Consider the following system under equilibrium: \[ \underbrace{\ce{Fe^{3+}(aq)}}_{\text{colorless}} + \underbrace{ \ce{SCN^{-}(aq)}}_{\text{colorless}} \rightleftharpoons \underbrace{\ce{FeSCN^{2+}(aq)}}_{\text{red}}\] If more \(Fe^{3+}\) is added to the reaction, what will happen? According to Le Chatelier's Principle, the system will react to minimize the stress. Since Fe 3+ is on the reactant side of this reaction, the rate of the forward reaction will increase in order to "use up" the additional reactant. This will cause the equilibrium to shift to the right , producing more FeSCN 2+ . For this particular reaction, we will be able to see that this has happened, as the solution will become a darker red color. There are a few different ways to state what happens here when more Fe 3+ is added, all of which have the same meaning: equilibrium shifts to the right equilibrium shifts to the product side the forward reaction is favored What changes does this cause in the concentrations of the reaction participants? 0 1 \(\ce{Fe^{3+}}\) Since this is what was added to cause the stress, the concentration of \(\ce{Fe^{3+}}\) will increase. (A shorthand way to indicate this: \(\ce{[Fe]^{3+}\: \uparrow}\) (Reminder: the square brackets represent "concentration") \(\ce{SCN^{-}(aq)}\) Equilibrium will shift to the right, which will use up the reactants. The concentration of \(\ce{SCN^{-}(aq)}\) will decrease \(\ce{[SCN]^{-}\: \downarrow}\) as the rate of the forward reaction increases. \(\ce{FeSCN^{2+}}\) When the forward reaction rate increases, more products are produced, and the concentration of \(\ce{FeSCN^{2+}}\) will increase. \(\ce{[FeSCN]^{2+}} \uparrow \) How about the value of K eq ? Notice that the concentration of some reaction participants have increased, while others have decreased. Once equilibrium has re-established itself, the value of K eq will be unchanged. The value of K eq does not change when changes in concentration cause a shift in equilibrium. What if more FeSCN 2+ is added? Again, equilibrium will shift to use up the added substance. In this case, equilibrium will shift to favor the reverse reaction, since the reverse reaction will use up the additional FeSCN 2+ . equilibrium shifts to the left equilibrium shifts to the reactant side the reverse reaction is favored How do the concentrations of reaction participants change? 0 1 \(\ce{Fe^{3+}}\) \(\ce{[Fe]^{3+}\: \uparrow}\) as the reverse reaction is favored \(\ce{SCN^{-}(aq)}\) \(\ce{[SCN]^{-}\: \uparrow}\) as the reverse reaction is favored \(\ce{FeSCN^{2+}}\) \(\ce{[FeSCN]^{2+}} \uparrow \) because this is the substance that was added Concentration can also be changed by removing a substance from the reaction. This is often accomplished by adding another substance that reacts (in a side reaction) with something already in the reaction. Let's remove SCN - from the system (perhaps by adding some Pb 2+ ions—the lead(II) ions will form a precipitate with SCN - , removing them from the solution). What will happen now? Equilibrium will shift to replace SCN - —the reverse reaction will be favored because that is the direction that produces more SCN - . equilibrium shifts to the left equilibrium shifts to the reactant side the reverse reaction is favored How do the concentrations of reaction participants change? 0 1 \(\ce{Fe^{3+}}\) \(\ce{[Fe]^{3+}\: \uparrow}\) as the reverse reaction is favored \(\ce{SCN^{-}}\) \(\ce{[SCN]^{-}\: \uparrow}\) as the reverse reaction is favored (but also ↓ because it was removed) \(\ce{FeSCN^{2+}}\) \(\ce{[FeSCN]^{2+}} \uparrow \) because this is the substance that was added