The initial rates of reaction $3 \mathrm{~A}+2 \mathrm{~B}+\mathrm{C} \longrightarrow$ Products, at…

The initial rates of reaction
$3 \mathrm{~A}+2 \mathrm{~B}+\mathrm{C} \longrightarrow$ Products, at different initial concentrations are given below:
$\begin{array}{llll}\text { Initial rate, } & {[\mathbf{A}]_{0}, \mathbf{M}} & {[\mathbf{B}]_{0}, \mathbf{M}} & {[\mathbf{C}]_{0}, \mathbf{M}} \\ \mathbf{M s}^{-1} & & & \\ 5.0 \times 10^{-3} & 0.010 & 0.005 & 0.010 \\ 5.0 \times 10^{-3} & 0.010 & 0.005 & 0.015 \\ 1.0 \times 10^{-2} & 0.010 & 0.010 & 0.010 \\ 1.25 \times 10^{-3} & 0.005 & 0.005 & 0.010\end{array}$
The order with respect to the reactants, $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are respectively
  1. $3,2,0$
  2. $3,2,1$
  3. $2,2,0$
  4. $2,1,0$

Solution

From $1^{\text {st }}$ and $2^{\text {nd }}$ sets of data - no change in rate is observed with the change in concentration of ' $\mathrm{C}^{\prime}$. So the order with respect to ' $\mathrm{C}$ ' is zero.
From $1^{\text {st }}$ and $4^{\text {th }}$ sets of data
Dividing eq. (4) by eq. (1)
$\frac{1.25 \times 10^{-3}}{5.0 \times 10^{-3}}=\left[\frac{0.005}{0.010}ight]^{x}$
or $0.25=(0.5)^{x}$ or $(0.5)^{2}=(0.5)^{x}$
$\therefore \mathrm{x}=2$
The order with respect to ' $\mathrm{A}$ ' is 2 from the $1^{\text {st }}$ and $3^{\text {rd }}$ sets of data dividing eq. (1) by eq. (3)
$\frac{5.0 \times 10^{-3}}{1.0 \times 10^{-2}}=\left[\frac{0.005}{0.010}ight]^{y}$
or $(0.5)^{1}=(0.5)^{y} \Rightarrow y=1$
The order with respect to ' $\mathrm{B}$ ' is 1
So the order with respective the reactants A, B and C is 2,1 and $0 .$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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