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  • Calculate the enthalpy change on freezing of 1.0 mol of water at10.0°C to ice at –10.0°C. delta fus H = 6.03 kJ mol–1 at 0°C.

6.10  Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 \degree C  to ice at  - 10.0 \degree C.

       \Delta _{ fus }H = 6.03 KJ mol ^{-1} at 0 \degree C.

Cp [H_2O(l)] = 75.3 J mol^{-1} K^{-1}\\\\.\: \: \: \: Cp [H_2O(s)] = 36.8 J mol^ { -1}K^{-1}

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Total enthalpy change is equal to the summation of all the energy required at three different stages-
(i) from 10^o C to 0^o C of 1 mol of water (water \rightarrowwater )
(ii) from 0^o C to 0^o C of 1 mol of ice (water \rightarrowice)
(iii) from 0^o C to -10^oC of 1 mole of ice (ice \rightarrowice)

So, the total enthalpy change 
\Delta H = C_p[H_2O(l)](\Delta T)+\Delta H_(freezing)+C_p[H_2O(s)]\Delta T 

         \\ = (75.3)(-10)+(-6.03\times 10^{3})+(36.8)(-10)\\ =-753-6030 - 368\\ =-7151 Jmol^{-1}\\ = 7.151kJ/mol
Hence the total enthalpy change in the transformation process is 7.151 kJ/mol

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manish

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