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KVPY Sample Paper KVPY Stream-SX Model Paper-11

  • question_answer
    Two \[{{l}^{st}}\] order reactions have half-lives in the ratio 3 : 2. Then the ratio of time intervals \[{{t}_{1}}:{{t}_{2}},\] will be? Where \[{{t}_{1}}\] is the time period for 25% completion of the first reaction and \[{{t}_{2}}\] is time required for 75% completion of the second reaction, \[[\log 2=0.3,\,\,\log 3=0.48]\]

    A) 0.2 : 1              

    B) 0.42 : 1

    C) 0.28 : 1            

    D) 0.3 : 1

    Correct Answer: D

    Solution :

    Time taken for 25% completion
    \[{{C}_{t}}={{C}_{0}}{{e}^{-1}}{{k}_{1}}\,{{t}_{1}}\]                   \[\Rightarrow \]\[\frac{3}{4}={{e}^{-}}{{k}_{1}}\,{{t}_{1}}\]
    As \[\ln \,(3/4)=-\,{{k}_{1}}\,{{b}_{1}}\] so \[{{t}_{1}}=\frac{1}{{{k}_{1}}}\ln \,\,\left( \frac{4}{3} \right).\]
    time taken for 75% completion \[{{t}_{2}}=2.\]\[\frac{\ln 2}{{{k}_{2}}}\]
    so required ration \[\frac{{{t}_{1}}}{{{t}_{2}}}=\frac{\ln \,(4/3){{k}_{2}}}{{{k}_{1}}.2{{\ln }_{2}}}=\frac{3}{2}\times \]\[\frac{(ln4-ln3)}{\ln 4}\]
    \[=0.3:1\]


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