2. Solved Papers
  3. NEET

NEET AIPMT SOLVED PAPER MAINS 2011

  • question_answer
    The rate of reaction \[2{{N}_{2}}{{O}_{5}}\xrightarrow[{}]{{}}4N{{O}_{2}}+{{O}_{2}}\]can be written in three ways \[\frac{-d[{{N}_{2}}{{O}_{5}}]}{dt}=k[{{N}_{2}}{{O}_{5}}]\] \[\frac{d[NO_{7}^{2}]}{dt}=k'[{{N}_{2}}{{O}_{5}}]\]\[\frac{d[O_{2}^{{}}]}{dt}=k''[{{N}_{2}}{{O}_{5}}]\]                                                   The relationship between k and k' and between k?? and k'' are

    A)  k? = 2 k; k? = k   

    B)  k? = 2k; k??=k/2

    C)  k? = 2k; k?? = 2k

    D)  k? = k; k?? = k

    Correct Answer: B

    Solution :

    Rate \[=-\frac{1}{2}\frac{d[{{N}_{2}}{{O}_{5}}]}{dt}=\frac{1}{4}\frac{d[N{{O}_{2}}]}{dt}=\frac{d[{{O}_{2}}]}{dt}\] \[\Rightarrow \]\[\frac{1}{2}k[{{N}_{2}}{{O}_{5}}]=\frac{1}{4}k'[{{N}_{2}}{{O}_{5}}]=k'\,'[{{N}_{2}}{{O}_{5}}]\] \[\Rightarrow \]               \[\frac{k}{2}=\frac{k'}{4}=k'\,'\] \[\therefore \]  \[k'=2k;k'\,'=\frac{k}{2}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner