The results given in the below table were obt | vedclass

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

Question

(A) From the rate law expression: $Rate = k [A]^x [B]^y$Using the data from experiments $I, II,$ and $III$:$6.00 	imes 10^{-3} = k (0.1)^x (0.1)^y \d

Vedclass · Accepted Answer

$0.3, 0.4$

Vedclass · Answer

(A) From the rate law expression: $Rate = k [A]^x [B]^y$Using the data from experiments $I, II,$ and $III$:$6.00 	imes 10^{-3} = k (0.1)^x (0.1)^y \dots(1)$$2.40 	imes 10^{-2} = k (0.1)^x (0.2)^y \dots(2)$$1.20 	imes 10^{-2} = k (0.2)^x (0.1)^y \dots(3)$Dividing $(3)$ by $(1)$: $2^x = 2 \implies x = 1$Dividing $(2)$ by $(1)$: $2^y = 4 \implies y = 2$Thus,the rate law is $Rate = k [A]^1 [B]^2$.For experiment $IV$:$7.20 	imes 10^{-2} = k (X)^1 (0.2)^2$Using $k$ from experiment $I$: $k = \frac{6.00 	imes 10^{-3}}{(0.1)(0.1)^2} = 6 \ L^2 \ mol^{-2} \ min^{-1}$$7.20 	imes 10^{-2} = 6 	imes X 	imes 0.04 \implies X = \frac{7.20 	imes 10^{-2}}{0.24} = 0.3 \ M$For experiment $V$:$2.88 	imes 10^{-1} = 6 	imes (0.3) 	imes Y^2$$Y^2 = \frac{2.88 	imes 10^{-1}}{1.8} = 0.16 \implies Y = 0.4 \ M$Therefore,$X = 0.3$ and $Y = 0.4$.

Experiment	$[A] / mol \ L^{-1}$	$[B] / mol \ L^{-1}$	Initial rate / $mol \ L^{-1} \ min^{-1}$
$I$	$0.1$	$0.1$	$6.00 \times 10^{-3}$
$II$	$0.1$	$0.2$	$2.40 \times 10^{-2}$
$III$	$0.2$	$0.1$	$1.20 \times 10^{-2}$
$IV$	$X$	$0.2$	$7.20 \times 10^{-2}$
$V$	$0.3$	$Y$	$2.88 \times 10^{-1}$

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

Similar Questions

What is the molecularity of a reaction? Explain its types with examples.

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 $(M/s)$
$1$ $0.02$ $0.02$ $2 \times 10^{-2}$
$2$ $0.02$ $0.04$ $4 \times 10^{-2}$
$3$ $0.04$ $0.04$ $8 \times 10^{-2}$

Identify the order of the following reaction: $2 NO_{2(g)} \longrightarrow 2 NO_{(g)} + O_{2(g)}$

For the reaction $3 \, A_{(g)} \xrightarrow{K} B_{(g)} + C_{(g)}$,$K$ is $10^{-14} \, L/mol \cdot min$. If $[A] = 0.5 \, M$,then the value of $-\frac{d[A]}{dt}$ (in $M/sec$) is:

For the reaction $2NO_{(g)} + O_{2_{(g)}} \rightarrow 2NO_{2_{(g)}}$,the volume of the reaction vessel is suddenly reduced to half of its original volume. If the reaction is first order with respect to $O_2$ and second order with respect to $NO$,the rate of the reaction will:

Vedclass Test Series

Exam Paper Generator

Online Exam Module