# Using the following data, calculate DeltaS_fus and DeltaS_vap for Li.

## Question:

Using the following data, calculate {eq}\Delta S_{fus} {/eq} and {eq}\Delta S_{vap} {/eq} for Li.

Metal Tm (K) Delta H fus (kJ/mol) Tb (K) Delta H vap (kJ/mol)
Li 454 2.99 1615 134.7
Na 371 2.60 1156 89.6
K 336 2.33 1033 77.1
Rb 312 2.34 956 69
Cs 302 2.10 942 66

## Gibb's Free Energy and Phase Change

Two phases exits at a phase change in equilibrium. Since the system is in equilibrium, {eq}\Delta G {/eq} of the process would equal zero. Because {eq}\Delta G = \Delta H - T \Delta S {/eq}, at equilibrium we can write

$$\Delta H = T \Delta S$$ which allows us to solve for {eq}\Delta S {/eq} as

$$\Delta S = \dfrac { \Delta H}{T}$$

Since we have the temperature and the {eq}\Delta H {/eq} for both the fusion and vaporization of Li, we can apply the equation

$$\Delta S = \dfrac { \Delta H}{T}$$

to calculate the entropy change of both process.

$$\Delta S_{fus} = \dfrac { 2.99 \dfrac {kJ}{mole}}{454 \ K}\\ \Delta S_{fus} = 0.006585 \dfrac {kJ}{mole \cdot K} = \boxed {6.59 \dfrac {J}{mole \cdot K}}$$

$$\Delta S_{vap} = \dfrac { 134.7 \dfrac {kJ}{mole}}{1615 \ K}\\ \Delta S_{vap} = 0.0834 \dfrac {kJ}{mole \cdot K} = \boxed {83.4 \dfrac {J}{mole \cdot K}}$$