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JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Afternoon)

  • question_answer
    For the following reactions, equilibrium constants are given : \[S(s)+{{O}_{2}}(g)\rightleftharpoons S{{O}_{2}}(g);{{K}_{1}}={{10}^{52}}\] \[2S(s)+3{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g);{{K}_{2}}={{10}^{129}}\] The equilibrium constant for the reaction, \[2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g)\]is             [JEE Main 8-4-2019 Afternoon]

    A) \[{{10}^{181}}\]                 

    B) \[{{10}^{154}}\]

    C) \[{{10}^{25}}\]       

    D)   \[{{10}^{77}}\]

    Correct Answer: C

    Solution :

    \[S(s)+{{O}_{2}}(g)\rightleftharpoons S{{O}_{2}}(g)\,\,\,\,\,\,\,{{K}_{1}}={{10}^{52}}\]?..(1)           \[2S(s)+3{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g)\,\,\,\,\,\,{{K}_{2}}={{10}^{129}}\]?..(2)           \[2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g)\,\,\,\,\,\,{{K}_{3}}=x\] multiplying equation (1) by 2; \[2SO(s)+2{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{2}}(g)\,\,\,\,\,\,K{{'}_{1}}={{10}^{104}}\] ...(3) \[\Rightarrow \]Substracting (3) from (2); we get \[2S{{O}_{2}}(g)+{{O}_{2}}(g)\rightleftharpoons 2S{{O}_{3}}(g);\] \[{{K}_{eq}}={{10}^{(129-104)}}={{10}^{25}}\]     


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