For an elementary reaction like $$\ce{A + B -> AB},$$ $$K_{c} = \frac{\ce{[AB]}}{\ce{[A][B]}}$$ and $$\frac{\mathrm{d}\ce{[AB]}}{\mathrm{d}t} = k\ce{[A][B]}.$$ If I let the reaction reach equilibrium, and at this point, without altering temperature, suck away $\ce{A}$ and add $\ce{B}$ at the same rate (so $-\mathrm{d}\ce{[A]}/\mathrm{d}t = \mathrm{d}\ce{[B]}/\mathrm{d}t$) and alter both concentrations slightly, once I am finished, will the system still be in equilibrium or try to shift?

What if the reaction is not an elementary reaction such that $\frac{\mathrm{d}\ce{[AB]}}{\mathrm{d}t} = k\ce{[A]}^n\ce{[B]}^m$?


A single "elementary reaction" does not involve an equilibrium.

There must be a forward reaction and a reverse reaction to have an equilibrium. This is at least two elementary reactions.

Considering an equilbrium reaction: Just as a square encloses the most area for a rectangle of given perimeter, if the sum [A] + [B] is kept constant, [A][B] is maximal when [A] = [B]. Deviation from this maximum will shift the equilibrium away from the product and toward the reactants. In other words, if the system is at equilibrium with [A] = [B], and [A] is decreased and [B] increased, keeping [A] + [B] constant, there will be net reverse reaction until equilibrium is again reached, and [AB] at equilibrium will be less than orginally.

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  • $\begingroup$ What if I maintain the product [A][B] to be constant while decreasing [A] and increasing [B]. After I stop, will the final state be at equilibrium? In both cases (maintaining [A][B] and [A]+[B] constant), does the new equilibrium state differ than the original? $\endgroup$ – Yandle Nov 26 '14 at 6:21
  • $\begingroup$ [A][B]=constant1; [A] + [B] = constant2. There are two equations and two unknowns, only two solutions at most (the intersections of a hyperbola and a line). In other words, you cannot decrease [A] and increase [B] while maintaining [A][B] and [A]+[B] constant. It is mathematically impossible. $\endgroup$ – DavePhD Nov 26 '14 at 12:03
  • $\begingroup$ @Yandle Instead, if you just keep [A][B] constant, decrease [A] and increase [B], without keeping [A]+[B] constant, the system will still be in equilibrium as long as [AB] is not changed. The decrease in [A] and increase in [B] would need to be in a manner that avoids changing the volume of the system, because changing the volume of the system with a constant amount of AB would change [AB]. $\endgroup$ – DavePhD Nov 26 '14 at 15:34

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