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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    One \[\text{d}{{\text{m}}^{\text{3}}}\]of 2M ethanoic acid is mixed with one \[\text{d}{{\text{m}}^{\text{3}}}\]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\[=\frac{1}{2}\times \frac{1}{2}\times \text{rate before dilution}\] Therefore, it is clear that the rate decreases four times.


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