1. Tardigrade
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  3. Chemistry
  4. At 0° C, ice and water are in equilibrium and Δ S and Δ G for the conversion of ice to liquid water are

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Q. At $0^{\circ} C$, ice and water are in equilibrium and $\Delta S$ and $\Delta G$ for the conversion of ice to liquid water are

Thermodynamics

Solution:

Since the given process is in equilibrium, i.e., $\Delta G=0$

$\Delta G=\Delta H-T \Delta S$

or, $0=\Delta H-T \Delta S$ or $\Delta H=T \Delta S$ or, $\Delta S=\frac{\Delta H }{T}=\frac{6}{273} \times 10^{3}\, JK ^{-1}$

$mol ^{-1}=21.98 \,JK ^{-1}\, mol ^{-1}$