## Example 4J.1 in 7th ed textbook

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Harmony Becerra
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### Example 4J.1 in 7th ed textbook

For under the "anticipate" section they state the that solid and liquid phases are in equilibrium. How can you tell just from the problem that they're in equilibrium?

Chem_Mod
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### Re: Example 4J.1 in 7th ed textbook

It states that the solid and liquid phases are in equilibrium at the melting point. This is just the definition of melting point: the temperature at which solid and liquid phases are in equilibrium. We know the melting point of water is 0 °C, so we would expect the reaction to be at equilibrium at this temperature and therefore ∆G to be 0.