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JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry
Given \((I)\) \({H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l);\) \(\Delta {H^o_{298\,K}} = - 285.9\,kJ\,mo{l^{ - 1}}\) \((II)\) \({H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);\) \(\Delta {H^o_{298\,K}} = - 241.8\,kJ\,mo{l^{ - 1}}\) The molar enthalpy of vapourisation of water will be.....\(kJ\,mol^{-1}\)
- A \(241.8\)
- B \(22\)
- C \(44.1\)
- D \(527.7\)
Answer & Solution
Correct Answer
(C) \(44.1\)
Step-by-step Solution
Detailed explanation
Given \({H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l);\) \(\Delta {H^o} = - 285.9\,kJ\,mo{l^{ - 1}}........(1)\) \({H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);\) \(\Delta {H^o}= - 241.8\,kJ\,mo{l^{ - 1}}........(2)\) We have to calculate…
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