Chemical Kinetics

To determine the time required for 99.9% completion of a first-order reaction with a half-life of 1 second, we can demonstrate that this time is 10 times the reaction's half-life (

t_%7B1%2F2%7D

).When a reaction is 99.9% complete, the remaining concentration is only 0.1% of the initial concentration. This can be expressed as

%5BR%5D_t%20%3D%20%5BR%5D_0%20-%200.999%5BR%5D_0

.Using the first-order kinetics equation:

k%20%3D%20%5Cfrac%7B2.303%7D%7Bt%7D%20%5Clog%20%5Cfrac%7B%5BR%5D_0%7D%7B%5BR%5D%7D

To find the time at which the reaction is 99.9% complete:

k%20%3D%20%5Cfrac%7B2.303%7D%7Bt%7D%20%5Clog%20%5Cfrac%7B%5BR%5D_0%7D%7B0.001%5BR%5D_0%7D%20%3D%20%5Cfrac%7B2.303%7D%7Bt%7D%20%5Clog%2010%5E3

Thus:

t%20%3D%20%5Cfrac%7B6.909%7D%7Bk%7D

For the half-life of a first-order reaction:

t_%7B1%2F2%7D%20%3D%20%5Cfrac%7B0.693%7D%7Bk%7D

Comparing the times:

%5Cfrac%7Bt%7D%7Bt_%7B1%2F2%7D%7D%20%3D%20%5Cfrac%7B6.909%7D%7Bk%7D%20%5Ctimes%20%5Cfrac%7Bk%7D%7B0.693%7D%20%3D%2010

Therefore, the time required for 99.9% completion is approximately 10 times the half-life, or 10 seconds. This coincides with option C, 10 seconds.

On the basis of the unit of the reaction we can see that the reaction is a zero order reaction.$

%5Cbegin%7Baligned%7D%20k%20%26%20%3D%5Cfrac%7B%5Ctext%20%7B%20Rate%20%7D%7D%7B%5BA%5D%5Ex%5BB%5D%5Ev%7D%20%5C%5C%20%26%20%3D%5Cfrac%7B%5Ctext%20%7B%20concentration%20%7D%7D%7B%5Ctext%20%7B%20time%20%7D%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7B(%5Ctext%20%7B%20concentration%20%7D)%5E%7B%5Cprime%20%5Cprime%7D%7D%20%5Cquad(%5Ctext%20%7B%20where%20%7D%7CA%7C%3D%7CB%7C)%20%5Cend%7Baligned%7D

Taking%20SI%20units%20of%20concentration%2C%20mol

\mathrm{L}^{-1}

and%20time%2C%20s%2C%20the%20units%20of

for%20different%20reaction%20order%20are%20listed%20in%20Table%203.3

%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5Ctext%20%7B%20Table%203.3%3A%20Units%20of%20rate%20constant%20%7D

$Reactienorderthits of rete conatantZero order reaction0molL−1s×1(molL−1)6=molL−1s−1molL−1s×1molL−16=molL−1s−1(molL^(-1))/((s))xx(1)/(((mol)L^(-1))^(6))=molL^(-1)s^(-1)First order reaction1molL−1s×1(molL−1)1=s−1molL−1s×1molL−11=s−1(molL^(-1))/((s))xx(1)/(((mol)L^(-1))^(1))=s^(-1)Second order reaction2molL−1×1s×mol−1Ls−1molL−1×1s×mol−1Ls−1(molL)/()^(-1)xx(1)/(s)xxmol^(-1)Ls^(-1)Fig. 3.3: Variation in the concentration vs time plot for a zero order reaction

High-Yield Trend

Chapter Questions
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Chemical Kinetics

High-Yield Trend

Chapter Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Chapter Questions
3 MCQs