For a zero order reaction,the plot of concent | vedclass

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

Question

(B) For a zero order reaction,the rate law is given by: $Rate = k[A]^0 = k$. Integrating this with respect to time,we get the integrated rate equation

Vedclass · Accepted Answer

linear with $-ve$ slope and non zero $+ve$ intercept

Vedclass · Answer

(B) For a zero order reaction,the rate law is given by: $Rate = k[A]^0 = k$. Integrating this with respect to time,we get the integrated rate equation: $[A] = -kt + [A]_0$. This equation follows the form of a straight line equation $y = mx + c$,where $y = [A]$,$x = t$,$m = -k$ (slope),and $c = [A]_0$ (intercept). Therefore,the plot of concentration $[A]$ versus time $t$ is a straight line with a negative slope equal to $-k$ and a non-zero positive intercept equal to $[A]_0$ on the concentration axis.

Time $(s)$	Concentration of the reactant $(mol \ L^{-1})$
$0$	$0.1$
$15$	$0.05$
$20$	$0.1 - x$

For a zero order reaction,the plot of concentration of reactant $vs$ time is (Hint: Consider the intercept on the concentration axis)

Similar Questions

For a zero-order reaction,the correct expression for rate constant $(k)$ at half-life time $(t_{1/2})$ is ($[R_0] =$ initial concentration of reactant).

For the reaction $A \longrightarrow 3 B$,the rate is given by $rate = k[A]^0$. Based on the data provided in the table,what is the concentration of the product $B$ (in $mol \ L^{-1}$) after $20 \ s$?
Time $(s)$ Concentration of the reactant $(mol \ L^{-1})$
$0$ $0.1$
$15$ $0.05$
$20$ $0.1 - x$

In a reaction,reactant $A$ decomposes $10 \%$ in $1 \ hour$,$20 \%$ in $2 \ hours$ and $30 \%$ in $3 \ hours$. The unit of rate constant of this reaction is

Which of the following statements about zero order reaction is not true?

The half-life of a zero order reaction $A \rightarrow \text{products}$ is $0.5 \ hr$. The initial concentration of $A$ is $4 \ mol \ L^{-1}$. How much time (in $hr$) does it take for its concentration to come from $2.0 \ mol \ L^{-1}$ to $1.0 \ mol \ L^{-1}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module