For the reaction A + B to text{products},what | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The rate law is given by: $	ext{Rate} = K[A]^{x}[B]^{y}$From experiment $(1)$: $5 	imes 10^{-4} = K[2.5 	imes 10^{-4}]^{x}[3 	imes 10^{-5}]^{y

Vedclass · Accepted Answer

$2$ with respect to $A$ and $1$ with respect to $B$

Vedclass · Answer

(B) The rate law is given by: $	ext{Rate} = K[A]^{x}[B]^{y}$From experiment $(1)$: $5 	imes 10^{-4} = K[2.5 	imes 10^{-4}]^{x}[3 	imes 10^{-5}]^{y} \quad \dots (i)$From experiment $(2)$: $4 	imes 10^{-3} = K[5 	imes 10^{-4}]^{x}[6 	imes 10^{-5}]^{y} \quad \dots (ii)$From experiment $(3)$: $1.6 	imes 10^{-2} = K[1 	imes 10^{-3}]^{x}[6 	imes 10^{-5}]^{y} \quad \dots (iii)$Dividing $(iii)$ by $(ii)$:$\frac{1.6 	imes 10^{-2}}{4 	imes 10^{-3}} = \left(\frac{1 	imes 10^{-3}}{5 	imes 10^{-4}}ight)^{x} \implies 4 = 2^{x} \implies x = 2$Dividing $(ii)$ by $(i)$:$\frac{4 	imes 10^{-3}}{5 	imes 10^{-4}} = \left(\frac{5 	imes 10^{-4}}{2.5 	imes 10^{-4}}ight)^{2} \cdot \left(\frac{6 	imes 10^{-5}}{3 	imes 10^{-5}}ight)^{y}$$8 = 2^{2} \cdot 2^{y} \implies 8 = 4 \cdot 2^{y} \implies 2 = 2^{y} \implies y = 1$Thus,the order with respect to $A$ is $2$ and with respect to $B$ is $1$.

$Exp.$	$[A] \ (mol \ L^{-1})$	$[B] \ (mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ s^{-1})$
$1$	$2.5 \times 10^{-4}$	$3 \times 10^{-5}$	$5 \times 10^{-4}$
$2$	$5 \times 10^{-4}$	$6 \times 10^{-5}$	$4 \times 10^{-3}$
$3$	$1 \times 10^{-3}$	$6 \times 10^{-5}$	$1.6 \times 10^{-2}$

Similar Questions

Define the following terms:
$(1)$ Rate law / Rate equation / Rate expression
$(2)$ Unimolecular reaction

For the reaction $2NOBr(g) \rightarrow 2NO(g) + Br_2(g)$, the rate law is $r = k[NOBr]^2$. If the rate constant $k = 1.62 \text{ M}^{-1} \text{ s}^{-1}$ and the concentration of $NOBr$ is $5 \times 10^{-3} \text{ M}$, what is the rate of the reaction?

For a reaction,$AB_5 \to AB + 4B$,the rate can be expressed in the following ways:
$-\frac{d[AB_5]}{dt} = K[AB_5]$ ; $\frac{d[B]}{dt} = K_1[AB_5]$
What is the correct relation between $K$ and $K_1$?

For a chemical reaction,$A + 2B \to C + D$,the rate of reaction increases $3$ times when the concentration of $A$ only is increased $9$ times. While when the concentration of $B$ only is increased $2$ times,the rate of reaction also increases $2$ times. The order of this reaction is:

The mechanism of the reaction $A + 2B \to D$ is
$2B \xrightarrow{k} B_2$ $[Slow]$
$B_2 + A \to D$ $[Fast]$
The rate law expression,order with respect to $A$,order with respect to $B$ and overall order of reaction are respectively

Vedclass Test Series

Exam Paper Generator

Online Exam Module