Question

Define enthalpy of sublimation. How is it related to enthalpy of fusion and enthalpy of vaporization?

Answer 1

Hint: You should know that heat of fusion, vaporization and sublimation are all examples of latent heat. So, in order to solve this question try to recall the concept of latent heat.

Complete step by step answer:
So, in simple words let us understand about sublimation.
Sublimation is the process of changing a solid into a gas without passing through the liquid phase. In order to sublimate a substance, we need a certain amount of energy which must be transferred to the substance in the form of heat or work.
The enthalpy of sublimation is the amount of energy that must be added to a mole of solid at constant pressure to turn it directly into a gas.
The heat of sublimation is generally expressed as $\Delta H$ in units of joules per mole.

Let us know a bit about the enthalpy of fusion and enthalpy of vaporization.
Enthalpy of fusion: The heat when a solid absorbs when it melts is called the enthalpy of fusion or heat of fusion generally quoted on a molar mass.
Enthalpy of vaporization: The amount of energy that must be added to a liquid substance, to transform a quantity of that substance into a gas. The enthalpy of vaporization is often quoted for the normal boiling temperature.

Let us see the relation of enthalpy of sublimation with enthalpy of fusion and enthalpy of vaporization:
In simple words, when solid is converted to vapour, the solid first gets converted to the liquid state and then to the vapour state. The enthalpy change remains the same.
Because,
\[{H_2}{O_{(s)}} \to {H_2}{O_{(l)}}{\text{ }}\Delta {{\text{H}}_{vap}} = + 6.01kJmo{l^{ - 1}}{\text{ at }}{{\text{0}}^ \circ }C\]
\[{H_2}{O_{(l)}} \to {H_2}{O_{(g)}}{\text{ }}\Delta {{\text{H}}_{vap}} = + 45.07kJmo{l^{ - 1}}{\text{ at }}{{\text{0}}^ \circ }C\]
\[{H_2}{O_{(s)}} \to {H_2}{O_{(g)}}{\text{ }}\Delta {\text{H = 51}}{\text{.08kJmo}}{{\text{l}}^{ - 1}}{\text{ at }}{{\text{0}}^ \circ }C\]
Therefore, $\Delta {H_{sub}} = \Delta {H_{fus}} + \Delta {H_{vap}}$ .

Note: As we know in sublimation we are transitioning from a solid directly to gas, which requires the system to absorb heat and therefore the reaction is endothermic. Hence, the enthalpy change is positive.