Given: (I) \(\mathrm{H_2(g)} + \tfrac12\mathrm{O_2(g)} \rightarrow \mathrm{H_2O}(l)\); \(\Delta H^\circ_{298\,K} = -285.9\ \mathrm{kJ\,mol^{-1}}\)
(II) \(\mathrm{H_2(g)} + \tfrac12\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(g)}\); \(\Delta H^\circ_{298\,K} = -241.8\ \mathrm{kJ\,mol^{-1}}\)
The molar enthalpy of vapourisation of water will be :
(II) \(\mathrm{H_2(g)} + \tfrac12\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(g)}\); \(\Delta H^\circ_{298\,K} = -241.8\ \mathrm{kJ\,mol^{-1}}\)
The molar enthalpy of vapourisation of water will be :
Show Hint
- \(241.8\ \mathrm{kJ\,mol^{-1}}\)
- \(22.0\ \mathrm{kJ\,mol^{-1}}\)
- \(44.1\ \mathrm{kJ\,mol^{-1}}\)
- \(527.7\ \mathrm{kJ\,mol^{-1}}\)
The Correct Option is C
Solution and Explanation
Reverse (I): \(\mathrm{H_2O}(l)\rightarrow \mathrm{H_2(g)}+\tfrac12\mathrm{O_2(g)}\), \(\Delta H = +285.9\ \mathrm{kJ}\).
Add (II): \(\mathrm{H_2(g)}+\tfrac12\mathrm{O_2(g)}\rightarrow\mathrm{H_2O(g)}\), \(\Delta H = -241.8\ \mathrm{kJ}\).
Sum: \(\mathrm{H_2O}(l)\rightarrow\mathrm{H_2O(g)}\), \(\Delta H_{vap} = 285.9 - 241.8 = +44.1\ \mathrm{kJ\,mol^{-1}}\).
Answer: (3) \(44.1\ \mathrm{kJ\,mol^{-1}}\)
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