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JEE Mains · Chemistry · STD 11 - 5. Thermodynamics and thermochemistry

આપેલ: \((I)\) \({H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(l);\) \(\Delta {H^o_{298\,K}} =  - 285.9\,kJ\,mo{l^{ - 1}}\) \((II)\) \({H_2}(g) + \frac{1}{2}{O_2}(g) \to {H_2}O(g);\) \(\Delta {H^o_{298\,K}} =  - 241.8\,kJ\,mo{l^{ - 1}}\)  તો પાણીની મોલર બાષ્પાયન એન્થાલ્પી .....\(kJ\,mol^{-1}\)

  1. A \(241.8\)
  2. B \(22\)
  3. C \(44.1\)
  4. D \(527.7\)
Verified Solution

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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