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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2009

  • question_answer
    The initial concentration of sugar solution is 0.12 M. On    doing   fermentation   the concentration of sugar decreases to 0.06 M in 10 h and to 0.045 M in 15 h. The order of the reaction is

    A)  0.5                  

    B)  1.0

    C)  1.5                  

    D)  2.0

    Correct Answer: B

    Solution :

     [b] \[0.12M\xrightarrow[{}]{10\,h}0.06M;{{t}_{50%}}=10\,h\] \[0.12M\xrightarrow[{}]{15\,h}0.045M;{{t}_{62.5%}}=15\,h\] \[0.12M\xrightarrow[{}]{20\,h}0.03M;{{t}_{75%}}=20\,h\] For first order reaction, \[{{t}_{75%}}={{t}_{50%}}\times 2\] Hence, this is a first order reaction. Further on putting the values in first order reaction, the value of k remains constant, hence it is a first order reaction.


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