[Shadow]

In a reaction:
A+B  C
B+C D
D+E C+F

Steady state condition can be applied to D and somehow to C. Why to C if it is a product in the reaction?

[Corribus]

Anything that is consumed and destroyed during the course of a reaction can be subject to the steady-state approximation. The only reason you are calling it a product is because it is generated along with F.

[Shadow]

C is not an intermediate, it appears in the sum reaction: A+2B+E C+F

[Corribus]

If you're going to add them like that, you have to include the reverse of the first reaction, since it's expressed as an equilibrium.
But I wouldn't add them like that, anyway. It's not helpful to look at such a complicated mechanism as a straightforward combination of three linear reactions.

[curiouscat]

C is not an intermediate, it appears in the sum reaction: A+2B+E C+F

I think you are right. SSA can only be applied to D here (if at all) but not to C.

[curiouscat]

If you're going to add them like that, you have to include the reverse of the first reaction, since it's expressed as an equilibrium.
But I wouldn't add them like that, anyway. It's not helpful to look at such a complicated mechanism as a straightforward combination of three linear reactions.

@Corribus

I think what he means is that this reaction mechanism is a given. As well as the overall stoichiometry.

If so, the SSA can be applied to D at best but not to C.

[Shadow]

Actually it can be applied to C, but that's why I asked my original question.

[Corribus]

You can apply it to anything you want. The question is whether it is kosher to do so. All the steady state approximation is, essentially, is assuming that the rate of change of the species over time is constant. It doesn't make sense to apply this to a species that is ONLY consumed or ONLY produced during the course of a reaction. However any species that is both consumed and produced during the course of the reaction could, in principle, have an invariant concentration as a function of time, provided the total rate of production is equal to the total rate of consumption. In the case of C, it is being consumed and produced during the reaction. Any "C" that is being produced along with F as a "final step" (although I discourage you from thinking about it that way), will subsequently be re-consumed by earlier steps in the mechanism, until equilibrium has been reached or you run out of reactants. There is no problem applying the SSA here - whether or not doing so will yield a quality prediction of the course of the reaction will need to be verified experimentally.

[curiouscat]

However any species that is both consumed and produced during the course of the reaction could, in principle, have an invariant concentration as a function of time, provided the total rate of production is equal to the total rate of consumption.

I disagree. I think any species that appears in the overall equation, either as a reactant or as a product, can never be applied the SSA.

I may be wrong. But if so I'd love to see an example where it makes sense to apply the SSA to a species that is a reactant / product and not strictly an intermediate.

[Shadow]

In this year IChO: (NiL)

[rwiew]

Yep, good example above. Also, where is this idea that C IS a product coming from? What in that reaction scheme is telling you that you will see C in the reaction mixture after full consumption of A, B and E? The problem with this question is the sum equation must be established experimentally by looking at what's going in and out and that is not given here, the mechanism must be made to fit the sum equation, not the other way around. With the information we have I might as well say that I will take the second elementary step as 2B + 2C  2D, the sum equation then becomes A + 3B + E  D + F and now you have D as the product and you'll say you can't use the SSA for it. Do you see what I mean? All these are, are letters on paper (or screen...) and you can do a lot of things with them, of which all but one (or zero if you're proposing a wrong mechanism) will not be representative of the experimental outcome. In conclusion two things: a) if we don't know the experimental sum equation, we can't say whether C or D is an observed product of the reaction and hence argue for SSA based on that b) SSA is just an approximation, it needs to be verified experimentally as was said already - but with the information we have here you can use it for C as much as you can for D.

[Shadow]

Nicely said. I thought that the equations can be summed up to an overall equation.