The rate constant for a first order reaction | vedclass

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

Question

(D) For a first order reaction,the rate equation is given by: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$ Given: $k = 4.606 	imes 10^{-3} \ s

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(D) For a first order reaction,the rate equation is given by: $k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$ Given: $k = 4.606 	imes 10^{-3} \ s^{-1}$,$[A]_0 = 2.0 \ g$,$[A]_t = 0.2 \ g$ Substituting the values: $4.606 	imes 10^{-3} = \frac{2.303}{t} \log_{10} \frac{2.0}{0.2}$ $4.606 	imes 10^{-3} = \frac{2.303}{t} \log_{10} (10)$ Since $\log_{10} (10) = 1$,we have: $4.606 	imes 10^{-3} = \frac{2.303}{t}$ $t = \frac{2.303}{4.606 	imes 10^{-3}} = \frac{1}{2} 	imes 10^3 = 500 \ s$

The rate constant for a first order reaction is $4.606 \times 10^{-3} \ s^{-1}$. The time required to reduce $2.0 \ g$ of the reactant to $0.2 \ g$ is......... $s$

Similar Questions

The time required for a first-order reaction to be $90\%$ complete is approximately .........

For a first-order reaction,the ratio of the time required for $99.9\%$ completion to the time required for $50\%$ completion is:

The decomposition of dinitrogen pentoxide $(N_2O_5)$ follows first order rate law. What will be the rate constant from the given data?
At $t = 800 \ s$,$[N_2O_5] = 1.45 \ mol \ L^{-1}$
At $t = 1600 \ s$,$[N_2O_5] = 0.88 \ mol \ L^{-1}$

Half-life period of a first-order reaction is $10 \ min$. Starting with an initial concentration of $12 \ M$,the rate after $20 \ min$ is:

For a first-order reaction,if the value of $a/(a - x)$ is $8$ after $10 \text{ minutes}$,what will be the rate constant $K$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module