available.

**Example #4:** Using the heat of vaporization for water in J/g, calculate the energy needed to boil 50.0 g of water at its boiling point of 100 °C.

**Solution:**

multiply the heat of vaporization (expressed in J/g) by the mass of the water involved.(2259 J/g) (50.0 g) = 112950 J = 113 kJ

**Example #5:** By what factor is the energy requirement to evaporate 75 g of water at 100 °C greater than the energy required to melt 75 g of ice at 0 °C?

**Solution:**

Notice how the amounts of water are the same. This is deliberate. The equality is important, not the amount.Change the 75 g to one mole and solve:

40.7 kJ / 6.02 kJ = 6.76Change the amount to 1 gram of water and solve:

2259.23 J / 334.166 J = 6.76If you insisted that you must do it for 75 g, then we have this:

(75 g * 2259.23 J/g) / (75 g * 334.166 J/g) = ???You can see that the 75 cancels out, leaving 6.76 for the answer.