Question

The standard molar entropy of liquid water at 273.15 K is 65 J K−1 mol−1, and that of ice at the same temperature is 43 J K−1 mol−1. Calculate the change in chemical potential of liquid water and of ice when the temperature is increased by 1 K from the normal melting point. Giving your reasons, explain which phase is thermodynamically the more stable at the new temperature.

Answer 1

Answer:

$\Delta \mu=-65\frac{J}{mol}$

$\Delta \mu=-43\frac{J}{mol}$

Liquid water is more stable.

Explanation:

Hello.

In this case, we can represent the change in chemical potential when the temperature changes via:

$\frac{\Delta \mu}{\Delta T}=-S^o$

In such a way, for the liquid water, by increasing 1 K the temperature, the change in the chemical potential is:

$\Delta \mu=-S^o*\Delta T=-65\frac{J}{mol*K}*1K=-65\frac{J}{mol}$

So for ice:

$\Delta \mu=-43\frac{J}{mol*K}*1K =-43\frac{J}{mol}$

Finally, since the phase with the lowest chemical potential is the most stable one, we infer that since liquid water's chemical potential is less than ice's chemical potential, liquid water is the most stable substance at that new temperature.

Best regards!