COMEDK2025ChemistryThermodynamics (C)Actual
The enthalpy change accompanying the freezing of 18 g of water at 5^ C to ice at -5^ C is: [Given: H_ ( fusion ) =6 ~kJ ~mol ⁻¹ at 0^ C ; C_p (H₂ O , l )=75.3 ~J ~K ⁻¹ ~mol ⁻¹ ; C_p (H₂ O , s )=36.8 ~J ~K ⁻¹ ~mol ⁻¹ ]
Options
- A-6.56 ~kJ ~mol ⁻¹
- B-4.2 ~kJ ~mol ⁻¹
- C+4.2 ~kJ ~mol ⁻¹
- D-5.68 ~kJ ~mol ⁻¹
Correct answer
A. -6.56 ~kJ ~mol ⁻¹
Step-by-step solution
The process of freezing 18 g (1 mole) of water at 5^ C to ice at -5^ C can be broken down into three steps: Cooling 1 mole of liquid water from 5^ C to 0^ C . H₁ = n C_p(l) T = 1 75.3 (0 - 5) 10⁻³ ~kJ = -0.3765 ~kJ Phase change of 1 mole of liquid water at 0^ C to ice at 0^ C . H₂ = - H_ fusion = -6.0 ~kJ Cooling 1 mole of ice from 0^ C to -5^ C . H₃ = n C_p(s) T = 1 36.8 (-5 - 0) 10⁻³ ~kJ = -0.184 ~kJ Total enthalpy change H = H₁ + H₂ + H₃ = -0.3765 - 6.0 - 0.184 = -6.5605 ~kJ ~mol ⁻¹ . Answer: -6.56 ~kJ ~mol ⁻¹