The rate constant k = 2.303 min^{-1} for a p | vedclass

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

Question

(B) The reaction is a $1^{st}$ order reaction because the unit of the rate constant is $min^{-1}$.For a $1^{st}$ order reaction,the integrated rate la

Vedclass · Accepted Answer

$0.2303 \ M \ min^{-1}$

Vedclass · Answer

(B) The reaction is a $1^{st}$ order reaction because the unit of the rate constant is $min^{-1}$.For a $1^{st}$ order reaction,the integrated rate law is $k = \frac{2.303}{t} \log \frac{[A_0]}{[A]}$.Given: $k = 2.303 \ min^{-1}$,$[A_0] = 1 \ mol/L$,and $t = 1 \ min$.Substituting the values: $2.303 = \frac{2.303}{1} \log \frac{1}{[A]}$.This simplifies to $1 = \log \frac{1}{[A]}$,which means $\frac{1}{[A]} = 10^1 = 10$.Therefore,$[A] = 0.1 \ mol/L$.The rate of reaction is given by $Rate = k 	imes [A]$.$Rate = 2.303 \ min^{-1} 	imes 0.1 \ M = 0.2303 \ M \ min^{-1}$.

The rate constant $k = 2.303 \ min^{-1}$ for a particular reaction,and the initial concentration of the reactant is $1 \ mol/L$. What is the rate of reaction after $1 \ minute$?

Similar Questions

For a first order gas phase elementary reaction $A_{(g)} \xrightarrow{k} 2B_{(g)}$,which of the following is $INCORRECT$?

Find the rate constant of a first-order reaction in $sec^{-1}$ having a half-life of $2.5 \ hours$.

The reaction,$X \to$ product follows first order kinetics. In $40 \, min$ the concentration of $X$ changes from $0.1 \, M$ to $0.025 \, M$. Then the rate of reaction when concentration of $X$ is $0.01 \, M$ is:

Which equation is correct for first order reactions?

$A$ first order reaction is half completed in $45 \, \text{minutes}$. How long does it need for $99.9 \%$ of the reaction to be completed? (in $hr$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module