Given: (I) H2(g) + (1/2)O2(g) → H2O(l); Δ H°298K - Tardigrade App

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3. Chemistry
4. Given: (I) H2(g) + (1/2)O2(g) → H2O(l); Δ H°298K = -285.9kJ mol-1 (II) H2(g) + (1/2)O2(g) → H2O(g); Δ H° 298K = -244.8kJ mol-1 The molar enthalpy of vapourisation of water will be :

Q. Given : (I)H2​(g)+21​O2​(g)→H2​O(l); ΔH298K∘​=−285.9kJmol−1 (II)H2​(g)+21​O2​(g)→H2​O(g); ΔH298K∘​=−244.8kJmol−1 The molar enthalpy of vapourisation of water will be : 6974 155 JEE MainJEE Main 2013Thermodynamics Report Error A

241.8kJmol−1

10% B

22.0kJmol−1

5% C

44.1kJmol−1

52% D

527.7kJmol−1

32%

Solution:

Given H2​(g)+21​O2​(g)→H2​O(l); ΔH∘=−285.9kJmol−1...(1) H2​(g)+21​O2​(g)→H2​O(g); ΔH∘=−244.8kJmol−1...(2) We have to calculate H2​O(l)→H2​O(g);ΔH∘=? On substracting eqn. (2) from eqn. (1) we get H2​O(l)→H2​O(g); ΔH∘=−241.8−(−285.9) =44.1kJmol−1