Get out a piece of paper to take notes on. Turn in your Take Home Exams If you didn’t write this down yesterday, write it down now: Equilibrium SRR: Pgs #589-595, 598-601, 568-570…Write your own guided outline. Get out a piece of paper to take notes on.

Test and “F” Information Overall Average: 52.74%...Failing 2nd: 54.53% 4th: 51.26% 5th: 58.95% 6th: 46.24% High: 112% Low: 3% “F”’s Was 12 and 13% Now 30 and 32.6%

I SEE LOTS OF RETAKE’S IN YOUR FUTURES!!! Must be done by March 17th Test corrections required

What is Equilibrium?

This is not Equilibrium?

Chemical Equilibrium in Nature: (The formation of stalagmites and Stalactites)

Equilibrium=

Chemical Equilibrium Consider the following reactions: H2(g) + O2(g)  H2O(g)…(1) and H2O(g)  H2(g) + O2(g) …(2) Reaction (2) is the reverse of reaction (1). At equilibrium the two opposing reactions occur at the same rate. Concentrations of chemical species do not change once equilibrium is established.

Expression for Equilibrium Constant Consider the following equilibrium system: wA + xB ⇄ yC + zD Kc = The numerical value of Kc is calculated using the concentrations of reactants and products that exist at equilibrium. Include only substances in the gas or aqueous phase. Solid’s and water’s concentrations do not change during a chemical reaction.

Expressions for Equilibrium Constants Examples: N2(g) + 3H2(g) ⇄ 2NH3(g); Kc = PCl5(g) ⇄ PCl3(g) + Cl2(g); Kc =

Interpretation of the Equilibrium Constant For K>1, we can state that the amount of product is greater than the amount of reactant. The equilibrium is shifts to the right. Likewise, for K<1, the amount of reactants is much greater than the products. The equilibrium shifts to the left.

Calculate K for the following reaction given the following equilibrium concentrations of PCl5, PCl3, and Cl2. PCl5(g) ⇄ PCl3(g) + Cl2(g) K = ?? (M): .200 .040 .080

What is the equilibrium concentration of Br2 if [HBr] = 0 What is the equilibrium concentration of Br2 if [HBr] = 0.35 M and [H2] = 0.22 M at equilibrium? H2(g) + Br2(g) ⇄ 2HBr(g) K = 62.5

Equilibrium constant is used to predict the direction of net reaction For a reaction of known Kc value, the direction of net reaction can be predicted by calculating the reaction quotient, Qc. Qc is called the reaction quotient, where for a reaction such as: aA + bB ⇄ cC + dD; Qc has the same expression as Kc , but Qc is calculated using concentrations that are not necessarily at equilibrium.

What does the reaction quotient tell us? If Qc = Kc,  the reaction is at equilibrium; If Qc < Kc,  the reaction is not at equilibrium and there’s a net forward reaction; If Qc > Kc,  the reaction is not at equilibrium and there’s a net reverse reaction.

Chemical Kinetics Chemical kinetics: the study of reaction rate Factors affecting reaction rate: Concentrations of reactants Catalyst Temperature Surface area of solid reactants or catalyst

Factors that Affect Reaction Rate Temperature Kinetic Theory: Increasing temperature means the molecules move faster. Concentrations of reactants More reactants mean more collisions if enough energy is present Catalysts Speed up reactions by lowering activation energy Surface area of a solid reactant Bread and Butter theory: more area for reactants to be in contact

D[A] = change in concentration of A over time period Dt Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = - D[A] Dt D[A] = change in concentration of A over time period Dt rate = D[B] Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative.

Calculating reaction rate The concentrations of N2O5 are 0.0124M and 0.0093 M at 600 and 1200 s after the reactants are mixed at the appropriate temperature. Evaluate the reaction rates for 2 N2O5 = 4 NO2 + O2 Solution: (0.0093 – 0.0124) 0.31e-2 M Decomposition rate of N2O5 = – ———————— = – —————— 1200 – 600 600 s = 5.2e-6 M s-1. Note however, rate of formation of NO2 = 1.02e-5 M s-1. rate of formation of O2 = 2.6e-6 M s-1.

Reaction Rates and Stoichiometry To generalize, for the reaction aA + bB cC + dD Reactants (decrease) Products (increase)

Le Châtelier’s Principle The Le Châtelier's principle states that: when factors that influence an equilibrium are altered, the equilibrium will shift to a new position that tends to minimize those changes. Factors that influence equilibrium: Concentration, temperature, and addition of a catalyst

The Effect of Changes in Concentration Consider the reaction: N2(g) + 3H2(g) ⇄ 2 NH3(g); If [N2] and/or [H2] is increased, Qc < Kc  a net forward reaction will occur to reach new equilibrium position. If [NH3] is increased, Qc > Kc, and a net reverse reaction will occur to come to new equilibrium position.

Concentration Changes Add more reactant  Shift to products Remove reactants  Shift to reactants Add more product  Shift to reactants Remove products  Shift to products

Temperature Changes Exothermic Reactions Consider heat as a product in exothermic reactions. Add heat  Shift to reactants Remove heat  Shift to products A + B = AB + Heat

Temperature Changes Endothermic Reactions Consider heat as a reactant in endothermic reactions. Add heat  Shift to products Remove heat  Shift to reactants A + B + heat = AB

Catalysts Adding a catalyst increases the rate at which equilibrium is achieved, but it does not change the composition of the equilibrium mixture.

S8(g) + 12O2(g)  8 SO3(g) + 808 kcals Given: S8(g) + 12O2(g)  8 SO3(g) + 808 kcals What will happen when …… Oxygen gas is added? The reaction vessel is cooled? Sulfur trioxide is removed? A catalyst is added to make it faster? Shifts to prodcuts  Shifts to Products – to replace heat Shift to products to replace it! No change!