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JEE Main & Advanced JEE Main Paper (Held On 16 April 2018)

  • question_answer
    At 320K, a gas\[{{A}_{2}}\] is 20% dissociated to\[A(g).\]The standard free energy change at 320K and  1 atm in \[J\,mo{{l}^{-1}}\]approximately:  \[(R=8.314\,J{{K}^{-1}}\,mo{{l}^{-1}};\ln \,2=0.693;\,\ln \,3=1.098)\][JEE Main 16-4-2018]

    A)  4763              

    B)  2068

    C)  4281              

    D)  1844

    Correct Answer: C

    Solution :

     \[{{A}_{2}}\rightleftharpoons 2A\] Initially, suppose \[[{{A}_{2}}]=1M\]and\[[A]=0M\] After 20 % dissociation, 80% of \[{{A}_{2}}\]remains. \[[{{A}_{2}}]=1\times \frac{80}{100}=0.8\,M\] 20% of 1 M is \[1\times \frac{20}{100}=0.2.[A]=2\times 0.2=0.4M\] The equilibrium constant \[K=\frac{[{{A}^{2}}]}{[{{A}_{2}}]}\] \[K=\frac{{{[0.4]}^{2}}}{[0.8]}=0.2\] \[\Delta {{G}^{0}}=-RT\ln k=-8.314J{{K}^{-1}}\] \[mo{{l}^{-1}}\times 320K\times \ln 0.2=4281\,J/mol\]


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