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JEE Main & Advanced AIEEE Solved Paper-2006

  • question_answer
    The enthalpy changes for the following processes are listed below \[C{{l}_{2}}(g)=2Cl(g)\]  \[242.3\text{ }kJ\text{ }mo{{l}^{-1}}\] \[{{I}_{2}}(g)=2I(g),\]                     \[151.0\text{ }kJ\text{ }mo{{l}^{-1}}\] \[ICl(g)=I(g)+Cl(g),\]       \[211.3\text{ }kJ\text{ }mo{{l}^{-1}}\] \[{{I}_{2}}(s)={{I}_{2}}(g),\]         \[62.76\text{ }kJ\text{ }mo{{l}^{-1}}\] Given that the standard states for iodine and chlorine are\[{{I}_{2}}(s)\]and\[C{{l}_{2}}(g)\], the standard enthalpy of formation of \[ICl\,(g)\] is     AIEEE  Solved  Paper-2006

    A) \[-\text{ }14.6\text{ }kJ\text{ }mo{{l}^{-1}}\]     

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

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

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

    Correct Answer: C

    Solution :

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


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