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Question -

Calculate the enthalpy change on freezing of1.0 mol of water at 10.0┬░C to ice at тАУ10.0┬░C. ╬ФfusH┬а= 6.03 kJ molтАУ1┬аat 0┬░C.

Cp[H2O(l)] = 75.3 J molтАУ1┬аKтАУ1

Cp[H2O(s)] = 36.8 J molтАУ1┬аKтАУ1



Answer -

Total enthalpy changeinvolved in the transformation is the sum of the following changes:

(a) Energy change involvedin the transformation of 1 mol of water at 10┬░C to 1 mol of water at 0┬░C.

(b) Energy change involvedin the transformation of 1 mol of water at 0┬░ to 1 mol of ice at 0┬░C.

(c) Energy change involvedin the transformation of 1 mol of ice at 0┬░C to 1 mol of ice at тАУ10┬░C.

= (75.3 J molтАУ1┬аKтАУ1) (0 тАУ 10)K + (тАУ6.03 ├Ч 103┬аJ molтАУ1) + (36.8 J molтАУ1┬аKтАУ1) (тАУ10 тАУ 0)K

= тАУ753 J molтАУ1┬атАУ 6030 J molтАУ1┬атАУ 368 J molтАУ1

= тАУ7151 J molтАУ1

= тАУ7.151 kJ molтАУ1

Hence,the enthalpy change involved in the transformation is тАУ7.151 kJ molтАУ1.

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