In a reaction, A + B rightarrow product, rate | vedclass

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Question

Let the order of reaction with respect to $\mathrm{A}$ and $\mathrm{B}$ is $\mathrm{x}$ and $\mathrm{y}$ respectively.

So, the law can be given as

Vedclass · Accepted Answer

rate $= k[A]^2[B]$

Vedclass · Answer

Let the order of reaction with respect to $\mathrm{A}$ and $\mathrm{B}$ is $\mathrm{x}$ and $\mathrm{y}$ respectively.

So, the law can be given as

$R=k[A]^{x}[B]^{y}$ ...... $(i)$

When the concentration of only $\mathrm{B}$ 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 $R^{\prime \prime}=k[2 A]^{x}[2 B]^{y}=8 R$  .... $(i i i)$

$\Rightarrow k 2^{x} 2^{y}[A]^{x}[B]^{y}=8\, R$ ..... $(iv)$

From Eqs. $(i)$ and $(ii)$,

we get $\Rightarrow \quad \frac{2 R}{R}=\frac{|A|^{x}|2 B|^{y}}{|A|^{x}|B|^{y}}$

$2=2^{y}$

$	herefore y=1$

From Eqs. (i) and (iv), 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,

$8=2^{x} 2^{1}$

$4=2^{x}$

$(2)^{2}=(2)^{x}$

$x=2$

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 increases by a factor of $8$ when the concentration of both the reactants $(A$ and $B)$ are doubled, rate law for the reaction can be written as

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