The enthalpy of vaporization of water at $100^{o} C$ is $40.63 k J m o l^{- 1}$ . It's entropy change for the vaporization would be :

406.3 J K^{- 1} m o l^{- 1}

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108.9 J K^{- 1} m o l^{- 1}

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108.9 k J K^{- 1} m o l^{- 1}

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4063 k J K^{- 1} m o l^{- 1}

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Solution

The correct option is B $108.9 J K^{- 1} m o l^{- 1}$
We know that entropy change in phase transformation is given as:
$△ S_{v a p} = \frac{△ H_{v a p}}{T}$
According to given data
$△ S_{v a p} = \frac{40.63 \times 10^{3} J m o l^{- 1}}{373 K}$
so,
$△ S_{v a p} = 108.9 J K^{- 1} m o l^{- 1}$

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Similar questions

100^{\circ} C

40.63 {kJ mol}^{- 1}

Calculate entropy change for vaporization of 1 mole of liquid water to steam at $100^{\circ} C$ if $Δ H_{v}$ =40.8 kJ $m o l^{- 1}$ .

Enthalpy of vaporization for water is 186.5 kJ mol^-1.

The entropy change during vaporization is ........ kJK^-1mol^-1

373 K

△ H_{v a p} = 2.257 k J / g

373 K

= 2.257 k J g^{- 1})

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