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JCECE Medical JCECE Medical 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 \[\text{250}{{\,}^{\text{o}}}\text{C}\]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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