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  • The results given in the below table were obtained during kinetic studies of the following reaction: A+2B→C+D. <table border="1" cellpadding="0" cellspacing="0" style="width:500p

The results given in the below table were obtained during kinetic studies of the following reaction:
A+2B→C+D.

    Experiment         Initial     Concentration [A]     Initial     Concentration [B]

    Initial Rate of     Formation of D
    1

     0.1

     0.1     6\times10^{-3}
    2     0.3     0.2     7\times10^{-2}
    3     0.3     0.4     2.8\times10^{-1}
    4     0.4     0.1     2.4\times10^{-2}

What will be correct rate law expression for the given reaction?

 

Option: 1

Rate=k[A][B]


Option: 2

Rate=\mathrm{k[A][B]^{2}}


Option: 3

\mathrm{Rate=k[A]^{2}[B]^{2} }


Option: 4

\mathrm{Rate=k[A]^{2}[B]}


Answers (1)

best_answer

In experiment 2 and 3 (Keeping the concentration of A constant), when the concentration ' B ' is doubled, the rate gets quadrapled. So, by initial rate method,

\mathrm{ \frac{7 \times 10^{-2}}{2.8 \times 10^{-1}}=\left(\frac{0.2}{0.4}\right)^b \\ }

\mathrm{ \frac{1}{4}=\left(\frac{1}{2}\right)^b}

Thus, b=2
Thus, order w.r.t to B is 2 .

In experiment 1 and 4 (Keeping the concentration of B constant), when the concentration quadrapled the rate quadrapled.
So, by initial rate method,

\mathrm{ \frac{6 \times 10^{-3}}{2.4 \times 10^{-2}}=\left(\frac{0.1}{0.4}\right)^a \\ }

\mathrm{ \frac{1}{4}=\left(\frac{1}{4}\right)^a }

Thus, a=1
Thus, order w.r.t to A is 1 .
So, rate law expression is.
\mathrm{ \text { Rate }=k[A][B]^2 }

Posted by

Ritika Jonwal

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