Rate equation

The natural logarithm of the power-law rate equation is

${\displaystyle \ln v_{0}=\ln k+x\ln[{\ce {A}}]+y\ln[{\ce {B}}]+\cdots }$

This can be used to estimate the order of reaction of each reactant. For example, the initial rate can be measured in a series of experiments at different initial concentrations of reactant ${\displaystyle {\rm {A}}}$ with all other concentrations ${\displaystyle [{\rm {B],[{\rm {C],\dots }}}}}$ kept constant, so that

${\displaystyle \ln v_{0}=x\ln[{\ce {A}}]+{\textrm {constant}}.}$

The slope of a graph of ${\displaystyle \ln v}$ as a function of ${\displaystyle \ln[{\ce {A}}]}$ then corresponds to the order ${\displaystyle x}$ with respect to reactant ${\displaystyle {\rm {A}}}$.[9][10]

However, this method is not always reliable because

1. measurement of the initial rate requires accurate determination of small changes in concentration in short times (compared to the reaction half-life) and is sensitive to errors, and
2. the rate equation will not be completely determined if the rate also depends on substances not present at the beginning of the reaction, such as intermediates or products.

The tentative rate equation determined by the method of initial rates is therefore normally verified by comparing the concentrations measured over a longer time (several half-lives) with the integrated form of the rate equation; this assumes that the reaction goes to completion.

For example, the integrated rate law for a first-order reaction is

${\displaystyle \ln {[{\ce {A}}]}=-kt+\ln {[{\ce {A}}]_{0}},}$

where ${\displaystyle [{\rm {A]}}}$ is the concentration at time ${\displaystyle t}$ and ${\displaystyle [{\rm {A]_{0}}}}$ is the initial concentration at zero time. The first-order rate law is confirmed if ${\displaystyle \ln {[{\ce {A}}]}}$ is in fact a linear function of time. In this case the rate constant ${\displaystyle k}$ is equal to the slope with sign reversed.[11][12]

The partial order with respect to a given reactant can be evaluated by the method of flooding (or of isolation) of Ostwald. In this method, the concentration of one reactant is measured with all other reactants in large excess so that their concentration remains essentially constant. For a reaction a·A + b·B → c·C with rate law ${\displaystyle v_{0}=k\cdot [{\rm {A}}]^{x}\cdot [{\rm {B}}]^{y},}$ the partial order ${\displaystyle x}$ with respect to ${\displaystyle {\rm {A}}}$ is determined using a large excess of ${\displaystyle {\rm {B}}}$. In this case

${\displaystyle v_{0}=k'\cdot [{\rm {A}}]^{x}}$ with ${\displaystyle k'=k\cdot [{\rm {B}}]^{y},}$

and ${\displaystyle x}$ may be determined by the integral method. The order ${\displaystyle y}$ with respect to ${\displaystyle {\rm {B}}}$ under the same conditions (with ${\displaystyle {\rm {B}}}$ in excess) is determined by a series of similar experiments with a range of initial concentration ${\displaystyle [{\rm {B]_{0}}}}$ so that the variation of ${\displaystyle k'}$ can be measured.[13]

For zero-order reactions, the reaction rate is independent of the concentration of a reactant, so that changing its concentration has no effect on the rate of the reaction. Thus, the concentration changes linearly with time. The rate law for zero order reaction is

${\displaystyle -{d[A] \over dt}=k[A]^{0}=k,}$

The unit of k is mol dm^-3 s^-1.[14] This may occur when there is a bottleneck which limits the number of reactant molecules that can react at the same time, for example if the reaction requires contact with an enzyme or a catalytic surface.[15]

Many enzyme-catalyzed reactions are zero order, provided that the reactant concentration is much greater than the enzyme concentration which controls the rate, so that the enzyme is saturated. For example, the biological oxidation of ethanol to acetaldehyde by the enzyme liver alcohol dehydrogenase (LADH) is zero order in ethanol.[16]

Similarly reactions with heterogeneous catalysis can be zero order if the catalytic surface is saturated. For example, the decomposition of phosphine (PH₃) on a hot tungsten surface at high pressure is zero order in phosphine, which decomposes at a constant rate.[15]

In homogeneous catalysis zero order behavior can come about from reversible inhibition. For example, ring-opening metathesis polymerization using third-generation Grubbs catalyst exhibits zero order behavior in catalyst due to the reversible inhibition that occurs between pyridine and the ruthenium center.[17]

A first order reaction depends on the concentration of only one reactant (a unimolecular reaction). Other reactants can be present, but their concentration has no effect on the rate. The rate law for a first order reaction is

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

The unit of k is s^-1.[18] Although not affecting the above math, the majority of first order reactions proceed via intermolecular collisions. Such collisions, which contribute the energy to the reactant, are necessarily second order. The rate of these collisions is, however, masked by the fact that the rate determining step remains the unimolecular breakdown of the energized reactant.

The half-life is independent of the starting concentration and is given by ${\textstyle t_{1/2}={\frac {\ln {(2)}}{k}}}$. The mean lifetime is τ = 1/k.[19]

Examples of such reactions are:

• ${\displaystyle {\ce {2N2O5 -> 4NO2 + O2}}}$ [20][21]
• ${\displaystyle {\ce {[CoCl(NH3)5]^2+ + H2O -> [Co(H2O)(NH3)5]^3+ + Cl-}}}$
• ${\displaystyle {\ce {H2O2 -> H2O + 1/2O2}}}$ [22]

In organic chemistry, the class of S_N1 (nucleophilic substitution unimolecular) reactions consists of first-order reactions. For example, in the reaction of aryldiazonium ions with nucleophiles in aqueous solution, ArN+2 + X⁻ → ArX + N₂, the rate equation is ${\displaystyle v_{0}=k[{\ce {ArN2+}}],}$ where Ar indicates an aryl group.[23]

A reaction is said to be second order when the overall order is two. The rate of a second-order reaction may be proportional to one concentration squared, ${\displaystyle v_{0}=k[{\ce {A}}]^{2},}$ or (more commonly) to the product of two concentrations, ${\displaystyle v_{0}=k[{\ce {A}}][{\ce {B}}].}$ As an example of the first type, the reaction NO₂ + CO → NO + CO₂ is second-order in the reactant NO₂ and zero order in the reactant CO. The observed rate is given by ${\displaystyle v_{0}=k[{\ce {NO2}}]^{2},}$ and is independent of the concentration of CO.[24]

For the rate proportional to a single concentration squared, the time dependence of the concentration is given by

${\displaystyle {\frac {1}{{\ce {[A]}}}}={\frac {1}{{\ce {[A]0}}}}+kt.}$

The unit of k is mol^-1 dm³ s^-1. [25]

The time dependence for a rate proportional to two unequal concentrations is

${\displaystyle {\frac {{\ce {[A]}}}{{\ce {[B]}}}}={\frac {{\ce {[A]0}}}{{\ce {[B]0}}}}e^{\left({\ce {[A]0}}-{\ce {[B]0}}\right)kt};}$

if the concentrations are equal, they satisfy the previous equation.

The second type includes nucleophilic addition-elimination reactions, such as the alkaline hydrolysis of ethyl acetate:[23]

${\displaystyle {\ce {CH3COOC2H5 + OH- -> CH3COO- + C2H5OH}}}$

This reaction is first-order in each reactant and second-order overall:

${\displaystyle v_{0}=k[{\ce {CH3COOC2H5}}][{\ce {OH-}}]}$

If the same hydrolysis reaction is catalyzed by imidazole, the rate equation becomes[23]

${\displaystyle v_{0}=k[{\text{imidazole}}][{\ce {CH3COOC2H5}}].}$

The rate is first-order in one reactant (ethyl acetate), and also first-order in imidazole, which as a catalyst does not appear in the overall chemical equation.

Another well-known class of second-order reactions are the S_N2 (bimolecular nucleophilic substitution) reactions, such as the reaction of n-butyl bromide with sodium iodide in acetone:

${\displaystyle {\ce {CH3CH2CH2CH2Br + NaI -> CH3CH2CH2CH2I + NaBr(v)}}}$

This same compound can be made to undergo a bimolecular (E2) elimination reaction, another common type of second-order reaction, if the sodium iodide and acetone are replaced with sodium tert-butoxide as the salt and tert-butanol as the solvent:

${\displaystyle {\ce {CH3CH2CH2CH2Br+NaO{\mathit {t}}-Bu->CH3CH2CH=CH2+NaBr+HO{\mathit {t}}-Bu}}}$

If the concentration of a reactant remains constant (because it is a catalyst, or because it is in great excess with respect to the other reactants), its concentration can be included in the rate constant, leading to a pseudo–first-order (or occasionally pseudo–second-order) rate equation. For a typical second-order reaction with rate equation ${\displaystyle v_{0}=k[{\ce {A}}][{\ce {B}}],}$ if the concentration of reactant B is constant then ${\displaystyle v_{0}=k[{\ce {A}}][{\ce {B}}]=k'[{\ce {A}}],}$ where the pseudo–first-order rate constant ${\displaystyle k'=k[{\ce {B}}].}$ The second-order rate equation has been reduced to a pseudo–first-order rate equation, which makes the treatment to obtain an integrated rate equation much easier.

One way to obtain a pseudo-first order reaction is to use a large excess of one reactant (say, [B]≫[A]) so that, as the reaction progresses, only a small fraction of the reactant in excess (B) is consumed, and its concentration can be considered to stay constant. For example, the hydrolysis of esters by dilute mineral acids follows pseudo-first order kinetics, where the concentration of water is constant because it is present in large excess:

${\displaystyle {\ce {CH3COOCH3 + H2O -> CH3COOH + CH3OH}}}$

The hydrolysis of sucrose (C₁₂H₂₂O₁₁) in acid solution is often cited as a first-order reaction with rate ${\displaystyle v_{0}=k[{\ce {C12H22O11}}].}$ The true rate equation is third-order, ${\displaystyle v_{0}=k[{\ce {C12H22O11}}][{\ce {H+}}][{\ce {H2O}}];}$ however, the concentrations of both the catalyst H⁺ and the solvent H₂O are normally constant, so that the reaction is pseudo–first-order.[26]

Elementary reaction steps with order 3 (called ternary reactions) are rare and unlikely to occur. However, overall reactions composed of several elementary steps can, of course, be of any (including non-integer) order.

	Zero order	First order	Second order	nth order (g = 1−n)
Rate Law	${\displaystyle -{d[{\ce {A}}]}/{dt}=k}$	${\displaystyle -{d[{\ce {A}}]}/{dt}=k[{\ce {A}}]}$	${\displaystyle -{d[{\ce {A}}]}/{dt}=k[{\ce {A}}]^{2}}$[27]	${\displaystyle -{d[{\ce {A}}]}/{dt}=k[{\ce {A}}]^{n}}$
Integrated Rate Law	${\displaystyle {\ce {[A] = [A]0}}-kt}$	${\displaystyle {\ce {[A] = [A]0}}e^{-kt}}$	${\displaystyle {\frac {1}{{\ce {[A]}}}}={\frac {1}{{\ce {[A]0}}}}+kt}$[27]	${\displaystyle [{\ce {A}}]^{g}={{\ce {[A]0}}^{g}}-gkt}$ [Except first order]
Units of Rate Constant (k)	${\displaystyle {\rm {\frac {M}{s}}}}$	${\displaystyle {\rm {\frac {1}{s}}}}$	${\displaystyle {\rm {\frac {1}{M\cdot s}}}}$	${\displaystyle {\frac {{\rm {M}}^{g}}{\rm {s}}}}$
Linear Plot to determine k	[A] vs. t	${\displaystyle {\ce {\ln([A])}}}$ vs. t	${\displaystyle {\ce {{\frac {1}{[A]}}}}}$ vs. t	${\displaystyle {\ce {{\rm {[A]}}^{g}}}}$ vs. t [Except first order]
Half-life	${\displaystyle t_{\frac {1}{2}}={\frac {{\ce {[A]0}}}{2k}}}$	${\displaystyle t_{\frac {1}{2}}={\frac {\ln(2)}{k}}}$	${\displaystyle t_{\frac {1}{2}}={\frac {1}{k{\ce {[A]0}}}}}$[27]	${\displaystyle t_{\frac {1}{2}}={\frac {{\ce {[A]0}}^{g}(1-2^{-g})}{gk}}}$ [Limit is necessary for first order]

Here ${\displaystyle {\rm {M}}}$ stands for concentration in molarity (mol · L⁻¹), ${\displaystyle t}$ for time, and ${\displaystyle k}$ for the reaction rate constant. The half-life of a first-order reaction is often expressed as t_1/2 = 0.693/k (as ln(2)≈0.693).

In fractional order reactions, the order is a non-integer, which often indicates a chemical chain reaction or other complex reaction mechanism. For example, the pyrolysis of acetaldehyde (CH₃CHO) into methane and carbon monoxide proceeds with an order of 1.5 with respect to acetaldehyde: ${\displaystyle v_{0}=k[{\ce {CH3CHO}}]^{3/2}.}$[28] The decomposition of phosgene (COCl₂) to carbon monoxide and chlorine has order 1 with respect to phosgene itself and order 0.5 with respect to chlorine: ${\displaystyle v_{0}=k{\ce {[COCl2] [Cl2]}}^{1/2}.}$[29]

The order of a chain reaction can be rationalized using the steady state approximation for the concentration of reactive intermediates such as free radicals. For the pyrolysis of acetaldehyde, the Rice-Herzfeld mechanism is

Initiation

${\displaystyle {\ce {CH3CHO -> .CH3 + .CHO}}}$

Propagation

${\displaystyle {\ce {.CH3 + CH3CHO -> CH3CO. + CH4}}}$

${\displaystyle {\ce {CH3CO. -> .CH3 + CO}}}$

Termination

${\displaystyle {\ce {2 .CH3 -> C2H6}}}$

where • denotes a free radical.[28][30] To simplify the theory, the reactions of the *CHO to form a second *CH₃ are ignored.

In the steady state, the rates of formation and destruction of methyl radicals are equal, so that

${\displaystyle {\frac {d[{\ce {.CH3}}]}{dt}}=k_{i}[{\ce {CH3CHO}}]-k_{t}[{\ce {.CH3}}]^{2}=0,}$

so that the concentration of methyl radical satisfies

${\displaystyle {\ce {[.CH3]\quad \propto \quad [CH3CHO]^{1/2}.}}}$

The reaction rate equals the rate of the propagation steps which form the main reaction products CH₄ and CO:

${\displaystyle v_{0}={\frac {d[{\ce {CH4}}]}{dt}}|_{0}=k_{p}{\ce {[.CH3][CH3CHO]}}\quad \propto \quad {\ce {[CH3CHO]^{3/2}}}}$

in agreement with the experimental order of 3/2.[28][30]

More complex rate laws have been described as being mixed order if they approximate to the laws for more than one order at different concentrations of the chemical species involved. For example, a rate law of the form ${\displaystyle v_{0}=k_{1}[A]+k_{2}[A]^{2}}$ represents concurrent first order and second order reactions (or more often concurrent pseudo-first order and second order) reactions, and can be described as mixed first and second order.[31] For sufficiently large values of [A] such a reaction will approximate second order kinetics, but for smaller [A] the kinetics will approximate first order (or pseudo-first order). As the reaction progresses, the reaction can change from second order to first order as reactant is consumed.

Another type of mixed-order rate law has a denominator of two or more terms, often because the identity of the rate-determining step depends on the values of the concentrations. An example is the oxidation of an alcohol to a ketone by hexacyanoferrate (III) ion [Fe(CN)₆³⁻] with ruthenate (VI) ion (RuO₄²⁻) as catalyst.[32] For this reaction, the rate of disappearance of hexacyanoferrate (III) is ${\displaystyle v_{0}={\frac {{\ce {[Fe(CN)6]^2-}}}{k_{\alpha }+k_{\beta }{\ce {[Fe(CN)6]^2-}}}}}$

This is zero-order with respect to hexacyanoferrate (III) at the onset of the reaction (when its concentration is high and the ruthenium catalyst is quickly regenerated), but changes to first-order when its concentration decreases and the regeneration of catalyst becomes rate-determining.

Notable mechanisms with mixed-order rate laws with two-term denominators include:

• Michaelis–Menten kinetics for enzyme-catalysis: first-order in substrate (second-order overall) at low substrate concentrations, zero order in substrate (first-order overall) at higher substrate concentrations; and
• the Lindemann mechanism for unimolecular reactions: second-order at low pressures, first-order at high pressures.

A reaction rate can have a negative partial order with respect to a substance. For example, the conversion of ozone (O₃) to oxygen follows the rate equation ${\displaystyle v_{0}=k{\ce {[O_3]^2}}{\ce {[O_2]^{-1}}}}$ in an excess of oxygen. This corresponds to second order in ozone and order (−1) with respect to oxygen.[33]

When a partial order is negative, the overall order is usually considered as undefined. In the above example, for instance, the reaction is not described as first order even though the sum of the partial orders is ${\displaystyle 2+(-1)=1}$, because the rate equation is more complex than that of a simple first-order reaction.

In a simple equilibrium between two species:

${\displaystyle {\ce {A <=> P}}}$

where the reaction starts with an initial concentration of reactant A, ${\displaystyle {\ce {[A]0}}}$, and an initial concentration of 0 for product P at time t=0.

Then the equilibrium constant K is expressed as:

${\displaystyle K\ {\stackrel {\mathrm {def} }{=}}\ {\frac {k_{1}}{k_{-1}}}={\frac {\left[{\ce {P}}\right]_{e}}{\left[{\ce {A}}\right]_{e}}}}$

where ${\displaystyle [{\ce {A}}]_{e}}$ and ${\displaystyle [{\ce {P}}]_{e}}$ are the concentrations of A and P at equilibrium, respectively.

The concentration of A at time t, ${\displaystyle [{\ce {A}}]_{t}}$, is related to the concentration of P at time t, ${\displaystyle [{\ce {P}}]_{t}}$, by the equilibrium reaction equation:

${\displaystyle {\ce {[A]_{\mathit {t}}=[A]0-[P]_{\mathit {t}}}}}$

The term ${\displaystyle {\ce {[P]0}}}$ is not present because, in this simple example, the initial concentration of P is 0.

This applies even when time t is at infinity; i.e., equilibrium has been reached:

${\displaystyle {\ce {[A]_{\mathit {e}}=[A]0-[P]_{\mathit {e}}}}}$

then it follows, by the definition of K, that

${\displaystyle [{\ce {P}}]_{e}={\frac {k_{1}}{k_{1}+k_{-1}}}{\ce {[A]0}}}$

and, therefore,

${\displaystyle \ [{\ce {A}}]_{e}={\ce {[A]0}}-[{\ce {P}}]_{e}={\frac {k_{-1}}{k_{1}+k_{-1}}}{\ce {[A]0}}}$

These equations allow us to uncouple the system of differential equations, and allow us to solve for the concentration of A alone.

The reaction equation was given previously as:

${\displaystyle v=k_{1}[{\ce {A}}]^{a}[{\ce {B}}]^{b}-k_{-1}[{\ce {P}}]^{p}[{\ce {Q}}]^{q}}$

For ${\displaystyle {\ce {A <=> P}}}$ this is simply

${\displaystyle -{\frac {d[{\ce {A}}]}{dt}}=k_{1}[{\ce {A}}]_{t}-k_{-1}[{\ce {P}}]_{t}}$

The derivative is negative because this is the rate of the reaction going from A to P, and therefore the concentration of A is decreasing. To simplify notation, let x be ${\displaystyle [{\ce {A}}]_{t}}$, the concentration of A at time t. Let ${\displaystyle x_{e}}$ be the concentration of A at equilibrium. Then:

${\displaystyle {\begin{aligned}-{\frac {d[{\ce {A}}]}{dt}}&={k_{1}[{\ce {A}}]_{t}}-{k_{-1}[{\ce {P}}]_{t}}\\[8pt]-{\frac {dx}{dt}}&={k_{1}x}-{k_{-1}[{\ce {P}}]_{t}}\\[8pt]&={k_{1}x}-{k_{-1}({\ce {[A]0}}-x)}\\[8pt]&={(k_{1}+k_{-1})x}-{k_{-1}{\ce {[A]0}}}\end{aligned}}}$

Since:

${\displaystyle k_{1}+k_{-1}=k_{-1}{\frac {{\ce {[A]0}}}{x_{e}}}}$

the reaction rate becomes:

${\displaystyle {\frac {dx}{dt}}={\frac {k_{-1}{\ce {[A]0}}}{x_{e}}}(x_{e}-x)}$

which results in:

${\displaystyle \ln \left({\frac {{\ce {[A]0}}-[{\ce {A}}]_{e}}{[{\ce {A}}]_{t}-[{\ce {A}}]_{e}}}\right)=(k_{1}+k_{-1})t}$.

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 k₁ + k₋₁. By measurement of [A]_e and [P]_e the values of K and the two reaction rate constants will be known.[34]

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:

${\displaystyle {\begin{aligned}&\left[{\ce {A}}\right]={\ce {[A]0}}{\frac {1}{k_{1}+k_{-1}}}\left(k_{-1}+k_{1}e^{-\left(k_{1}+k_{-1}\right)t}\right)+{\ce {[P]0}}{\frac {k_{-1}}{k_{1}+k_{-1}}}\left(1-e^{-\left(k_{1}+k_{-1}\right)t}\right)\\[8pt]&\left[{\ce {P}}\right]={\ce {[A]0}}{\frac {k_{1}}{k_{1}+k_{-1}}}\left(1-e^{-\left(k_{1}+k_{-1}\right)t}\right)+{\ce {[P]0}}{\frac {1}{k_{1}+k_{-1}}}\left(k_{1}+k_{-1}e^{-\left(k_{1}+k_{-1}\right)t}\right)\end{aligned}}}$

When the equilibrium constant is close to unity and the reaction rates very fast for instance in conformational analysis of molecules, other methods are required for the determination of rate constants for instance by complete lineshape analysis in NMR spectroscopy.

${\displaystyle {\ce {A -> B}}}$ and ${\displaystyle {\ce {A -> C}}}$, with constants ${\displaystyle k_{1}}$ and ${\displaystyle k_{2}}$ and rate equations ${\displaystyle -{\frac {d[{\ce {A}}]}{dt}}=(k_{1}+k_{2})[{\ce {A}}]}$; ${\displaystyle {\frac {d[{\ce {B}}]}{dt}}=k_{1}[{\ce {A}}]}$ and ${\displaystyle {\frac {d[{\ce {C}}]}{dt}}=k_{2}[{\ce {A}}]}$

The integrated rate equations are then ${\displaystyle [{\ce {A}}]={\ce {[A]0}}e^{-(k_{1}+k_{2})t}}$; ${\displaystyle [{\ce {B}}]={\frac {k_{1}}{k_{1}+k_{2}}}{\ce {[A]0}}\left(1-e^{-(k_{1}+k_{2})t}\right)}$ and ${\displaystyle [{\ce {C}}]={\frac {k_{2}}{k_{1}+k_{2}}}{\ce {[A]0}}\left(1-e^{-(k_{1}+k_{2})t}\right)}$.

One important relationship in this case is ${\displaystyle {\frac {{\ce {[B]}}}{{\ce {[C]}}}}={\frac {k_{1}}{k_{2}}}}$

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: ${\displaystyle {\ce {A + H2O -> B}}}$ and ${\displaystyle {\ce {A + R -> C}}}$. The rate equations are: ${\displaystyle {\frac {d[{\ce {B}}]}{dt}}=k_{1}{\ce {[A][H2O]}}=k_{1}'[{\ce {A}}]}$ and ${\displaystyle {\frac {d[{\ce {C}}]}{dt}}=k_{2}{\ce {[A][R]}}}$, where ${\displaystyle k_{1}'}$ is the pseudo first order constant.[35]

The integrated rate equation for the main product [C] is ${\displaystyle {\ce {[C]=[R]0}}\left[1-e^{-{\frac {k_{2}}{k_{1}'}}{\ce {[A]0}}\left(1-e^{-k_{1}'t}\right)}\right]}$, which is equivalent to ${\displaystyle \ln {\frac {{\ce {[R]0}}}{{\ce {[R]0-[C]}}}}={\frac {k_{2}{\ce {[A]0}}}{k_{1}'}}\left(1-e^{-k_{1}'t}\right)}$. Concentration of B is related to that of C through ${\displaystyle [{\ce {B}}]=-{\frac {k_{1}'}{k_{2}}}\ln \left(1-{\frac {\ce {[C]}}{\ce {[R]0}}}\right)}$

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