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Consider the reaction A → products at 311 K. The concentration of A was monitored over time and the data was analyzed by plotting. It was found that a plot of 1/[A] vs time gave a straight line relationship. It was also observed that it took 24.5 s for the concentration of A to decrease from 0.757 M to 0.107 M. What is the half life for this reaction when [A]o = 0.757 M?

Answer :

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The half life for this reaction when [A]o = 0.757 M is 4.033 s

integrated rate law for 2nd order reaction:

1/[A]o = 1/[A] - k*t

so, for 2nd order reaction, 1/[A] vs t will be straight line

integrated rate law for 1st order reaction:

ln [A] = –kt + ln [A]o

So, for 1st order reaction, ln[A] vs t will be straight line

integrated rate law for zero order reaction:

[A] = [A]o – k*t

So, for zero order reaction, [A] vs t will be straight line

Here,

1/[A] vs t is straight line

so, order of A is 2

use integrated rate law for 2nd order reaction:

1/[A] = 1/[A]o + k*t

1/(0.107) = 1/(0.757) + k*24.5

9.346 = 1.321 +k*24.5

k*24.5 = 8.025

k = 0.328 M-1.s-1

Given:

rate constant, k = 0.328 M-1.s-1

use relation between rate constant and half life of 2nd order reaction

t1/2 = 1/([A]o*k)

= 1/(0.757*0.328)

= 4.033 s

The half-life of a chemical reaction can be defined as the time it takes for the concentration of a given reactant to reach 50% of its initial concentration (that is, the time it takes for the concentration of a reactant to reach half its initial value). This is represented by the symbol "t1/2" and is usually expressed in seconds.

Learn more about half life of a chemical reaction:

https://brainly.in/question/15482143

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