EAMCET Medical EAMCET Medical Solved Paper-2007

  • question_answer
    \[\text{9}\text{.2 g }{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\]is heated in a 1 L vessel till equilibrium state is established\[{{N}_{2}}{{O}_{4}}(g)2N{{O}_{2}}(g)\] In equilibrium state 50% \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{4}}}\]was dissociated equilibrium constant will be(mol. wt. of \[{{\text{N}}_{2}}{{O}_{4}}=92\])

    A)  0.1                                        

    B)  0.4

    C)   0.3                                       

    D)  0.2

    Correct Answer: D

    Solution :

                     \[{{N}_{2}}{{O}_{4}}(g)2N{{O}_{2}}(g)\] Molar concentration of \[[{{N}_{2}}{{O}_{4}}]=\frac{9.2}{92}=0.1\,mol/L\] In equilibrium state when it 50% dissociates. \[[{{N}_{2}}{{O}_{4}}]=0.05\,M\] \[[N{{O}_{2}}]=0.1\,M\] \[{{K}_{c}}=\frac{{{[N{{O}_{2}}]}^{2}}}{[{{N}_{2}}{{O}_{4}}]}=\frac{0.1\times 0.1}{0.05}=0.2\]


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