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JEE Main & Advanced Sample Paper JEE Main Sample Paper-42

  • question_answer
    For the reaction \[A+BC+D,\]equilibrium  concentration of [C] =[D] = 0.5M if we start with 1 mole each of A and B. Percentage of A converted into C if we start with 2 moles of A and 1 mole of B, is

    A)  25%                           

    B)  40%

    C)  66.66%                      

    D)  33.33%

    Correct Answer: D

    Solution :

    \[A+BC+D\] \[\left. \begin{matrix}    1 & 1 & 0 & 0  \\    (1-x) & (1-x) & x & 0  \\ \end{matrix} \right]\text{Set}\,\text{I}\]\[\left. \begin{matrix}    2 & 1 & 0 & 0  \\    (2-y) & (1-y) & y & y  \\ \end{matrix} \right]\text{Set}\,\text{II}\] \[{{K}_{c}}=\frac{{{x}^{2}}}{{{(1-x)}^{2}}}=\frac{{{y}^{2}}}{(2-y)(1-y)}=1\]since x = 0.5 \[\therefore \]\[y=\frac{2}{3}\] Fraction of A converted into products \[=\frac{2/3}{2}=\frac{1}{3}=0.33\]


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