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| The '''rate law''' or '''rate equation''' for a [[chemical reaction]] is an equation that links the [[reaction rate]] with concentrations or pressures of reactants and constant parameters (normally rate coefficients and partial [[reaction order]]s).<ref>[http://goldbook.iupac.org/R05141.html IUPAC Gold Book definition of rate law]. See also: According to [[IUPAC]] [[Compendium of Chemical Terminology]].</ref> To determine the rate equation for a particular system one combines the reaction rate with a [[mass balance]] for the system.<ref>Kenneth A. Connors ''Chemical Kinetics, the study of reaction rates in solution'', 1991, VCH Publishers. This book''' contains most of the rate equations in this article and their derivation.</ref> For a generic reaction {{nowrap|''a''A + ''b''B → C}} with no intermediate steps in its [[reaction mechanism]] (that is, an [[elementary reaction]]), the rate is given by
| | == smiled and shook his head == |
| :<math>r\; =\; k[\mathrm{A}]^x[\mathrm{B}]^y</math>
| |
| where [A] and [B] express the concentration of the species A and B, respectively (usually in moles per liter ([[molarity]], M)); ''x'' and ''y'' must be determined experimentally (a common mistake is assuming they represent stoichiometric coefficients but this is not the case). ''k'' is the ''rate coefficient'' or ''rate constant'' of the reaction. The value of this coefficient ''k'' depends on conditions such as temperature, ionic strength, surface area of the [[adsorbent]] or light irradiation. For elementary reactions, the rate equation can be derived from first principles using [[collision theory]] under well-stirred conditions.
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|
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| The rate equation is a [[differential equation]], and it can be [[integral|integrated]] to obtain an '''integrated rate equation''' that links concentrations of reactants or products with time.
| | Method, the fastest way [http://www.nrcil.net/fancybox/lib/rakuten_LV_34.html ルイヴィトンの新作バッグ] is to improve practice in the [http://www.nrcil.net/fancybox/lib/rakuten_LV_138.html ルイヴィトン 正規店] conduct of life and death? 'Burly man asked puzzled.<br><br>'That [http://www.nrcil.net/fancybox/lib/rakuten_LV_48.html ルイヴィトン 肩掛けバッグ] I understand.' Thor said.<br><br>flood, Raytheon certainly understand.<br><br>on Earth strong gush fastest times, is a big Nirvana era, and the [http://www.nrcil.net/fancybox/lib/rakuten_LV_60.html ルイヴィトン タイガ 財布] use of wood grain ya era. Nirvana had great times, just a few years, the weak point of the human from the original, evolving [http://www.nrcil.net/fancybox/lib/rakuten_LV_66.html ルイヴィトン 店舗] out of a lot of strong moments, and even the strong line star [http://www.nrcil.net/fancybox/lib/rakuten_LV_34.html ルイヴィトンの新作バッグ] out. And after a few decades, wood ya crystal was found to [http://www.nrcil.net/fancybox/lib/rakuten_LV_137.html ルイヴィトン バック] also catalyze a number of strong.<br><br>'understand, you still remain in the killing fields?' burly man asked.<br><br>'This is the battle of life and death, but also very dangerous fight ah.' Thor said.<br><br>'wrong, wrong is ridiculous.' burly man laughed, smiled and shook his head, 'It seems you do not have a teacher, or that [http://www.nrcil.net/fancybox/lib/rakuten_LV_94.html ルイヴィトン 財布 価格] your teacher level is low.'<br>Luo Feng<br>all three listened.<br><br>a domain master talk to three of them say something strong |
| | 相关的主题文章： |
| | <ul> |
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| | <li>[http://www.shijiewang.net/home.php?mod=space&uid=73184 http://www.shijiewang.net/home.php?mod=space&uid=73184]</li> |
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| | <li>[http://www.qi-chem.com/plus/feedback.php?aid=5 http://www.qi-chem.com/plus/feedback.php?aid=5]</li> |
| | |
| | <li>[http://cgi.educities.edu.tw/cgi-bin/cgiwrap/haojas/f4ubook/yybbs.cgi http://cgi.educities.edu.tw/cgi-bin/cgiwrap/haojas/f4ubook/yybbs.cgi]</li> |
| | |
| | </ul> |
|
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| ==Stoichiometric reaction networks== | | == what does he get the message == |
| The most general description of a chemical reaction network considers a number <math>N</math> of distinct chemical species reacting via <math>R</math> reactions.<ref>Heinrich, R. and Schuster, S. (1996) The regulation of cellular systems. Chapman & Hall, New York.</ref>
| |
| <ref>Chen, L. and Wang, R. and Li, C. and Aihara, K. (2010) Modeling biomolecular networks in cells: structures and dynamics. Springer.</ref> The chemical equation of the <math>j</math>-th reaction can then be written in the generic form
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| :<math>
| | A frame of reference to explore the area for a long time? 'Luo Feng suddenly raised wariness,' Why explore for a long time, what does he get the message? '<br><br>which light wing, Vaillant endless.<br>Luo Feng afraid of what appeared midway<br>twists and [http://www.nrcil.net/fancybox/lib/rakuten_LV_65.html ルイヴィトン 財布 モノグラム] turns, all the unexpected factors must be [http://www.nrcil.net/fancybox/lib/rakuten_LV_61.html ルイヴィトン財布ランキング] minimized.<br><br>'ah, it seems he did not seem to know, just busy in exploring everywhere, the entire bottom waters foes are endless celestial rocks, there are numerous caves, canyons, mountains and [http://www.nrcil.net/fancybox/lib/rakuten_LV_140.html ルイヴィトン バック] so on the day the body rocks, so the fear is going to explore thousands era are unlikely to find my cave. 'Luo Feng secretly nod.<br>Before<br>, yourself out from that cave, Ernst \u0026 Young, the Lord seems no doubt.<br><br>actually do [http://www.nrcil.net/fancybox/lib/rakuten_LV_109.html ルイヴィトン ダミエ 新作] not doubt is normal, [http://www.nrcil.net/fancybox/lib/rakuten_LV_80.html メンズ ルイヴィトン 財布] because the waters that countless caves, [http://www.nrcil.net/fancybox/lib/rakuten_LV_14.html ルイヴィトン 新作] many [http://www.nrcil.net/fancybox/lib/rakuten_LV_17.html ルイヴィトン キャップ] of which are in the waters looking for treasure in the drilling of various caves. Ernst \u0026 Young found that when the Lord ...... Luo Feng Luo Feng in the cave, the Lord only when [http://www.nrcil.net/fancybox/lib/rakuten_LV_105.html ベルト ルイヴィトン] Luo Feng Ernst \u0026 Young is also a treasure hunt, and many did not realize that the hole |
| s_{1j} X_1 + s_{2j} X_2 \ldots + s_{Nj} X_{N} \xrightarrow{k_j} \ r_{1j} X_{1} + \ r_{2j} X_{2} + \ldots + r_{Nj} X_{N},
| | 相关的主题文章： |
| </math>
| | <ul> |
| | | |
| which is often written in the equivalent form
| | <li>[http://bbs.dcgzcy.com/home.php?mod=space&uid=26924 http://bbs.dcgzcy.com/home.php?mod=space&uid=26924]</li> |
| | | |
| :<math>
| | <li>[http://www.hanban.com/plus/feedback.php?aid=304 http://www.hanban.com/plus/feedback.php?aid=304]</li> |
| \sum_{i=1}^{N} s_{ij} X_i \xrightarrow{k_j} \sum_{i=1}^{N}\ r_{ij} X_{i}.
| | |
| </math>
| | <li>[http://www.libertad.org.ar/news.cgi http://www.libertad.org.ar/news.cgi]</li> |
| | | |
| Here
| | </ul> |
| | |
| : <math>j</math> is the reaction index running from 1 to <math>R</math>,
| |
| : <math>X_i</math> denotes the <math>i</math>-th chemical species,
| |
| : <math>k_j</math> is the [[Reaction rate constant|rate constant]] of the <math>j</math>-th reaction and
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| : <math>s_{ij}</math> and <math>r_{ij}</math> are the stoichiometric coefficients of reactants and products, respectively.
| |
| | |
| The rate of such reaction can be inferred by the [[law of mass action]]
| |
| | |
| :<math>
| |
| f_j([\vec{X}])= k_j \prod_{z=1}^N [X_z]^{s_{zj}}
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| </math>
| |
| | |
| which denotes the flux of molecules per unit time and unit volume. Here <math>[\vec{X}]=([X_1], [X_2], ... ,[X_N])</math> is the vector of concentrations. Note that this definition includes the [[elementary reaction]]s:
| |
| | |
| * '''zero-order reactions'''
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| for which <math>s_{zj}=0</math> for all <math>z</math>,
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| * '''first-order reactions'''
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| for which <math>s_{zj}=1</math> for a single <math>z</math>,
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| * '''second-order reactions'''
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| for which <math>s_{zj}=1</math> for exactly two <math>z</math>, i.e., a bimolecular reaction, or <math>s_{zj}=2</math> for a single <math>z</math>, i.e., a dimerization reaction.
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| | |
| Each of which are discussed in detail below. One can define the [[Stoichiometry#Stoichiometry_matrix|stoichiometric matrix]]
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| : <math>S_{ij}=r_{ij}-s_{ij},</math>
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| denoting the net extend of molecules of <math>i</math> in reaction <math>j</math>. The reaction rate equations can then be written in the general form
| |
| | |
| :<math>
| |
| \frac{d [X_i]}{dt} =\sum_{j=1}^{R} S_{ij} f_j([\vec{X}]).
| |
| </math> | |
| | |
| Note that this is the product of the stochiometric matrix and the vector of reaction rate functions.
| |
| Particular simple solutions exist in equilibrium, <math>\frac{d [X_i]}{dt}=0</math>, for systems composed of merely reversible reactions. In this case the rate of the forward and backward reactions are equal, a principle called [[detailed balance]]. Note that detailed balance is a property of the stoichiometric matrix <math>S_{ij}</math> alone and does not depend on the particular form of the rate functions <math>f_j</math>. All other cases where detailed balance is violated are commonly studied by [[flux balance analysis]] which has been developed to understand [[metabolic pathway]]s.<ref>Szallasi, Z. and Stelling, J. and Periwal, V. (2006) System modeling in cell biology: from concepts to nuts and bolts. MIT Press Cambridge.</ref><ref>Iglesias, P.A. and Ingalls, B.P. (2010) Control theory and systems biology. MIT Press Cambridge.</ref>
| |
| | |
| The rate equation of a reaction with a multi-step mechanism cannot, in general, be deduced from the stoichiometric coefficients of the overall reaction; it must be derived theoretically using [[Steady State theory|quasi-steady state assumptions]] from the underlying elementary reactions or determined experimentally. The equation may involve fractions, or it may depend on the concentration of an intermediate species.
| |
| | |
| ==Zero-order reactions==
| |
| A '''Zero-order reaction''' has a rate that is independent of the concentration of the reactant(s). Increasing the concentration of the reacting species will not speed up the rate of the reaction i.e. the amount of substance reacted is proportional to the time. Zero-order reactions are typically found when a material that is required for the reaction to proceed, such as a surface or a [[catalyst]], is saturated by the reactants. The rate law for a zero-order reaction is
| |
| | |
| :<math>\ r = k</math>
| |
| | |
| where r is the reaction rate and k is the reaction rate coefficient with units of concentration or time. If, and only if, this zeroth-order reaction 1) occurs in a closed system, 2) there is no net build-up of intermediates, and 3) there are no other reactions occurring, it can be shown by solving a [[mass balance]] equation for the system that:
| |
| | |
| :<math> r = -\frac{d[A]}{dt}=k</math>
| |
| | |
| If this [[differential equation]] is [[integral|integrated]] it gives an equation often called the '''integrated zero-order rate law'''.
| |
| | |
| :<math>\ [A]_t = -kt + [A]_0</math>
| |
| | |
| where <math>\ [A]_t</math> represents the concentration of the chemical of interest at a particular time, and <math>\ [A]_0</math> represents the initial concentration.
| |
| | |
| A reaction is zero order if concentration data are plotted versus time and the result is a straight line. A plot of <math>\ [A]_t</math> vs. time t gives a straight line with a slope of <math> -k </math>.
| |
| | |
| The half-life of a reaction describes the time needed for half of the reactant to be depleted (same as the [[half-life]] involved in [[nuclear decay]], which is a first-order reaction). For a zero-order reaction the half-life is given by
| |
| | |
| : <math>\ t_ \frac{1}{2} = \frac{[A]_0}{2k}</math> | |
| | |
| ;Example of a zero-order reaction
| |
| * Reversed [[Haber process]]: <math>2NH_3 (g) \rightarrow \; 3H_2 (g) + N_2 (g)</math>
| |
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| The order of a reaction cannot be deduced from the chemical equation of the reaction.
| |
| | |
| ==First-order reactions==
| |
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| {{see also|Order of reaction}}
| |
| | |
| A '''first-order reaction''' depends on the concentration of only one reactant (a '''unimolecular reaction'''). Other reactants can be present, but each will be zero-order. The rate law for a reaction that is first order with respect to a reactant A is
| |
| :<math>\frac{-d[A]}{dt} \equiv r = k[A]</math>
| |
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| ''k'' is the first order rate constant, which has units of 1/s.
| |
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| The '''integrated first-order rate law''' is
| |
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| :<math>\ \ln{[A]} = -kt + \ln{[A]_0}</math>
| |
| | |
| A plot of <math>\ln{[A]}</math> vs. time ''t'' gives a straight line with a slope of <math>-k</math>.
| |
| | |
| The half-life of a first-order reaction is independent of the starting concentration and is given by <math>\ t_ \frac{1}{2} = \frac{\ln{(2)}}{k}</math>.
| |
| | |
| Examples of reactions that are first-order with respect to the reactant:
| |
| | |
| * <math>\mbox{H}_2 \mbox{O}_2 (l) \rightarrow \; \mbox{H}_2\mbox{O} (l) + \frac{1}{2}\mbox{O}_2 (g)</math>
| |
| * <math>\mbox{SO}_2 \mbox{Cl}_2 (l) \rightarrow \; \mbox{SO}_2 (g) + \mbox{Cl}_2 (g)</math>
| |
| * <math>2\mbox{N}_2 \mbox{O}_5 (g) \rightarrow \; 4\mbox{NO}_2 (g) + \mbox{O}_2 (g)</math>
| |
| | |
| ===Further properties of first-order reaction kinetics===
| |
| The integrated first-order rate law
| |
| :<math>\ \ln{[A]} = -kt + \ln{[A]_0}</math>
| |
| is usually written in the form of the exponential decay equation
| |
| :<math>A=A_0e^{-kt}\,</math>
| |
| A different (but equivalent) way of considering first order kinetics is as follows: The exponential decay equation can be rewritten as:
| |
| :<math>A=A_{0}\left( e^{-k\Delta t_{p}} \right)^{n}</math>
| |
| where <math>\Delta t_{p}</math> corresponds to a specific time period and <math>n</math> is an integer corresponding to the number of time periods. At the end of each time period, the fraction of the reactant population remaining relative to the amount present at the start of the time period, <math>f_{RP}</math>, will be:
| |
| :<math>\frac{A_{n}}{A_{n-1}} =f_{RP}=e^{-k\Delta t_{p}}</math>
| |
| Such that after <math>n</math> time periods, the fraction of the original reactant population will be:
| |
| :<math>\frac{A}{A_{0}}\equiv \frac{A_{n}}{A_{0}}=\left( e^{-k\Delta t_{p}} \right)^{n}=\left( f_{RP} \right)^{n}=\left( 1-f_{BP} \right)^{n}</math>
| |
| where: <math>f_{BP}</math> corresponds to the fraction of the reactant population that will break down in each time period.
| |
| This equation indicates that the fraction of the total amount of reactant population that will break down in each time period is independent of the initial amount present. When the chosen time period corresponds to <math>\Delta t_{p}=\frac{\ln \left( 2 \right)}{k}</math>, the fraction of the population that will break down in each time period will be exactly ½ the amount present at the start of the time period (i.e. the time period corresponds to the half-life of the first-order reaction).
| |
| | |
| The average rate of the reaction for the n<sup>th</sup> time period is given by:
| |
| :<math>r_{avg,n}=-\frac{\Delta A}{\Delta t_{p}}=\frac{A_{n-1}-A_{n}}{\Delta t_{p}}</math>
| |
| Therefore, the amount remaining at the end of each time period will be related to the average rate of that time period and the reactant population at the start of the time period by:
| |
| :<math>A_{n}=A_{n-1}-r_{avg,n}\Delta t_{p}</math>
| |
| Since the fraction of the reactant population that will break down in each time period can be expressed as:
| |
| :<math>f_{BP}=1-\frac{A_{n}}{A_{n-1}}</math>
| |
| The amount of reactant that will break down in each time period can be related to the average rate over that time period by:
| |
| :<math>f_{BP}=\frac{r_{avg,n}\Delta t_{p}}{A_{n-1}}</math>
| |
| Such that the amount that remains at the end of each time period will be related to the amount present at the start of the time period according to:
| |
| :<math>A_{n}=A_{n-1}\left( 1-\frac{r_{avg,n}\Delta t_{p}}{A_{n-1}} \right)</math>
| |
| This equation is a recursion allowing for the calculation of the amount present after any number of time periods, without need of the rate constant, provided that the average rate for each time period is known.
| |
| <ref>Walsh R, Martin E, Darvesh S. A method to describe enzyme-catalyzed reactions by combining steady state and time course enzyme kinetic parameters... Biochim Biophys Acta. 2010 Jan;1800:1-5</ref>
| |
| | |
| ==Second-order reactions==
| |
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| A '''second-order reaction''' depends on the concentrations of one second-order reactant, or two first-order reactants.
| |
| | |
| For a second order reaction, its reaction rate is given by:
| |
| | |
| :<math>\ -\frac{d[A]}{dt} = 2k[A]^2</math> or <math>\ -\frac{d[A]}{dt} = k[A][B]</math> or <math>\ -\frac{d[A]}{dt} = 2k[B]^2</math>
| |
| | |
| In several popular kinetics books, the definition of the rate law for second-order reactions is written instead as<math>-\frac{d[A]}{dt} = k[A]^2</math>. This effectively conflates the 2 inside the constant, k, whose numerical meaning then becomes different. This simplifying convention is followed in the integrated rate laws provided below. However, this simplification leads to potentially problematic inconsistencies, i.e. if the reaction rate is described in terms of product formation vs reactant disappearance. Instead, the option of keeping the 2 in the rate law (rather than absorbing it into a rate constant with an altered meaning) maintains a consistent meaning for k and is considered more correct technically. This more technically consistent convention is almost always used in peer-reviewed literature, tables of rate constants, and simulation software.<ref name="2nd-order">[http://www.rcdc.nd.edu/compilations/Ali/Ali.htm NDRL Radiation Chemistry Data Center]. See also: [http://www.getcited.org/puba/101600761 Christos Capellos and Bennon H. Bielski ''"Kinetic systems: mathematical description of chemical kinetics in solution"'' 1972, Wiley-Interscience (New York)].</ref>
| |
| | |
| The '''integrated second-order rate laws''' are respectively
| |
| | |
| :<math>\frac{1}{[A]} = \frac{1}{[A]_0} + kt </math>
| |
| | |
| or
| |
| | |
| :<math>\frac{[A]}{[B]} = \frac{[A]_0}{[B]_0} e^{([A]_0 - [B]_0)kt}</math>
| |
| | |
| [A]<sub>0</sub> and [B]<sub>0</sub> must be different to obtain that integrated equation.
| |
| | |
| The half-life equation for a second-order reaction dependent on one second-order reactant is <math>\ t_ \frac{1}{2} = \frac{1}{k[A]_0}</math>. For such a reaction, the half-life progressively doubles as the concentration of the reactant falls to half its initial value.
| |
| | |
| Another way to present the above rate laws is to take the log of both sides:
| |
| <math>\ln{}r = \ln{}k + 2\ln\left[A\right] </math>
| |
| | |
| ;Examples of a Second-order reaction:
| |
| * <math>2\mbox{NO}_2(g) \rightarrow \; 2\mbox{NO}(g) + \mbox{O}_2(g)</math>
| |
| | |
| ===Pseudo-first-order===
| |
| | |
| Measuring a second-order reaction rate with reactants A and B can be problematic: The concentrations of the two reactants must be followed simultaneously, which is more difficult; or measure one of them and calculate the other as a difference, which is less precise. A common solution for that problem is the '''pseudo-first-order approximation'''.
| |
| | |
| If the concentration of one of a reactants remains constant because it is supplied in great excess, its concentration can be absorbed within the rate constant, obtaining a '''pseudo''' first-order reaction constant, because in fact, it depends on the concentration of only one reactant. If, for example, [B] remains constant, then:
| |
| | |
| <math>\ r = k[A][B] = k'[A]</math>
| |
| | |
| where <math>k'=k[B]_0</math> (k' or k<sub>obs</sub> with units s<sup>−1</sup>) and an expression is obtained identical to the first order expression above.
| |
| | |
| One way to obtain a pseudo-first-order reaction is to use a large excess of one of the reactants ([B]>>[A]) would work for the previous example) so that, as the reaction progresses, only a small amount of the reactant is consumed, and its concentration can be considered to stay constant. By collecting <math>k'</math> for many reactions with different (but excess) concentrations of [B], a plot of <math>k'</math> versus [B] gives <math>k</math> (the regular second order rate constant) as the slope.
| |
| | |
| Example:
| |
| The hydrolysis of esters by dilute mineral acids follows pseudo-first-order kinetics where the concentration of water is present in large excess.
| |
| :CH<sub>3</sub>COOCH<sub>3</sub> + H<sub>2</sub>O → CH<sub>3</sub>COOH + CH<sub>3</sub>OH
| |
| | |
| ==Summary for reaction orders 0, 1, 2, and ''n''==
| |
| | |
| Elementary reaction steps with order 3 (called '''ternary reactions''') are [[Elementary reaction|rare and unlikely]] to occur. However, overall reactions composed of several elementary steps can, of course, be of any (including non-integer) order.
| |
| | |
| {| class="wikitable"
| |
| !
| |
| !Zero-Order
| |
| !First-Order
| |
| !Second-Order
| |
| !''n''th-Order
| |
| |-
| |
| |Rate Law
| |
| |<math>-\frac{d[A]}{dt} = k</math>
| |
| |<math>-\frac{d[A]}{dt} = k[A]</math>
| |
| |<math>-\frac{d[A]}{dt} = k[A]^2</math><ref name="2nd-order"/>
| |
| |<math>-\frac{d[A]}{dt} = k[A]^n</math>
| |
| |-
| |
| |Integrated Rate Law
| |
| |<math>\ [A] = [A]_0 - kt</math>
| |
| |<math>\ [A] = [A]_0 e^{-kt}</math>
| |
| |<math>\frac{1}{[A]} = \frac{1}{[A]_0} + kt</math><ref name="2nd-order"/>
| |
| |<math>\frac{1}{[A]^{n-1}} = \frac{1}{{[A]_0}^{n-1}} + (n-1)kt</math>
| |
| <small>[Except first order]</small>
| |
| |-
| |
| |Units of Rate Constant (''k'')
| |
| |<math>\rm\frac{M}{s}</math>
| |
| |<math>\rm\frac{1}{s}</math>
| |
| |<math>\rm\frac{1}{M \cdot s}</math>
| |
| |<math>\frac{1}{{\rm M}^{n-1} \cdot \rm s}</math>
| |
| |-
| |
| |Linear Plot to determine ''k''
| |
| |<math>[A] \ \mbox{vs.} \ t</math>
| |
| |<math>\ln ([A]) \ \mbox{vs.} \ t </math>
| |
| |<math>\frac{1}{[A]} \ \mbox{vs.} \ t</math>
| |
| |<math>\frac{1}{[A]^{n-1}} \ \mbox{vs.} \ t</math>
| |
| <small>[Except first order]</small>
| |
| |-
| |
| |Half-life
| |
| |<math>t_{1/2} = \frac{[A]_0}{2k}</math>
| |
| |<math>t_{1/2} = \frac{\ln (2)}{k}</math>
| |
| |<math>t_{1/2} = \frac{1}{k[A]_0}</math><ref name="2nd-order"/>
| |
| |<math>t_{1/2} = \frac{2^{n-1}-1}{(n-1)k{[A]_0}^{n-1}}</math>
| |
| <small>[Except first order]
| |
| |}
| |
| | |
| Where M stands for concentration in [[molarity]] (mol · L<sup>−1</sup>), ''t'' for time, and ''k'' for the reaction rate constant. The half-life of a first-order reaction is often expressed as ''t''<sub>1/2</sub> = 0.693/''k'' (as ln2 = 0.693).
| |
| | |
| ==Equilibrium reactions or opposed reactions==
| |
| | |
| A pair of forward and reverse reactions may define an [[Chemical equilibrium|equilibrium]] process. For example, A and B react into X and Y and vice versa (s, t, u, and v are the [[stoichiometric coefficient]]s):
| |
| | |
| :<math>\ sA + tB \rightleftharpoons uX + vY</math> | |
| | |
| The reaction rate expression for the above reactions (assuming each one is elementary) can be expressed as:
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| :<math> r = {k_1 [A]^s[B]^t} - {k_2 [X]^u[Y]^v}\,</math>
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| where: k<sub>1</sub> is the rate coefficient for the reaction that consumes A and B; k<sub>2</sub> is the rate coefficient for the backwards reaction, which consumes X and Y and produces A and B.
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| The constants k<sub>1</sub> and k<sub>2</sub> are related to the equilibrium coefficient for the reaction (K) by the following relationship (set r=0 in balance):
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| :<math> {k_1 [A]^s[B]^t = k_2 [X]^u[Y]^v}\,</math>
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| :<math> K = \frac{[X]^u[Y]^v}{[A]^s[B]^t} = \frac{k_1}{k_2}</math>
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| [[Image:ChemicalEquilibrium.svg|thumb|300px|right|Concentration of A (A<sub>0</sub> = 0.25 mole/l) and B versus time reaching equilibrium k<sub>f</sub> = 2 min<sup>-1</sup> and k<sub>r</sub> = 1 min<sup>-1</sup>]]
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| ===Simple example===
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| In a simple equilibrium between two species:
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| :<math> A \rightleftharpoons B </math>
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| Where the reactions starts with an initial concentration of A, <math>[A]_0</math>, with an initial concentration of 0 for B at time t=0.
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| Then the constant K at equilibrium is expressed as:
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| :<math>K \ \stackrel{\mathrm{def}}{=}\ \frac{k_{f}}{k_{b}} = \frac{\left[B\right]_e} {\left[A\right]_e}</math>
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| Where <math>[A]_e</math> and <math>[B]_e</math> are the concentrations of A and B at equilibrium, respectively.
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| The concentration of A at time t, <math>[A]_t</math>, is related to the concentration of B at time t, <math>[B]_t</math>, by the equilibrium reaction equation:
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| :<math>\ [A]_t = [A]_0 - [B]_t </math> | |
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| Note that the term <math>[B]_0</math> is not present because, in this simple example, the initial concentration of B is 0.
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| This applies even when time t is at infinity; i.e., equilibrium has been reached:
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| :<math>\ [A]_e = [A]_0 - [B]_e </math>
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| then it follows, by the definition of K, that
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| :<math>\ [B]_e = x = \frac{k_{f}}{k_f+k_b}[A]_0 </math>
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| and, therefore,
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| :<math>\ [A]_e = [A]_0 - x = \frac{k_{b}}{k_f+k_b}[A]_0 </math>
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| These equations allow us to uncouple the [[system of equations|system of differential equations]], and allow us to solve for the concentration of A alone.
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| The reaction equation, given previously as:
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| :<math> r = {k_1 [A]^s[B]^t} - {k_2 [X]^u[Y]^v}\,</math>
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| :<math> -\frac{d[A]}{dt} = {k_f [A]_t} - {k_b [B]_t}\,</math> | |
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| The derivative is negative because this is the rate of the reaction going from A to B, and therefore the concentration of A is decreasing. To simplify annotation, let x be <math>[A]_t</math>, the concentration of A at time t. Let <math>x_e</math> be the concentration of A at equilibrium. Then:
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| :<math> -\frac{d[A]}{dt} = {k_f [A]_t} - {k_b [B]_t}\,</math>
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| :<math> -\frac{dx}{dt} = {k_f x} - {k_b [B]_t}\,</math>
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| :<math> -\frac{dx}{dt} = {k_f x} - {k_b ([A]_0 - x)}\,</math>
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| :<math> -\frac{dx}{dt} = {(k_f + k_b)x} - {k_b [A]_0}\,</math>
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| Since:
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| :<math> k_f + k_b = {k_b \frac{[A]_0}{x_e}} </math>
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| The [[reaction rate]] becomes:
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| :<math>\ \frac{dx}{dt} = \frac{k_b[A]_0}{x_e} (x_e - x) </math>
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| which results in: | |
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| :<math> \ln \left(\frac{[A]_0 - [A]_e}{[A]_t-[A]_e}\right) = (k_f + k_b)t </math>
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| A plot of the negative [[natural logarithm]] of the concentration of A in time minus the concentration at equilibrium versus time t gives a straight line with slope k<sub>f</sub> + k<sub>b</sub>. By measurement of A<sub>e</sub> and B<sub>e</sub> the values of K and the two [[reaction rate constant]]s will be known.<ref>For a worked out example see: ''Determination of the Rotational Barrier for Kinetically Stable Conformational Isomers via NMR and 2D TLC An Introductory Organic Chemistry Experiment'' Gregory T. Rushton, William G. Burns, Judi M. Lavin, Yong S. Chong, Perry Pellechia, and Ken D. Shimizu [[J. Chem. Educ.]] '''2007''', 84, 1499. [http://jchemed.chem.wisc.edu/Journal/Issues/2007/Sep/abs1499.html Abstract]</ref>
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| ===Generalization of simple example===
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| If the concentration at the time t = 0 is different from above, the simplifications above are invalid, and a system of differential equations must be solved. However, this system can also be solved exactly to yield the following generalized expressions:
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| <math>\left[ A \right]=\left[ A \right]_{0}\frac{1}{k_{f}+k_{b}}\left( k_{b}+k_{f}e^{-\left( k_{f}+k_{b} \right)t} \right)+\left[ B \right]_{0}\frac{k_{b}}{k_{f}+k_{b}}\left( 1-e^{-\left( k_{f}+k_{b} \right)t} \right)</math>
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| <math>\left[ B \right]=\left[ A \right]_{0}\frac{k_{f}}{k_{f}+k_{b}}\left( 1-e^{-\left( k_{f}+k_{b} \right)t} \right)+\left[ B \right]_{0}\frac{1}{k_{f}+k_{b}}\left( k_{f}+k_{b}e^{-\left( k_{f}+k_{b} \right)t} \right)</math>
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| When the equilibrium constant is close to unity and the reaction rates very fast for instance in [[Conformational isomerism|conformational analysis]] of molecules, other methods are required for the determination of rate constants for instance by complete lineshape analysis in [[NMR spectroscopy]].
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| ==Consecutive reactions==
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| If the rate constants for the following reaction are <math>k_1</math> and <math>k_2</math>; <math> A \rightarrow \; B \rightarrow \; C </math>, then the rate equation is:
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| For reactant A: <math> \frac{d[A]}{dt} = -k_1 [A] </math>
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| For reactant B: <math> \frac{d[B]}{dt} = k_1 [A] - k_2 [B]</math>
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| For product C: <math> \frac{d[C]}{dt} = k_2 [B]</math>
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| With the individual concentrations scaled by the total population of reactants to become probabilities, linear systems of differential equations such as these can be formulated as a [[master equation]]. The differential equations can be solved analytically and the integrated rate equations are
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| <math>[A]=[A]_0 e^{-k_1 t}</math>
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| <math>\left[ B \right]=\left\{ \begin{array}{*{35}l}
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| \left[ A \right]_{0}\frac{k_{1}}{k_{2}-k_{1}}\left( e^{-k_{1}t}-e^{-k_{2}t} \right) & k_{1}\ne k_{2} \\
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| \left[ A \right]_{0}k_{1}te^{-k_{1}t}+\left[ B \right]_{0}e^{-k_{1}t} & \text{otherwise} \\
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| \end{array} \right.</math>
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| <math>\left[ C \right]=\left\{ \begin{array}{*{35}l}
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| \left[ A \right]_{0}\left( 1+\frac{k_{1}e^{-k_{2}t}-k_{2}e^{-k_{1}t}}{k_{2}-k_{1}} \right)+\left[ B \right]_{0}\left( 1-e^{-k_{2}t} \right)+\left[ C \right]_{0} & k_{1}\ne k_{2} \\
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| \left[ A \right]_{0}\left( 1-e^{-k_{1}t}-k_{1}te^{-k_{1}t} \right)+\left[ B \right]_{0}\left( 1-e^{-k_{1}t} \right)+\left[ C \right]_{0} & \text{otherwise} \\
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| \end{array} \right.</math>
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| The [[steady state (chemistry)|steady state]] approximation leads to very similar results in an easier way.
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| ==Parallel or competitive reactions== | |
| When a substance reacts simultaneously to give two different products, a parallel or competitive reaction is said to take place.
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| *''Two first order reactions'':
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| <math> A \rightarrow \; B </math> and <math> A \rightarrow \; C </math>, with constants <math> k_1</math> and <math> k_2</math> and rate equations <math>-\frac{d[A]}{dt}=(k_1+k_2)[A]</math>, <math> \frac{d[B]}{dt}=k_1[A]</math> and <math> \frac{d[C]}{dt}=k_2[A]</math>
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| The integrated rate equations are then <math>\ [A] = [A]_0 e^{-(k_1+k_2)t}</math>; <math>[B] = \frac{k_1}{k_1+k_2}[A]_0 (1-e^{-(k_1+k_2)t})</math> and
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| <math>[C] = \frac{k_2}{k_1+k_2}[A]_0 (1-e^{-(k_1+k_2)t})</math>.
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| One important relationship in this case is <math> \frac{[B]}{[C]}=\frac{k_1}{k_2}</math>
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| *''One first order and one second order reaction'':<ref>José A. Manso et al."A Kinetic Approach to the Alkylating Potential of Carcinogenic Lactones" Chem. Res. Toxicol. 2005, 18, (7) 1161-1166</ref>
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| This can be the case when studying a bimolecular reaction and a simultaneous hydrolysis (which can be treated as pseudo order one) takes place: the hydrolysis complicates the study of the reaction kinetics, because some reactant is being "spent" in a parallel reaction. For example A reacts with R to give our product C, but meanwhile the hydrolysis reaction takes away an amount of A to give B, a byproduct: <math> A + H_2O \rightarrow \ B </math> and <math> A + R \rightarrow \ C </math>. The rate equations are: <math> \frac{d[B]}{dt}=k_1[A][H_2O]=k_1'[A]</math> and <math> \frac{d[C]}{dt}=k_2[A][R]</math>. Where <math>k_1'</math> is the pseudo first order constant.
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| The integrated rate equation for the main product [C] is <math> [C]=[R]_0 \left [ 1-e^{-\frac{k_2}{k_1'}[A]_0(1-e^{-k_1't})} \right ] </math>, which is equivalent to <math> ln \frac{[R]_0}{[R]_0-[C]}=\frac{k_2[A]_0}{k_1'}(1-e^{-k_1't})</math>. Concentration of B is related to that of C through <math> [B]=-\frac{k_1'}{k_2} ln \left ( 1 - \frac{[C]}{[R]_0} \right )</math>
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| The integrated equations were analytically obtained but during the process it was assumed that <math>[A]_0-[C]\approx \;[A]_0</math> therefeore, previous equation for [C] can only be used for low concentrations of [C] compared to [A]<sub>0</sub>
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| ==General dynamics of unimolecular conversion==
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| For a general unimolecular reaction involving interconversion of <math>N</math> different species, whose concentrations at time <math>t</math> are denoted by <math>X_1(t)</math> through <math>X_N(t)</math>, an analytic form for the time-evolution of the species can be found. Let the rate constant of conversion from species <math>X_i</math> to species <math>X_j</math> be denoted as <math>k_{ij}</math>, and construct a rate-constant matrix <math>K</math> whose entries are the <math>k_{ij}</math>.
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| Also, let <math>X(t)=(X_1(t),X_2(t),...,X_N(t))^T</math> be the vector of concentrations as a function of time.
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| Let <math>J=(1,1,1,...,1)^T</math> be the vector of ones.
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| Let <math>I</math> be the <math>N</math>×<math>N</math> identity matrix.
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| Let <math>Diag</math> be the function that takes a vector and constructs a diagonal matrix whose on-diagonal entries are those of the vector.
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| Let <math> \displaystyle\mathcal{L}^{-1}</math> be the inverse Laplace transform from <math>s</math> to <math>t</math>.
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| Then the time-evolved state <math>X(t)</math> is given by
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| :<math>X(t)=\displaystyle\mathcal{L}^{-1}[(sI+Diag(KJ)-K^T)^{-1}X(0)]</math>,
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| thus providing the relation between the initial conditions of the system and its state at time <math>t</math>.
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| ==See also==
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| *[[Michaelis–Menten kinetics]]
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| *[[Petersen matrix]]
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| *[[Reaction-diffusion equation]]
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| *[[Reactions on surfaces]]: rate equations for reactions where at least one of the reactants [[adsorption|adsorbs]] onto a surface
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| *[[Reaction progress kinetic analysis]]
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| *[[Reaction rate]]
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| *[[Reaction rate constant]]
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| *[[Steady state (chemistry)|Steady state approximation]]
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| ==References==
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| {{reflist}}
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| {{Reaction mechanisms}}
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| {{DEFAULTSORT:Rate Equation}}
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| [[Category:Chemical kinetics]]
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| [[Category:Chemical engineering]]
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| [[cy:Cyfradd adwaith#Hafaliadau cyfradd]]
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