Question 6.10:
Calculate the enthalpy change on freezing of 1.0 mol of waterat 10.0C to ice at 10.0C. ΔfusH= 6.03 kJ mol1at 0C.
Cp[H2O(l)] = 75.3 J mol1K1
Cp[H2O(s)] = 36.8 J mol1K1
Answer:Total enthalpy change involved in the transformation is the sum of the following changes:
(a) Energy change involved in the transformation of 1 mol of water at 10C to 1 mol of water at 0C.
(b) Energy change involved in the transformation of 1 mol of water at 0 to 1 mol of ice at 0C.
(c) Energy change involved in the transformation of 1 mol of ice at 0C to 1 mol of ice at 10C.
= (75.3 J mol1K1) (0 10)K + (6.03 × 103J mol1) + (36.8 J mol1K1) (10 0)K
= 753 J mol1 6030 J mol1 368 J mol1
= 7151 J mol1
= 7.151 kJ mol1
Hence, the enthalpy change involved in the transformation is 7.151 kJ mol1.
Pls explain this question i antfigureout wny we have to arry this method plsexplain
In order to calculate the enthalpy change for water at 10 degree Celsius to ice at -10 degree Celsius, we need to calculate the enthalpy change of all the transformations involved in the process.
The three transformations are:
- Water at 10 degree Celsius to water at 0 degree Celsius.This process requires removal of heat which can be calculated from specific heat of water.
- Water at 0 degree Celsius to ice at ice at 0 degree Celsius. This process involves heat of fusion as the state is changing at constant temperature.
- Ice at 0 degree Celsius to ice at -10 degree Celsius.This process requires removal of heat which can be calculated from specific heat of ice.
The overall enthalpy change will be sum of the enthalpy changes in these transformations.