The
steady state approximation, 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 {{chem2| ^{239}U -> ^{239}Np -> ^{239}Pu}}. If the rate constants for the following reaction are and ; , combining the
rate equations with a
mass balance for the system yields three coupled differential equations:
Reaction rates For species A: \frac{d[\ce A]}{dt} = -k_1 [\ce A] For species B: \frac{d[\ce B]}{dt} = k_1 [\ce A] - k_2 [\ce B] :Here the first (positive) term represents the formation of B by the first step , whose rate depends on the initial reactant A. The second (negative) term represents the consumption of B by the second step , whose rate depends on B as the reactant in that step. For species C: \frac{d[\ce C]}{dt} = k_2 [\ce B]
Analytical solutions The analytical solutions for these equations (supposing that initial concentrations of every substance except for A are zero) are: : [\ce A]=[\ce A]_0 e^{-k_1 t} : \left[ \ce B \right]= \begin{cases} \left[ \ce 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} \\\\ \left[ \ce A \right]_{0}k_{1}te^{-k_{1}t};&k_{1} = k_{2} \\ \end{cases} : \left[ \ce C \right]= \begin{cases} \left[ \ce A \right]_{0}\left( 1+\frac{k_{1}e^{-k_{2}t}-k_{2}e^{-k_{1}t}}{k_{2}-k_{1}} \right);&k_{1}\ne k_{2} \\\\ \left[ \ce A \right]_{0}\left( 1-e^{-k_{1}t}-k_{1}te^{-k_{1}t} \right);&k_{1} = k_{2} \\ \end{cases}
Steady state If the steady state approximation is applied, then the derivative of the concentration of the intermediate is set to zero. This reduces the second differential equation to an algebraic equation which is much easier to solve. : \frac{d[\ce B]}{dt} = 0 = k_1 [\ce A] - k_2 [\ce B] \Rightarrow \; [\ce B] = \frac{k_1}{k_2} [\ce A]. Therefore, \tfrac{d[\ce C]}{dt} = k_1 [\ce A], so that [\ce C]=[\ce A]_0 \left (1- e^{-k_1 t} \right ). Since [\ce B] = \tfrac{k_1}{k_2} [\ce A] = \tfrac{k_1}{k_2}[\ce A]_0 e^{-k_1 t} , the concentration of the reaction intermediate B changes with the same time constant as [A] and is not in a steady state in that sense.
Validity The analytical and approximated solutions should now be compared in order to decide when it is valid to use the steady state approximation. The analytical solution transforms into the approximate one when k_2 \gg k_1 , because then e^{-k_2t} \ll e^{-k_1t} and k_2-k_1 \approx \; k_2. Therefore, it is valid to apply the steady state approximation only if the second reaction is much faster than the first ( is a common criterion), because that means that the intermediate forms slowly and reacts readily so its concentration stays low. The graphs show concentrations of A (red), B (green) and C (blue) in two cases, calculated from the analytical solution. When the first reaction is faster it is not valid to assume that the variation of [B] is very small, because [B] is neither low or close to constant: first A transforms into B rapidly and B accumulates because it disappears slowly. As the concentration of A decreases its rate of transformation decreases, at the same time the rate of reaction of B into C increases as more B is formed, so a maximum is reached when t=\begin{cases} \frac{\ln \left( \frac{k_{1}}{k_{2}} \right)}{k_{1}-k_{2}} & \, k_{1}\ne k_{2} \\\\ \frac{1}{k_{1}} & \, k_{1} = k_{2} \\ \end{cases}From then on the concentration of B decreases. When the second reaction is faster, after a short
induction period during which the steady state approximation does not apply, the concentration of B remains low (and more or less constant in an absolute sense) because its rates of formation and disappearance are almost equal and the steady state approximation can be used. The equilibrium approximation can sometimes be used in chemical kinetics to yield similar results to the steady state approximation. It consists in assuming that the intermediate arrives rapidly at chemical equilibrium with the reactants. For example,
Michaelis-Menten kinetics can be derived assuming equilibrium instead of steady state. Normally the requirements for applying the steady state approximation are laxer: the concentration of the intermediate is only needed to be low and more or less constant (as seen, this has to do only with the rates at which it appears and disappears) but it is not required to be at equilibrium. == Example ==