A first order reaction was started with a dec | 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:$K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$Given:Initial concentration $[A]_0 = \f

Vedclass · Accepted Answer

$4.606 	imes 10^{-3} \; \sec^{-1}$

Vedclass · Answer

(C) For a first order reaction,the rate constant $K$ is given by:$K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$Given:Initial concentration $[A]_0 = \frac{1}{10} \; M = 0.1 \; M$Final concentration $[A]_t = \frac{1}{100} \; M = 0.01 \; M$Time $t = 8 \; 	ext{minutes} + 20 \; 	ext{seconds} = (8 	imes 60) + 20 = 500 \; \sec$Substituting the values:$K = \frac{2.303}{500} \log \frac{0.1}{0.01}$$K = \frac{2.303}{500} \log 10$Since $\log 10 = 1$:$K = \frac{2.303}{500} = 0.004606 \; \sec^{-1}$$K = 4.606 	imes 10^{-3} \; \sec^{-1}$

Experiment	$[A]$ $(mol \ L^{-1})$	$[B]$ $(mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ min^{-1})$
$I$	$0.10$	$0.20$	$6.93 \times 10^{-3}$
$II$	$0.10$	$0.25$	$6.93 \times 10^{-3}$
$III$	$0.20$	$0.30$	$1.386 \times 10^{-2}$

$A$ 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:

Similar Questions

What is the value of the rate constant for a first-order reaction,if it takes $15 \ min$ for the consumption of $20 \%$ of the reactants?

The half-life period for a first-order reaction is . . . . . . .

The time of completion of $90\%$ of a first order reaction is approximately

Sucrose hydrolyses in acidic medium into glucose and fructose by first order rate law with $t_{1/2} = 3 \text{ hours}$. The percentage of sucrose remaining after $6 \text{ hours}$ is . . . . . . . (Nearest integer) (Given: $\log 2 = 0.3010$ and $\log 3 = 0.4771$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module