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The enthalpy of vaporization of liquid water using the data: (i). H 2 (g) + 1 2 O 2 g → H 2 O ( l ) ∆ H = - 285.77 k J / m o l (ii). H 2 g + 1 2 O 2 g → H 2 O ( g ) , ∆ H = - 241.84 k J / m o l in kJ/mol is

Options

  1. A+43.93
  2. B-43.93
  3. C+527.61
  4. D-527.61

Correct answer

A. +43.93

Step-by-step solution

Enthalpy of vapourisation of H 2 O is H 2 O l → H 2 O g i.e. (ii)-(i) Enthalpy of vaporization of liquid water = 285.77 - 241.84 = + 43.93 k J m o l - 1

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