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BCECE Medical BCECE Medical Solved Papers-2002

  • question_answer
    In the reaction, \[2C(s)+{{O}_{2}}(g)2CO\,\,(g)\] the partial pressure of CO and \[{{O}_{2}}\] is 8 atm and 4 arm respectively, then find its equilibrium constant:

    A)  16             

    B)  24

    C)  8              

    D)  2

    Correct Answer: A

    Solution :

    Key Idea: Only gaseous products have partial pressure. Given    \[{{p}_{CO}}=8\,\,atm,\,\,{{p}_{{{O}_{2}}}}=4\,\,atm\]                 \[2C\,\,(s)+{{O}_{2}}(g)2CO\,\,(g)\] For above equilibrium \[{{K}_{p}}=\frac{{{({{p}_{CO}})}^{2}}}{{{p}_{{{O}_{2}}}}}=\frac{8\times 8}{4}=16\,\,atm\]


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