Calculate the enthalpy of fusion of 1 mole of ice at -5c to water at 5c given
?H fusion =6.03 kilo joule mole ^-1
Cp H2O liquid =75.3 joule mole^-1
CpH2O solid = 36.8 joule mole^-1
Dear Student
Using formula
Total ∆H = [Cp (H2O) (l)] ∆t + ∆Hfreezing + [Cp (H2O) (s)] ∆t
= 75.3 x (0-5) + (- 6.03 x 103 ) + 36.8 (-5-0) [Convert ∆H fusion =6.03 kilo joule mole-1 to J/mol-1]
= -376.5 +(- 6.03 x 103 ) - 184
= - 560.5 - 6030
= - 6590.5 J/mol
= - 6.590 kJ/mol
Using formula
Total ∆H = [Cp (H2O) (l)] ∆t + ∆Hfreezing + [Cp (H2O) (s)] ∆t
= 75.3 x (0-5) + (- 6.03 x 103 ) + 36.8 (-5-0) [Convert ∆H fusion =6.03 kilo joule mole-1 to J/mol-1]
= -376.5 +(- 6.03 x 103 ) - 184
= - 560.5 - 6030
= - 6590.5 J/mol
= - 6.590 kJ/mol
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