JEE Main Chemistry Class 11 Chemical Thermodynamics Part 1 Questions

In order to calculate the enthalpy change for H2O at 5‌∘C to ice at −5‌∘C, we need to calculate the enthalpy change of all the transformation involved in the process.
(a) Energy change of 1mol,H2O(l), at 5∘C
→1mol,H2O(I),0∘C
(b) Energy change of 1mol,H2O(l), at 0‌∘C
→1mol,H2O(s)(ice),0∘C
(c) Energy change of 1mol, ice (s), at 0‌∘C
→1mol‌ , ice ‌(s),−5∘C
Total ∆H
=Cp[H2O(l)]∆T+∆H‌ freezing ‌+Cp[H2O(s)]∆T
=(75.3Jmol−1K−1)(0−5)K+(−6×103Jmol−1)
+(36.8Jmol−1K−1)(−5−0)K
∆H=−6.56kJmol−1 (cxothermic process)
‌ So, ‌∆H=6.56kJmol−1‌ . ‌