Question

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C.
∆_fusH = 6.03 kJ mol^-1 at 0^oC
C_p [H₂O_(l)] = 75.3 J mol^-1K^-1
C_p [H₂O_(s)] = 36.8 J mol^-1K^-1

Answer 1

Total enthalpy change involved in the transformation is the sum of the following changes and can be represented as :

According to Hess’ law:
(∆H) = (∆H₁) + (∆H₂) + (∆H₃)
(∆H₁) = (75.3 J mol^-1K^-1) (0 - 10) K = -753 J mol^-1
(∆H₂) (solidification) = (-6.03 × 10³ J mol^-1) or 60630 J mol^-1
(∆H₃) = (36.8 J mol^-1K^-1) (-10 - 0) K = -368 J mol^-1
Substituting the values we get,
= -753 J mol^-1 - 6030 J mol^-1 - 368 J mol^-1
= -56451 J mol^-1
= -5.645 kJ mol^-1
Hence, the enthalpy change involved in the transformation is -5.645 kJ mol^-1.