JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

  • question_answer
                    Given                 (A) \[{{\operatorname{H}}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\to {{\operatorname{H}}_{2}}O(l):\] \[\Delta {{\operatorname{H}}^{0}}_{298K}=-285.9\operatorname{k}\operatorname{J}\operatorname{m}\operatorname{o}{{1}^{-1}}\]                 (B)\[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(g);\] \[\Delta {{H}^{0}}_{298K}=-241.8\operatorname{k}\operatorname{j}\operatorname{mo}{{1}^{-1}}\]                 The molar enthalpy of vaporization of water will be:                  JEE Main Online Paper (Held On 09 April 2013) 

    A)                 \[241.8\,\,kJ\,\,mo{{l}^{-1}}\]                   

    B)                 \[22.0\,\,kJ\,\,mo{{l}^{-1}}\]                

    C)                 \[44.1\,\,kJ\,\,mo{{l}^{-1}}\]                

    D)                        \[527.7\,\,kJ\,\,mo{{l}^{-1}}\]

    Correct Answer: C

    Solution :

                    Molar enthalpy of vaporization                 \[=\text{ }285.9-241.8\]                 \[=44.1\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\]                


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