10.3
反応機構:定常状態近似
定常状態近似は、真の定常状態と区別するために準定常状態近似とも呼ばれ、複雑な反応機構の計算を簡素化するために広く使われている手法です。このアプローチは、逆反応や複数のステップを含む多段階反応を扱う際に特に有用であり、数学的複雑さを大幅に増やし、解析的にほぼ解けない反応にします。
定常状態近似は、反応中…
多くの反応は反応性中間体を含む複数の基本的なステップを経て進行します。これらのステップを加えることで、中間体なしの全体的な反応が得られます。
反応物Rが中間体Iを介して生成物Pを形成する反応を考えます。
ここでは、最初のステップは速く、かつ可逆的です。第二段階はゆっくりとした速度を決定するステップで、全体の速度を制御します。
最初はIの濃度が急速に上昇します。わずかな最大値に達した後、減少し、ほぼ一定に近い低い値で安定します。したがって、その濃度はRやPに比べて無視できるほど低いままです。
Iは最初のステップで形成されることに注意してください。しかし、最初のステップと反応の2番目のステップの逆に消費されます。
定常状態近似では、Iの濃度が一定であると仮定します。したがって、Iの形成速度は消費速度に等しい。
Iを解いて全体のレート法則に代入すると、中間数を含まない最終的なレート式が得られます。この速度法則は、複数の基本的なステップを経て進行する反応に有用です。
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Q1: What is the steady-state approximation and why is it useful for complex reactions?
The steady-state approximation assumes that reactive intermediates maintain a low, nearly constant concentration during a reaction. This simplifies calculations for multi-step reactions involving reverse reactions, making them analytically solvable. It reduces mathematical complexity significantly compared to solving full differential equations for each elementary step.
Q2: How does an intermediate's concentration change during a reaction?
Initially, intermediate concentration rises quickly to a small maximum, then decreases and stabilizes at a low, nearly constant value. This occurs because the intermediate is formed in the first step but consumed by both the reverse reaction and subsequent steps. After the induction period, its concentration remains negligible compared to reactants and products.
Q3: What does it mean when d[I]/dt equals zero in the steady-state approximation?
Setting d[I]/dt = 0 means the rate of intermediate formation equals its rate of destruction. The intermediate reaches a quasi-steady state where its concentration changes negligibly. This mathematical assumption allows you to solve for the intermediate's concentration and substitute it into the overall rate law to eliminate intermediates.
Q4: How do you derive the overall rate law using the steady-state approximation?
Set the rate of formation of the intermediate equal to its rate of consumption, then solve for the intermediate's concentration. Substitute this expression into the rate law for the rate-determining step. The result is a final rate expression containing only reactants and products, with no intermediates appearing in the equation.
Q5: Why is the steady-state approximation called quasi-steady-state?
The term quasi-steady-state distinguishes this approximation from a true steady state. The intermediate concentration is not truly constant throughout the entire reaction; it rises during an initial induction period before stabilizing. The approximation assumes negligible change only after this induction period, when the intermediate reaches its low, stable concentration.
Q6: How does the steady-state approximation relate to equilibrium constants?
The steady-state approximation aligns with using equilibrium constants of the first elementary process. When you solve for intermediate concentration under steady-state conditions, the resulting rate law incorporates the equilibrium constant from the fast, reversible first step. This consistency shows that both approaches yield equivalent rate expressions for multi-step reactions.
Q7: What role does the rate-determining step play in the steady-state approximation?
The rate-determining step controls the overall reaction rate and determines which rate law you substitute the intermediate concentration into. The slow step's rate law, combined with the steady-state expression for the intermediate, produces the final overall rate law. This approach works for reaction mechanisms where the rate-determining step follows one or more fast, reversible steps.