Given: (I) H_2(g) + 12O_2(g) H_2O(l); H^ _ 298 K = -285.9 kJ mol^ -1 (II) H_2(g) + 12O_2(g) H_2O(g); H^ _ 298...

Question

Given: (I) H_2(g) + 12O_2(g) H_2O(l); H^ _ 298 K = -285.9 kJ mol^ -1 (II) H_2(g) + 12O_2(g) H_2O(g); H^ _ 298 K = -241.8 kJ mol^ -1 The molar enthalpy of vapourisation of water will be :

Accepted Answer

(3)

JEE Main 2013 Chemistry question

Solution

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