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CMC Medical CMC-Medical Ludhiana Solved Paper-2015

  • question_answer
    The rate of a gaseous reaction is equal to k[A] [B]. The volume of the reaction vessel containing these gases, is reduced by \[\left( \frac{1}{2} \right)\]of the initial volume. The new rate of the reaction would be

    A)  \[\frac{1}{8}\]                                  

    B)  \[\frac{8}{1}\]

    C)  \[\frac{1}{4}\]                                  

    D)  \[\frac{4}{1}\]

    Correct Answer: D

    Solution :

                    Rate\[=k\,[A]\,[B]=kab\] \[r=kab\]                             ?(i) Volume is reduced to \[\frac{1}{2}\]then concentration becomes double. \[\therefore \]  \[r=k(2a)\,(2b)\]                 \[r=4kab\]                           ?(ii) From Eqs. (i) and (ii), we get                 \[\frac{r}{r}=4\]\[\Rightarrow \]\[4r=r\]


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