Enthalpy change for the process, H(2)O...

Enthalpy change for the process,
`H_(2)O"(ice")hArr H_(2)O"(water)"`
is `"6.01 kJ mol"^(-1)`. The entropy change of 1 mole of ice at its melting point will be

A

`"12 J K"^(-1)"mol"^(-1)`

B

`"22 J K"^(-1)"mol"^(-1)`

C

`"100 J K"^(-1)"mol"^(-1)`

D

`"30 J K"^(-1)"mol"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:

B

`DeltaS_("fusion")=(DeltaH_("fusion"))/(T_("fusion"))=(6.01xx1000)/(273)`
`="22 J K"^(-1)"mol"^(-1)`

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