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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Self Evaluation Test - Equilibrium

  • question_answer
    Gaseous \[{{N}_{2}}{{O}_{4}}\] dissociates into gaseous \[N{{O}_{2}}\]according to the reaction    \[[{{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)]\] At 300 K and 1 atm pressure, the degree of dissociation of \[{{N}_{2}}{{O}_{4}}\] is 0.2. If one mole of \[{{N}_{2}}{{O}_{4}}\] gas is contained in a vessel, then the density of the equilibrium mixture is:

    A) 1.56 g/L          

    B) 6.22 g/L

    C) 3.11g/L            

    D) 4.56 g/L

    Correct Answer: C

    Solution :

    [c]  \begin{align}   & \begin{matrix}    {} & {} & {{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)  \\ \end{matrix} \\  & \begin{matrix}    t=0 & {} & 1  \\ \end{matrix}\begin{matrix}    {} & {} & {} & 0  \\ \end{matrix} \\  & \begin{matrix}    t=\text{eq}\text{.} & 1-\alpha  & {}  \\ \end{matrix}\begin{matrix}    {} & 2\alpha   \\ \end{matrix} \\ \end{align}\] Where \[\alpha \] = Degree of dissociation. Mol. wt. of mixture \[=\frac{(1-\alpha )\times {{M}_{{{N}_{2}}{{O}_{4}}}}+2\alpha \times {{M}_{N{{O}_{2}}}}}{(1+\alpha )}\] \[=\frac{(1-0.2)92+2\times 0.2\times 46}{(1+0.2)}=76.66\] Now, As per ideal gas equation \[PV=nRT\] \[P{{M}_{mixture}}=dRT\] \[\therefore d=\frac{P{{M}_{mix}}}{RT}=\frac{1\times 76.66}{0.0821\times 300}=3.11g/L\]


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