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VIT Engineering VIT Engineering Solved Paper-2015

  • question_answer
    The concentrations of the reactant A in the reaction \[A\to B\] at different times are given Below
    Concentration (M) Time (Minutes)
              0.069            0
              0.052            17
              0.035            34
              0.018            51
    The rate constant of the reaction according to the correct order of reaction is

    A) 0.001 M/min     

    B) \[0.001{{\min }^{-1}}\]

    C) \[0.001\min /M\]            

    D) \[0.001{{M}^{-1}}{{\min }^{-1}}\]

    Correct Answer: A

    Solution :

    Interval Change in conc. Rate
    0-17min 0.069-0.0532 = 0.017M \[\text{ }\!\!\Delta\!\!\text{  }\phi \text{ =}\frac{\text{2 }\!\!\pi\!\!\text{ }}{\text{ }\!\!\lambda\!\!\text{ }}\text{ }\!\!\times\!\!\text{ path difference}\]
    17-34 min 0.052-0.035=0.017M \[=\frac{2\pi }{\lambda }\times \Delta x\]
    34-51 min 0.035-0.018M \[\Rightarrow \]
    As the rate remains constant i.e. independent of concentration, the reaction is of zero order. According to rate law Rate \[\Delta \phi =\frac{2\pi }{\lambda }\left( {{\mu }_{1}}{{L}_{1}}-{{\mu }_{2}}{{L}_{2}} \right)\]


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