J & K CET Engineering J and K - CET Engineering Solved Paper-2009

  • question_answer
    The rate constant for a first order reaction is \[6.909\,{{\min }^{-1}},\]Therefore, the time required, in minute, for the participation of 75% of the initial reactant is

    A) \[2/3\text{ }log2\]     

    B) \[~2/3\text{ }log\text{ }4\]

    C) \[~3/2\text{ }log\,2\]       

    D) \[~3/2\text{ }log\text{ }4\]

    Correct Answer: A

    Solution :

     For first order reaction, \[k=\frac{2.303}{t}\log \left( \frac{a}{a-x} \right)\] \[6.909=\frac{2.303}{t}\log \frac{100}{100-75}\] \[t=\frac{1}{3}\log \frac{100}{25}\] \[=\frac{1}{3}\log 4=\frac{1}{3}\log {{2}^{2}}\] \[=\frac{2}{3}\log \,2\]


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