Q4 / 120 · VITEEE 2010 · Chemistry
For a reaction of type $\mathrm{A}+\mathrm{B} \rightarrow$ products, it is observed that doubling concentration of $\mathrm{A}$ causes the reaction rate to be four times as great, but doubling amount of $\mathrm{B}$ does not affect the rate. The unit of rate constant is
- A$\mathrm{s}^{-1}$
- B$\mathrm{s}^{-1} \mathrm{~mol} \mathrm{~L}^{-1}$
- Correct answerC$\mathrm{s}^{-1} \mathrm{~mol}^{-1} \mathrm{~L}$
- D$\mathrm{s} \mathrm{s}^{-1} \mathrm{~mol}^{-2} \mathrm{~L}^{2}$
Solution
Let the initial rate be $R$ and order with respect to $A$ be $x$ and $B$ be $y$. Thus, rate lawcan be written as, rate, $R=[A]^{x}[B]^{y}$...(i)
After doubling the concentration of $A$, rate becomes $4 R$
$$
4 R=[2 A]^{x}[B]^{y}...(ii)
$$
After doubling the concentration of $B$, rate remains $R$,
$R=[A]^{x}[2 B\}$...(iii)
From Eq. (i) and (ii), we get
$$
\frac{R}{4 R}=\left(\frac{1}{2}\right)^{x} \Rightarrow\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{x}
$$
So, $x=2$
From Eq. (i) and (iii), we get
$$
\frac{R}{R}=\left[\frac{1}{2}\right]^{y} \Rightarrow\left(\frac{1}{1}\right)^{0}=\left(\frac{1}{2}\right)^{y}
$$
So, $Y=0$
Hence, the rate lawis, rate $\mathrm{R}=[A]^{2}[B]^{0}$ This clearly shows that the order of this reaction is 2 and for second order reaction units of rate constant are $\mathrm{mol}^{-1} \mathrm{Ls}^{-1}$.
Hence, the correct answer is option: 'C'.