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 For the reaction,3A+2B\rightarrow C+D,  the differential rate law can be written as :

Option: 1

\frac{1}{3}\frac{d\left [ A \right ]}{dt}=\frac{d\left [ C \right ]}{dt}=k\left [ A \right ]^{n}\left [ B \right ]^{m}

Option: 2

-\frac{d\left [ A \right ]}{dt}=\frac{d\left [ C \right ]}{dt}=k\left [ A \right ]^{n}\left [ B \right ]^{m}

Option: 3

+\frac{1}{3}\frac{d\left [ A \right ]}{dt}=\frac{d\left [ C \right ]}{dt}=k\left [ A \right ]^{n}\left [ B \right ]^{m}

Option: 4

-\frac{1}{3}\frac{d\left [ A \right ]}{dt}=\frac{d\left [ C \right ]}{dt}=k\left [ A \right ]^{n}\left [ B \right ]^{m}

Answers (1)


Let's consider a general reaction:

aA+bB\rightleftharpoons cC+dD

The differential rote law for the above reaction is:

\frac{-1}{a}\:\frac{d\left ( A \right )}{dt}=\frac{-1}{b}\:\frac{d\left ( B \right )}{dt}=\frac{1}{c}\:\frac{d\left ( c \right )}{dt} = \frac{1}{d}\:\frac{d\left ( D \right )}{dt} = k\left [ A \right ]^{3}\left [ B \right ]^{2}

Similarly for 3A+2B\rightarrow C+D, we have 

\frac{-1}{3}\:\frac{d\left ( A \right )}{dt}= \frac{-1}{2}\:\frac{d\left ( B \right )}{dt}= \:\frac{d\left ( C \right )}{dt}= \:\frac{d\left ( D \right )}{dt} = k\left [ A \right ]^{3}\left [ B \right ]^{2}

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Divya Prakash Singh

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