Steady state (chemistry)

The steady state approximation,[1] occasionally called the stationary-state approximation or Bodenstein's quasi-steady state approximation, involves setting the rate of change of a reaction intermediate in a reaction mechanism equal to zero so that the kinetic equations can be simplified by setting the rate of formation of the intermediate equal to the rate of its destruction.

In practice it is sufficient that the rates of formation and destruction are approximately equal, which means that the net rate of variation of the concentration of the intermediate is small compared to the formation and destruction, and the concentration of the intermediate varies only slowly, similar to the reactants and products (see the equations and the green traces in the figures below).

Its use facilitates the resolution of the differential equations that arise from rate equations, which lack an analytical solution for most mechanisms beyond the simplest ones. The steady state approximation is applied, for example, in Michaelis-Menten kinetics.

As an example, the steady state approximation will be applied to two consecutive, irreversible, homogeneous first order reactions in a closed system. (For heterogeneous reactions, see reactions on surfaces.) This model corresponds, for example, to a series of nuclear decompositions like ²³⁹U → ²³⁹Np → ²³⁹Pu.

If the rate constants for the following reaction are k₁ and k₂; A → B → C, combining the rate equations with a mass balance for the system yields three coupled differential equations:

The reaction H₂ + Br₂ → 2 HBr has the following mechanism:

The rate of each species are:

${\displaystyle {\frac {d[{\ce {H}}Br]}{dt}}=k_{2}[{\ce {B}}r][{\ce {H}}_{2}]+k_{3}[{\ce {H}}][{\ce {B}}r_{2}]-k_{4}[{\ce {H}}][{\ce {H}}Br]}$

${\displaystyle {\frac {d[{\ce {H}}]}{dt}}=k_{2}[{\ce {B}}r][{\ce {H}}_{2}]-k_{3}[{\ce {H}}][{\ce {B}}r_{2}]-k_{4}[{\ce {H}}][{\ce {H}}Br]}$

${\displaystyle {\frac {d[{\ce {B}}r]}{dt}}=2k_{1}[{\ce {B}}r_{2}]+k_{3}[{\ce {H}}][{\ce {B}}r_{2}]+k_{4}[{\ce {H}}][{\ce {H}}Br]-2k_{5}[{\ce {B}}r]^{2}}$

${\displaystyle {\frac {d[{\ce {B}}r_{2}]}{dt}}=-k_{1}[{\ce {B}}r_{2}]-k_{3}[{\ce {H}}][{\ce {B}}r_{2}]+k_{5}[{\ce {B}}r]^{2}}$

${\displaystyle {\frac {d[{\ce {H}}_{2}]}{dt}}=-k_{2}[{\ce {B}}r][{\ce {H}}_{2}]+k_{4}[{\ce {H}}][{\ce {H}}Br]}$

These equations cannot be solved, because each one has values that change with time. For example, the first equation contains the concentrations of [Br], [H₂] and [Br₂], which depend on time, as can be seen in their respective equations.

To solve the rate equations the steady state approximation can be used. The reactants of this reaction are H₂ and Br₂, the intermediates are H and Br, and the product is HBr.

For solving the equations, the rates of the intermediates are set to 0 in the steady state approximation:

${\displaystyle {\begin{aligned}&{\frac {d[{\ce {H}}]}{dt}}=k_{2}[{\ce {B}}r][{\ce {H}}_{2}]-k_{3}[{\ce {H}}][{\ce {H}}Br]-k_{4}[{\ce {H}}][{\ce {H}}Br]=0\\&\qquad \longrightarrow k_{2}[{\ce {B}}r][{\ce {H}}_{2}]=k_{3}[{\ce {H}}][{\ce {H}}Br_{2}]\end{aligned}}}$

${\displaystyle {\frac {d[{\ce {B}}r]}{dt}}=2k_{1}[{\ce {B}}r_{2}]-k_{2}[{\ce {B}}r][{\ce {H}}_{2}]+k_{3}[{\ce {H}}][{\ce {B}}r_{2}]+k_{4}[{\ce {H}}][{\ce {H}}Br]-2k_{5}[{\ce {B}}r]^{2}}$

From the reaction rate of H, k₂[Br][H₂] − k₃[H][Br₂] − k₄[H][HBr] = 0 , so the reaction rate of Br can be simplified:

${\displaystyle {\begin{aligned}&2k_{1}[{\ce {B}}r_{2}]-2k_{5}[{\ce {B}}r]^{2}=0\\&\qquad \longrightarrow [{\ce {B}}r]={\frac {k_{1}}{k_{5}}}^{\frac {1}{2}}[{\ce {B}}r]^{\frac {1}{2}}\end{aligned}}}$

${\displaystyle {\frac {d[{\ce {H}}Br]}{dt}}=k_{2}[{\ce {B}}r][{\ce {H}}_{2}]+k_{3}[{\ce {H}}][{\ce {B}}r_{2}]-k_{4}[{\ce {H}}][{\ce {H}}Br]}$

${\displaystyle {\begin{aligned}&k_{2}[{\ce {B}}r][{\ce {H}}_{2}]-k_{4}[{\ce {H}}][{\ce {B}}r]=k_{3}[{\ce {H}}][{\ce {H}}Br_{2}]\\&\qquad \longrightarrow {\frac {d[{\ce {H}}Br]}{dt}}=2k_{3}[{\ce {H}}][{\ce {B}}r_{2}]\end{aligned}}}$

The reaction rate of HBr can also be simplifed, changing k₂[Br][H₂] − k₄[H][Br] to k₃[H][Br₂], since both values are equal.

The concentration of H from equation 1 can be isolated:

${\displaystyle [{\ce {H}}]}={\frac {k_{2}[{\ce {B}}r][{\ce {H}}_{2}]}{k_{3}[{\ce {B}}r_{2}]+k_{4}[{\ce {H}}][{\ce {H}}Br]}}={\frac {k_{2}\left({\frac {k_{1}}{k_{5}}}\right)^{\frac {1}{2}}[{\ce {B}}r_{2}]^{\frac {1}{2}}[{\ce {H}}_{2}]}{k_{3}[{\ce {B}}r_{2}]+k_{4}[{\ce {H}}Br]}}$

The concentration of this intermediate is small and changes with time like the concentrations of reactants and product. It is inserted into the last differential equation to give

${\displaystyle {\frac {d[{\ce {H}}Br]}{dt}}=2k_{3}[{\ce {H}}][{\ce {B}}r_{2}]=2k_{3}\left\lfloor {\frac {k_{2}\left({\frac {k_{1}}{k_{5}}}\right)^{\frac {1}{2}}[{\ce {B}}r_{2}]^{\frac {1}{2}}[{\ce {H}}_{2}]}{k_{3}[{\ce {B}}r_{2}]+k_{4}[{\ce {H}}Br]}}\right\rfloor [{\ce {B}}r_{2}].}$

Simplifying the equation leads to

${\displaystyle {\frac {d[{\ce {H}}Br]}{dt}}={\frac {2k_{3}k_{2}\left({\frac {k_{1}}{k_{5}}}\right)^{\frac {1}{2}}[{\ce {B}}r_{2}]^{\frac {1}{2}}[{\ce {H}}_{2}]}{k_{3}+{\frac {k_{4}[{\ce {H}}Br]}{[{\ce {B}}r_{2}]}}}}.}$

The experimentally observed rate is

${\displaystyle v={\frac {k'[{\ce {H}}_{2}][{\ce {B}}r_{2}]^{\tfrac {1}{2}}}{1+k''{\frac {[{\ce {H}}Br]}{[{\ce {B}}r_{2}]}}}}.}$

The experimental rate law is very similar to the rate obtained with the steady state approximation, if ${\displaystyle k'}$ is ${\textstyle 2k_{3}k_{2}{\sqrt {\frac {k_{1}}{k_{5}}}}}$ and ${\displaystyle 1+k''}$ is ${\displaystyle k_{3}+k_{4}}$.

• Lindemann mechanism
• Reaction progress kinetic analysis
• Steady state (biochemistry)

• https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Kinetics/Reaction_Mechanisms/Steady-State_Approximation