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EAMCET Medical EAMCET Medical Solved Paper-2003

  • question_answer
    In the reaction\[PC{{l}_{5}}(g)\rightleftharpoons PC{{l}_{3}}(g)+C{{l}_{2}}(g).\]the equilibrium concentrations of \[\text{PC}{{\text{l}}_{\text{5}}}\]and \[\text{PC}{{\text{l}}_{\text{3}}}\]are 0.4 and 0.2 mol/L respectively. If the value of \[{{\text{K}}_{c}}\]is 0.5, what is the concentration of \[\text{C}{{\text{l}}_{\text{2}}}\]in mol/L?

    A)  2.0                                        

    B)  1.5

    C)  1.0                                        

    D)  0.5

    Correct Answer: C

    Solution :

                     \[\underset{\begin{smallmatrix}  \text{0}\,\text{time}\,\text{1}\,\text{mole} \\  \text{atequili}\,1-x\,  \\  \text{brium}\,=0.4 \end{smallmatrix}}{\mathop{PC{{l}_{5}}}}\,\rightleftharpoons \underset{\begin{smallmatrix}  0\,mole \\  \,\,\,\,x  \\  \,\,\,0.2 \end{smallmatrix}}{\mathop{PC{{l}_{3}}}}\,+\underset{\begin{smallmatrix}  0\,mole \\  \,\,\,x  \\  \,\,\,\,- \end{smallmatrix}}{\mathop{C{{l}_{2}}}}\,\] \[{{K}_{c}}=\frac{[PC{{l}_{3}}][C{{l}_{2}}]}{[PC{{l}_{5}}]}{{K}_{c}}=0.5\] \[0.5=\frac{0.2\times [C{{l}_{2}}]}{0.4}\] \[[C{{l}_{2}}]=\frac{0.4\times 0.5}{0.2}=\text{1}\,\text{mol/L}\]


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