One mole of ice is converted into water at `273K`. The entropies of `H_2O(s)` and `H_2O(l)` are `38.20` and `60.01Jmol^(-1)K^(-1)` respectively. The enthalpy change for the conversion is:

Options

A. 59.59 J/mol
B. 595.59 J/mol
C. 5959.5 J/mol
D. 59.0 J/mol

Answer

C

Explanation

`H_(2)O_((s)) hArr H_(2)O_((l))`
`Delta S= S_(l)^(@)- S_(s)^(@)=60.03 -38.20 = 21.83`
`Delta H = T Delta S=273 xx21.83 =5959.5 J mol^(-1)`