2. Solved Papers
  3. Haryana PMT

Haryana PMT Haryana PMT Solved Paper-1999

  • question_answer
    The rate constant for the reaction\[2{{N}_{2}}{{O}_{5}}\to 4N{{O}_{2}}+{{O}_{2}}\] is \[3.0\times {{10}^{-5}}\,{{\sec }^{-1}}\]. If the rate is \[2.4\times {{10}^{-5}}mol\]\[2.4\times {{10}^{-5}}mol\,litr{{e}^{-1}}\,{{\sec }^{-1}}\], then the concentration of \[{{N}_{2}}{{O}_{5}}\] (in \[mol\text{ }litr{{e}^{-1}}\]) is:

    A)  \[1.4\]                                 

    B)  \[1.2\]

    C)  \[0.8\]                                 

    D)  \[0.04\]

    Correct Answer: C

    Solution :

                    For the first order reaction, \[2{{N}_{2}}{{O}_{5}}\xrightarrow{{}}4N{{O}_{2}}+{{O}_{2}}\] Rate  \[\frac{dx}{dt}=K[{{N}_{2}}{{O}_{5}}]\]    ?...(1) Given, \[\frac{dx}{dt}=2.40\times {{10}^{-5}}mol\,litr{{e}^{-1}}{{\sec }^{-1}}\]                 \[K=3.0\times {{10}^{-5}}{{\sec }^{-1}}\]                 \[[{{N}_{2}}{{O}_{5}}]=?\] Substituting these value in equation (1)                 \[2.4\times {{10}^{-5}}mol\,litr{{e}^{-1}}{{\sec }^{-1}}\]                 \[=3\times {{10}^{-5}}{{\sec }^{-1}}\times [{{N}_{2}}{{O}_{5}}]\] or \[[{{N}_{2}}{{O}_{5}}]=\frac{2.4\times {{10}^{-5}}\,mol\,litr{{e}^{-1}}\,{{\sec }^{-1}}}{3.0\times {{10}^{-5}}{{\sec }^{-1}}}\]                 \[=0.8\,mol\,litr{{e}^{-1}}\]


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