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NEET CHEMISTRYMedium

If the enthalpy change for the transition of liquid water to steam is 30 kJ mol130 \text{ kJ mol}^{-1} at 27C27^{\circ}\text{C}, the entropy change for the process would be:

A

1.0 J mol1 K11.0 \text{ J mol}^{-1} \text{ K}^{-1}

B

0.1 J mol1 K10.1 \text{ J mol}^{-1}\text{ K}^{-1}

C

100 J mol1 K1100 \text{ J mol}^{-1}\text{ K}^{-1}

D

10 J mol1 K110 \text{ J mol}^{-1}\text{ K}^{-1}

Step-by-Step Solution

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

Given data: Enthalpy change, ΔH=30 kJ mol1=30000 J mol1\Delta H = 30 \text{ kJ mol}^{-1} = 30000 \text{ J mol}^{-1} Temperature, T=27C=27+273 K=300 KT = 27^{\circ}\text{C} = 27 + 273 \text{ K} = 300 \text{ K}

Substituting the values into the formula: ΔS=30000 J mol1300 K\Delta S = \frac{30000 \text{ J mol}^{-1}}{300 \text{ K}} ΔS=100 J mol1 K1\Delta S = 100 \text{ J mol}^{-1}\text{ K}^{-1} .

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