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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2001

  • question_answer
    One \[d{{m}^{3}}\]of 2M ethanoic acid is mixed with one \[d{{m}^{3}}\] of 3M ethanol to form an ester \[C{{H}_{3}}COOH+{{C}_{2}}{{H}_{5}}OH\xrightarrow{{}}\] \[C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O\] The decrease in the initial rate if each solution is diluted with an equal volume of water would be:

    A)  2 times      

    B)  4 times

    C)  \[0.25\]times    

    D)  \[0.5\] times

    Correct Answer: B

    Solution :

    \[C{{H}_{3}}COOH+{{C}_{2}}{{H}_{5}}OH\]\[\xrightarrow{{}}C{{H}_{3}}COO{{C}_{2}}{{H}_{5}}+{{H}_{2}}O\] \[\therefore \] Rate \[\propto [C{{H}_{3}}COOH][{{C}_{2}}{{H}_{5}}OH]\] As per question each solution is diluted with an equal volume of water. Hence the concentration becomes half. So, rate \[=\frac{1}{2}\times \frac{1}{2}\times \] rate before dilution Therefore, it is clear that the rate decreases four times.


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