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

Similar Questions

Write the unit of the rate constant for the following reactions:
$1.$ $5/2$ order
$2.$ $n$ order

For three reactions of first,second,and third order,the rate constants are numerically equal. If the concentration of the reactant is the same and greater than $1 \ M$,which of the following is true for the rates of these three reactions?

The half-life for a second-order reaction is $30 \ min$. If the initial concentration is $0.1 \ M$,then the value of the rate constant will be $............ \ M^{-1} \ min^{-1}$.

$A + B \rightarrow \text{product}$. The rate of this reaction is given by $v = K[A]^2[B]^0$. What is the change in the rate of reaction when the concentration of $A$ is doubled and the concentration of $B$ is doubled?

Which among the following reactions can be an example of a pseudo first order reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module