Question

67
Calculate the enthalpy change
water at 10°C to ice at
10°C.
freezing of 1.0 mol of
6.03 kJ (nol
at 0°C​

Answer 1

Answer:

you are a brilliant student you can do it by yourself

Answer 2

Answer:

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

According to Hess's law :

(ΔH)=(ΔH1)+(ΔH2)+(ΔH2)

(ΔH1)=(75.3Jmol−1K−1)(0−10)K=−753Jmol−1

(ΔH2)(solidification)=(−6.03×103Jmol−1)or60630Jmol−1

(ΔH3)−(36.8Jmol−1K−1)(−10−0)K=−368Jmol−1

Substituting the values we get 

=−753Jmol−1=6030Jmol−1−368Jmol−1

=−56451Jmol−1

=−5.645kJmol−1

hence the enthalpy change involved in the transformation is −5.645kJmol−1