CHEMICAL EQUILIBRIUM

 Up to this point we have mostly been considering reactions “to completion”, where all the reactants change into product. reversible  However, most reactions are reversible = occurs in both the forward and the reverse directions.

 N 2 (g) + 3H 2 (g)  2NH 3 (g) forward  N 2 (g) + 3H 2 (g)  2NH 3 (g) reverse N 2 (g) + 3H 2 (g) 2NH 3 (g) or N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)   Combined in one equation using double arrows

Chemical Equilibrium  A state in which the forward and reverse reactions balance each other and when the forward reaction proceeds at the same rate as the reverse reaction.  Concentrations of reactants and products are constant at equilibrium. (Constant ≠ equal)

Rate vs. Time

Question  Can we change the equilibrium position thereby increasing the amount of products in a reaction? adding stress  Yes – by adding stress to a system in equilibrium.

LE CHÂTELIER’S PRINCIPLE

Le Châtelier’s Principle stress is applied relieves the stress.  If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress.  Stress is anything that upsets equilibrium – changes in concentration, pressure, or temperature.

Le Châtelier’s Principle  If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress.  Concentration  Pressure  Temperature Note: Only temperature affects K. The larger the value of K, the more product at equilibrium.

Concentration  Measure of molarity (moles/L)  If you ↑ concentration of a reactant, equilibrium will shift toward the products.  If you ↓ concentration of a reactant, equilibrium will shift toward the reactants.

 CO(g) + 3H 2 (g) ↔ CH 4 (g) + H 2 O(g) CO(g) H 2 O(g)

Changes in Concentration A + B C + D  Increasing the concentration of “A” will shift the reaction to the right  we need to get rid of excess “A”  Decreasing the concentration of “C” will shift the reaction to the right  there is a deficit, so more “C” needs to be made

Temperature  Increasing the temperature shifts the reaction away from the side that contains the “heat”  Decreasing the temperature shifts the reaction toward the side that contains the “heat”

heat CO(g) + 3H 2 (g) ↔ CH 4 (g) + H 2 O(g) + heat

Sample Reaction heat + NH 4 Cl (s) NH 3 (g) + HCl (g) endothermic  In the above endothermic reaction, increasing the temperature will drive the reaction to the right (in other words, forward)

Pressure PV=nRT  Ideal Gas Law: PV=nRT  If ↑ P then ↑ n, which means more number of atoms.  If ↑ P, then the equilibrium will shift toward the side with fewer moles of gas.

 CO(g) + 3H 2 (g) ↔ CH 4 (g) + H 2 O(g) 4 moles of gas 2 moles of gas Note: If moles of gaseous reactant = moles of gaseous product, then no shift in equilibrium will occur from a change in pressure

Equilibrium constant  Kequilibrium constant  K is called the equilibrium constant. It is a ratio of the concentrations of products to the concentration of reactants.

Equilibrium Constant Expression aA + bB ↔ cC + dD K eq = [C] c [D] d [A] a [B] b A & B = molar [ ] of reactants C & D = molar [ ] of products Exponents a, b, c, and d = coefficients in the balanced equation.

Equilibrium Constant K eq > 1  If K eq > 1: products are favored at equilibrium  If K eq < 1  If K eq < 1: reactants are favored at equilibrium

Important!  Only substances that are gases and aqueous get factored into the equilibrium expression  Pure liquids and solids do not appear in the expression.

Example #1  Write the equilibrium expression for the following reaction: 2 CO (g) + O 2 (g) ↔ 2 CO 2 (g)

Answer  K eq = [CO 2 ] 2 / ([CO] 2 [O 2 ])

Example #2  Write the equilibrium expression for the following reaction: CO (g) + 3 H 2 (g) ↔ CH 4 (g) + H 2 O (g)

Answer  K eq = [CH 4 ][H 2 O] / ([CO][H 2 ] 3 )

Drill #14May 19 & 20, 2014  What are the 3 types of stress that can affect the equilibrium of a system?  Which states of matter get factored into an equilibrium expression? K eq > 1  If K eq > 1: are favored at equilibrium  If K eq < 1  If K eq < 1: are favored at equilibrium

Drill #16May 27 & 28, 2014  An equilibrium mixture of N 2, O 2, and NO gases at 1500 K is determined to consits of 6.4 x mole/L of N 2, 1.7 x mol/L of O 2 and 1.1 x mo/L of NO. What is the equilibrium constant for the system at this temperature?  Given: the molarity of the three gases  Unknown: K  Write the balanced equation  Substitute the given values for the [ ]s into the equilibrium expression.

Answer to Drill N 2 + O 2 2NO K = [NO] 2 [N 2 ][O 2 ] K = (1.1 x mo/L) 2 (6.4 x mole/L )(1.7 x mol/L ) K = 1.1 x The value of K is < 1 therefore reactants are favored at equilibrium.

Equilibrium Constant K eq > 1  If K eq > 1: products are favored at equilibrium  If K eq < 1  If K eq < 1: reactants are favored at equilibrium

Temperature Think of heat as a reactant or a product. CO(g) + 3H 2 (g) ↔ CH 4 (g) + H 2 O(g) + heat Is this Exothermic or Endothermic? Reversible reactions are exothermic (energy as heat shown on product side) in one direction and endothermic (energy as heat shown on reactant side) in the other.

Drill #17May 28 & 29, Concentration Changed 2A (g) + 3B (g) 2C (g) + D (g) Action Effect?  Increase [A] Shift to the ?  Increase [C] Shift to the ?  Decrease [B] Shift to the ?  Decrease [C] Shift to the ?

Drill #17May 28 & 29, Concentration Changed 2A (g) + 3B (g) 2C (g) + D (g) Action Effect?  Increase [A] Shift to the right  Increase [C] Shift to the left  Decrease [B] Shift to the left  Decrease [C] Shift to the right

Drill #17May 28 & 29, Volume Changed 2A (g) + 3B (g) 2C (g) + D (g) Action Effect?  Volume decreased Shifts to the ?  Volume increased Shifts to the ?

Drill #17May 28 & 29, Volume Changed 2A (g) + 3B (g) 2C (g) + D (g) Action Effect?  Volume decreased Shifts to right (side with fewest moles)  Volume increased Shifts to left (side with most moles)

Drill #17May 28 & 29, Temp. Changed for Endothermic Rxn. A (g) + B (g) C (g) + D (g) Action Effect?  Increase temp Shift to the ?  Decrease temp Shift to the ?

Drill #17May 28 & 29, Temp. Changed for Endothermic Rxn. A (g) + B (g) C (g) + D (g) Action Effect?  Increase temp Shifts to right (endothermic rxn.)  Decrease temp Shifts to left (exothermic rxn.)

Agenda  Quiz  Finish Notes on Ice Table  Finish Worksheet and new worksheet on Ice Table  Chemical Energy notes

INTRO TO ICE BOXES

Equilibrium Problems  The equilibrium constant can also be used to calculate equilibrium concentrations from initial concentrations  Use an ICE box!  To start with, we will use a generic equation  A (g) + B (g)  2 C (g)  [A] =.100 M and [B] =.100 M

A (g) + B (g)  2 C (g) AB 2 C Initial.100 M 0 Change -x-x+2x Equilibrium.100 – x 2x A and B are assigned –x values because they are being reacted, while C gets a +x because products are being formed. The 2x comes from the coefficient in front of the C

ICE BOX PRACTICE PROBLEMS We will complete in class examples and then you will try on your own!

Drill #18June 2 & 3, 2014  What is energy?  What are the two different types of energy?

What is Energy? The ability to do work or produce heat. Potential Energy – stored energy. Kinetic Energy – energy of motion.

Agenda  Equilibrium Constant for Acids and Bases  Worksheets  Chemical Energy Diagrams  Final Exam Review Project

Strong Acids/Bases vs. Weak Acids/Bases  STRONG  Ionize almost 100% in water (react to completion)  WEAK  Do not ionize completely, achieve an equilibrium

HOW CAN I TELL THE DIFFERENCE BETWEEN STRONG AND WEAK ACIDS?

Two different ways:  Study the list of strong acids and bases  Look at the arrow in an acid/base balanced equation  If arrow is one direction  it is STRONG  If arrow is two directions   it is WEAK

Introduction  Strong Acid Example  HNO 3(aq) + H 2 O (l)  H 3 O + (aq) + NO 3 - (aq)  Weak Acid Example  CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO - (aq)  Strong Base Example:  NaOH (aq) + H 2 O (l)  Na + (aq) + OH - (aq)  Weak Base Example:  NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq)

INSTEAD OF USING KEQ WE USE A DIFFERENT CONSTANT FOR WEAK ACIDS AND BASES

Ionization Constant of a Weak Acid  K a – acid ionization constant  Ka varies at different temperatures  Generic Example:

Weak Acid Example  CH 3 COOH (aq) + H 2 O (l)  H 3 O + (aq) + CH 3 COO - (aq)  Weak acids have a Ka < 1  Leads to small [H 3 O + ]  pH of weak acids is calculated by solving the expression for [H 3 O + ] at equilibrium and then taking the –log[H 3 O + ]

Ionization Constant of a Weak Base  K b – base ionization constant  Weak bases = K b < 1  Leads to a small [OH - ]  To find pOH take -log[OH - ]

Weak Acid Example  HA (aq) + H 2 O (l)  H 3 O + (aq) + A - (aq)  Initially, you have 1.00 M HA. Calculate the equilibrium concentrations of HA, H 3 O +, A -, and calculate the pH. K a = 1.80 x  Use an ICE table!

Weak Acid Example HA H3O+H3O+H3O+H3O+ A-A-A-A- I C-x+x+x E 1.00 – x +x+x

Weak Acid Example  Write K a expression:  Assume x is small because K a is very small, therefore x can be ignored K a = 1.80 x = [H 3 O + ][A - ] = x 2 [HA] 1.00-x K a = 1.80 x = x

Weak Acid Example  x = [H 3 O + ] = [A - ] = 4.20 x M  pH = - log [H 3 O + ] = 2.37 K a = 1.80 x = x

Weak Base Example  NH 3(aq) + H 2 O (l)  NH 4 + (aq) + OH - (aq)  Initially, you have a concentration of M NH 3. Find the concentrations at equilibrium. Calculate the pH. K b = 1.80 x  Use an ICE table!

Weak Base Example NH 3 NH 4 + OH - I C-x+x+x E – x +x+x

Weak Base Example  Write K b expression:  Assume x is small because K b is very small, therefore x can be ignored K b = 1.80 x = [NH 4 ][OH - ] = x 2 [NH 3 ] x K b = 1.80 x = x

Weak Base Example  x = [NH 4 + ] = [OH - ] = 4.20 x M  pOH = - log [OH - ] = 3.37  pH + pOH = 14  Therefore, = 10.6 = pH K b = 1.80 x = x

DIFFERENT TYPES OF EQUILIBRIUM

Concentration Equilibrium K c (or K eq )  nA + mB ↔ xC + yD  K c = [C] x [D] y [A] n [B] m

 Remember…equilibrium is where the rates of forward and reverse reactions are the same. It means that the concentrations do not change, NOT that they are identical.

 Because equilibrium expressions have to do with concentration (in molarity) we do not include items that are not in solution so NO LIQUID or SOLID states! They are in excess so can be ignored.

Acid Equilibrium  Acid + H 2 O ↔ H 3 O + + Acid Ion or HA + H 2 O ↔ H 3 O + + A - K a = [H 3 O + ][A - ] [HA][H 2 O] [HA] Because water is a solvent (liquid) and its conc. greatly exceeds the acid, we can assume that the conc. of water does not change.

Base Equilibrium  Base + H 2 O ↔ OH - + Base Ion or B + H 2 O ↔ OH - + HB + K b = [OH - ][HB + ] [B][H 2 O] [B]

Acid Strength A stronger acid will react completely to form ions (strong electrolytes) and hydronium ions (H 3 O + ) in water (dissociation) Hydrochloric acid (all HCL molecules are ionized into hydronium and chloride ions) HCl + H 2 O  H 3 O + + Cl - Acetic Acid (not all A.A. molecules are ionized into hydronium and ions) CH 3 COOH + H 2 O  H 3 O + + CH 3 COO -

Examples of Strong & Weak Acids StrongWeak HCl CH 3 COOH HBr H 2 CO 3 H 2 SO 4 HClO

Base Strength  A strong base has the strongest affinity for H + ions and dissociates entirely into metal ions and OH -.  For ex. Calcium oxide (CaO); oxygen strongly attracts H + ions.  Strong bases: CaO, NaOH, KOH  Weak base: ammonia (NH 3 )

Sample Problem  Write an equation to show the dissociation of hydrochloric acid in water.

Answer HCl + H 2 O → H 3 O + + Cl −  HCl is a STRONG ACID

PropertyStrong AcidWeak Acid K a valueK a is largeK a is small Position of equilibrium Far to the right (a lot of dissociation) Far to the left (little dissociation) [H+] compared to original [HA] [H+]≈[HA] o [H+]<<[HA] o Strength of conjugate base A- is much weaker A- is much stronger

Graphic Representation of the Behavior of Acids of Different Strengths in Aqueous Solution

Solubility Equilibrium  Salt (s) ↔ Cation (aq) + Anion (aq)  Solids are not included in equilibrium equations! So… K sp = [Cation][Anion]

Solubility Equilibrium Example  CaF 2 (s) ↔ Ca +2 (aq) + 2F - (aq)  Solids are not included in equilibrium equations! So…  K sp = [Ca +2 ][F - ] 2

Drill #15May22 & 23, Write the equilibrium expression for the reaction: H 2 (g) + I 2 (s)  2HI (g) 2. How would the following shift the equilibrium in the equation (forward, reverse or no change): H + (aq) + Cl - (aq) ↔ HCl (aq) kJ a) Increasing temperature b) Increasing pressure c) Adding Cl - d) Removing HCl

Answers:May22 & 23, Write the equilibrium expression for the reaction: H 2 (g) + I 2 (s)  2HI (g) K eq = [HI] 2 [H 2 ] 2. How would the following shift the equilibrium in the equation (forward, reverse or no change): H + (aq) + Cl - (aq) ↔ HCl (aq) kJ a) Increasing temperature (reverse) b) Increasing pressure (forward) c) Adding Cl - (forward) d) Remove HCl (forward)