10.3
Many reactions proceed through multiple elementary steps involving reactive intermediates. Adding these steps gives the overall reaction without intermediates.
Consider a reaction where reactant, R, forms product, P, through an intermediate, I.
Here, the first step is fast and reversible. The second step is a slow and rate-determining step that controls the overall rate.
Initially, the concentration of I rises quickly. It reaches a small maximum, then decreases and stabilizes at a low, nearly constant value. So, its concentration remains negligible compared to R and P.
Note that I is formed during the first step. However, it is consumed during the reverse of the first step and the second step of the reaction.
The steady-state approximation assumes that the concentration of I remains constant. So, the rate of formation of I equals its rate of consumption.
Solving for I and substituting it into the overall rate law yields the final rate expression with no intermediates. This rate law is useful for reactions that proceed through multiple elementary steps.
The steady-state approximation, also referred to as the quasi-steady-state approximation to differentiate it from a true steady state, is a widely used method for simplifying calculations in complex reaction mechanisms. This approach is particularly useful when dealing with multi-step reactions that involve reverse reactions or several steps, which can significantly increase mathematical complexity and make the reactions nearly unsolvable analytically.
The steady-state approximation operates on the assumption that the intermediate species (I) in a reaction maintain a low, stable concentration throughout the bulk of the reaction, following an initial induction period. During this induction period, the concentration of intermediates rises from zero, but their rates of change become negligibly small for the majority of the reaction.
Therefore, the steady-state approximation often assumes d[I]/dt = 0 for each reactive intermediate. This means that, the rate at which an intermediate is formed is roughly equal to its rate of destruction, maintaining a near-constant steady-state concentration. Using calculus, the rate of increase of intermediate concentration due to the forward reaction is balanced by decreases from the reverse reaction and the rate-determining step's forward progress.
For a reaction with a rate-determining step yielding the rate law, substituting the steady-state concentration of I into the equations results in a rate law with constants combined into a new constant 'k.' Interestingly, this approach aligns with using the equilibrium constant of the first elementary process, showcasing consistency between the two applications of the steady-state approximation. The choice of mathematical approach depends on available information and the specific goals in understanding a chemical reaction.
Many reactions proceed through multiple elementary steps involving reactive intermediates. Adding these steps gives the overall reaction without intermediates.
Consider a reaction where reactant, R, forms product, P, through an intermediate, I.
Here, the first step is fast and reversible. The second step is a slow and rate-determining step that controls the overall rate.
Initially, the concentration of I rises quickly. It reaches a small maximum, then decreases and stabilizes at a low, nearly constant value. So, its concentration remains negligible compared to R and P.
Note that I is formed during the first step. However, it is consumed during the reverse of the first step and the second step of the reaction.
The steady-state approximation assumes that the concentration of I remains constant. So, the rate of formation of I equals its rate of consumption.
Solving for I and substituting it into the overall rate law yields the final rate expression with no intermediates. This rate law is useful for reactions that proceed through multiple elementary steps.
From Chapter 10:
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