BHU PMT BHU PMT (Screening) Solved Paper-2009

  • question_answer
    One mole of\[{{N}_{2}}{{O}_{4}}\]is heated in a flask with a volume of\[10\text{ }d{{m}^{3}}\]. At equilibrium 1.708 mole of\[N{{O}^{2}}\]and 0.146 mole of\[{{N}_{2}}{{O}_{4}}\]were found at\[134{}^\circ C\]. The equilibrium constant will be

    A)  \[250\,mol\,d{{m}^{-3}}\]

    B) \[300\,mol\,d{{m}^{-3}}\]

    C) \[200\,mol\,d{{m}^{-3}}\]

    D) \[230\,mol\,d{{m}^{-3}}\]

    E)  none of the above

    Correct Answer: E

    Solution :

                    \[\underset{\begin{smallmatrix}  \,\,\,\,\,1 \\  0.146 \end{smallmatrix}}{\mathop{{{N}_{2}}{{O}_{4}}}}\,\underset{\begin{smallmatrix}  \,\,\,\,\,0 \\  1.708 \end{smallmatrix}}{\mathop{2N{{O}_{2}}}}\,\,\,\,\,\,\,\underset{\begin{smallmatrix}  at\,\operatorname{int}ial \\  at\,equilibrium \end{smallmatrix}}{\mathop{{}}}\,\] Volume\[=10\text{ }d{{m}^{3}}\] \[K=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}=\frac{{{\left[ \frac{1.708}{10} \right]}^{2}}}{\left[ \frac{0.146}{10} \right]}\] \[K=1.99\,mol\,d{{m}^{-3}}\]


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