For a second-order reaction where both reacta | vedclass

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

Question

(NONE) The rate constant for a second-order reaction with equal initial concentrations $a$ is given by $k = \frac{1}{t} 	imes \frac{x}{a(a-x)}$.For $

Vedclass · Accepted Answer

$3000$

Vedclass · Answer

(NONE) The rate constant for a second-order reaction with equal initial concentrations $a$ is given by $k = \frac{1}{t} 	imes \frac{x}{a(a-x)}$.For $20\%$ completion,$x = 0.2a$,so $a-x = 0.8a$. Thus,$k = \frac{1}{500} 	imes \frac{0.2a}{a(0.8a)} = \frac{1}{500} 	imes \frac{0.2}{0.8a} = \frac{1}{500} 	imes \frac{1}{4a} = \frac{1}{2000a}$.For $80\%$ completion,$x = 0.8a$,so $a-x = 0.2a$. Thus,$k = \frac{1}{t} 	imes \frac{0.8a}{a(0.2a)} = \frac{1}{t} 	imes \frac{0.8}{0.2a} = \frac{1}{t} 	imes \frac{4}{a} = \frac{4}{ta}$.Equating the two expressions for $k$: $\frac{1}{2000a} = \frac{4}{ta}$.Solving for $t$: $t = 2000 	imes 4 = 8000 \ s$. Wait,re-evaluating the formula: The standard form for $2nd$ order is $k = \frac{1}{t} \frac{x}{a(a-x)}$.At $20\%$,$x=0.2a$,$t=500$: $k = \frac{1}{500} \frac{0.2a}{a(0.8a)} = \frac{1}{500} \frac{0.2}{0.8a} = \frac{1}{2000a}$.At $80\%$,$x=0.8a$: $k = \frac{1}{t} \frac{0.8a}{a(0.2a)} = \frac{1}{t} \frac{0.8}{0.2a} = \frac{4}{ta}$.$\frac{1}{2000a} = \frac{4}{ta} \implies t = 8000 \ s$.

Experiment	Initial $[I_2]$ $(mol \ L^{-1})$	Initial $[H^{+}]$ $(mol \ L^{-1})$	Initial $[CH_3COCH_3]$ $(mol \ L^{-1})$	Initial rate $(mol \ L^{-1} \ s^{-1})$
$1$	$0.01$	$0.1$	$0.1$	$0.096$
$2$	$0.01$	$0.2$	$0.1$	$0.192$
$3$	$0.02$	$0.2$	$0.1$	$0.192$
$4$	$0.01$	$0.2$	$0.2$	$0.384$

For a second-order reaction where both reactants have the same initial concentration,it takes $500 \ s$ for the reaction to be $20\%$ complete. How many seconds will it take for the reaction to be $80\%$ complete (in $s$)?

Similar Questions

For a given reaction $t_{1/2} = \frac{1}{Ka}$. The order of the reaction is

Which among the following statements is $NOT$ true about rate constant?

For the $2^{nd}$ order reaction $R \rightarrow P$,the following graph holds true. Identify the incorrect statement.

$2 \ FeCl_3 + SnCl_2 \rightarrow 2 \ FeCl_2 + SnCl_4$. This reaction is an example of:

Vedclass Test Series

Exam Paper Generator

Online Exam Module