The decomposition of benzene diazonium chlori | vedclass

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

Question

(A) For a first order reaction,$k = \frac{2.303}{t} \log \frac{a}{a-x}$.For $t_{1/4}$,$x = \frac{a}{4}$,so $t_{1/4} = \frac{2.303}{k} \log \frac{a}{3a

Vedclass · Accepted Answer

$272$

Vedclass · Answer

(A) For a first order reaction,$k = \frac{2.303}{t} \log \frac{a}{a-x}$.For $t_{1/4}$,$x = \frac{a}{4}$,so $t_{1/4} = \frac{2.303}{k} \log \frac{a}{3a/4} = \frac{2.303}{k} \log \frac{4}{3}$.For $t_{1/10}$,$x = \frac{a}{10}$,so $t_{1/10} = \frac{2.303}{k} \log \frac{a}{9a/10} = \frac{2.303}{k} \log \frac{10}{9}$.Taking the ratio: $\frac{t_{1/4}}{t_{1/10}} = \frac{\log(4/3)}{\log(10/9)} = \frac{\log 4 - \log 3}{\log 10 - \log 9} = \frac{2 \log 2 - \log 3}{1 - 2 \log 3}$.Using $\log 2 = 0.3$ and $\log 3 = 0.477$: $\frac{t_{1/4}}{t_{1/10}} = \frac{2(0.3) - 0.477}{1 - 2(0.477)} = \frac{0.6 - 0.477}{1 - 0.954} = \frac{0.123}{0.046} \approx 2.67$.Thus,$\frac{t_{1/4}}{t_{1/10}} 	imes 100 \approx 267$. The closest option is $272$.

Similar Questions

$t_{1/4}$ can be taken as the time taken for the concentration of a reactant to drop to $3/4$ of its initial value. If the rate constant for a first order reaction is $K$,the $t_{1/4}$ can be written as (in $/K$)

For a first order reaction,the graph between $\log \frac{a}{(a-x)}$ (on $y$-axis) and time (in $min$,on $x$-axis) gave a straight line passing through origin. The slope is $2 \times 10^{-3} \ min^{-1}$. What is the rate constant (in $min^{-1}$)?

The value of the velocity constant for a first-order reaction is $3.46 \times 10^{-3} \ min^{-1}$. The time for half-change is ........ $min$.

For a reaction $A \rightarrow$ product,the rate constant is $2 \times 10^{-2} \ s^{-1}$. The initial concentration of $A$ is $1.0 \ mol \ dm^{-3}$. What is the value of $\log \frac{1}{[A]_{t}}$ after $100 \ s$?

$20 \%$ of a first order reaction was found to be completed at $10:00 \ am$. At $11:30 \ am$ on the same day,$20 \%$ of the reaction was found to be remaining. The half-life period in minutes of the reaction is

Vedclass Test Series

Exam Paper Generator

Online Exam Module