Calculate the enthalpy change on freezin | Class 11 Chemistry Chapter Thermodynamics, Thermodynamics NCERT Solutions

Question:

Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. ΔfusH = 6.03 kJ mol-1 at 0°C.

Cp[H2O(l)] = 75.3 J mol-1 K-1

Cp[H2O(s)] = 36.8 J mol-1 K-1

Answer:

Total enthalpy change involved in the transformation is the sum of the following changes:

(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.

(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.

(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at -10°C.

Total ΔH = Cp [H2OCI] ΔT  + ΔHfreezing + Cp[H2O(s)] ΔH

= (75.3 J mol-1 K-1) (0 - 10)K + (-6.03 × 103 J mol-1) + (36.8 J mol-1 K-1) (-10 - 0)K

= -753 J mol-1 - 6030 J mol-1 - 368 J mol-1

= -7151 J mol-1

= -7.151 kJ mol-1

Hence, the enthalpy change involved in the transformation is -7.151 kJ mol-1.


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Comments

  • Sanjeev
  • 2018-12-03 22:08:29

Not clear solutions


  • Siddharth
  • 2018-08-18 15:57:32

But ncert answer didn't match


  • Hero
  • 2018-05-09 11:34:38

Excellent


Comment(s) on this Question

Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 10: Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at -10.0°C. &....