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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2011

  • question_answer
    The enthalpy changes for the following processes are listed below \[C{{l}_{2}}(g)2Cl(g),242.3kJmo{{l}^{-1}}\] \[{{I}_{2}}(g)2I(g),151.0\,kJmo{{l}^{-1}}\] \[ICl(g)I(g)+Cl(g),211.3\,kJmo{{l}^{-1}}\] \[{{I}_{2}}(s){{I}_{2}}(g),62.76\,kJmo{{l}^{-1}}\] Given that the standard states for iodine and chlorine are\[{{I}_{2}}(s)\]and\[C{{l}_{2}}(g)\]the standard enthalpy of formation of\[ICI(g)\]is

    A)  \[-14.6kJmo{{l}^{-1}}\]  

    B)  \[-16.8\text{ }kJ\text{ }mo{{l}^{-1}}\]

    C)  \[+\text{ }16.8\text{ }kJ\text{ }mo{{l}^{-1}}\]  

    D)  \[+\text{ }244.8\text{ }kJ\text{ }mo{{l}^{-1}}\]

    Correct Answer: C

    Solution :

     \[\frac{1}{2}{{I}_{2}}(s)+\frac{1}{2}C{{l}_{2}}(g)ICl(g)\] \[\Delta H=\left[ \frac{1}{2}\Delta {{H}_{s\to g}}{{I}_{2}}+\frac{1}{2}\Delta {{H}_{diss}}C{{l}_{2}}+\frac{1}{2}\Delta {{H}_{diss}}{{I}_{2}} \right]\]        \[-\Delta {{H}_{ICl}}\] \[=\left[ \frac{1}{2}\times 62.76+\frac{1}{2}\times 242.3+\frac{1}{2}\times 151.0 \right]-211.3\] \[=228.03-211.3\] \[\Delta H=16.73\text{ }kJ\,mo{{l}^{-1}}\]


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