Why is$∆{\text{S}}_{\text{vap}}$of a substance always larger than $∆{\text{S}}_{\text{fus}}$.

The mechanism through which the state of a liquid converts into the condition of a vapour is known as vaporisation. As the temperature rises, the kinetic energy of the molecules rises as well.

When two light atoms combine to form a heavier one, this is called fusion. The new atom's total mass is less than the two that generated it; the "missing" mass is released as energy.

Upon vaporization, the substance changes its physical state from liquid to gas, for example water –

${\text{H}}_{\text{2}}\text{O(l)}\to {\text{H}}_{\text{2}}\text{O(g)}$

The entropy of vaporization can be represented as –

$∆{\text{S}}_{\text{vaporization}}{\text{=S}}_{\text{gas}}{\text{-S}}_{\text{liquid}}$

Similarly, upon fusion the substance changes its physical state from liquid to solid, for example water –

${\text{H}}_{\text{2}}\text{O(l)}\to {\text{H}}_{\text{2}}\text{O(s)}$

The entropy of fusion can be represented as –

$∆{\text{S}}_{\text{fusion}}\text{=}{\text{S}}_{\text{liquid}}{\text{-S}}_{\text{solid}}$

The difference in entropy is –

$\begin{array}{rcl}∆\mathrm{S}& =& ∆{\mathrm{S}}_{\mathrm{vaporization}}-∆{\mathrm{S}}_{\mathrm{fusion}}\\ & =& {\text{(S}}_{\text{gas}}{\text{-S}}_{\text{liquid}}{\text{) - (S}}_{\text{liquid}}{\text{-S}}_{\text{solid}}\text{)}\\ & =& {\text{S}}_{\text{gas}}{\text{+S}}_{\text{solid}}\end{array}$

The vaporization is characterized by the energy gain from heat and the volume expansion, characteristic to entropy increase. Whereas fusion involves volume contraction - as the substance is solidified, so the total volume is reduced.

Therefore, the vaporization entropy is larger than the fusion entropy.