Rate equation

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For a chemical reaction, the rate law or rate equation is an equation which links the reaction rate with concentrations or pressures of reactants. To determine the rate equation for a particular system one combines the reaction rate with a mass balance for the system. For a reaction s A + t B → C + D, in a closed system a simple rate equation is of the form

$r = \frac{d[C]}{dt} = k(T)[A]^{n}[B]^{m}$

In this equation, [X] expresses the concentration of a given X, usually in mol/litre and k(T) is known as the reaction rate coefficient or rate constant, although it is not really a constant, because it includes everything that affects reaction rate outside concentration: mainly temperature but also ionic strength, surface area of the adsorbent or light irradiation.

The exponents n and m are called reaction orders and depend on the reaction mechanism. The stoichiometric coefficients and reaction orders are very often equal, but only in one step reactions, molecularity (number of molecules or atoms actually colliding), stoichiometry and reaction order must be the same.

Complicated rate equations are not of the form above, and they can be a sum of terms like it or have quantities in the denominator (see further sections)

The rate equation is a differential equation, and it can be integrated in order to obtain an integrated rate equation that links concentrations of reactants or products with time.

If the concentration of one of the reactants remains constant (because it is a catalyst or it is in great excess with respect to the other reactants) its concentration can be included in the rate constant, obtaining a pseudo constant: if B is the reactant whose concentration is constant then $r = k [A][B] = k'[A]$ . The second order rate equation has been reduced to a pseudo first order rate equation. This makes the treatment to obtain an integrated rate equation much easier.

1 Zero-order reactions
2 First-order reactions
3 Second-order reactions
- 3.1 Pseudo first order
4 Summary for reaction orders 0, 1, 2 and n
5 Equilibrium reactions or opposed reactions
6 Consecutive reactions
7 Parallel or competitive reactions
8 See also
9 References

[edit] Zero-order reactions

A zero-order reaction has a rate which 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. Zero-order reactions are typically found when a material 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

$\ r = k$

Where r is the reaction rate and k is the reaction rate coefficient, k has units of concentration/time. If, and only if, this zero-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 for the system that:

$r = -\frac{d[A]}{dt}=k$

If this differential equation is integrated it gives an equation which is often called the integrated zero-order rate law

$\ [A] = -kt + [A]_0$

where $\ [A]$ represents the concentration of the chemical of interest at a particular time and $\ [A]_0$ represents the initial concentration.

A reaction is zero order if concentration data are plotted versus time and the result is a straight line. The slope of this resulting line is the zero order rate constant k.

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

$\ t_ \frac{1}{2} = \frac{[A]_0}{2k}$ .

Example of a Zeroth-order reaction

Reversed Haber process: $2NH_3 (g) \rightarrow \; 3H_2 (g) + N_2 (g)$

[edit] First-order reactions

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 first-order reaction is

$\ r = k[A]$

k is the first order rate constant that has units of 1/time

If, and only if, this first-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 for the system that

$-\frac{d[A]}{dt} = k[A]$

The integrated first-order rate law is

$\ \ln{[A]} = -kt + \ln{[A]_0}$

A plot of time $\ t$ vs. $\ -ln{ [A]}$ gives a straight line with a slope equal to the reaction rate constant. The half life of a first-order reaction can be determined using the equation $\ t_ \frac{1}{2} = \frac{\ln{(2)}}{k}$ . Charactaristic of a first-order reaction is that all the half-lives are equal.

Examples of First-order reactions

$2\mbox{H}_2 \mbox{O}_2 (l) \rightarrow \; 2\mbox{H}_2\mbox{O} (l) + \mbox{O}_2 (g)$
$2\mbox{SO}_2 \mbox{Cl}_2 (l) \rightarrow \; 2\mbox{SO}_2 (g) + \mbox{Cl}_2 (g)$
$2\mbox{N}_2 \mbox{O}_5 (g) \rightarrow \; 4\mbox{NO}_2 (g) + \mbox{O}_2 (g)$

These are all first-order with respect to the reactant.

[edit] Second-order reactions

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:

$\ r = k[A]^2$ or $\ r = k[A][B]$

The integrated second-order rate laws are respectively

$\frac{1}{[A]} = kt + \frac{1}{[A]_0}$ or

$\frac{[A]}{[B]} = \frac{[A]_0}{[B]_0} e^{([A]_0 - [B]_0)kt}$

[A]₀ and [B]₀ must be different, in order to obtain that integrated equation.

The half-life equation for a second-order reaction dependent on one second-order reactant is $\ t_ \frac{1}{2} = \frac{1}{k[A]_0}$ . For a second-order reaction half-lives progressively double.

Another way to present the above rate laws is to take the log of both sides: $\ln{}r = \ln{}k + 2\ln\left[A\right]$

Examples of a Second-order reaction

$2\mbox{NO}_2(g) \rightarrow \; 2\mbox{NO}(g) + \mbox{O}_2(g)$

[edit] Pseudo first order

Measuring a second order reaction rate 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 either [A] or [B] remain constant as the reaction proceeds, then the reaction can be considered pseudo first order because in fact it only depends on the concentration of one reactant. If for example [B] remains constant then:

$\ r = k[A][B] = k'[A]$

where $k' = k [B] 0$ (k' or k_obs with units s^-1) and we have an expression 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 $k'$ for many reactions with different (but excess) concentrations of [B]; a plot of $k'$ versus [B] gives $k$ (the regular second order rate constant) as the slope.

[edit] Summary for reaction orders 0, 1, 2 and n

Reactions with order 3 are very rare, to the point that some chemists think they don't exist, and (almost) always involve dinitrogen pentoxide N₂O₅.

	Zeroth Order	First Order	Second Order	n-th Order
Rate Law	$-\frac{d[A]}{dt} = k$	$-\frac{d[A]}{dt} = k[A]$	$-\frac{d[A]}{dt} = k[A]^2$	$-\frac{d[A]}{dt} = k [A]^n$
Integrated Rate Law	$\ [A] = [A]_0 - kt$	$\ [A] = [A]_0 e^{-kt}$	$\frac{1}{[A]} = \frac{1}{[A]_0} + kt$	$\frac{1}{[A]^{n-1}} = \frac{1}{{[A]_0}^{n-1}} + (n-1)kt$ [Except first order]
Units of Rate Constant $\ k$	$\frac{M}{s}$	$\frac{1}{s}$	$\frac{1}{M \cdot s}$	$\frac{1}{M^{n-1} \cdot s}$
Linear Plot to determine $\ k$	$[A] \ \mbox{vs.} \ t$	$\ln ([A]) \ \mbox{vs.} \ t$	$\frac{1}{[A]} \ \mbox{vs.} \ t$	$\frac{1}{[A]^{n-1}} \ \mbox{vs.} \ t$ [Except first order]
Half-life	$t_{1/2} = \frac{[A]_0}{2k}$	$t_{1/2} = \frac{\ln (2)}{k}$	$t_{1/2} = \frac{1}{[A]_0 k}$	$t_{1/2} = \frac{2^{n-1}-1}{(n-1)k[A_0]^{n-1}}$ [Except first order]

[edit] Equilibrium reactions or opposed reactions

A pair of forward and reverse reactions may define an equilibrium process. For example A and B react into X and Y and vice versa (s, t, u and v are the stoichiometric coefficients):

sA + tB ⇌ uX + vY

The reaction rate expression for the above reactions (assuming they each are elementary) can be expressed as:

$r = {k_1 [A]^s[B]^t} - {k_2 [X]^u[Y]^v}\,$

where: k₁ is the rate coefficient for the reaction which consumes A and B; k₂ is the rate coefficient for the backwards reaction, which consumes X and Y and produces A and B.

The constants k₁ and k₂ are related to the equilibrium coefficient for the reaction (K) by the following relationship (set r=0 in balance):

${k_1 [A]^s[B]^t = k_2 [X]^u[Y]^v}\,$

$K = \frac{[X]^u[Y]^v}{[A]^s[B]^t} = \frac{k_1}{k_2}$

[edit] Consecutive reactions

If the rate constants for the following reaction are $k 1$ and $k 2$ ; $A \rightarrow \; B \rightarrow \; C$ , then the rate equation is:

For reactant A: $\frac{d[A]}{dt} = -k_1 [A]$

For reactant B: $\frac{d[B]}{dt} = k_1 [A] - k_2 [B]$

For product C: $\frac{d[C]}{dt} = k_2 [B]$

These differential equations can be solved analytically and the integrated rate equations (supposing that initial concentrations of every substance except A are zero) are

$[A]=[A]_0 e^{-k_1 t}$

$[B]=[A]_0 \frac{k_1}{k_2 - k_1}\left ( e^{-k_1t}-e^{-k_2t} \right )$

$[C] = \frac{[A]_0}{k_2-k_1} \left [ k_2 \left ( 1- e^{-k_1t} \right ) - k_1 \left (1- e^{-k_2t} \right ) \right ] \quad = [A]_0 \left (1 + \frac{k_1 e^{-k_2t}-k_2e^{-k_1t}}{k_2-k_1} \right )$

The steady state approximation leads to very similar results in an easier way.

[edit] Parallel or competitive reactions

When a substance reacts simultaneously to give two different products, a parallel or competitive reaction is said to have place.

Two first order reactions:

$A \rightarrow \; B$ and $A \rightarrow \; C$ , with constants $k 1$ and $k 2$ and rate equations $-\frac{d[A]}{dt}=(k_1+k_2)[A]$ , $\frac{d[B]}{dt}=k_1[A]$ and $\frac{d[C]}{dt}=k_2[A]$

The integrated rate equations are then $\ [A] = [A]_0 e^{-(k_1+k_2)t}$ ; $[B] = \frac{k_1}{k_1+k_2}[A]_0 (1-e^{-(k_1+k_2)t})$ and $[C] = \frac{k_2}{k_1+k_2}[A]_0 (1-e^{-(k_1+k_2)t})$ .

One important relationship in this case is $\frac{[B]}{[C]}=\frac{k_1}{k_2}$

One first order and one second order reaction:

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: $A + H_2O \rightarrow \ B$ and $A + R \rightarrow \ C$ . The rate equations are: $\frac{d[B]}{dt}=k_1[A][H_2O]=k_1'[A]$ and $\frac{d[C]}{dt}=k_2[A][R]$ . Where $k 1'$ is the pseudo first order constant.

The integrated rate equation for the main product [C] is $[C]=[R]_0 \left [ 1-e^{-\frac{k_2}{k_1'}[A]_0(1-e^{-k_1't})} \right ]$ , which is equivalent to $ln \frac{[R]_0}{[R]_0-[C]}=\frac{k_2[A]_0}{k_1'}(1-e^{-k_1't})$ . Concentration of B is related to that of C through $[B]=-\frac{k_1'}{k_2} ln \left ( 1 - \frac{[C]}{[R]_0} \right )$

The integrated equations were analytically obtained but during the process it was assumed that $[A]_0-[C]\approx \;[A]_0$ therefeore, previous equation for [C] can only be used for low concentrations of [C] compared to [A]₀

[edit] See also

Reactions on surfaces: rate equations for reactions where, at least one of the reactants adsorbs onto a surface
Steady state approximation

[edit] References

Kenneth A. Connors Chemical Kinetics, the study of raction rates in solution, 1990, VCH Publishers. This book contains all the rate equations in this article and their derivation.

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Categories: Chemical kinetics | Physical chemistry | Chemical engineering