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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2000

  • question_answer
    \[{{H}_{2}}(g)+C{{l}_{2}}(g)\to 2HCl(g).\] \[\Delta H=-44k.cal\] \[2Na(s)+2HCl(g)\to 2NaCl(s)+{{H}_{2}}(g),\] \[\Delta H=-152k.cal\] For the reaction, \[Na(s)+\frac{1}{2}C{{l}_{2}}(g)\xrightarrow{{}}NaCl(s),\Delta H=?\]

    A)  \[-108k,\,cal\]

    B)  \[-196k,\,cal\]

    C)  \[-98k,\,cal\]

    D)  \[54k,\,cal\]

    Correct Answer: C

    Solution :

    \[{{H}_{2}}(g)+C{{l}_{2}}(g)\xrightarrow{{}}2HCl(g)\] \[\Delta H=-44k.\,cal\]      ??(i) \[2Na(s)+2HCl(g)\xrightarrow{{}}2NaCl(s)+{{H}_{2}}(g)\] \[\Delta H=-152k.cal\]     ??.(ii) \[Na(s)+\frac{1}{2}C{{l}_{2}}(g)\xrightarrow{{}}NaCl\,(s)\,\Delta H=?\] By adding eq. (i) and (ii) we have \[Na(s)+\frac{1}{2}C{{l}_{2}}(g)\to NaCl\,\Delta H=-98k.cal\] \[2Na(s)+C{{l}_{2}}(g)\xrightarrow{{}}2NaCl,\] \[\Delta H=-196k.\,\,cal\]    ?..(iii) dividing eq. (iii) by 2 we have \[Na(s)+\frac{1}{2}C{{l}_{2}}(g)\to NaCl\Delta H=-98k.cal\]


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