# How would you find the molar enthalpy of vaporization for a substance, given 3.21 mol of the substance absorbs 28.4 kJ of energy as heat when the substance changes from a liquid to a gas?

Nov 2, 2015

$\text{8.85 kJ/mol}$

#### Explanation:

The molar enthalpy of vaporization is simply the enthalpy change needed in order for one mole of a given substance to undergo a liquid $\to$ vapor phase change.

Now, the molar enthalpy of vaporization will carry a positive sign, since it's used with the notion of energy needed, or energy absorbed, by a system in order to cause its molecules to go from the liquid state into the gaseous state.

Phase changes always occur at constant temperature, which is why you'll sometimes see the molar enthlpy of vaporization being referred to as the latent molar enthalpy of vaporization.

So, you know that you have $3.21$ moles of a substance and that $\text{28.4 kJ}$ of energy must be absorbed in order for the phase change to take place.

Mathematically, this can be expressed as

color(blue)(q = n * DeltaH_"vap") " ", where

$q$ - the amount of heat absorbed
$n$ - the number of moles of the substance
$\Delta {H}_{\text{vap}}$ - the molar enthalpy change of vaporization

Plug your values into this equation and rearrange to solve for $\Delta {H}_{\text{vap}}$

$q = n \cdot \Delta {H}_{\text{vap" implies DeltaH_"vap}} = \frac{q}{n}$

$\Delta {H}_{\text{vap" = "28.4 kJ"/"3.21 moles" = "8.8474 kJ/mol}}$

Rounded to thre sig figs, the answer will be

DeltaH_"vap" = color(green)("8.85 kJ/mol")