 # AP Chapter 15 Equilibrium *Chapters 15, 16 and 17 are all EQUILIBRIUM chapters* HW: 4 13 15 17 21 27 29 31 37 41 55 69.

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AP Chapter 15 Equilibrium *Chapters 15, 16 and 17 are all EQUILIBRIUM chapters* HW: 4 13 15 17 21 27 29 31 37 41 55 69

15.1 - The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. N 2 O 4  2 NO 2 Reactants and Products CANNOT escape the container!

Depicting Equilibrium In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow N 2 O 4 (g) 2 NO 2 (g)

At equilibrium the forward and reverse directions of the reaction are occurring at equal rate. Rate f = Rate r

The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate.

A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant. The amounts are not equal!!

K = The Equilibrium Constant 15.2 -

The Equilibrium Constant Consider the reaction: The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD Multiplication of Products to their “coefficient powers” Multiplication of Reactants to their “coefficient powers” concentration “reactants” “products”

The Equilibrium Constant K c = [C] c [D] d [A] a [B] b aA + bBcC + dD K values are usually UNITLESS! (even if they don’t truly cancel)

Examples: K c = [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = [N 2 O 4 ] [NO 2 ] 2 N2O4 (g)N2O4 (g) 2 NO 2 (g)

Example: K c = [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = ?? [NO 2 ] 2 [N 2 O 4 ]

15.2 – Writing Equilibrium Constant Expressions Homogeneous Equilibrium – Equilibrium when the species are in the same phase N 2 O 4 (g) 2 NO 2 (g) K c = [NO 2 ] 2 [N 2 O 4 ] K p = P NO2 2 P N2O4 Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written based on pressures.

Relationship between K c and K p From the ideal gas law we know that Rearranging it, we get PV = nRT P = RT nVnV Using the Kp expression, Kp becomes: K p = RT nBVnBV b nAVnAV a For the reaction: aA(g)  bB(g) K p = PBbPAaPBbPAa

Relationship between K c and K p n/V is concentration and the RT terms can be pulled out to give: Where K p = K c (RT)  n  n = b-a = moles of gaseous product − moles of gaseous reactant K p = b a (RT)  n [A] [B] R = 0.08206

Relationship between K c and K p Example: H 2(g) + Br 2(g) = 2HBr (g)  n = 2 – 2 = 0,  Kc = Kp  Example:  n = 2 – 1 = 1 K p = K c (RT)  n N 2 O 4 (g) 2 NO 2 (g)

Relationship between K c and K p N 2 (g) + 3H 2 (g) = 2 NH 3 (g) If Kc = 9.60 at 300 o C, what is Kp? (Kp = 4.34 x 10 -3 )

15.3 – Interpreting the Equilibrium Constant If K >> 1, the reaction is product-favored; product predominates at equilibrium. If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

15. 3 - Form of K -How you write the reaction determines the value of K The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction. 1 0.212 = K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = 4.72 at 100  C [N 2 O 4 ] [NO 2 ] 2 N2O4 (g)N2O4 (g) 2 NO 2 (g)

Form of K -K values depend on the way the reaction is balanced. -The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. K c = = 0.212 at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2 2 N 2 O 4 (g) 4 NO 2 (g)

Multiple Equilibria A + BC + D E + F K = Conclusion: The overall reaction is the product of the equilibrium constants of the individual reactions! Summary Pg 638

CH 3 COOH(aq) + H 2 O(l)  H 3 O + (aq) + CH 3 COO - (aq) K = ??? = But – the [H2O] is VERY high (~55.5 M) so it is essentially constant, so K becomes: K = ??? = (Liquids are USUALLY not in K expressions) 15.4 – Heterogeneous Equilibrium

The Concentrations of Solids and Liquids Are Essentially Constant d = g/L MM = g/mol d/MM = = = mol/L For solids and liquids, density and MM are constants at constant temperature  concentration is a constant.

The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression K c = [Pb 2+ ] [Cl − ] 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq)

Kc = [CO 2 ]OR Kp = P CO2 CaCO 3 (s) CO 2 (g) + CaO (s)

15.5 – Calculating Equilibrium Constants Example 15.9 – Pg 643 A closed system initially containing 1.000 x 10 -3 M H 2 and 2.000 x 10 -3 M I 2 at 448  C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 -3 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (s) 2 HI (g)

What do we know: [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change -x +2x At equilibrium1.87 x 10 -3 H 2 (g) + I 2 (s) 2 HI (g)

Now we know the change in [HI] [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change+1.87 x 10 -3 At equilibrium1.87 x 10 -3

Can fill in the remainder of the changes based on stoichiometry… [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium1.87 x 10 -3 H 2 (g) + I 2 (s) 2 HI (g)

Now fill in the Equilibrium Values… [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium6.5 x 10 -5 1.065 x 10 -3 1.87 x 10 -3

And finally calculate K… K c = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x 10 -3 ) 2 (6.5 x 10 -5 )(1.065 x 10 -3 )

15.6 - Calculating Equilibrium Concentrations The ICE Method

ICE Initial Change Equilibrium *Make a chart for these calculations*

Example: Cis-stilbene  trans-stilbene Kc = 24.0 at 200 o C [cis] i = 0.850 M [trans] i = 0 M WORK:

More Examples: 15.11

15.12

% Yield % =

On Page 13 of notes!!

15.6 – What does K tell us? Predicting the direction of a reaction: Calculate a “Reaction Quotient” (Qc). It has the same FORM as Kc, but can be used at ANY time. If Q>K, then products are too high, and reaction will shift LEFT If Q<K, then products are too low, and reaction will shift RIGHT If Q = K, then the reaction is at equilibrium

Example – In what direction will the reaction proceed? H 2 + I 2 2HI K = 54.3 at 430 o C Initial concentrations in a 1 L container: H 2 = 0.243 mol I 2 = 0.146 mol HI = 1.98 mol

15.7 – Factors that Affect Chemical Equilibrium LeChatlier’s Principle – When an external stress is applied to a system at equilibrium, the system adjusts to offset the stress.

LeChatlier’s – Changes in Concentration Does NOT change the value of K FeSCN +2  Fe +3 + SCN -1 -Add NaSCN -Add Fe(NO 3 ) 3 -Add H 2 C 2 O 4

Changes in Volume and Pressure Only affects gases Does not change the value of K Lowering volume, creates an increase in concentration (mol / L) Lowering volume, increasing concentration, causes a shift to the side of the reaction with FEWER MOLES OF GAS Increasing volume, decreasing concentration, causes a shift to the side of the reaction with MORE MOLES OF GAS If there is no change in the moles of gas, changing volume and pressure create NO SHIFT

Changes in Volume and Pressure Adding “an inert gas” does NOT effect equilibrium, because the partial pressures of all of the gases in the reaction do not change

Changes in Temperature ALTERS K – this is the only change that changes the Equilibrium Constant! If heat is added, the reaction shifts to the ENDOTHERMIC reaction If heat is removed, the reaction shifts to the EXOTHERMIC reaction

Catalyst Does not change the value of K Lowers the activation energy (for both the forward and the reverse reaction) Makes the reaction reach equilibrium faster

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