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Reactions are reversibleReactions are reversible A + B C + D ( forward) C + D A + B (reverse) Initially there is only A and B so only the forward reaction is possible As C and D build up, the reverse reaction speeds up while the forward reaction slows down. Eventually the rates are equal
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Reaction Rate Time Forward Reaction Reverse reaction Equilibrium
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What is equal at Equilibrium? Rates are equal Concentrations are not. Rates are determined by concentrations and activation energy. The concentrations do not change at equilibrium. or if the reaction is verrrry slooooow.
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The units for KThe units for K Are determined by the various powers and units of concentrations. They depend on the reaction.
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K is CONSTANTK is CONSTANT At any temperature. Temperature affects rate. The equilibrium concentrations don’t have to be the same only K. Equilibrium position is a set of concentrations at equilibrium. There are an unlimited number.
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Calculate KCalculate K N 2 + 3H 2 3NH 3 Initial At Equilibrium [N 2 ] 0 =1.000 M [N 2 ] = 0.921M [H 2 ] 0 =1.000 M [H 2 ] = 0.763M [NH 3 ] 0 =0 M [NH 3 ] = 0.157M
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Equilibrium and PressureEquilibrium and Pressure Some reactions are gaseous PV = nRT P = (n/V)RT P = CRT C is a concentration in moles/Liter C = P/RT
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Equilibrium and PressureEquilibrium and Pressure 2SO 2 (g) + O 2 (g) 2SO 3 (g) Kp = (P SO3 ) 2 (P SO2 ) 2 (P O2 ) K = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]
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The Reaction QuotientThe Reaction Quotient Tells you the directing the reaction will go to reach equilibrium Calculated the same as the equilibrium constant, but for a system not at equilibrium Q = [Products] coefficient [Reactants] coefficient Compare value to equilibrium constant
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What Q tells usWhat Q tells us If Q<K : Not enough products : Shift to right If Q>K : Too many products : Shift to left If Q=K system is at equilibrium
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Checking the assumptionChecking the assumption The rule of thumb is that if the value of X is less than 5% of all the other concentrations, our assumption was valid. If not we would have had to use the quadratic equation More on this later. Our assumption was valid.
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Using K C to determine [products] and [reactants] at equilibrium Only known K C and initial concentrations of reactants Changes in concentrations ( Δ C) of chemical compounds related to STOICHIOMETRIC rations in chemical equation.
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Steps for Complex Problems 1)Make an ICE Chart 2)Set up the equilibrium expression 3)Solve for Δ C
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Ice ChartIce Chart “ICE”ReactantsProducts Initial Concentrations (I): Change in Concentration ( Δ C or x): Equilibrium Concentrations (E):
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Example 3:Example 3: H 2 O (g) is present in a rigid container at 25°C with an initial partial pressure of 0.784 atm. What are the partial pressures of H 2(g) and O 2(g) at equilibrium? (K P = 2.0 x 10 -42 )
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Example 3: continuedExample 3: continued If K eq < 1x10 -4, remove the “x or Δ C” value in denominator. Δ C is very small compared to initial concentration so subtraction would not be a huge difference. Only works when adding or subtracting Δ C If concentrations or partial pressures are very small where their magnitude is approximately equal to K eq, CANNOT discount Δ C value.
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Example 4:Example 4: Sulfur trioxide decomposes to form sulfur dioxide and oxygen at 300°C°. Calculate the concentrations of all chemical compounds at equilibrium with an initial SO 3 concentration of 0.100M and K C = 1.6 x 10 -10.
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Example 5:Example 5: 0.194 mol of COCl 2 comes to equilibrium in a 5.8L container at 25°C (K C = 7.27 x 10 -38 ). Find the equilibrium concentrations of all chemical compounds in the following equation. ( Hint: first find the initial [COCl 2 ] ) COCl 2 (g) CO (g) + Cl 2 (g)
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% Ionization% Ionization
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Le Chatelier’s PrincipleLe Chatelier’s Principle If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress. 3 Types of stress
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Change amounts of reactants and/or products Adding product makes Q>K Removing reactant makes Q>K Adding reactant makes Q<K Removing product makes Q<K Determine the effect on Q, will tell you the direction of shift
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Variables disrupting equilibrium 1) Addition or removal of chemical compounds from the reaction. 2) Pressure 3) Temperature/Heat
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Basic ConceptBasic Concept Main concept in chemical equilibrium Change in a variable that alters the equilibrium of a system (chemical reaction) products a shift in the OPPOSITE direction Reaction shifts to counteract the variable’s influence Reaction tries to get back to equilibrium state-----SO reaction shifts
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Basic Concept (cont.)Basic Concept (cont.) What happens to a chemical reaction when equilibrium shifts When one side of a chemical reaction is stressed, the reaction shifts to the side of LEAST stress !
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Using K C to determine [products] and [reactants] at equilibrium Only known K C and initial concentrations of reactants Changes in concentrations ( Δ C) of chemical compounds related to STOICHIOMETRIC rations in chemical equation.
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Change PressureChange Pressure By changing volume System will move in the direction that has the least moles of gas. Because partial pressures (and concentrations) change a new equilibrium must be reached. System tries to minimize the moles of gas.
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Change in PressureChange in Pressure By adding an inert gas Partial pressures of reactants and product are not changed No effect on equilibrium position
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Change in TemperatureChange in Temperature Affects the rates of both the forward and reverse reactions. Doesn’t just change the equilibrium position, changes the equilibrium constant. The direction of the shift depends on whether it is exo- or endothermic
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Example 5:Example 5: Predict the influence of certain changes to the following chemical reaction. 2SO 3(g) 2SO 2(g) + O 2(g) Δ H°= 197.84 kJ/mol A) increasing reaction’s temperature B) increasing reaction’s pressure C) adding more O 2 at equilibrium D) Removing O 2 at equilibrium
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Δ G° Δ G Concentration and partial pressures can change for reactants and products. Δ G = Δ G° + RTlnQ What happens when we leave the standard state conditons?
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At equilibrium, Δ G = 0, so reaction quotient (Q) = equilibrium constant (K) At equilibrium Δ G° = - RTlnK Enables the reaction’s equilibrium constant (K) to be calculated from the change in free energy ( Δ G°) Δ G = Δ G° + RTlnQ
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The magnitude of Δ G° indicates how far the chemical reaction in its standard state is from equilibrium. Δ G° = 0, equilibrium Δ G°= large value, far from equilibrium Δ G° = small value, close to equilibrium The sign (+, - ) indicates which direction the reaction needs to shift to achieve equilibrium Positive (+) -------- shift to left, no reaction Negative (-) -------- shift to right, reaction goes to completion What is the relationship between free energy( Δ G) and K?
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Calculate the equilibrium constant for the following reaction given Δ G° = -33 kJ/mol. N 2 + 3H 2 2NH 3 Example 1Example 1
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Reaction between H 2 O and Acid HA + H 2 O H 3 O + + A - Write the equilibrium expression
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Acid Dissociation Constant (K a ) Quantifies how much an ACID dissociates in water Describes the acid strength K a = [H 3 O + ] [A - ] [HA]
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Acid Dissociation Constant (K a ) Strong Acids > 1 Weak Acids <<<< 1 Examples: HCl Ka = 1x10 6 CH 3 CO 2 H Ka = 1.8 x 10 -5
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Base Ionization Constant (K b ) Quantifies how much a BASE dissociates in water Describes base strength B + H 2 O BH + + OH - K b = [BH + ] [OH - ] [B]
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Other helpful equationsOther helpful equations pKa = -logKa pK b = -logK b
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Weak Acid pH CalculationsWeak Acid pH Calculations Weak acids only partially dissociate Calculating: Use ICE method Compare reaction quotient (Q a ) to equilibrium constant (Ka)----- % Ionization When can we ignore change in initial acid concentration ( Δ C) ?
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When can we ignore change in initial acid concentration ( Δ C)? If Δ C < 5% of initial acid concentration Ka < 1x10 -4 THEN, Δ C can be disregarded for equilibrium acid concentration
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Example 1:Example 1: Determine the H 3 O + ion concentration and pH of a 1.0M acetic acid solution with a Ka = 1.8 x 10 -5
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Example 2:Example 2: Calculate the pH of a 0.10M NH 3 solution with a pK b of 4.74
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K a and K b MUST determine K b from K a value Use relationship between an acid and its conjugate base K a K b = K w
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Example 5:Example 5: Find the pH of a 3.5M NH 3 solution with a K b value of 1.8 x 10 -5
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NaOAc HOAc Now we would have 2 sources for OAc - ion 1) Acetic Acid Dissociation 2) Salt Hydrolysis/Dissociation **Acetate ion is COMMON to both reactions** What would happen to the pH if we added….
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Common ion Ion present in both acid/base dissociation and salt ionization Common ion effect Seen with weak acids and bases Hinders the ionization of a weak acid or base due to the presence of a common ion between the acid/base and a salt Common Ion EffectCommon Ion Effect
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HF + H 2 O (l) H 3 O + + F - Addition of KF KF K + + F - Result is less [H 3 O + ] so pH ----more BASIC Example 1:Example 1:
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NH 3 + H 2 O (l) OH - + NH 4 + Addition of NH 4 Cl NH 4 Cl NH 4 + + Cl - Result is less [OH - ] so pH ----more ACIDIC Example 2:Example 2:
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450ml of 1.0M HCO 2 H is mixed with 250ml of 0.40M NaHCO 2 at 25°C. Find the pH of the solution (Ka = 1.8 x 10 -4 ). Example 3:Example 3:
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Salts Ionic compound made up of CATION and ANION Has acidic and basic properties Based on ions produced when salts dissociate No acid/base properties—group I/II cations (ex. Na +, Li +, K +, Ca +2 ) No basic properties—conjugate bases from monoprotic acids (ex. Cl -, Br -, NO 3 - ) Ex. NaCl, CaBr 2
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1. Salt Formation from Strong Base and Weak Acid Salt forms a BASIC solution. Conjugate base ion reacts with water to give hydroxide (OH - ) ions. Ex. Potassium fluoride (KF) KF K + + F - F - + H 2 O HF + OH -
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2. Salt Formation from a Strong Acid and Weak Base Salt forms an ACIDIC solution Conjugate acid reactions with water to give hydronium ion (H 3 O + ) Ex. Ammonium nitrate (NH 4 NO 3 )
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3. Salt Formation from Strong Acid and Strong Base Salt forms a NEUTRAL solution Conjugate base resulting from salt dissociation is weak Ex. Sodium chloride (NaCl)
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Example 1:Example 1: Calculate the concentration of HOAc, OAc - and OH - at equilibrium in a 0.10M NaOAc solution (Ka for HOAc = 1.8 x 10 -5 ).
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Titration Lab technique commonly utilized to determine an UNKNOWN concentration of a chemical compound with a KNOWN concentration of another chemical compound. Chemical compounds combine with exact stoichiometric proportions Analyte— Chemical compound with unknown concentration Titrant— Chemical compound with known concentration Measured with volume and concentration Added to chemical compound with unknown concentration in titration
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2 Types of Acid-Base Titrations 1)Strong Acid/Strong Base Titrations 2)Weak Acid/Strong Base Titrations
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1. Strong Acid/Strong Base Titrations Low initial pH value Sharp increase in pH before equivalence point Equivalence point is pH = 7 Rapid pH increase after equivalence point **Indicators with pH range 4-10 helpful for these titrations **Neutralization reactions
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Strong Acid with Strong Base Titrant Strong Base with Strong Acid Titrant
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2. Weak Acid/Strong Base Titrations High initial pH value pH = pKa at half-neutralization [weak acid] = [conjugate base] Ka = [H 3 O + ] [A - ] SO Ka = [A - ]/[HA] is 1:1 [HA] Ka = [H 3 O + ], SO pH = pKa
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2. Weak Acid/Strong Base Titrations Equivalence point > 7 on pH scale **Indicators with pH range > 7 helpful as pH equivalence point is basic
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Example 1: Weak Acid/Strong Base Titration Calculations 30 ml of 0.5M HC 2 H 3 O 2 titrated with 0.50M NaOH. Ka for acetic acid is 1.8x10 -5. a) Find the initial pH of 0.5M acetic acid. HC 2 H 3 O 2 + OH - C 2 H 3 O 2 - + H 2 O (titration view) HC 2 H 3 O 2 H + + C 2 H 3 O 2 - (in detail)
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Example 1: Weak Acid/Strong Base Titration Calculations b) Find the pH after 15ml of NaOH were added pH = pKa + log [C 2 H 3 O 2 - ] / [HC 2 H 3 O 2 ] -used when dealing with buffer solutions - buffer —mixture of weak acid/conjugate base -more on this concept later
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Example 1: Weak Acid/Strong Base Titration Calculations c) Find the pH at the equivalence point. C 2 H 3 O 2 - + H 2 O HC 2 H 3 O 2 + OH -
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Weak acid/conjugate base mixtures OR weak base/conjugate acid mixtures “buffers” or reduces the affect of a change in the pH of a solution Absorbs slight changes in pH resulting from the addition of small acid/base amounts to water. Buffers
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HOAc + H 2 O (l) H 3 O + + OAc – What happens if an acid is added??? Reacts with OAc ion [HOAc] increases slight, [OAc] decreases slightly, ratio mostly the same No pH change Example 1: (cont.)Example 1: (cont.)
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HOAc + H 2 O (l) H 3 O + + OAc – What happens if a base is added??? Reacts with HOAc ion More OAc ion formed, removes excess OH - from solution No pH change Example 1: (cont.)Example 1: (cont.)
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1)Acidic Buffers Formed from mixing a weak acid and its conjugate base pH < 7 Ex. HOAc and OAc – 2)Basic Buffers Formed from mixing a weak base and its conjugate acid pH > 7 Ex. NH 3 and NH 4 + Types of buffersTypes of buffers
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Compare Ka and Kb from acid-conjugate base pair Ka > Kb, (generally > 1x10 -7 ) -------- acidic buffer Ka 1x10 -7 ) -------- basic buffer How do we tell acidic vs. basic buffers?
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1)Method applying “Common Ion Effect” 2)Henderson-Hasselbalch equation Calculating the pH of buffers
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Easier method pH = pKa + log[A - ]/[HA] pOH = pKb + log[HB + ]/[B] [B] = molarity of weak base [HB + ] = molarity of conjugate acid Assumption: weak acids and conjugate bases do NOT change concentration with equilibrium. Henderson-Hasselbalch Equation
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Find the pH of a buffered solution created by mixing 0.15mol NH 4 NO 3 with 0.65L of a 0.25M NH 3 solution. Assume that the volume change is negligible. (Kb = 1.8x10 -5 ) Example 1:Example 1:
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