Chemistry · Ch 7 — Chemical Kinetics

Half Life Period of a Reaction

7.6

Half Life Period of a Reaction

Definition. The half life of a reaction, $t_{1/2}$ , is the time it takes for the reactant's concentration to fall to exactly HALF its initial value.

First order half life -- and why it is a constant. Start from the first order rate-constant formula, $k=\dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$ , and substitute the half-life condition $t=t_{1/2}$ , $[A]=[A]_0/2$ :

k=\frac{2.303}{t_{1/2}}\log\frac{[A]_0}{[A]_0/2}=\frac{2.303}{t_{1/2}}\log2=\frac{2.303\times0.3010}{t_{1/2}}\quad\Rightarrow\quad t_{1/2}=\frac{0.693}{k}

The crucial feature: $[A]_0$ has completely CANCELLED OUT of the algebra along the way (it appeared in both numerator and denominator of the log argument as a ratio of 2, independent of its actual value). So for a first order reaction, the half life is a FIXED NUMBER, the same no matter whether you start with a large or a tiny amount of reactant -- a genuinely distinguishing signature of first order kinetics, and the reason radioactive decay (always first order) is quoted with one single half life regardless of sample size.

Zero order half life -- and why it is NOT a constant. Start instead from the zero order formula, $k=\dfrac{[A]_0-[A]}{t}$ , and make the same substitution:

k=\frac{[A]_0-[A]_0/2}{t_{1/2}}=\frac{[A]_0/2}{t_{1/2}}\quad\Rightarrow\quad t_{1/2}=\frac{[A]_0}{2k}

Here $[A]_0$ survives in the final formula: the zero order half life is DIRECTLY PROPORTIONAL to the starting concentration -- double the initial amount, and the half life itself doubles, in sharp contrast to first order behaviour.

The general nth-order half life ( $n\neq1$ ) is $t_{1/2}=\dfrac{2^{n-1}-1}{(n-1)k[A]_0^{n-1}}$ , which specialises to the $n=0$ result above; the $n=1$ (first order) result is obtained separately as the limiting case, since the $(n-1)$ in the denominator itself vanishes exactly at $n=1$ .

A worked feel for the numbers. If a first order reaction takes 8 hours for 90% completion, it takes only 5.59 hours for 80% completion (Example 4) -- LESS time for less conversion, as expected, but note the times are not simply proportional to percentage because of the logarithmic relationship. Given a half life, the percentage decomposed after any elapsed time follows directly (Example 5); and completing 99.9% of a first order reaction always takes almost exactly TEN half lives (Example 6), since $\log1000=3=10\times0.3$ roughly matches $10\times\log2\approx10\times0.301=3.01$ . …

Misc more-to-know-nth-order-halflifeMore to know -- half life for an nth order reaction

Worked out. For a reactant A with order $n\neq1$ , the general half life formula is $t_{1/2}=\dfrac{2^{n-1}-1}{(n-1)k[A]_0^{n-1}}$ -- setting $n=0$ recovers $t_{1/2}=[A]_0/2k$ (half life proportional to $[A]_0$ ), and the first order case $t_{1/2}=0.693/k$ is recovered separately by taking the limit $n\rightarrow1$ of the differential rate law, since the $(n-1)$ denominator itself vanishes at $n=1$ . …

Misc example-4Example 4 -- time for 80% completion given the time for 90% completion (first order)

Worked out. A first order reaction takes 8 hours for 90% completion. Calculate the time required for 80% completion. ( $\log5=0.6989$ , $\log10=1$ ). Book's solution: let $[A]_0=100$ M. At $t_{90\%}=8$ h, $[A]=10$ M, so $k=\dfrac{2.303}{8}\log\dfrac{100}{10}=\dfrac{2.303}{8}\log10=\dfrac{2.303}{8}(1)$ . At $t_{80\%}$ , $[A]=20$ M, so $t_{80\%}=\dfrac{2.303}{k}\log\dfrac{100}{20}=\dfrac{2.303}{k}\log5$ . Substituting k: $t_{80\%}=\dfrac{8\times\log5}{\log10}=8\times0.6989=5.59$ hours. …

Misc example-5Example 5 -- percentage decomposed after a given time, from the half life (first order)

Worked out. The half life of a first order reaction $x\rightarrow\text{products}$ is $6.932\times10^4$ s at 500 K. What percentage of x would be decomposed on heating at 500 K for 100 min? ( $e^{0.06}=1.06$ ). Book's solution: $k=\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{6.932\times10^4}=10^{-5}\ \text{s}^{-1}$ . Using $k=\dfrac{1}{t}\ln\dfrac{[A]_0}{[A]}$ with $t=100\times60=6000$ s: $10^{-5}\times6000=0.06=\ln\dfrac{[A]_0}{[A]}\Rightarrow\dfrac{[A]_0}{[A]}=e^{0.06}=1.06$ . Fraction decomposed $=\dfrac{[A]_0-[A]}{[A]_0}\times100=\left(1-\dfrac{1}{1.06}\right)\times100\approx5.6\%$ …

Misc example-6Example 6 -- time for 99.9% completion is about ten half lives (first order)

Worked out. Show that for a first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion. Book's solution: let $[A]_0=100$ ; at $t_{99.9\%}$ , $[A]=100-99.9=0.1$ . $t_{99.9\%}=\dfrac{2.303}{k}\log\dfrac{100}{0.1}=\dfrac{2.303}{k}\log1000=\dfrac{2.303}{k}(3)=\dfrac{6.909}{k}\approx\dfrac{10\times0.6932}{k}=10\times t_{1/2}$ (since $t_{1/2}=0.6932/k$ ). …

Misc evaluate-yourself-3Evaluate Yourself 3 -- three half-life and first-order-kinetics practice problems

Worked out. Book's practice box (no printed solutions). (1) In a first order reaction $A\rightarrow\text{products}$ , 60% of a sample of A decomposes in 40 min. What is the half life? (2) The rate constant for a first order reaction is $2.3\times10^{-4}\ \text{s}^{-1}$ . If the initial concentration is 0.01 M, what concentration remains after 1 hour? (3) An ester's hydrolysis in aqueous solution was studied by titrating the liberated carboxylic acid against NaOH; the ester concentration at t = 0, 30, 60, 90 min is 0.85, 0.80, 0.754, 0.71 mol L $^{-1}$ . Show that the reaction follows first order kinetics. Working through: (1) $k=\dfrac{2.303}{40}\log\dfrac{100}{40}=\dfrac{2.303}{40}(0.398)=0.02292\ \text{min}^{-1}\Rightarrow t_{1/2}=0.693/0.02292\approx30.2$ min. (2) $t=3600$ s, $k\times t=2.3\times10^{-4}\times3600=0.828=\ln([A]_0/[A])\Rightarrow[A]_0/[A]=e^{0.828}\approx2.289\Rightarrow[A]=0.01/2.289\approx4.37\times10^{-3}$ M. (3) Computing $k=\dfrac{2.303}{t}\log([A]_0/[A])$ at each interval gives $k_{30}\approx2.03\times10^{-3}$ , $k_{60}\approx1.98\times10^{-3}$ , $k_{90}\approx1.99\times10^{-3}\ \text{min}^{-1}$ -- …

Misc pharmacokinetics-applicationApplication -- chemical kinetics in pharmaceuticals (pharmacokinetics)

Worked out. Pharmacokinetics -- the study of drug lifetimes and bioavailability in the body -- uses the half life concept directly to decide a drug's prescribed dosage and dosing frequency. Paracetamol, a common anti-pyretic/analgesic, has a plasma half life of 2.5 hours, meaning its blood concentration halves every 2.5 hours; after 10 hours (exactly 4 half lives, since $10/2.5=4$ ), only $(1/2)^4=1/16=6.25\%$ of the original dose remains in the body. This half-life knowledge is exactly why paracetamol is typically prescribed once every 6 hours -- close enough to a couple of half lives that the drug does not fully clear before the next dose, keeping plasma concentration within an …

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