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Liquid water freezes at 273K under exter...

Liquid water freezes at 273K under external pressure of 1 atm. The process is at equilibrium `H_(2)O (l) hArr H_(2)O (s)` at 273 K & 1 atm. However it was required to calculate the thermodynamic parameters of the fusion process occuring at same pressure & different temperature.Using the following data, answer the question that follow. `d_("ice") = 0.9` gm/cc, `d_(H_(2)O(l))` = 1 gm//c, `C_(P) [H_(2)O(s))] = 36.4 JK^(-1) mol^(-1)`
`C_(P) [H_(2)O(l)] = 75.3 JK^(-1) mol^(-1), Delta H_("fusion") = 6008.2 J mol^(-1)`
The value of ''`Delta H_("fusion")`'' at 263K & 1 atm will be

A

`+6008.2 J "mole"^(-1)`

B

`5619.2 J "mole"^(-1)`

C

`-5619.2 J "mole"^(-1)`

D

`6619.2 J "mole"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
B

`Delta H_("fusion")` at 263and 1 atm will be occurred as follows: `"ice"_(263K) rarr "ice"_(273K) rarr H_(2)O_((l)_(273)) rarr H_(2)O_((l) 263)`
So `Delta H_("fusion")` at 263K = `Cp_("ice") xx Delta T + Delta H_("fusion")`
at `273- CP_(H_(2)O). Delta T`
`= 36.4 xx 10 + 6008.2 - 75.3 xx 10`
=5619.2 J/mole
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