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JCECE Engineering JCECE Engineering Solved Paper-2003

  • question_answer
    For reaction,\[PC{{l}_{3}}(g)+C{{l}_{2}}(g)PC{{l}_{5}}(g)\]the value of\[{{K}_{c}}\]at\[{{250}^{o}}C\]is\[26\]. At the same temperature, the value of \[{{K}_{p}}\] is:

    A) \[0.46\]                               

    B) \[0.61\]

    C) \[0.95\]                               

    D)  \[0.73\]

    Correct Answer: B

    Solution :

    Key Idea:\[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\]                 \[PC{{l}_{3}}(g)+C{{l}_{2}}(g)PC{{l}_{5}}(g)\] \[\therefore \]  \[\Delta {{n}_{g}}=1-2=-1\] Given    \[{{K}_{c}}=26\]                 \[T={{250}^{o}}C=250+273=523\,\,K\]                 \[R=0.0821\] \[\therefore \]  \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\]                 \[=26{{(0.0821\times 523)}^{-1}}\]                 \[=26\times \frac{1}{0.0821\times 523}\]                 \[=0.61\]


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