Question

About the steady state approximation ((BIO)chemical kinetics)
@shrutipande9

Answer 1

ohh yes please....see i just need to clear d concept....i dnt want to go into d "chemistry" details...

Answer 2

Ok I am going to summarize the principle in one sentence:
"The steady-state principle assumes a constant but dynamic concentration of reaction intermediates"

Answer 3

Ok what does mean:
Consider a reaction
\[\huge \sf A \overset{k_{1}}{\rightarrow} B \overset{k_{2}}{\rightarrow} C\]
From a avg reaction we could imagine the cures to look like this:

Ok so what we see is that the intermediate is being created and then faints away as C is being produced. Lets now assume that \(\Large \sf k_{2}>k_{1} \) we then get to see for a enzymatic reaction:

We see that B remains constant after a little time (we call that pre-steady-state). Mathematical we get to say:
\[\Large \frac{ d[B] }{ dt }=k_{1} \times [A]-k_{2} \times [B]=0\]
Where the differential is an expression for the rate.
By this approximation we get to see that:
\[\Large k_{1} \times [A]=k_{2} \times [B]\]
This expression is very useful to evaluate the rate equations!