What is the numerical value of the rate const | vedclass

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

Question

(A) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given that the reaction i

Vedclass · Accepted Answer

$0.0223$

Vedclass · Answer

(A) For a first-order reaction,the rate constant $k$ is given by the formula: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given that the reaction is $20 \%$ completed,the remaining concentration $[A]_t$ is $80 \%$ of the initial concentration $[A]_0$. So,$[A]_t = 0.80 [A]_0$ and $t = 10 \ min$. Substituting these values: $k = \frac{2.303}{10} \log \frac{1}{0.8} = \frac{2.303}{10} \log(1.25)$. Since $\log(1.25) \approx 0.0969$,we get $k = \frac{2.303 	imes 0.0969}{10} \approx 0.0223 \ min^{-1}$.

What is the numerical value of the rate constant of a first-order reaction that is $20 \%$ completed in $10 \ min$ (in $min^{-1}$)?

Similar Questions

The rate constant for a first order reaction is $60 \ s^{-1}$. How much time will it take to reduce the initial concentration of the reactant to its $1/16^{th}$ value?

What is the rate constant of a first-order reaction if $60 \%$ of the reactant decomposes in $45 \ minutes$?

The rate constant for a first order reaction is $60 \text{ s}^{-1}$. How much time will it take to reduce the concentration of the reactant to $1/20^{th}$ of its initial value (in $\text{ s}$)?

Which of the following is correct regarding $t_{1/2}$ of a reaction if we double the initial concentration of a reactant in a first-order reaction?

In a first order reaction,the concentration of the reactant is reduced to $(1/8)^{th}$ of its initial concentration in $23.03 \ min$. What is the half-life period of the reaction (in $min$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module