10.3
반응 기전: 정상 상태 근사
정상 상태 근사는 진정한 정상 상태와 구분하기 위해 준정상 상태 근사라고도 불리며, 복잡한 반응 기전에서 계산을 단순화하기 위해 널리 사용되는 방법입니다. 이 접근법은 역반응이나 여러 단계를 포함하는 다단계 반응을 다룰 때 특히 유용하며, 이는 수학적 복잡성을 크게 증…
많은 반응은 반응성 중간체를 포함하는 여러 기본 단계를 거칩니다. 이 단계를 추가하면 중간 성분 없이 전체 반응이 가능합니다.
반응물 R이 중간 I를 통해 생성물 P를 형성하는 반응을 생각해 봅시다.
여기서는 첫 번째 단계가 빠르고 되돌릴 수 있습니다. 두 번째 단계는 전체 속도를 제어하는 느리고 속도를 결정하는 단계입니다.
처음에는 I 집중도가 빠르게 상승합니다. 이 수치는 작은 최대치에 도달한 후 감소하여 낮고 거의 일정한 값으로 안정화됩니다. 따라서 R과 P에 비해 농도는 무시할 만합니다.
I는 첫 번째 단계에서 형성된다는 점을 유의하세요. 하지만 첫 번째 단계와 두 번째 단계의 반대 단계에서 소모됩니다.
정상 상태 근사는 I의 농도가 일정하다고 가정합니다. 따라서 I의 형성 속도는 소비 속도와 같다.
I를 풀고 전체 요율 법칙에 대입하면 중간 계량이 없는 최종 계율 식이 나옵니다. 이 속도 법칙은 여러 기본 단계를 거치는 반응에 유용합니다.
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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.