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Question: Enthalpy change for the process \[H_{2}O(ice) \rightleftharpoons H_{2}O(water)\] is \(6.01kJmol^{-...

Enthalpy change for the process

H2O(ice)⇌H2O(water)H_{2}O(ice) \rightleftharpoons H_{2}O(water)

is 6.01kJmol−16.01kJmol^{- 1} The entropy change of 1 mole of ice at its melting point will be.

A

12JK−1mol−112JK^{- 1}mol^{- 1}

B

22JK−1mol−122JK^{- 1}mol^{- 1}

C

100JK−1mol−1100JK^{- 1}mol^{- 1}

D

30JK−1mol−130JK^{- 1}mol^{- 1}

Answer

22JK−1mol−122JK^{- 1}mol^{- 1}

Explanation

Solution

:

}{= \frac{6.01 \times 1000}{273} = 22JK^{- 1}mol^{- 1}}$$