What does “thermodynamic equilibrium” mean for an enzyme-substrate complex?

From Fersht, Enzyme Structure and Mechanism p. 87:

The Michaelis-Menten mechanism assumes that the enzyme-substrate complex is in thermodynamic equilibrium with free enzyme and substrate.

In my understanding what this means is that the (E-, S- and ES-concentration dependent) rates of association and dissociation have equated. So we're kind of in this situation: where "product" would be referring to [ES] and "reactant" to [E] and [S]. Does that make sense?

• Product is still $P$; $ES$ is the transient Enzyme-Substrate complex. If you mean product in the sense of product of a reaction then yes but it is a confusing usage. So stick to calling $ES$ as a transient state – WYSIWYG Jun 3 '14 at 9:41
• @tmottm Can you please accept answers, which answer your question satisfactory? This is a nice gesture towards the people who write the answers. You can find all your questions in your User menu. – Chris Jun 13 '14 at 10:27

Thermodynamic equilibrium means that:

Rate of forward reaction = Rate of backward reaction

In this case:

$E + S\xrightleftharpoons[k2]{k1} ES\\ \ \\ \ \\ \\ at\ equilibrium:\\ \ \ \\ \ k1.[E][S] \tiny{\ (forward\ rate)}\normalsize= k2.[ES] \tiny\ (backward\ rate)$

This was the initial assumption in the Michaelis-Menten model.

Later on this was improvised by assuming pseudo-steady state of ES complex. This means that $[ES]$ does not change over time, which is both as a result of its production by the reversible reaction: $E + S \leftrightharpoons\ ES$ and consumption by the irreversible reaction $\ ES\ \xrightarrow{k3} E+P$

that is:

$k1[E][S]\tiny\ (production)\normalsize=(k2+k3)[ES]\tiny\ (consumption)$