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(D) Let the order of reaction with respect to $A$ and $B$ be $x$ and $y$ respectively.So,the rate law can be given as:$R = k[A]^{x}[B]^{y} \dots (i)$W

Vedclass · Accepted Answer

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

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(D) Let the order of reaction with respect to $A$ and $B$ be $x$ and $y$ respectively.So,the rate law can be given as:$R = k[A]^{x}[B]^{y} \dots (i)$When the concentration of only $B$ is doubled,the rate is doubled:$2R = k[A]^{x}[2B]^{y} \dots (ii)$Dividing $(ii)$ by $(i)$:$2 = 2^{y} \Rightarrow y = 1$When concentrations of both $A$ and $B$ are doubled,the rate increases by a factor of $8$:$8R = k[2A]^{x}[2B]^{y} \dots (iii)$Dividing $(iii)$ by $(i)$:$8 = 2^{x} 	imes 2^{y}$Substituting $y = 1$:$8 = 2^{x} 	imes 2^{1}$ $\Rightarrow 4 = 2^{x}$ $\Rightarrow x = 2$Thus,the rate law is $R = k[A]^{2}[B]$.

Entry	$[A]$ in $M$	$[B]$ in $M$	Initial rate $(M/s)$
$1$	$0.02$	$0.02$	$2 \times 10^{-2}$
$2$	$0.02$	$0.04$	$4 \times 10^{-2}$
$3$	$0.04$	$0.04$	$8 \times 10^{-2}$

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

Similar Questions

For a reaction $A + B \longrightarrow P$,the following data are provided. The rate constant for this reaction in standard units is:
Entry $[A]$ in $M$ $[B]$ in $M$ Initial rate $(M/s)$
$1$ $0.02$ $0.02$ $2 \times 10^{-2}$
$2$ $0.02$ $0.04$ $4 \times 10^{-2}$
$3$ $0.04$ $0.04$ $8 \times 10^{-2}$

For the first order decomposition reaction of $N_2O_5$,it is found that -
$(a)$ $2N_2O_5 \rightarrow 4NO_{2(g)} + O_{2(g)}$ ; $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$
$(b)$ $N_2O_5 \rightarrow 2NO_{2(g)} + 1/2 O_{2(g)}$ ; $-\frac{d[N_2O_5]}{dt} = k'[N_2O_5]$
Which of the following is true?

For a reaction $A + B \rightarrow \text{product}$,if $[A]$ is doubled keeping $[B]$ constant,the rate of reaction doubles. Calculate the order of reaction with respect to $A$.

Rate of reaction is $r_1 = k[A]^a[B]^b$. If the concentration of $A$ is doubled and $B$ is halved,the new rate is $r_2$. What is the value of $\frac{r_2}{r_1}$?

The value of $\frac{t_{0.875}}{t_{0.50}}$ for $n^{th}$ order reaction is

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