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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 41

  • question_answer
    Given that for a reaction of order n. the integrated form of the rate equation is \[K=\frac{1}{t(n-1)}\left[ \frac{1}{{{C}^{n-1}}}-\frac{1}{C_{0}^{n-1}} \right]\] where \[{{C}_{0}}\] and C are the values of the reactant concentration at the start and after time t. What is the relationship between \[{{t}_{3/4}}\] and \[{{t}_{1/2}}\] where \[{{t}_{3/4}}\] is the time required for C to become \[1/4\text{ }{{C}_{0}}\]-

    A) \[{{t}_{3/4}}={{t}_{1/2}}\left[ {{2}^{n-1}}+1 \right]\]

    B) \[{{t}_{3/4}}={{t}_{1/2}}\left[ {{2}^{n-1}}-1 \right]\]

    C) \[{{t}_{3/4}}={{t}_{1/2}}\left[ {{2}^{n+1}}-1 \right]\]            

    D) \[{{t}_{3/4}}={{t}_{1/2}}\left[ {{2}^{n+1}}+1 \right]\]

    Correct Answer: A

    Solution :

    [a] \[{{t}_{3/4}}=\frac{1}{K(n-1)}\left[ \frac{1}{{{\left( \frac{{{C}_{0}}}{4} \right)}^{n-1}}}-\frac{1}{C_{0}^{n-1}} \right]\] \[=\frac{1}{K(n-1)}({{4}^{n-1}}-1)C_{0}^{n-1}\] \[\therefore {{t}_{1/2}}=\frac{1}{K(n-1)}({{2}^{n-1}}-1)\] So \[\frac{{{t}_{3/4}}}{{{t}_{1/2}}}={{2}^{n-1}}+1\]


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