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JEE Main & Advanced JEE Main Paper (Held On 11 April 2014)

  • question_answer
    For the reaction, \[2{{N}_{2}}{{O}_{5}}\to 4N{{O}_{2}}+{{O}_{2}},\]the rate equation can be expressed in two ways\[-\frac{d\left[ {{N}_{2}}{{O}_{5}} \right]}{dt}=k\left[ {{N}_{2}}{{O}_{5}} \right]\] and \[+\frac{d\left[ N{{O}_{2}} \right]}{dt}=k'\left[ {{N}_{2}}{{O}_{5}} \right]\]k and k' are related as:   [JEE Main Online Paper ( Held On 11 Apirl  2014 )

    A) k = k'                   

    B) 2k = k'

    C) k = 2k'                 

    D) k = 4k'

    Correct Answer: B

    Solution :

    Rate of disappearance of reactant = Rate of appearance of products \[-\frac{1}{2}\frac{d\left[ {{N}_{2}}{{O}_{5}} \right]}{dt}=\frac{1}{4}\frac{d\left[ N{{O}_{2}} \right]}{dt}\] \[\frac{1}{2}k\left[ {{N}_{2}}{{O}_{5}} \right]\frac{1}{4}k'\left[ {{N}_{2}}{{O}_{5}} \right]\]\[\frac{k}{2}=\frac{k'}{4}\]\[\therefore \]\[k'=2k\]


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