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J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    The standard enthalpy of formation of\[{{\text{H}}_{\text{2}}}\text{(g)}\] and\[C{{l}_{2}}(g)\]and\[HCl(g)\]are \[218\text{ }kJ/mol,\]\[\text{ }121.68\text{ }kJ/mol\]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 these

    Correct Answer: E

    Solution :

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


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