2. Solved Papers
  3. RAJASTHAN PMT

RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • question_answer
    \[9.2\,\,g\,\,{{N}_{2}}{{O}_{4}}\]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% N^4 was dissociated, equilibrium constant will be                 \[(\text{mol}.\text{wt}\text{.}\,\,\text{of}\,\,{{N}_{2}}{{O}_{4}}=92)\]

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


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