19 gm of ice is converted into water at `0^(@)C` and 1 atm. The entropies of `H_(2)O(s)` and `H_(2)O(l)` are 38.2 and `60 J//"mol " K` respectively. The enthalpy change for this conversion is :
A
`5951.4 J//"mol"`
B
`595.14 J//"mol"`
C
`-5951.4 J//"mol"`
D
None of these
Text Solution
Verified by Experts
The correct Answer is:
A
At equilibrium `therefore " "DeltaH=T.DeltaS` `H_(2)O(s)rarrH_(2)O(l)` `DeltaS=S_(H_(2)O(l))-S_(H_(2)O(s))=21.8 J//"mol-K"` `DeltaH=273 xx (21.8)=5951.4 J//"mol"`
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18g of ice is converted into water at 0^(@)C and 1 atm. The entropies of H_(2)O (s) and H_(2)O (l) are 38.2 and 60J/mol K respectively. The enthalpy cange for this conversion is:
One mole of ice is converted into water at 273 K. The entropies of H_(2)O(s) and H_(2)O(l) are 38.20 and 60.01 J mol^(-1)K^(-1) respectively. Calculate the enthalpy change for this conversion a ?
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