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AIIMS AIIMS Solved Paper-2011

  • 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 :

    \[A\text{ }\xrightarrow{{}}Product\]
    initially a 0.
    after time t \[(a-x)\] \[x\]
    After \[{{t}_{1/4}}\] \[\left( a-\frac{a}{4} \right)\] \[\frac{a}{4}\]
    For first-order kinetics \[k=\frac{2.303}{t}\log \left( \frac{a}{a-x} \right)\] \[\therefore \]   \[k=\frac{2.303}{{{t}_{1/4}}}\log \frac{a}{\frac{3a}{4}}\] \[{{t}_{1/4}}=\frac{2.303\log \frac{4}{3}}{k}\] \[=\frac{2.303(\log 4-\log 3)}{k}\] \[=\frac{2.303(0.6020-0.4771)}{k}\] \[=\frac{2.303\times 0.1249}{k}\] \[=\frac{0.2876}{k}\] \[=\frac{0.29}{k}\]


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