Mathematical representation for t_{1/4} life | vedclass

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

Question

(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right)$.At $t_{1/4}$,the amount

Vedclass · Accepted Answer

$t_{1/4} = \frac{2.303}{k} \log \left( \frac{4}{3} \right)$

Vedclass · Answer

(C) For a first-order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right)$.At $t_{1/4}$,the amount of reactant consumed is $\frac{1}{4}$ of the initial concentration $[A]_0$.Therefore,the remaining concentration $[A]_t = [A]_0 - \frac{1}{4}[A]_0 = \frac{3}{4}[A]_0$.Substituting these values into the rate equation:$k = \frac{2.303}{t_{1/4}} \log \left( \frac{[A]_0}{\frac{3}{4}[A]_0} \right)$$k = \frac{2.303}{t_{1/4}} \log \left( \frac{4}{3} \right)$Rearranging for $t_{1/4}$ gives $t_{1/4} = \frac{2.303}{k} \log \left( \frac{4}{3} \right)$.

Mathematical representation for $t_{1/4}$ life for a first-order reaction is given by:

Similar Questions

Consider a certain reaction $A \rightarrow \text{Products}$ with $k = 2.0 \times 10^{-2} \ s^{-1}$. Calculate the concentration of $A$ remaining after $100 \ s$ if the initial concentration of $A$ is $1.0 \ mol \ L^{-1}$.

What is the rate constant of a first-order reaction if $0.08 \ mol$ of reactant reduces to $0.02 \ mol$ in $23.03 \ min$ (in $min^{-1}$)?

$A$ first order reaction was started with a decimolar solution of the reactant. $8$ minutes and $20$ seconds later,its concentration was found to be $M/100$. The rate constant of the reaction is:

Which one of the following formulas represents a first-order reaction?

If the rate law is $r = K[A]$,then the concentration of the reactant remaining after time $t = 1/k$ is: ($[A]_0$ is the concentration of the reactant at $t = 0$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module