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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2004

  • question_answer
    The enthalpy of reaction,\[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}O(g)\] is \[\Delta {{H}_{1}}\] and that of\[{{H}_{{{2}_{(g)}}}}+\frac{1}{2}{{O}_{2}}(g)\to {{H}_{2}}{{O}_{(l)}}\] is \[\Delta {{H}_{2}}\]. Then:

    A)  \[\Delta {{H}_{1}}<\Delta {{H}_{2}}\]                     

    B)  \[\Delta {{H}_{1}}+\Delta {{H}_{2}}=0\]

    C)  \[\Delta {{H}_{1}}>\Delta {{H}_{2}}\]                     

    D)  \[\Delta {{H}_{1}}=\Delta {{H}_{2}}\]

    Correct Answer: A

    Solution :

    The enthalpy of \[{{H}_{2}}O\,(l)\] is less than that of\[{{H}_{2}}O\,(g)\], hence more energy will be released when \[{{H}_{2}}O\,(l)\] is formed, therefore \[\Delta {{H}_{1}}<\Delta {{H}_{2}}\].


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