If the enthalpy change for the transition of liquid water to steam is $30 ext{ kJ mol}^{-1}$ at $27^{\circ}\

Question

If the enthalpy change for the transition of liquid water to steam is $30 	ext{ kJ mol}^{-1}$ at $27^{\circ}	ext{C}$, the entropy change for the process would be:

Accepted Answer

For a phase transition process (such as the conversion of liquid water to steam) occurring at a constant temperature, the change in entropy ($\Delta S$) is related to the enthalpy change of the transition ($\Delta H$) and the absolute temperature ($T$) by the equation:
$\Delta S = \frac{\Delta H}{T}$

Given data:
Enthalpy change, $\Delta H = 30 	ext{ kJ mol}^{-1} = 30000 	ext{ J mol}^{-1}$
Temperature, $T = 27^{\circ}	ext{C} = 27 + 273 	ext{ K} = 300 	ext{ K}$

Substituting the values into the formula:
$\Delta S = \frac{30000 	ext{ J mol}^{-1}}{300 	ext{ K}}$
$\Delta S = 100 	ext{ J mol}^{-1}	ext{ K}^{-1}$ .

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