Which function of [X] plotted against time wi | vedclass

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

Question

(D) For a second order reaction,the rate law is given by: $-\frac{d[X]}{dt} = k[X]^2$Integrating this expression: $\int_{[X]_0}^{[X]} -\frac{d[X]}{[X]

Vedclass · Accepted Answer

$\frac{1}{[X]}$

Vedclass · Answer

(D) For a second order reaction,the rate law is given by: $-\frac{d[X]}{dt} = k[X]^2$Integrating this expression: $\int_{[X]_0}^{[X]} -\frac{d[X]}{[X]^2} = \int_{0}^{t} k dt$This results in: $\frac{1}{[X]} - \frac{1}{[X]_0} = kt$Rearranging into the equation of a straight line $(y = mx + c)$: $\frac{1}{[X]} = kt + \frac{1}{[X]_0}$Here,$y = \frac{1}{[X]}$,$m = k$,$x = t$,and $c = \frac{1}{[X]_0}$.Thus,a plot of $\frac{1}{[X]}$ against time $t$ yields a straight line.

Which function of $[X]$ plotted against time will give a straight line for a second order reaction? $X \to \text{Product}$

Similar Questions

For an elementary chemical reaction,$A_2 \underset{k_{-1}}{\overset{k_1}{\longleftrightarrow}} 2A$,the expression for $\frac{d[A]}{dt}$ is

For the following graphs,choose the correct option regarding the order of reaction:

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?

For the following rate law,determine the unit of the rate constant: Rate $= -\frac{d[R]}{dt} = k[A]^{\frac{1}{2}}[B]^{2}$

In the hydrolysis of an organic chloride in the presence of a large excess of water; $RCl + H_2O \rightarrow ROH + HCl$. What are the molecularity and order of the reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module