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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 38

  • question_answer
    Plotting a graph of log \[{{t}_{1/2}}\] against \[log{{\left[ A \right]}_{0}}\] of a reactant for a first order reaction, the slope will be

    A) \[-1\]                

    B)        \[-2\]   

    C) 0         

    D)        +1

    Correct Answer: C

    Solution :

    [c] \[{{t}_{1/2}}=\frac{0.693}{k}\] \[log\text{ }{{t}_{1}}/2=log0.693-logk=constant\]          


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