Find the Enthalpy change-

Asked by singh_mp65 | 20th Feb, 2010, 05:05: PM

Expert Answer:

The enthalpy change on freezing from 10 degree celcius to -10 degree celcius may be expressed as:

                             ΔH1                          ΔH2                            ΔH3

Liquid (10οC) Liquid (0οC) Solid (0οC) Solid (-10οC)

 

ΔH1= n Cp [H2O(l)] x ΔT = 1 x 75.03 x 10 = 0.7503 kJ mol-1

ΔH2= n (-ΔfusHo) = -1 x 6.03 = -6.03 kJ mol-1

ΔH3= n Cp [H2O(s)] x ΔT = -1 x 36.8 x 10 = - 0.368 kJ mol-1

 

ΔH = ΔH1 +  ΔH2 +  ΔH3

ΔH = 0.7503 -6.03 - 0.368 = -5.6477 kJ mol-1

 

 

Answered by  | 24th Feb, 2010, 04:15: PM

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