Consider the kinetic data given in the follow | vedclass

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

Question

(A) The rate law is given by $r = K [A]^{n_1} [B]^{n_2} [C]^{n_3}$.Comparing experiments $1$ and $2$,$[A]$ and $[C]$ are constant,but $[B]$ changes an

Vedclass · Accepted Answer

$6.75$

Vedclass · Answer

(A) The rate law is given by $r = K [A]^{n_1} [B]^{n_2} [C]^{n_3}$.Comparing experiments $1$ and $2$,$[A]$ and $[C]$ are constant,but $[B]$ changes and the rate remains constant,so $n_2 = 0$.Comparing experiments $1$ and $3$,$[A]$ and $[B]$ are constant,$[C]$ doubles,and the rate doubles,so $n_3 = 1$.Comparing experiments $1$ and $4$,$[B]$ and $[C]$ are constant,$[A]$ increases by $1.5$ times,and the rate increases by $1.5$ times,so $n_1 = 1$.Thus,the rate law is $r = K [A] [C]$.Using experiment $1$: $6.0 	imes 10^{-5} = K 	imes 0.2 	imes 0.1$,which gives $K = 3.0 	imes 10^{-3} \ L \ mol^{-1} \ s^{-1}$.Now,for $[A] = 0.15 \ mol \ dm^{-3}, [B] = 0.25 \ mol \ dm^{-3}, [C] = 0.15 \ mol \ dm^{-3}$:$r = (3.0 	imes 10^{-3}) 	imes 0.15 	imes 0.15 = 6.75 	imes 10^{-5} \ mol \ dm^{-3} \ s^{-1}$.Comparing this with $Y 	imes 10^{-5}$,we get $Y = 6.75$.

Experiment No.	$[A] \ (mol \ dm^{-3})$	$[B] \ (mol \ dm^{-3})$	$[C] \ (mol \ dm^{-3})$	Rate of reaction $(mol \ dm^{-3} \ s^{-1})$
$1$	$0.2$	$0.1$	$0.1$	$6.0 \times 10^{-5}$
$2$	$0.2$	$0.2$	$0.1$	$6.0 \times 10^{-5}$
$3$	$0.2$	$0.1$	$0.2$	$1.2 \times 10^{-4}$
$4$	$0.3$	$0.1$	$0.1$	$9.0 \times 10^{-5}$

Similar Questions

If the half-life period of a reaction is inversely proportional to the initial concentration,the order of the reaction is:

The hydrolysis of an ester is catalyzed by dilute acids $A$ and $B$. The rate constants for the two processes are $K_A$ and $K_B$ respectively. If $K_A > K_B$,which of the following statements is correct?

Differential form of the rate equation is $\frac{dx}{dt} = k[P][Q]^{0.5}[R]^{0.5}$. Which statement about the above equation is wrong?

$1 \ L$ of $2 \ M$ $CH_3COOH$ is mixed with $1 \ L$ of $3 \ M$ $C_2H_5OH$ to form an ester. The rate of the reaction with respect to the initial rate when each solution is diluted with an equal volume of water will be (in times).

For the reaction: $2A + B \to A_2B$; the rate $= K[A][B]^2$ with $K = 2.0 \times 10^{-6} \ L^2 \ mol^{-2} \ s^{-1}$. Initial concentrations of $A$ and $B$ are $[A]_0 = 0.2 \ mol/L$ and $[B]_0 = 0.4 \ mol/L$ respectively. Calculate the rate of reaction after $[A]$ is reduced to $0.12 \ mol/L$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module