In a reaction, $A+B \rightarrow$ Product, rate is doubled when the concentration of $B$ is doubled, and rate increases by a factor of 8 when the concentrations of both the reactants $(A$ and $B$ ) are doubled. Rate law for the reaction can be written as

Question

Aicharya · Accepted Answer

Let the order of reaction with respect to $A$ and $B$ is $x$ and $y$ respectively. So, the rate law can be given as
$R=k[A]^x[B]^y$ ...(i)
When the concentration of only $B$ is doubled, the rate is doubled, so
$R_1=k[A]^x[2 B]^y=2 R$ ...(ii)
If concentrations of both the reactants $A$ and $B$ are doubled, the rate increases by a factor of 8 , so
$\begin{aligned}
R^{\prime \prime}=k[2 A]^x[2 B]^y & =8 R \\
\Rightarrow k 2^x 2^y[A]^x[B]^y & =8 R \quad \ldots (iii)
\end{aligned}$
From Eq. (i) and (ii), we get
$\begin{array}{rlrl}
& \Rightarrow & \frac{2 R}{R} & =\frac{[A]^x[2 B]^y}{[A]^x[B]^y} \\
& \therefore & =2^y \\
& y & =1
\end{array}$
From Eq. (i) and (iii), we get
$\Rightarrow \frac{8 R}{R}=\frac{2^x 2^y[A]^x[B]^y}{[A]^x[B]^y}$
or $8=2^x 2^y$
Substitution of the value of $y$ gives,
$\begin{aligned}
& 8=2^x 2^1 \\
& 4=2^x
\end{aligned}$
$\begin{aligned}
& & (2)^2 & =(2)^x \\
\therefore & & x & =2
\end{aligned}$
Substitution of the value of $x$ and $y$ in Eq. (i) gives,
$R=k[A]^2[B]$

In a reaction, $A+B \rightarrow$ Product, rate is doubled when the concentration of $B$ is doubled, and rate…

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