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J & K CET Medical J & K - CET Medical Solved Paper-2009

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

    A)  2/3 log 2                             

    B)  2/3 log 4

    C)  3/2 log 2                             

    D)  3/2 log 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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