પ્રક્રિયા $2 A + B \rightarrow A _{2} B $ માટે વેગ $=k[ A ][ B ]^{2}$ છે જેમાં $k =2.0 \times 10^{-6} \,mol ^{-2}\, L ^{2} \,s ^{-1}$ છે. જ્યારે $[ A ]=0.1 \,mol \,L ^{-1},[ B ]=0.2\, mol \,L ^{-1}$ હોય ત્યારે પ્રક્રિયાનો પ્રારંભિક વેગ ગણો. $[A] $ ઘટીને $0.06 \,mol\, L ^{-1}$ થાય પછી પ્રક્રિયાનો વેગ ગણો.

The initial rate of the reaction is

Rate $=k[ A ][ B ]^{2}$

$=\left(2.0 \times 10^{-6}\, mol ^{-2} \,L ^{2}\, s ^{-1}\right)\left(0.1\, mol\, L ^{-1}\right)\left(0.2 \,mol\, L ^{-1}\right)^{2}$

$=8.0 \times 10^{-9}\, mol ^{-2}\, L ^{2}\, s ^{-1}$

When $[A]$ is reduced from $0.1$ $mol$ $L_{1}$ to $0.06$ $mol ^{-1}$, the concentration of $A$ reacted $=$ $(0.1-0.06) \,mol \,L ^{-1}=0.04 \,mol \,L ^{-1}$

Therefore, concentration of $B$ reacted $=\frac{1}{2} \times 0.04 \,mol\, L ^{-1}=0.02 \,mol\, L ^{-1}$

Then, concentration of $B$ available, $[ B ]=(0.2-0.02)\, mol\, L ^{-1}$
$=0.18 \,mol\, L ^{-1}$

After $[A]$ is reduced to $0.06 \,mol\, L ^{-1}$, the rate of the reaction is given by,

Rate $=k[ A ][ B ]^{2}$

$=\left(2.0 \times 10^{-6}\, mol ^{-2}\, L ^{2} \,s ^{-1}\right)\left(0.06 \,mol \,L ^{-1}\right)\left(0.18 \,mol\, L ^{-1}\right)^{2}$

$=3.89 \,mol\, L ^{-1}\, s ^{-1}$