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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Sample Paper Topic Test - Chemical Equilibrium

  • question_answer
    At a certain temperature the equilibrium constant \[{{K}_{c}}\] is 0.25 for the reaction \[{{A}_{2}}(g)+{{B}_{2}}(g){{C}_{2}}(g)+{{D}_{2}}(g),\] If we take 1 mole of each of the four gases in a 10 litre container, what would be equilibrium concentration of \[{{A}_{2}}(g)\]?

    A) 0.331 M

    B) 0.033 M          

    C) 0.133 M

    D) 1.33 M

    Correct Answer: C

    Solution :

    [c] \[{{Q}_{C}}=\frac{1\times 1}{1\times 1}=1\]
    \[HN{{O}_{3}}\] \[{{Q}_{C}}>{{K}_{C}}\] so reaction will proceed in backward direction:
    \[{{A}_{2}}(g)+{{B}_{2}}(g)\rightleftharpoons {{C}_{2}}(g)+{{D}_{2}}(g)\]
    Conc. Eqm
    \[\frac{1+x}{10}\]        \[\frac{1+x}{10}\]          \[\frac{1-x}{10}\]                                   \[\frac{1-x}{10}\]
    \[0.25=\frac{{{\left( \frac{1-x}{10} \right)}^{2}}}{{{\left( \frac{1+x}{10} \right)}^{2}}}\] \[\Rightarrow \]   \[x=0.333\]
    \[[A{{  }_{2}}(g)]=\frac{1+x}{10}=\frac{1.333}{10}=0.1333\]


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