For a reaction: 3X to 4Y,calculate the rate o | vedclass

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

Question

(C) The rate law for the reaction $3X 	o 4Y$ is given by $-\frac{1}{3} \frac{d[X]}{dt} = k[X]^n$.For an $n$-th order reaction where $n 
eq 1$,the in

Vedclass · Accepted Answer

$\frac{1}{9\sqrt{3}} 	imes 10^{-4} \ M \ \min^{-1}$

Vedclass · Answer

(C) The rate law for the reaction $3X 	o 4Y$ is given by $-\frac{1}{3} \frac{d[X]}{dt} = k[X]^n$.For an $n$-th order reaction where $n 
eq 1$,the integrated rate law is $\frac{1}{[X]^{n-1}} - \frac{1}{[X]_0^{n-1}} = 3(n-1)kt$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \frac{1}{[X]^3}$,we identify $n-1 = 3$,so $n = 4$.The slope of the graph is $3(n-1)k = 	an 30^{\circ} = \frac{1}{\sqrt{3}}$.Substituting $n-1 = 3$,we get $3(3)k = \frac{1}{\sqrt{3}}$,which implies $9k = \frac{1}{\sqrt{3}}$,or $k = \frac{1}{9\sqrt{3}}$.The rate of reaction $r$ is given by $r = k[X]^n = \frac{1}{9\sqrt{3}} 	imes (0.1)^4$.$r = \frac{1}{9\sqrt{3}} 	imes 10^{-4} \ M \ \min^{-1}$.

For a reaction: $3X \to 4Y$,calculate the rate of reaction based on the following graph when the concentration of $X$ is $0.1 \ M$.

Similar Questions

For a reaction of order $(n - 1)$,what is the relationship between the half-life period $(t_{1/2})$ and the initial concentration of the reactant $([R]_0)$?

For the reaction $2A + B \rightarrow \text{product}$,the rate of the reaction is $15 \times 10^{-2} \ mol \ dm^{-3} \ sec^{-1}$ when $[A] = 0.3 \ mol \ dm^{-3}$ and $[B] = 0.05 \ mol \ dm^{-3}$. What is the value of the rate constant if the reaction is first order in both the reactants?

Consider the following reaction: $A \longrightarrow \text{Products}$. This reaction is completed in $100 \ min$. The rate constant of this reaction at $t_1 = 10 \ min$ is $10^{-2} \ min^{-1}$. What is the rate constant (in $min^{-1}$) at $t_2 = 20 \ min$?

The rate law for a reaction between the substances $A$ and $B$ is given by,$rate = k[A]^n[B]^m$. On doubling the concentration of $A$ and halving the concentration of $B$,the ratio of the new rate to the earlier rate of the reaction will be as

Vedclass Test Series

Exam Paper Generator

Online Exam Module