For a first order reaction A rightarrow text{ | 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:$K = \frac{2.303}{t} \log \left(\frac{[A]_0}{[A]_t}\right)$Given:Initial

Vedclass · Accepted Answer

$0.9212$

Vedclass · Answer

(A) For a first order reaction,the rate constant $K$ is given by the formula:$K = \frac{2.303}{t} \log \left(\frac{[A]_0}{[A]_t}ight)$Given:Initial concentration $[A]_0 = 0.1 \, M$Concentration after time $t$,$[A]_t = 0.001 \, M$Time $t = 5 \, \min$Substituting the values:$K = \frac{2.303}{5} \log \left(\frac{0.1}{0.001}ight)$$K = \frac{2.303}{5} \log(100)$Since $\log(100) = 2$:$K = \frac{2.303 	imes 2}{5} = \frac{4.606}{5} = 0.9212 \, \min^{-1}$

For a first order reaction $A \rightarrow \text{Products}$,initial concentration of $A$ is $0.1 \, M$,which becomes $0.001 \, M$ after $5 \, \min$. Rate constant for the reaction in $\min^{-1}$ is .... .

Similar Questions

If the rate constant for a first order reaction is $k$,then find the time required for completion of $80 \%$ of the reaction.

Calculate the half-life of a first-order reaction if the rate constant of the reaction is $2.772 \times 10^{-3} \ s^{-1}$. (in $s$)

If the half-life $(t_{1/2})$ of a reaction is $69.3 \ s$ and the rate constant $(k)$ is $10^{-2} \ s^{-1}$,then the order of the reaction is:

The degree of dissociation of a first-order reaction is

For the reaction $A \to B$,the rate law expression is: $\text{Rate} = k[A]$. Which of the following statements is incorrect?

Vedclass Test Series

Exam Paper Generator

Online Exam Module