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In a chemical reaction i.e., $A+2 B \longrightarrow$ products, when concentration of $A$ is doubled, rate of the reaction increases 4 times and when concentration of $B$ alone is doubled rate continues to be the same. The order of the reaction is
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The correct answer is:
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Let the order of reaction w.r.t. $A$ is $x$ and w.r.t. $B$ is $y$.
$r_{1}=k[A]^{x}[B]^{y} \quad \text{...(i)}$
$r_{2}=k[2 A]^{x}[B]^{y} \quad \text{...(ii)}$
$r_{3}=k[A]^{x}[2 B]^{y} \quad \text{...(iii)}$
$\begin{aligned} \frac{r_{1}}{r_{2}} &=\frac{k[A]^{x}[B]^{y}}{k[2 A]^{x}[B]^{y}} \Rightarrow \frac{1}{4}=\left(\frac{1}{2}\right)^{x} \\ \Rightarrow \quad\left(\frac{1}{2}\right)^{2} &=\left(\frac{1}{2}\right)^{x} \\ \Rightarrow \quad x &=2 \end{aligned}$
Similarly,
$$
\begin{aligned}
& \frac{r_{1}}{r_{3}}=\frac{k[A]^{x}[B]^{y}}{k[A]^{x}[2 B]^{y}} \Rightarrow 1=\left(\frac{1}{2}\right)^{y} \\
\Rightarrow \quad\left(\frac{1}{2}\right)^{0} &=\left(\frac{1}{2}\right)^{y} \\
y &=0
\end{aligned}
$$
Hence, the rate of equation is
$$
\text { Rate }=k[A]^{2}[B]^{0}
$$
Order of reaction $=2$
$r_{1}=k[A]^{x}[B]^{y} \quad \text{...(i)}$
$r_{2}=k[2 A]^{x}[B]^{y} \quad \text{...(ii)}$
$r_{3}=k[A]^{x}[2 B]^{y} \quad \text{...(iii)}$
$\begin{aligned} \frac{r_{1}}{r_{2}} &=\frac{k[A]^{x}[B]^{y}}{k[2 A]^{x}[B]^{y}} \Rightarrow \frac{1}{4}=\left(\frac{1}{2}\right)^{x} \\ \Rightarrow \quad\left(\frac{1}{2}\right)^{2} &=\left(\frac{1}{2}\right)^{x} \\ \Rightarrow \quad x &=2 \end{aligned}$
Similarly,
$$
\begin{aligned}
& \frac{r_{1}}{r_{3}}=\frac{k[A]^{x}[B]^{y}}{k[A]^{x}[2 B]^{y}} \Rightarrow 1=\left(\frac{1}{2}\right)^{y} \\
\Rightarrow \quad\left(\frac{1}{2}\right)^{0} &=\left(\frac{1}{2}\right)^{y} \\
y &=0
\end{aligned}
$$
Hence, the rate of equation is
$$
\text { Rate }=k[A]^{2}[B]^{0}
$$
Order of reaction $=2$
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