Exams › JEE Main › Chemistry
Given: (I) H2(g) + 1/2 O2(g) → H2O(l); ΔH°298 K = −285.9 kJ mol−1 (II) H2(g) + 1/2 O2(g) → H2O(g); ΔH°298 K = −241.8 kJ mol−1 The molar enthalpy of vapourisation of water will be :
- 241.8 kJ mol−1
- 22.0 kJ mol−1
- 44.1 kJ mol−1
- 527.7 kJ mol−1
Correct answer: 241.8 kJ mol−1
Solution
The molar enthalpy of vaporization of water can be determined by the difference in enthalpy between the formation of liquid water and gaseous water. Since the enthalpy change for forming liquid water is -285.9 kJ/mol and for gaseous water is -241.8 kJ/mol, the difference of 241.8 kJ/mol indicates the energy required to vaporize liquid water.
Related JEE Main Chemistry questions
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