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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 45

  • question_answer
    \[{{t}_{1/4}}\] can be taken as the time taken for the concentration of a reactant to drop to \[\frac{3}{4}\] of its initial value. If the rate constant for a first order reaction is k, the \[{{t}_{1/4}}\] can be written as

    A) \[0.75/k\]

    B)        \[0.69/k\]

    C) \[0.29/k\]

    D)        \[0.10/k\]

    Correct Answer: C

    Solution :

    \[{{t}_{1/4}}=\frac{2.303}{k}\log \frac{1}{3/4}=\frac{2.303}{k}\log \frac{4}{3}\] \[=\frac{2.303}{k}(\log 4-\log 3)=\frac{2.303}{k}(2\log 2-\log 3)\] \[=\frac{2.303}{k}(2\times 0.301-0.4771)=\frac{0.29}{k}\]


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