$K = \frac{k_{forward}}{k_{reverse}}$

Aarthi_Lakshmi_2A
Posts: 14
Joined: Wed Sep 21, 2016 2:57 pm

Does the Steady-State method require integration? I don't quite understand how it works. And how does the Steady-State method work if there are multiple intermediates?

Chem_Mod
Posts: 18891
Joined: Thu Aug 04, 2011 1:53 pm
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The steady-state approximation assumes the concentration of intermediates is constant (after an initial induction period). For each intermediate, the rate of formation is set equal to the rate of consumption. If an intermediate is formed or consumed in more than one elementary reaction, the sums are set equal to each other. This can be done for every intermediate. These equations can be solved to express each intermediate concentration in terms of concentrations of reactants and products. Then the rate law for a product-forming elementary reaction can be used to solve for the overall rate law. This method is typically used when it is not clear which step is rate-determining. The resulting rate law could be integrated if needed.

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