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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Critical Thinking

  • question_answer
    At 700 K, the equilibrium constant \[{{K}_{p}}\] for the reaction \[2S{{O}_{3(g)}}\]⇌\[2S{{O}_{2(g)}}+{{O}_{2(g)}}\] is \[1.80\times {{10}^{-3}}\] and kPa is 14, (R = 8.314 Jk-1 mol?1). The numerical value in moles per litre of \[{{K}_{c}}\] for this reaction at the same temperature will be [AFMC 2001]

    A)                 \[3.09\ \times \ {{10}^{-7}}\]mol-litre     

    B)                         \[5.07\ \times \ {{10}^{-8}}\]mol-litre

    C)                 \[8.18\ \times \ {{10}^{-9}}\]mol-litre     

    D)                 \[9.24\ \times \ {{10}^{-10}}\]mol-litre

    Correct Answer: A

    Solution :

               \[\underset{2}{\mathop{2S{{O}_{3}}}}\,\]⇌ \[\underset{\,\,\,\,\,3}{\mathop{2S{{O}_{2}}+{{O}_{2}}}}\,\]         \[\Delta n=3-2=+1\];  \[{{K}_{p}}=1.80\times {{10}^{-3}}\]         \[{{[RT]}^{\Delta n}}={{(8.314\times 700)}^{1}}\]         \[{{K}_{c}}=\frac{{{K}_{p}}}{{{(RT)}^{\Delta n}}}=\frac{1.8\times {{10}^{-3}}}{{{(8.314\times 700)}^{1}}}\]                 \[=3.09\times {{10}^{-7}}\] mole-litre.


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