KVPY Sample Paper KVPY Stream-SX Model Paper-3

  • question_answer
    For a general  \[{{n}^{th}}\] order process \[A\to ~P\] with initial concentration of reactant \[''a''\] and rate constant k, the expression for time for 75% completion of reaction is -

    A) \[\frac{1}{n-1}\left( \frac{{{2}^{n-1}}-2}{{{a}^{n-1}}} \right)\frac{1}{k}\]       

    B) \[\frac{1}{n-1}\left( \frac{{{2}^{n-2}}-2}{{{a}^{n-1}}} \right)\frac{1}{k}\]

    C) \[\frac{1}{n-1}\left( \frac{{{2}^{2n-2}}-1}{{{a}^{n-1}}} \right)\frac{1}{k}\]     

    D) \[\frac{1}{n-2}\left( \frac{{{2}^{2n-2}}-1}{{{a}^{2n-1}}} \right)\frac{1}{k}\]

    Correct Answer: C

    Solution :

    for a nth order reaction :
    \[kt=\frac{1}{n-1}\left[ \frac{1}{{{[A]}^{n-1}}}-\frac{1}{[A]_{0}^{n-1}} \right]\]
    For 75% completion of reaction \[[A]=\frac{{{[A]}_{0}}}{4}\]
    \[\Rightarrow t=\frac{1}{k(n-1)}\left[ \frac{{{4}^{n-1}}-1}{[A]_{0}^{n-1}} \right]\]
          \[=\frac{1}{k\,(n-1)}\left( \frac{{{2}^{2n-2}}-1}{{{a}^{n-1}}} \right)\]

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