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

$A$ first order reaction with respect to the reactant $A$ has a rate constant of $6 \; sec^{-1}$. If we start with $[A] = 0.5 \; mol/L$,then in what time the concentration of $A$ becomes $0.05 \; mol/L$ ........... $sec$

For a reaction,$r = k[A]$. If $50 \%$ of $A$ is decomposed in $120 \ \text{minutes}$,the time taken for $90 \%$ decomposition of $A$ is $............ \ \text{minutes}$.

In a first order reaction,if the concentration of reactant drops from $0.8 \ mol \ L^{-1}$ to $0.4 \ mol \ L^{-1}$ in $15$ minutes,what is the time required to drop the concentration from $0.1 \ mol \ L^{-1}$ to $0.025 \ mol \ L^{-1}$?

$99 \%$ of a first order reaction was completed in $64 \text{ min}$. In how many minutes will $99.9 \%$ of the reaction be complete?

For the first order reaction $A \rightarrow B$,the rate constant is $0.25 \ s^{-1}$. If the concentration of $A$ is reduced to half,the value of the rate constant will be: (in $s^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module