Chemical Equilibrium UNIT 11 (PART 2) Pb 2+ (aq) + 2 Cl - (aq)  PbCl 2 (s)

Warm Up: 1.What is a forward reaction? What is a reverse reaction? 2.What does it mean for a reaction to be “reversible”? 3.Are all reactions reversible? 2

What is equilibrium? Definition (dictionary.com): a state of rest or balance due to the equal action of opposing forces Chemical Equilibrium: A process where a forward and reverse reaction occur at equal rates ***Not all chemical reactions are reversible!!!

General Characteristics of Equilibrium 1.DYNAMIC (molecules in constant motion) 2.REVERSIBLE 3.Can be approached from either direction (reaction can run in the forward direction or the reverse direction) 1.DYNAMIC (molecules in constant motion) 2.REVERSIBLE 3.Can be approached from either direction (reaction can run in the forward direction or the reverse direction)

Characteristics of Dynamic Equilibrium After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained.

6 Equilibrium Activity Directions Purpose: take a closer look at equilibrium systems in order to understand how equilibrium is reached and the meaning of the equilibrium constant. Work in Groups of 3 or4 Member 1 and 2 – In charge of “Reaction” Member 3 and 4 – In charge of Data Collection/Calculations Be sure to count the RED and GREEN chips – You need 30 of each! Pull 3 chips at a time:R + G  B +Y At least 1 R and 1G… replace with 1 B and 1 Y (put other back in) At least 1 B and 1 Y… replace with 1 R and 1G Neither combination… Record as no reaction Continue until equilibrium is reached (at least 8-10 reactions at equilibrium) We will complete the DATA ANALYSIS tomorrow!

Closing: How do you know when equilibrium was reached in the activity? (Think about: How did you know to stop picking up chips?) 7

Graphing Dynamic Equilibrium Equilibrium achieved when product and reactant concentrations remain constant!!

Warm Up: Predict what the graph of the reaction rate of the reactants and products would look like if plotted vs. time. 9

The Equilibrium Expression, K eq K eq = equilibrium constant (for a given T) Brackets "[ ]" = concentration (molarity) "a, b, c, and d" = coefficients from balanced equation The "c" in K c = concentration (K c = a special K eq based on concentration)

There are two cases when a species is not shown in the equilibrium expression: #1: SOLIDS – (s) after the formula #2: pure LIQUIDS – (l) after the formula

Equilibrium Position Reaction is product favored (Equilibrium [P] is greater than equilibrium [R]) Reaction is reactant favored (Equilibrium [R] is greater than equilibrium [P]) 12 At equilibrium, the rate of the forward and reverse reactions are equal… NOT the concentration of the reactants vs. products!

Example #7 Write the equilibrium expression for the oxidation- reduction reaction occurring between iron(III) chloride and tin(II) chloride: 2 FeCl 3 (aq) + SnCl 2 (aq)  2 FeCl 2 (aq) + SnCl 4 (aq)

Example #8 Write the equilibrium expression for the replacement of silver ions by copper: Cu (s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag (s)

15 Warm Up: 2. Derive the chemical equation from the following equilibrium constant expression.

“If a system at equilibrium is stressed, the system tends to shift its equilibrium position to counter the effect of the stress.” Le Chatelier’s Principle

How does a “stress” influence equilibrium? The impact of addition of reactants on reaction rate

The Seesaw Analogy

“Stresses” to a System 1)Changes in amount of species 2)Changes in pressure or volume 3)Changes in temperature 4)Adding an inert substance 5)Adding a catalyst

“Stresses” (factors) that can cause changes at equilibrium 1) Changes in amount of species o Add reactant; system shifts to theRIGHT! (produces more products) o Add product; system shifts to theLEFT! (produces more reactants) o Remove reactant; system shifts toLEFT! o Remove product; system shifts toRIGHT!

Example #9: Predict the direction of shift based on the following concentration changes for the reaction: CH 4 (g) + 2S 2 (g)  CS 2 (g) + 2H 2 S(g) (A) Some S 2 (g) is added. (B) Some CS 2 (g) is added. (C) Some H 2 S(g) is removed. (D) Some argon gas (an inert gas) is added.

“Stresses” (factors) that can cause changes at equilibrium 2) Changes in pressure or volume o If P goes down (same as V goes up), system shifts to increased # of moles of gas o If P goes up (same as V goes down), system shifts to decreased # of moles of gas

Example #10: Predict the effect of increasing pressure (decreasing volume) on each of the following reactions. (a) CH 4 (g) + 2S 2 (g)  CS 2 (g) + 2H 2 S(g) (b) H 2 (g) + Br 2 (g)  2HBr(g) (c) CO 2 (g) + C(s)  2CO(g) (d) PCl 5 (g)  PCl 3 (g) + Cl 2 (g)

“Stresses” (factors) that can cause changes at equilibrium 3) Changes in temperature o Write heat as a product (exothermic) or reactant (endothermic) o System shifts to get rid of added heat: o System shifts LEFT for exo. reactions as T goes up o System shifts RIGHT for endo. reactions as T goes up

Example #11: Predict the effect of increasing temperature on each of the following reactions: (a) CO (g) + 3 H 2 (g)  CH 4 (g) + H 2 O (g)ΔH < 0 (b) CO 2 (g) + C (s)  2 CO (g) ΔH > 0 (c) 4 NH 3 (g) + 5 O 2 (g)  4 NO (g) + 6 H 2 O (g) EXO. (d) 2 H 2 O (g)  2 H 2 (g) + O 2 (g) ENDO.

“Stresses” (factors) that can cause changes at equilibrium 4) Adding an inert substance o If a substance is NOT in the reaction (or in the K eq expression) any changes will have NO EFFECT on equilibrium! Ex.) If CO 2 is added to the system below, there would be no effect on the equilibrium. H 2 (g) + Br 2 (g)  2 HBr (g)

“Stresses” (factors) that can cause changes at equilibrium 5) Adding a catalyst o What do catalysts do? They increase the rate of the reaction! o Adding a catalyst will not affect equilibrium. It only changes the rate at which you reach equilibrium.

Example #12: How can the reaction below be shifted to the right? List all possibilities! CO(g) + 2 H 2 (g) CH 3 OH (g) ΔH = +

Example #13: How can the reaction below be shifted to the right? This process (the Haber process) is used in the industry to produce ammonia. List all possibilities! N 2 (g) + 3H 2 (g)  2NH 3 (g) ∆H = negative

N 2 (g) + 3 H 2 (g)  2 NH 3 (g) + heat N 2 (g) + 3 H 2 (g)  2 NH 3 (g) + heat K = 3.5 x 10 8 at 298 K K = 3.5 x 10 8 at 298 K Haber-Bosch Process for NH 3