In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:
What is the order of the reaction with respect to A and B?
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore,
$\mathrm{r}_{0}=k[\mathrm{~A}]^{x}[\mathrm{~B}]^{y}$
$5.07 \times 10^{-5}=k[0.20]^{x}[0.30]^{y}$ ...(i)
$5.07 \times 10^{-5}=k[0.20]^{x}[0.10]^{y}$ ...(ii)
$1.43 \times 10^{-4}=k[0.40]^{x}[0.05]^{y}$ ....(iii)
Dividing equation (i) by (ii), we obtain
$\frac{5.07 \times 10^{-5}}{5.07 \times 10^{-5}}=\frac{k[0.20]^{x}[0.30]^{y}}{k[0.20]^{x}[0.10]^{y}}$
$\Rightarrow 1=\frac{[0.30]^{y}}{[0.10]^{y}}$
$\Rightarrow\left(\frac{0.30}{0.10}\right)^{0}=\left(\frac{0.30}{0.10}\right)^{y}$
$\Rightarrow y=0$
Dividing equation (iii) by (ii), we obtain
$\frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{k[0.40]^{x}[0.05]^{y}}{k[0.20]^{x}[0.30]^{y}}$
$\Rightarrow \frac{1.43 \times 10^{-4}}{5.07 \times 10^{-5}}=\frac{[0.40]^{x}}{[0.20]^{x}} \quad\left[\begin{array}{l}\text { Since } y=0, \\ {[0.05]^{y}=[0.30]^{y}=1}\end{array}\right]$
$\Rightarrow 2.821=2^{x}$
$\Rightarrow \log 2.821=x \log 2$ (Taking log on both sides)
$\Rightarrow x=\frac{\log 2.821}{\log 2}$
= 1.496
= 1.5 (approximately)
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is zero.
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