J & K CET Engineering J and K - CET Engineering Solved Paper-2003

  • question_answer
    The enthalpy of vaporisation of liquid water using the data \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}O(l)\] \[\Delta H=-285.77\,kJ/mol\] \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}O(g);\] \[\Delta H=-241.84\,kJ/mol\]is

    A)  \[+\,43.93\,kJ/mol\]

    B)  \[-43.93\,kJ/mol\]

    C)  \[527.61\,kJ/mol\]

    D)  \[-527.61\,kJ/mol\]

    Correct Answer: A

    Solution :

     (I) \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}O(l);\] \[\Delta \Eta =-285.77\,kJ/mol\] (II) \[{{H}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}O(g);\] \[\Delta \Eta =-241.84\,kJ/mol\] \[{{H}_{2}}O(l)\xrightarrow{{}}{{H}_{2}}O(g);\] \[\Delta \Eta =?\] Subtract the Eq. (I) from (II) \[{{H}_{2}}O(l)\xrightarrow{{}}{{H}_{2}}O(g)\] \[\Delta \Eta =-241.84-(-285.77)\] \[=-241.84+285.77\] \[\Delta H=+\,43.93\,kJ/mol\]


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