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Date: 30-9-2018
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Date: 23-9-2018
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The half-life of a reaction, t1/2, is the duration of time required for the concentration of a reactant to drop to one-half of its initial concentration.
[latex] [A]{}_{t1/2}{}_{ } [/latex]= [latex] frac{1}{2} [A]{}_{0} [/latex]
Half-life is typically used to describe first-order reactions and serves as a metric to discuss the relative speeds of reactions. A slower reaction will have a longer half-life, while a faster reaction will have a shorter half-life.
To determine the half-life of a first-order reaction, we can manipulate the integrated rate law by substituting t1/2 for t and [A]t1/2 = [A]0 for [A]t, then solve for t1/2:
ln = –kt + ln (integrated rate law for a first-order reaction)
ln [latex] frac{1}{2} [A]{}_{0} [/latex]= –[latex] extit{k} t{}_{1/2 }+ ln {[A]}_0 [/latex]
ln [latex] frac{frac{1}{2}{ m }{{ m [A]}}_0{ m }}{{[A]}_0} [/latex]= –[latex] extit{k} t{}_{1/2 } [/latex]
ln [latex] frac{1}{2} [/latex]= –[latex] extit{k} t{}_{1/2 } [/latex]
[latex] t{}_{1/2 } [/latex] = – [latex] frac{{ m ln }frac{1}{2} }{k} [/latex] = [latex] frac{0.693}{k} [/latex]
Since the half-life equation of a first-order reaction does not include a reactant concentration term, it does not rely on the concentration of reactant present. In other words, a half-life is independent of concentration and remains constant throughout the duration of the reaction. Consequently, plots of kinetic data for first-order reactions exhibit a series of regularly spaced t1/2 intervals (Figure 17.10 “Generic First-Order Reaction Kinetics Plot”).
Figure 17.10. Generic First-Order Reaction Kinetics Plot
This graph shows repeating half-lives on a kinetics plot of a generic first-order reaction.
Example 1
A reaction having a first-order rate has a rate constant of 4.00 x 10-3 s-1.
1. Determine the half-life.
2. How long will it take for a sample of reactant at 1.0 M to decrease to 0.25 M?
3. What concentration of the 1.0 M sample of reactant would you expect to be present after it has reacted for 500 s?
Solution
1. [latex] t{}_{1/2 } [/latex] = [latex] frac{0.693}{k} [/latex] = [latex] frac{0.693}{{ m 4.00 x }{{ m 10}}^{-3}{ m }{{ m s}}^{-1}} [/latex] = 173 s
2. A simple way to calculate this is to determine how many half-lives it will take to go from 1.00 M to 0.250 M and use the half-life calculated in part 1.
1 half-life = 0.500 M
2 half-lives = 0.250 M
Therefore, it will take 2 x 173 s = 346 s.
3. We can use the rate-constant value in the integrated rate law to determine the concentration remaining.
ln [latex] frac{{[A]}_t}{{[A]}_0} [/latex]= –[latex] extit{k}t [/latex]
ln [latex] frac{{[A]}_t}{1.0 M} [/latex]= –[latex] (4.00 x 10{}^{-3} s{}^{-1})(500 s) [/latex]
ln [latex] frac{{[A]}_t}{1.0 M} [/latex] = -2
[latex] frac{{[A]}_t}{1.0 M} [/latex] = [latex] e{}^{-2 } [/latex]= 0.135
[latex] [A]{}_{t} [/latex] = 0.14 M
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