# 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

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

## Answer and Explanation:

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.

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

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

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