JEE Main & Advanced Sample Paper JEE Main Sample Paper-35

  • question_answer
    \[PC{{l}_{5}}(g)\overset{{}}{leftrightarrows}PC{{l}_{3}}(g)+C{{l}_{2}}(g)\]. In above reaction, at equilibrium condition mole fraction of \[PC{{l}_{5}}\] is \[0.4\] and mole fraction of\[C{{l}_{2}}\], is\[0.3\]. Then find out mole fraction of \[PC{{l}_{3}}\]

    A)  \[0.3\]                                 

    B)  \[0.7\]

    C)  \[0.4\]                                 

    D)  \[0.6\]

    Correct Answer: A

    Solution :

    \[\underset{t={{t}_{eq}}}{\mathop{t=0}}\,\underset{\begin{smallmatrix}  1 \\  1-x \end{smallmatrix}}{\mathop{PC{{l}_{5}}}}\,(g)\,\,\underset{\begin{smallmatrix}  0 \\  x \end{smallmatrix}}{\mathop{PC{{l}_{3}}(g)\,}}\,+\underset{\begin{smallmatrix}  0 \\  x \end{smallmatrix}}{\mathop{C{{l}_{2}}(g)}}\,\] Total moles \[=\text{ }1\text{ }+\text{ }x\] Given \[\frac{1-x}{1+x}=0.4\] \[x=\frac{3}{7}\] \[^{x}PC{{l}_{3}}\,=\frac{\frac{3}{7}}{1+\frac{3}{7}}\,=0.3\]

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