2. Solved Papers
  3. Chhattisgarh PMT

Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2008

  • question_answer
    \[9.2\text{ }g\]\[{{N}_{3}}{{O}_{4}}\] is heated in a \[1I,\] vessel till equilibrium state is established \[{{N}_{2}}{{O}_{4}}(g)2N{{O}_{2}}(g)\] In equilibrium state 50% \[{{N}_{2}}{{O}_{4}}\] was dissociated, equilibrium constant will be (mol. wt. of \[{{N}_{2}}{{O}_{4}}=92\])

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


You need to login to perform this action.
You will be redirected in 3 sec spinner