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JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Morning)

  • question_answer
                 In a chemical reaction, \[A+2B2C+D,\]the initial concentration of B was 1.5 times of the concentration of A, but the equilibrium   concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction is [JEE Main Online Paper Held On 12-Jan-2019 Morning]

    A) 1                                 

    B) \[\frac{1}{4}\]    

    C) 4                                 

    D)   16

    Correct Answer: C

    Solution :

    \[\begin{matrix}    {} & A & + & 2B &  & 2C+ & D  \\    Initial & a & {} & 1.5a & {} & 0 & 0  \\    At.eq. & a-x & {} & 1.5a-2x & {} & 2x & x  \\ \end{matrix}\] As given, at equilibrium, \[a-x=1.5a-2x\Rightarrow x=0.5a\] \[[C]=2x=0.5a\times 2=a,[D]=x=0.5a\] \[[B]=1.5a-2\times 0.5a=0.5a\] \[[A]=a-x=a-0.5a=0.5a\] \[k=\frac{{{[C]}^{2}}[D]}{[A]{{[B]}^{2}}}=\frac{{{a}^{2}}\times 0.5a}{{{(0.5a)}^{2}}\times 0.5a}=\frac{{{a}^{2}}}{0.25{{a}^{2}}}=4\]


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