For a first-order reaction,the ratio of the t | vedclass

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

Question

(C) For a first-order reaction,the time $t$ is given by $t = \frac{2.303}{K} \log \frac{a}{a-x}$.For $99.9\%$ completion,$x = 0.999a$,so $a-x = 0.001a

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) For a first-order reaction,the time $t$ is given by $t = \frac{2.303}{K} \log \frac{a}{a-x}$.For $99.9\%$ completion,$x = 0.999a$,so $a-x = 0.001a$. Thus,$t_{99.9\%} = \frac{2.303}{K} \log \frac{a}{0.001a} = \frac{2.303}{K} \log 10^3 = \frac{2.303 	imes 3}{K}$.For $50\%$ completion (half-life),$x = 0.5a$,so $a-x = 0.5a$. Thus,$t_{50\%} = \frac{2.303}{K} \log \frac{a}{0.5a} = \frac{2.303}{K} \log 2 \approx \frac{2.303 	imes 0.3010}{K}$.The ratio $\frac{t_{99.9\%}}{t_{50\%}} = \frac{3}{0.3010} \approx 10$.

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

Similar Questions

Show that in a first order reaction,the time required for completion of $99.9 \%$ is $10$ times the half-life $(t_{1/2})$ of the reaction.

In a first order reaction,the concentration of the reactant decreases from $20 \ mmol \ dm^{-3}$ to $8 \ mmol \ dm^{-3}$ in $40 \ minute$. Find the rate constant of the reaction.

The time of half change of a first order reaction is ....... initial concentration.

The reaction $2H_2O_2 \to 2H_2O + O_2$ is a

For a first order reaction $A \rightarrow P$,$t_{1/2}$ (half-life) is $10 \text{ days}$. The time required for $1/4^{th}$ conversion of $A$ (in days) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module