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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Law of equilibrium and Equilibrium constant

  • question_answer
    Calculate the partial pressure of carbon monoxide from the following    \[CaC{{O}_{3(s)}}\xrightarrow{\Delta }Ca{{O}_{(s)}}+C{{O}_{2}}\uparrow \]; \[{{K}_{p}}=8\times {{10}^{-2}}\] \[C{{O}_{2(g)}}+{{C}_{(s)}}\to 2C{{O}_{(g)}}\] ; \[{{K}_{p}}=2\]\[\]            [Orissa JEE 2004]

    A)                 0.2         

    B)                         0.4

    C)                 1.6         

    D)                 4

    Correct Answer: B

    Solution :

               Given, \[CaC{{O}_{3}}(s)\xrightarrow{\Delta }\,CaO(s)+C{{O}_{2}}(g)\uparrow \]                    \[C(s)+C{{O}_{2}}(g)\] ⇌ \[2CO(g)\]                    \[K{{p}_{2}}=\frac{{{[pCO]}^{2}}}{[pC{{O}_{2}}]}\] ;  \[pCO=\sqrt{[K{{p}_{1}}\times K{{p}_{2}}]}\]                                 \[pCO=\sqrt{[8\times {{10}^{-2}}\times 2]}=\sqrt{16\times {{10}^{-2}}}\]\[=4\times {{10}^{-1}}=0.4\]


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