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Question: At 373 K steam and water are in equilibrium and \(\Delta H = 40.98kJmol^{- 1}\) what will be \(\Delt...

At 373 K steam and water are in equilibrium and ΔH=40.98kJmol−1\Delta H = 40.98kJmol^{- 1} what will be ΔS\Delta S for conversion of water into steam?

H2O(l)→H2O(g)H_{2}O_{(l)} \rightarrow H_{2}O_{(g)}

A

109.8JK−1mol−1109.8JK^{- 1}mol^{- 1}

B

31JK−1mol−131JK^{- 1}mol^{- 1}

C

21.98JK−1mol−121.98JK^{- 1}mol^{- 1}

D

326JK−1mol−1326JK^{- 1}mol^{- 1}

Answer

109.8JK−1mol−1109.8JK^{- 1}mol^{- 1}

Explanation

Solution

: ΔSvap=ΔHvapTb=40.98×1000373=109.8JK−1mol−1\Delta S_{vap} = \frac{\Delta H_{vap}}{T_{b}} = \frac{40.98 \times 1000}{373} = 109.8JK^{- 1}mol^{- 1}