If 75 % of a first order reaction was complet | 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 constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $75 \%$ completion,$[A]_t = [A]_0

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(D) For a first order reaction,the rate constant $k$ is given by $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$.For $75 \%$ completion,$[A]_t = [A]_0 - 0.75[A]_0 = 0.25[A]_0$,so $k = \frac{2.303}{90} \log \frac{1}{0.25} = \frac{2.303}{90} \log 4 = \frac{2.303 	imes 2 \log 2}{90} = \frac{2.303 	imes 0.6}{90} \ min^{-1}$.For $60 \%$ completion,$[A]_t = [A]_0 - 0.60[A]_0 = 0.40[A]_0$,so $t = \frac{2.303}{k} \log \frac{1}{0.4} = \frac{2.303}{k} \log 2.5$.Substituting $k$: $t = \frac{2.303 	imes 90}{2.303 	imes 0.6} 	imes 0.4 = \frac{90 	imes 0.4}{0.6} = \frac{36}{0.6} = 60 \ minutes$.

If $75 \%$ of a first order reaction was completed in $90 \ minutes$,$60 \%$ of the same reaction would be completed in approximately (in minutes)..........
(Take: $\log 2=0.30 ; \log 2.5=0.40)$

Similar Questions

Which equation represents the $1/4^{th}$ life for a first-order reaction?

$A$ first order reaction is found to have a rate constant,$k = 5.5 \times 10^{-14} \ s^{-1}$. Find the half-life of the reaction.

The half-life of a first-order reaction is $69.35 \, \text{sec}$. The value of the rate constant of the reaction is .......... $\sec^{-1}$.

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 $t = 1600 \ s$,$[N_2O_5] = 0.88 \ mol \ L^{-1}$

The reaction,$X \to$ product follows first order kinetics. In $40 \, min$ the concentration of $X$ changes from $0.1 \, M$ to $0.025 \, M$. Then the rate of reaction when concentration of $X$ is $0.01 \, M$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

If $75 \%$ of a first order reaction was completed in $90 \ minutes$,$60 \%$ of the same reaction would be completed in approximately (in minutes)..........(Take: $\log 2=0.30 ; \log 2.5=0.40)$