Compounds $A$ and $B$ react according to the equation $2 A(g)+B(g) \longrightarrow 2 C(g)+D(g)$. The initial…
Compounds $A$ and $B$ react according to the equation $2 A(g)+B(g) \longrightarrow 2 C(g)+D(g)$.
The initial rate of formation was determined at different initial concentrations of $A$ and $B$. The following results were obtained. The rate law for this reaction may be [All concentrations are in $\mathrm{mol} / \mathrm{L}]$
Rate $=k[A]^2[B]$
Rate $=k[A][B]^2$
Rate $=k[A][B]$
Rate $=k[A]^2[B]^0$
Solution
$2 A+B \longrightarrow 2 C+D$
Let order with respect to $A$ and $B$ are $x$ and $y$ respectively.
$\therefore \quad$ Rate $=k[A]^x[B]^y$
$7.2 \times 10^{-2}=k(0.3)^x(0.2)^y$ ...(i)
$2.88 \times 10^{-1}=k(0.3)^x(0.4)^y$ ...(ii)
$6 \times 10^{-3}=k(0.1)^x(0.1)^y$ ...(iii)
$2.4 \times 10^{-2}=k(0.4)^x(0.1)^y$ ...(iv)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{7.2 \times 10^{-2}}{2.88 \times 10^{-1}}=\frac{(0.2)^y}{(0.4)^y}$
$\frac{1}{4}=\left(\frac{1}{2}\right)^y$
$\because \quad\left(\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^y$
$\therefore \quad y=2$
Dividing Eq. (iii) by Eq. (iv), we get
$\frac{6 \times 10^{-3}}{2.4 \times 10^{-2}}=\frac{(0.1)^x}{(0.4)^x}$
$\left(\frac{1}{4}\right)^1=\left(\frac{1}{4}\right)^x$
$\left(\frac{1}{4}\right)^1=\left(\frac{1}{4}\right)^x$
$\therefore \quad x=1$
$\begin{aligned} & \because \text { Rate }=k[A]^x[B]^y \\ & \therefore \text { Rate }=k[A][B]^2\end{aligned}$
