BCECE Engineering BCECE Engineering Solved Paper-2014

  • question_answer
    The enthalpy changes for the following processes are listed below        \[\text{C}{{\text{l}}_{\text{2}}}\text{(g)}\to \text{2Cl(g),242}\text{.3 kJ mo}{{\text{l}}^{\text{-1}}}\] \[{{\text{I}}_{\text{2}}}\text{(g)}\to \text{2I(g),151}\text{.0 kJ mo}{{\text{l}}^{\text{-1}}}\] \[\text{ICl(g)}\to \text{I(g)+Cl(g),211}\text{.3 kJ mo}{{\text{l}}^{\text{-1}}}\] \[{{\text{I}}_{\text{2}}}\text{(g)}\to {{\text{I}}_{\text{2}}}\text{(g),62}\text{.76 kJ mo}{{\text{l}}^{\text{-1}}}\] Given that the standard states for iodine and chlorine are\[{{I}_{2}}(s)\] and \[C{{l}_{2}}(g),\] the standard enthalpy for the formation of \[ICl(g)\] is

    A) \[-\text{14}.\text{6 kJ mo}{{\text{l}}^{\text{-1}}}\]      

    B) \[-\text{16}.\text{8 kJ mo}{{\text{l}}^{\text{-1}}}\]

    C) \[+\text{16}.\text{8 kJ mo}{{\text{l}}^{-}}^{\text{1}}\]    

    D) \[+\text{244}.\text{8 kJ mo}{{\text{l}}^{\text{-1}}}\]

    Correct Answer: C

    Solution :

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


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