J & K CET Medical J & K - CET Medical Solved Paper-2010

  • question_answer
    The standard enthalpy of formation of\[{{H}_{2}}(g)\] and\[C{{l}_{2}}(g)\]and\[HCl(g)\]are 218 kJ/mol, 121.68 kJ/mol and\[-92.31\text{ }kJ/mol\]respectively. Calculate standard enthalpy change in kJ for \[\frac{1}{2}{{H}_{2}}(g)+\frac{1}{2}C{{l}_{2}}(g)\xrightarrow{{}}HCl(g)\]

    A) \[+\,431.99\]                     

    B)  \[-\,246.37\]

    C) \[-\,431.99\]                      

    D)  \[+\,247.37\]

    E) None of the above

    Correct Answer: E

    Solution :

    (*) For reaction, \[\frac{1}{2}{{H}_{2}}(g)+\frac{1}{2}C{{l}_{2}}(g)\xrightarrow{{}}HCl(g)\] \[\Delta H=\Delta {{H}_{f}}(HCl)\]                 \[-\left[ \frac{1}{2}\times \Delta {{H}_{f}}({{H}_{2}})+\frac{1}{2}\times \Delta {{H}_{f}}(C{{l}_{2}}) \right]\] \[=-92.31-\left[ \frac{1}{2}\times (218)+\frac{1}{2}\times (121.68) \right]\] \[=-262.15\text{ }kJ\]


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