Total enthalpy change for freezing of 1 mol of water at 10^{\circ} {\C} to ice at -10^{\circ} {\C} is \_\_\_\_ (Given : ∆_{\text{fus }} {\H}=x {\kJ} \/ {\mol} {\C}_{ {\p}} ( {\H}_2 {\O}(ℓ)] = {\y} {\J} {\mol}^{-1} {\K}^{-1} {\C}_{ {\p}} ( {\H}_2 {\O}( { s})] = {\zJ} {\mol}^{-1} {\K}^{-1} [7 Apr 2025 Shift 1]

Correct Answer is : −10(100x+y+z)

Chemical Thermodynamics Part 3

Section: Chemistry

Question : 15 of 22

Marks: +1, -0

Total enthalpy change for freezing of 1 mol of water at 10∘C to ice at −10∘C is ____
(Given : ∆‌fus ‌H=x‌kJ∕mol Cp(H2O(ℓ)]=yJmol−1K−1 Cp(H2O( s)]=zJ‌mol−1‌K−1

[7 Apr 2025 Shift 1]

x−10y−10z
10(100x+y+z)
−x−10y−10z
−10(100x+y+z)

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