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2.4B: Steady-State and Pre-equilibrium Approximations

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    The steady state approximation, occasionally called the stationary-state approximation, involves setting the rate of change of a reaction intermediate in a reaction mechanism equal to zero. It is important to note that steady state approximation does not assume the reaction intermediate concentration to be constant (and therefore its time derivative being zero), it assumes that the variation in the concentration of the intermediate is almost zero: the concentration of the intermediate is very low, so even a big relative variation in its concentration is small, if considered quantitatively.

    Its use facilitates the resolution of the differential equations that arise from rate equations, which lack an analytical solution for most mechanisms beyond the most simple 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. If the rate constants for the following reaction are \(k_1\) and \(k_2\);

    \[A \rightarrow \; B \rightarrow \; C\]

    combining the rate equations with a mass balance for the system yields for following differential rate laws:

    For species A:

    \[\dfrac{d[A]}{dt} = -k_1 [A]\]

    For species B:

    \[\dfrac{d[B]}{dt} = k_1 [A] - k_2 [B]\]

    For species C:

    \[\dfrac{d[C]}{dt} = k_2 [B]\]

    The analytical solutions for these equations (supposing that initial concentrations of every substance except for A are zero) are:

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

    \[\left[ B \right]=\left\{ \begin{matrix} \left[ A \right]_{0}\dfrac{k_{1}}{k_{2}-k_{1}}\left( e^{-k_{1}t}-e^{-k_{2}t} \right);\,\,k_{1}\ne k_{2} \\ \left[ A \right]_{0}k_{1}te^{-k_{1}t}\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{otherwise} \\ \end{matrix} \right.\]

    \[\left[ C \right]=\left\{ \begin{matrix} \left[ A \right]_{0}\left( 1+\dfrac{k_{1}e^{-k_{2}t}-k_{2}e^{-k_{1}t}}{k_{2}-k_{1}} \right);\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,k_{1}\ne k_{2} \\ \left[ A \right]_{0}\left( 1-e^{-k_{1}t}-k_{1}te^{-k_{1}t} \right);\,\,\,\,\,\,\text{otherwise} \\ \end{matrix} \right.\]

    Steady State Approximation

    If the steady state approximation is applied, then the derivative of the concentration of the intermediate is set to zero.

    \[\dfrac{d[B]}{dt} = 0 = k_1 [A] - k_2 [B] \Rightarrow \; [B] = \dfrac{k_1}{k_2} [A]\]

    therefore

    \[\dfrac{d[C]}{dt} = k_1 [A] so [C]=[A]_0 \left (1- e^{-k_1 t} \right ).\]

    Validity of Approximation

    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 one (\(k_2/k_1 > 10\) is a right 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

    \[t=\left\{ \begin{matrix} \dfrac{\ln \left( \dfrac{k_{1}}{k_{2}} \right)}{k_{1}-k_{2}} & \, k_{1} \ne k_{2} \\ \dfrac{1}{k_{1}} & \, otherwise \\ \end{matrix} \right.. \]

    From then on the concentration of \(B\) decreases.

    • When the second reaction is faster, after a short induction period, concentration of \(B\) remains low (and more or less constant) because its rate of formation and disappearance are almost equal and the steady state approximation can be used.

    The equilibrium approximation can be used sometimes in chemical kinetics to yield similar results as the steady state approximation (Michaelis-Menten kinetics can be derived assuming equilibrium instead of steady state): it consists in assuming that the intermediate is at chemical equilibrium. 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 needed to be at equilibrium, which is usually difficult to prove and involves heavier assumptions.

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