Clear Notion
Login

Concepts

Read through concept notes with worked explanations and diagrams, chapter by chapter.

Concept 03 / 06Chemical Kinetics

Integrated Rate Equations and Half-Life for Zero and First Order Reactions

Derivation and use of integrated rate laws for zero-order and first-order reactions, their graphical forms, half-life expressions, examples of zero and first order reactions, and time relations such as t99.9 = 10 t1/2.

Figure 1
The rate law gives rate in terms of concentration; the integrated rate equation gives concentration as a function of time, which can be measured directly. Zero-order reactions. −d[A]dt=k-\frac{d[\mathrm{A}]}{dt} = k, so [A]=[A]0−kt[\mathrm{A}] = [\mathrm{A}]_{0} - kt and k=[A]0−[A]tk = \frac{[\mathrm{A}]_{0} - [\mathrm{A}]}{t}. A plot of [A][\mathrm{A}] against tt is a straight line of slope −k-k (see figure). Half-life t1/2=[A]02kt_{1/2} = \frac{[\mathrm{A}]_{0}}{2k} (proportional to initial concentration). The reaction is complete at t=[A]0kt = \frac{[\mathrm{A}]_{0}}{k}. Examples: decomposition of gaseous ammonia on a hot platinum surface at high pressure (2NH3→N2+3H2\mathrm{2NH_{3} \rightarrow N_{2} + 3H_{2}}), where the surface is saturated so the rate is independent of pressure; decomposition of HI on gold; some enzyme-catalysed reactions at high substrate concentration. First-order reactions. −d[A]dt=k[A]-\frac{d[\mathrm{A}]}{dt} = k[\mathrm{A}], so ln⁡[A]=ln⁡[A]0−kt\ln[\mathrm{A}] = \ln[\mathrm{A}]_{0} - kt, [A]=[A]0e−kt[\mathrm{A}] = [\mathrm{A}]_{0}e^{-kt}, k=2.303tlog⁡[A]0[A]k = \frac{2.303}{t}\log\frac{[\mathrm{A}]_{0}}{[\mathrm{A}]}. A plot of ln⁡[A]\ln[\mathrm{A}] (or log⁡[A]\log[\mathrm{A}]) against tt is a straight line of slope −k-k (or −k2.303-\frac{k}{2.303}). Half-life t1/2=0.693kt_{1/2} = \frac{0.693}{k}, independent of the initial concentration: every half-life halves what is left. After nn half-lives, the fraction remaining is (12)n\left(\frac{1}{2}\right)^{n}. Useful relations: t75%=2t1/2t_{75\%} = 2t_{1/2}; t87.5%=3t1/2t_{87.5\%} = 3t_{1/2}; t99%=2t90%t_{99\%} = 2t_{90\%}; t99.9%=10t1/2t_{99.9\%} = 10t_{1/2} (approximately). Examples: decomposition of N2O5\mathrm{N_{2}O_{5}} and N2O\mathrm{N_{2}O}, hydrogenation of ethene (r=k[C2H4]r = k[\mathrm{C_{2}H_{4}}] with excess hydrogen), all natural radioactive decay. General half-life relation. For an nnth order reaction, t1/2∝[A]01−nt_{1/2} \propto [\mathrm{A}]_{0}^{1-n}: zero order ∝[A]0\propto [\mathrm{A}]_{0}; first order independent; second order ∝1[A]0\propto \frac{1}{[\mathrm{A}]_{0}}. Key Takeaways: - Zero order: [A]=[A]0−kt[\mathrm{A}] = [\mathrm{A}]_{0} - kt; t1/2=[A]02kt_{1/2} = \frac{[\mathrm{A}]_{0}}{2k}. - First order: k=2.303tlog⁡[A]0[A]k = \frac{2.303}{t}\log\frac{[\mathrm{A}]_{0}}{[\mathrm{A}]}; t1/2=0.693kt_{1/2} = \frac{0.693}{k}. - Straight line: [A][\mathrm{A}] vs tt (zero), ln⁡[A]\ln[\mathrm{A}] vs tt (first). - t1/2∝[A]01−nt_{1/2} \propto [\mathrm{A}]_{0}^{1-n}. - Pitfall: a constant half-life is the signature of a first-order reaction only. Brain-Teaser 1: A first-order reaction has k=1.15×10−3 s−1k = 1.15 \times 10^{-3}\,\mathrm{s^{-1}}. How long will 5 g5\,\mathrm{g} of the reactant take to reduce to 3 g3\,\mathrm{g}? Solution: t=2.3031.15×10−3log⁡53=2003×0.2218=444 st = \frac{2.303}{1.15 \times 10^{-3}}\log\frac{5}{3} = 2003 \times 0.2218 = 444\,\mathrm{s}. Brain-Teaser 2: Show that for a first-order reaction, the time for 99%99\% completion is twice that for 90%90\% completion. Solution: t99=2.303klog⁡1001=2.303×2kt_{99} = \frac{2.303}{k}\log\frac{100}{1} = \frac{2.303 \times 2}{k}; t90=2.303klog⁡10010=2.303kt_{90} = \frac{2.303}{k}\log\frac{100}{10} = \frac{2.303}{k}. So t99=2t90t_{99} = 2t_{90}. Brain-Teaser 3: The half-life of a first-order reaction is 60 min60\,\mathrm{min}. What fraction remains after 4 h4\,\mathrm{h}, and how long does it take for 87.5%87.5\% to react? Solution: 4 h=44\,\mathrm{h} = 4 half-lives, so (12)4=116\left(\frac{1}{2}\right)^{4} = \frac{1}{16} remains. 87.5%87.5\% reacted means 18\frac{1}{8} remains, which takes 33 half-lives =180 min= 180\,\mathrm{min}.

Practice this concept

Free · no account required · instant solutions

Integrated Rate Equations and Half-Life for Zero and First Order Reactions — DPP

10 questionsSee the solution the moment you answer

Practice