E + S <-> ES<-> EP <-> E + P
In the case where the is none, or next to no product present initially, the chemical potential is a large negative number, resulting in a steep downhill slope for the third step, allowing us to write EP -> E + P
We now have:
E + S <-> ES <-> EP -> E + P
We shall also assume that E+S, ES, and E+P are separated by large barriers, allowing each transition to be treated independently. The transition involving binding of substrates from solution is characterised by a first order rate, proportional to the substrate concentration.
Throwing a single enzyme into a vat of substrate at initial concentration cs, and negligible product, will spend a fraction of time Pe unoccupied, and the rest (1-Pe) bound to substrate. These two times can be said to be nearly constant in time, so the enzyme converts the substrate at a constant rate.
We are now left with the final reaction of:
E + S <-> ES -> E + P
Forward rate of the first step = csk1
Reverse rate of the first step = k-1
Forward rate of the second step = k2
Michaelis constant = (k-1 + k2)/k1
vmax = k2Ce
v = vmax(cs / (Km + cs))