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The enthalpy change on freezing of 1 mol...

The enthalpy change on freezing of 1 mol of water at `5^(@)C` to ice at `-5^(@) C` is
(Given `Delta_("fus") H = 6 kJ mol^(-1)` at `0^(@) C , C_(p) (H_(2) O , l) = 75.3 J mol^(-1) K^(-1) , C_(p) (H_(2) O , s) = 36.8 J mol^(-1) K^(-1))`

A

`5.81 kJ mol^(-1)`

B

`5.44 kJ mol^(-1)`

C

`6.00 kJ mol^(-1)`

D

`6.56 kJ mol^(-1)`

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The correct Answer is:
D
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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) '' DeltaS_("fusion") '' at 263 K & 1 atm will be

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Consider the following reaction : CO_((g)) + 2H_(2(g)) hArr CH_(3)OH_((g)) Given : Delta_(r) H^(@) (CH_(3)OH, g) = -201 kJ/mol, Delta_(r) H^(@) (CO, g) = -114 kJ/mol S^(@) (CH_(3)OOH, g) = 240 J/K-mol , S^(@) (H_(2), g) = 29 JK^(-1)mol^(-1) S^(@) (CO, g) = 198 J/mol-K , C^(@)_(p,m) (H_(2)) = 28.8 J/mol-K C^(@)_(p,m) (CO) = 29.4 J/mol-K , C^(@)_(p,m) (CH_(3)OH) = 44 J/mol-K and ln ((320)/(300)) = 0.06 , all data at 300 K Delta_(r) S^(@) at 320 K is :

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