WB JEE Medical WB JEE Medical Solved Paper-2015

  • question_answer For the reaction, \[2S{{O}_{2}}(g)+{{O}_{2}}(g)2S{{O}_{3}}(g)\]at 300 K, the value of \[\Delta {{G}^{o}}\]is\[~-690.9\text{ }R.\] The equilibrium constant value for the reaction at that temperature is (R is gas constant)

    A)  \[10\,\text{at}{{\text{m}}^{-1}}\]

    B)  \[10\,\text{atm}\]

    C)  10

    D)  1

    Correct Answer: A

    Solution :

     For the reaction, \[2S{{O}_{2}}(g)+{{O}_{2}}(g)2S{{O}_{3}}(g)\] Given that, \[\Delta {{G}^{o}}=-690.9R,T=300\,K\] From the formula, \[\Delta {{G}^{o}}=-RT\,\ln \,K\] \[-690.9\,R=-R\times 300\times \ln \,K\] or \[\ln \,k=\frac{690.9}{300}=2.303\] or \[=2.303\,\log \,K=2.303\] or \[K={{10}^{1}}=10\] unit of \[K=(atm){{\Delta }^{n}}={{(atm)}^{2-3}}={{(atm)}^{-1}}\] \[\therefore \] \[K=10\,at{{m}^{-1}}\]


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