AP Chapter 15 Equilibrium *Chapters 15, 16 and 17 are all EQUILIBRIUM chapters* HW: 4 13 15 17 21 27...
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Transcript of AP Chapter 15 Equilibrium *Chapters 15, 16 and 17 are all EQUILIBRIUM chapters* HW: 4 13 15 17 21 27...
AP Chapter 15Equilibrium
*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.
N2O4 2 NO2
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
N2O4 (g) 2 NO2 (g)
At equilibrium the forward and reverse directions of the reaction are occurring at equal rate.
Ratef = Rater
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!!
The Equilibrium Constant• Consider the reaction:
• The equilibrium expression for this reaction would be
Kc = [C]c[D]d
[A]a[B]b
aA + bB cC + dD
Multiplication of Products to their “coefficient powers”
Multiplication of Reactants to their “coefficient powers”
concentration
“reactants” “products”
The Equilibrium Constant
Kc = [C]c[D]d
[A]a[B]b
aA + bB cC + dD
K values are usually UNITLESS!(even if they don’t truly cancel)
15.2 – Writing Equilibrium Constant Expressions
• Homogeneous Equilibrium – Equilibrium when the species are in the same phase
N2O4 (g) 2 NO2 (g)
Kc = [NO2]2
[N2O4]Kp = PNO2
2
PN2O4
Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written based on pressures.
Relationship between Kc and Kp
• From the ideal gas law we know that
• Rearranging it, we get
PV = nRT
P = RTnV
• Using the Kp expression, Kp becomes:
Kp = RT
nB
Vb
RTnA
Va
For the reaction: aA(g) bB(g)
Kp = PBb
PAa
Relationship between Kc and Kp
n/V is concentration and the RT terms can be pulled out to give:
Where
Kp = Kc (RT)n
n = b-a = moles of gaseous product − moles of gaseous reactant
Kp = b
a(RT)Dn
[A]
[B]
R = 0.08206
Relationship between Kc and Kp
• Example:H2(g) + Br2(g) = 2HBr (g)
Dn = 2 – 2 = 0, Kc = Kp Example:
Dn = 2 – 1 = 1
Kp = Kc (RT)n
N2O4 (g) 2 NO2 (g)
Relationship between Kc and Kp
N2(g) + 3H2(g) = 2 NH3(g)
If Kc = 9.60 at 300oC, 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.
10.212=
Kc = = 0.212 at 100C[NO2]2
[N2O4]N2O4 (g) 2 NO2 (g)
Kc = = 4.72 at 100C
[N2O4][NO2]2N2O4 (g)2 NO2 (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.
Kc = = 0.212 at 100C[NO2]2
[N2O4]N2O4 (g) 2 NO2 (g)
Kc = = (0.212)2 at 100C[NO2]4
[N2O4]22 N2O4 (g) 4 NO2 (g)
Multiple Equilibria
C + D E + F
K = K =
Conclusion: The overall reaction is the product of the equilibrium constants of the individual reactions!
Summary Pg 638
CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO-(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/LMM = g/mold/MM = = = mol/L
For solids and liquids, density and MM are constants at constant temperature concentration is a constant.
gramsLitergramsmole
grams molesx
Liter gram
The Concentrations of Solids and Liquids Are Essentially Constant
Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression
Kc = [Pb2+] [Cl−]2
PbCl2 (s) Pb2+ (aq) + 2 Cl−
(aq)
15.5 – Calculating Equilibrium ConstantsExample 15.9 – Pg 643
• A closed system initially containing 1.000 x 10-3 M H2 and 2.000 x 10-3 M I2 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 Kc at 448 C for the reaction taking place, which is
H2 (g) + I2 (s) 2 HI (g)
What do we know:
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -x -x +2x
At equilibrium 1.87 x 10-3
H2 (g) + I2 (s) 2 HI (g)
Now we know the change in [HI]
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change +1.87 x 10-3
At equilibrium 1.87 x 10-3
Can fill in the remainder of the changes based on stoichiometry…
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At equilibrium 1.87 x 10-3
H2 (g) + I2 (s) 2 HI (g)
Now fill in the Equilibrium Values…
[H2], M [I2], M [HI], M
Initially 1.000 x 10-3 2.000 x 10-3 0
Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3
At equilibrium 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3
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?
H2 + I2 2HIK = 54.3 at 430oCInitial concentrations in a 1 L container:H2 = 0.243 molI2 = 0.146 molHI = 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 KFeSCN+2 Fe+3 + SCN-1
-Add NaSCN
-Add Fe(NO3)3
-Add H2C2O4
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