CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    For the reaction \[2{{N}_{2}}{{O}_{5}}(g)\xrightarrow[{}]{{}}4N{{O}_{2}}(g)+{{O}_{2}}(g)\] if the concentration of\[N{{O}_{2}}\]increases by \[5.2\times {{10}^{-3}}\]M in 100 s then the rate of the reaction is:

    A) \[1.3\times {{10}^{-5}}M{{s}^{-1}}\]

    B)        \[0.5\times {{10}^{-4}}M{{s}^{-1}}\]

    C) \[7.6\times {{10}^{-4}}M{{s}^{-1}}\]

    D)        \[2\times {{10}^{-3}}M{{s}^{-1}}\]

    E) \[2.5\times {{10}^{-5}}M{{s}^{-1}}\]

    Correct Answer: A

    Solution :

    Rate of reaction\[=\frac{1}{4}\frac{d(N{{O}_{2}})}{dt}=\frac{5.2\times {{10}^{-3}}}{4\times 100}\] \[=1.3\times {{10}^{-5}}M{{s}^{-1}}\]


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