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

  • question_answer
    For the reaction \[N{{H}_{4}}HS(g)\rightleftharpoons N{{H}_{3}}(g)+{{H}_{2}}S(g)\] in a closed flask, the equilibrium pressure is P atm. The standard free energy of the reaction would be:

    A) \[-RT\] ln p        

    B) \[-RT\] (ln p - ln 2)

    C) \[-2RT\] ln p                   

    D) \[-2RT\] (ln p - ln 2)

    Correct Answer: D

    Solution :

    [d] \[{{P}_{N{{H}_{3}}}}={{P}_{{{H}_{2}}S}}=\frac{P}{2}atm\] \[{{K}_{p}}={{P}_{N{{H}_{3}}}}{{P}_{{{H}_{2}}S}}={{\left( \frac{P}{2} \right)}^{2}}=\frac{{{P}^{2}}}{4}\] \[\Delta G=-RT\,ln\,{{K}_{p}}=-RT\,ln\,{{\left( \frac{p}{2} \right)}^{2}}\] \[=-2RT\left[ ln\,p-ln\,2 \right]\]


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