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  • delta(H)of fusion= 6.03 kJ/mol at 0 degree Celsius. 
  • Cp[water(liquid)]=75.3 J/mol/K 
  • Cp[water(solid)]=36.8 J/mol/K

(Original Question: The enthalpy change on freezing of one mole of waterat 5 degree centigrate to ice at-5degrees centigrate)

 

1 Answer

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ΔH = Cp.ΔT  (for changing 1mol of liquid water from 5 to -0°)

      =75.3 x5 = .377kj

from (0 to -5)  

36.8 x-5 = - .184kj

Total enthalpy change 

= .377 +(-6.03) + (-.184)

= -5.837 kjmol-1

(p.s.: Kindly verify the solution, as it has been solved by a commerce guy .)

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