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NEET Sample Paper NEET Sample Test Paper-5

  • question_answer
    Decomposition of \[PC{{l}_{5}}\] at 323 K is represented by \[PC{{l}_{5(g)}}PC{{l}_{3(g)}}+C{{l}_{2(g)}}\]. It the degree of \[PC{{l}_{5}}\] is \[\alpha \] and the total pressure developed is P, then the partial pressure of \[PC{{l}_{3}}\] at the given temperature            

    A)  \[\frac{\alpha }{1+\alpha }P\]                             

    B)  \[\frac{1-\alpha }{1+\alpha }P\]

    C)                  \[P\alpha \]                                      

    D)  none of these

    Correct Answer: A

    Solution :

    \[\begin{align}   & \underset{1-\alpha }{\mathop{\underset{\begin{smallmatrix}  1 \\   \end{smallmatrix}}{\mathop{PC{{l}_{5(g)}}}}\,}}\,\underset{\begin{smallmatrix}   \\  \alpha  \end{smallmatrix}}{\mathop{\underset{0}{\mathop{PC{{l}_{3(g)}}}}\,}}\,+\underset{\alpha }{\mathop{\underset{{}}{\mathop{\underset{0}{\mathop{C{{l}_{2\,(g)}}}}\,}}\,}}\,initial\,\,conc. \\  & at\,\,equilibrium \\ \end{align}\] Total number of mole at equilibrium                         \[=1-\alpha +\alpha +\alpha \]                                 \[=1+\alpha \] Total pressure = P \[\because \]  Partial pressure \[=\frac{mole\,\,fraction}{total\,pressure}\] \[\therefore \]  Partial pressure of \[PC{{l}_{3}}=\frac{\alpha }{1+\alpha }P\]


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