JEE Main2017ChemistryThermodynamics (C)Actual
The enthalpy change on freezing of 1 mol of water at 5 ° C to ice at - 5 ° C is: (Given Δ 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 )
Options
- A6 .00   kJ   mol - 1
- B5 .81   kJ   mol - 1
- C5 .44   kJ   mol - 1
- D6 .56   kJ   mol - 1
Correct answer
D. 6 .56   kJ   mol - 1
Step-by-step solution
In order to calculate the enthalpy change for H 2 O at 5 ° C to ice at - 5 ° C , we need to calculate the enthalpy change of all the transformation involved in the process. (a) Energy change of 1   mol ,   H 2 O l , at 5 ° C → 1   mol ,   H 2 O   l ,   0 ° C (b) Energy change of 1   mol , H 2 O   l at 0 ° C → 1   mol ,   H 2 O   s   ice ,   0 ° C (c) Energy change of 1   mol , Ice s , at 0 ° C →