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BCECE Engineering BCECE Engineering Solved Paper-2012

  • question_answer
    For the reaction, \[2{{N}_{2}}{{O}_{5}}\to 4N{{O}_{2}}+{{O}_{2}}\] rate and rate constant are \[1.02\times {{10}^{-4}}\]and \[3.4\times {{10}^{-5}}{{s}^{-1}}\]     respectively, then concentration of \[{{\text{N}}_{\text{2}}}{{\text{O}}_{\text{5}}}\]at that time will be

    A) \[\text{1}\text{.732}\]                  

    B)  3                            

    C)  \[1.02\times {{10}^{-4}}\]           

    D)         \[3.4\times {{10}^{5}}\]

    Correct Answer: B

    Solution :

    \[2{{N}_{2}}{{O}_{5}}\xrightarrow{{}}4N{{O}_{2}}+{{O}_{2}}\] \[\frac{-d[{{N}_{2}}{{O}_{5}}]}{dt}=k[{{N}_{2}}{{O}_{5}}]\] \[1.02\times {{10}^{-4}}=3.4\times {{10}^{-5}}{{s}^{-1}}\times [{{N}_{2}}{{O}_{5}}]\] \[\therefore \]  \[[{{N}_{2}}{{O}_{5}}]=\frac{1.02\times {{10}^{-4}}}{3.4\times {{10}^{-5}}}=3\]


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