Chemical Equilibrium………..  Until now, we’ve treated reactions as though they can only go in one direction….with all of the reactants turning into products…  Most reactions that we have witnessed appear this way…  Are all reactions like this?  Is it possible to reverse a reaction after it is completed?  Can we turn CO 2 and H 2 O back into gasoline and O 2 after it burns?  2 C 8 H O 2 → 16 CO H 2 O  16 CO H 2 O → 2 C 8 H O 2 ?

Chemical equilibrium Most chemical reactions are reversible! Most chemical reactions are reversible! Reactions will continue to occur until they finally maintain a ratio of products versus reactants that will not change Reactions will continue to occur until they finally maintain a ratio of products versus reactants that will not change This ratio is called the Equilibrium Constant Expression, or K eq, or K c, and is equal to the final product concentration over the final reactant concentration This ratio is called the Equilibrium Constant Expression, or K eq, or K c, and is equal to the final product concentration over the final reactant concentration K = Products K = Products Reactants Reactants

Let’s take the following chemical reaction: Let’s take the following chemical reaction: aA + bB cC + dD When a chemical reaction reached equilibria, the Equilibrium Constant Expression, or K eq, is: K eq = [C] c [D] d [A] a [B] b [A] a [B] b

Chemical Equilibrium (continued) A, B, C, and D represent the concentration of each chemical at the end of the reaction This concentration is measured in Molarity, or M Once the reaction reaches equilibrium, the concentrations of these molecules WILL NOT CHANGE K eq = [C] c [D] d [A] a [B] b [A] a [B] b

Let’s examine the following reaction…… This reaction is reversible At equilibrium, the constant, or K eq, is equal to: K eq = [NO 2 ] 2 [N 2 O 4 ] 1 BUT WHAT DOES THIS TELL US? N 2 O 4(g) 2 NO 2(g) at 25 0 C N 2 O 4(g) 2 NO 2(g) at 25 0 C

It tells us whether we have more products or reactants at equilibrium!!!! Looking at the same reaction: N 2 O 4(g) 2 NO 2(g) at 25 0 C If the concentration of N 2 O 4(g) at equilibrium was.0445 M and the concentration of NO 2(g) was.0161 M, then: K eq = [NO 2 ] 2 = [.0161] 2 = = 582 [N 2 O 4 ] [.0445] 100,000 This is the same as: Is this a exothermic or endothermic reaction….and why?

Why is this important in the real world?

IT IS IMPORTANT BECAUSE: The K eq tells us whether we have MORE PRODUCTS OR REACTANTS AT EQUILIBRIUM! The K eq tells us whether we have MORE PRODUCTS OR REACTANTS AT EQUILIBRIUM! Remember, K eq are temperature specific - at higher or lower temperatures, the ratios will be different, because temperature affects a reaction! Remember, K eq are temperature specific - at higher or lower temperatures, the ratios will be different, because temperature affects a reaction!

Suppose we have this reaction: CO (g) + H 2 O (g) H 2(g) + CO 2(g) The K eq = 5.10 at C. 1 This means that products are favored over reactants almost 5 to 1!! Let’s say we are at some point in time, NOT AT EQUILIBRIUM and we measure the concentrations of each of the molecules in the reaction……

Is it possible to predict if a reaction is at equilibrium? CO (g) + H 2 O (g) H 2(g) + CO 2(g) The K eq = 5.10 at C. The concentrations are [CO] =.15 M, The concentrations are [CO] =.15 M, [H 2 O] =.25 M, [H 2 ] =.42 M, and [CO 2 ] =.37 M. Can you set up a ratio of concentrations? Can you set up a ratio of concentrations? DO IT NOW!!!!!

Does your ratio look like this? Q = [.42] 1 [.37] 1 [.15] 1 [.25] 1 [.15] 1 [.25] 1 This is called the REACTION QUOTIENT, OR Q This is called the REACTION QUOTIENT, OR Q It is set up the same way as K eq, but it is not K eq – It is set up the same way as K eq, but it is not K eq – Why is it not the same as K eq ? Why is it not the same as K eq ?

What does Q tell you??? The reaction quotient, or Q, tells you where you are in the reaction. You compare it to your K eq ! The reaction quotient, or Q, tells you where you are in the reaction. You compare it to your K eq ! Q = [.42] 1 [.37] 1 = 4.14 [.15] 1 [.25] 1 1 [.15] 1 [.25] 1 1 Q = 4.14, which is less than K eq = 5.10

Q = 4.14, which is less than K eq = If Q < K eq, then the reaction will proceed toward the products to balance the ratio! If Q < K eq, then the reaction will proceed toward the products to balance the ratio! If Q > K eq, then the reaction will proceed toward the reactants to balance the ratio! If Q > K eq, then the reaction will proceed toward the reactants to balance the ratio! It Q = K eq, then the reaction is…….. It Q = K eq, then the reaction is…….. AT EQUILIBRIUM!!!!! That means it is possible for a reaction to reverse itself and turn product back into reactant! That means it is possible for a reaction to reverse itself and turn product back into reactant!

AT EQUILIBRIUM… - The forward reaction, where molecules A and B are forming molecules C and D A + B > C + D - Occurs at the same rate as the reverse reaction, the reaction where molecules C and D are forming molecules A and B C + D > A + B That is how the ratio at equilibrium never changes!

Chemical Equilibrium does not mean: That there are equal concentrations of reactant and product when the reaction reaches equilibrium! That there are equal concentrations of reactant and product when the reaction reaches equilibrium! It simply means that there is eventually a ratio reached that will not change! It simply means that there is eventually a ratio reached that will not change!

Let’s watch a computer simulation of equilibrium in action!

Le Chatelier’s Principle: If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the molecules, the system will shift its equilibrium position to counteract the disturbance…… If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the molecules, the system will shift its equilibrium position to counteract the disturbance…… WHAT DOES THAT MEAN….?

Fritz Haber was a German Scientist who utilized LaChatelier’s Principle in making ammonia for explosives in World War I: N 2(g) + 3H 2(g) > 2 NH 3(g) K eq = at C and 200 atm At equilibrium, products are favored, but how do we get the reaction to continue to make NH 3, or ammonia?

N 2(g) + 3H 2(g) 2 NH 3(g) By adding more N 2(g) or H 2(g), the reaction would…..? By adding more N 2(g) or H 2(g), the reaction would…..? Shift to the right and make more ammonia - but this requires more reactant and more money! Shift to the right and make more ammonia - but this requires more reactant and more money! By adding more NH 3(g), the reaction would….? By adding more NH 3(g), the reaction would….? Shift to the left and form more N 2(g) and H 2(g), which is NOT what we want! Shift to the left and form more N 2(g) and H 2(g), which is NOT what we want!

N 2(g) + 3H 2(g) 2 NH 3(g) By removing NH 3(g)….. By removing NH 3(g)….. The reaction shifts to the right, N 2(g) and H 2(g) hurrying to create equilibrium again and create more NH 3(g). The reaction shifts to the right, N 2(g) and H 2(g) hurrying to create equilibrium again and create more NH 3(g). This causes even more NH 3(g) to be produced, driving the reaction to completion and producing the maximum amount of ammonia! This causes even more NH 3(g) to be produced, driving the reaction to completion and producing the maximum amount of ammonia! The ammonia is removed by liquefying it at C The ammonia is removed by liquefying it at C The reaction will always maintain its K eq = 9.60 The reaction will always maintain its K eq =

K c, or K eq, only applies to chemicals reacting in the solid, liquid, or gaseous state. It does not apply to aqueous solutions (reactions that occur in water). K sp deals with aqueous reactions, or reactions that occur in water!