[Peng Zhong]

Hi Dear chemical experts,

If species 1 and species 2 are ideal gases and are reacting with each other with slow to medium reaction rate,

A<---->B

such that the mole fraction x1 and x2 will try to reach the chemical equilibrium composition. If I were to separate this mixture that is already in chemical equilibrium by inputting mechanical work, is the required mimimum work governed still by the generic separation work equation (which can be derived by entropy of mixing),

W_rev = T*R*(x1*ln(x1)+x2*ln(x2))

Thank you very much.

Peng Zhong

[jeffmoonchop]

If they are reacting with each other wouldn't the equation be A + B  C? At equilibrium you would be separating 3 components, the reactants and the product. Or am I not understanding you?

[Peng Zhong]

A and B are NO2 and N2O4 for example, they form each other without the 3rd species

[mjc123]

That would be a reaction of the form 2A  B. Reaction would change the number of moles of gas, which would alter the pressure or volume. But if the reaction really is slow compared to the mechanical separation, your expression ought to be right. But is it slow? I have seen it demonstrated in the classroom; the equilibrium responds quite rapidly to changes of volume or pressure.

[Peng Zhong]

Let's just assume my mechanical separation process is fast compared to the equilibrium reaction. I think that separation work equation might still work in that case, but I am not sure.