Chemical Equilibrium Chapter 17 Chapter 17
Dec 26, 2015
Chemical Equilibrium
Chapter 17Chapter 17
Equilibrium vs. Kinetics
Kinetics:Kinetics: speed of a reaction or processspeed of a reaction or process
how fast?how fast?
Equilibrium:Equilibrium: extent of reaction or processextent of reaction or process
how much?how much?
Chemical Equilibrium
Reactant and product concentrations remain constantReactant and product concentrations remain constant
Molecular level:Molecular level: rapid activity (dynamic)rapid activity (dynamic)
Macroscopic level:Macroscopic level: unchangingunchanging
At equilibrium:At equilibrium: raterateforwardforward = rate= ratereversereverse
Does not limit timeDoes not limit time
Law of Mass Action
If:If:
Then:Then:
And:And:
reverseforward rate rate
nm ]k[products ]reactants[k
K]reactants[k
]k[products
rate
ratem
n
reverse
forward
K = equilibrium constant
m, n = coefficients in balanced equation
Law of Mass Action
For:For:
Equilibrium Equilibrium expression:expression:
mD lC kB jA
kj
ml
c[B][A]
[D][C] K
Value of K Favors
K < 0.01 Reactants
K > 100 Products
0.01 < K < 100 Neither
Direction of Reaction and Q
For:For:
Equilibrium Equilibrium expression:expression:
Reaction Reaction Quotient:Quotient:
mD lC kB jA
kj
ml
c[B][A]
[D][C] K
tat time, [B][A]
[D][C] Q
kj
ml
c
]O[N
][NO K
42
22
c
N2O4(g) 2 NO2(g)
Practice
1.1. Writing expression for KWriting expression for K
2. 2. Q vs. K and reaction directionQ vs. K and reaction direction
3.3. K for a multistep processK for a multistep process
4.4. K for reaction “multiples”K for reaction “multiples”
Write K
The decomposition of dinitrogen pentoxide:
N2O5(g) NO2(g) + O2(g)
The combustion of propane gas:
C3H8(g) + O2(g) CO2(g) + H2O(g)
Write K
The decomposition of dinitrogen pentoxide:
2 N2O5(g) 4 NO2(g) + O2(g)
The combustion of propane gas:
C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)
252
24
2c
]O[N
][O][NO K
5283
42
32
c]][OH[C
O][H][CO K
Write K
4 NH3(g) + O2(g) 4 NO(g) + 6 H2O(g)
2 NH3(g) + 5/2 O2(g) 2 NO(g) + 3 H2O
K vs. Q
For the reaction: N2O4(g) 2NO2(g)
Kc = 0.21 at 1000C.
At a point during the reaction, [N2O4] = 0.12M and [NO2] = 0.55M.
(a) Find Q. Is the reaction at equilibrium?
(b) If not, in which direction is it progressing?
K vs. Q
N2O4(g) 2NO2(g) Kc = 0.21 at 1000C.
At a point, [N2O4] = 0.12M and [NO2] = 0.55M.
(a) Find Q. Is the reaction at equilibrium?
(b) If not, in which direction is it progressing?
5.2(.12)
(0.55)
]O[N
][NO Q
2
42
22
c
left Right toso K Q cc
K for multistep reactions
Nitrogen dioxide, a toxic pollutant that contributes to photochemical smog, can develop in combustion engines from N2 and O2.
(1) N2 + O2 2NO Kc1 = 4.3 x 10-25
(2) 2NO + O2 2NO2 Kc2 = 6.4 x 109
(a) Show that Qc for the overall reaction is the same as the product of Qcs of the individual reactions.
(b) Calculate Kc for the overall reaction.
K for multistep reactions(1) N2 + O2 2NO Kc1 = 4.3 x 10-
25
(2) 2NO + O2 2NO2 Kc2 = 6.4 x 109
N2 + 2 O2 2 NO2
(a) Qc
(b) Kc,overall
222
22
22
22
22
2
c2c1c]][O[N
][NO
][O[NO]
][NO
]][O[N
[NO]QQ Q
15925c2c1c 108.2106.4 103.4K KK
K for multistep reactions
For the following
(1) Br2 2 Br
(2) Br + H2 HBr + H
(3) H + Br HBr
(a) Write the overall balanced reaction.
(b) Write out the individual expressions for Qc and show that their product is equivalent to the overall Qc.
Multiples of K
For the ammonia reaction:
N2(g) + 3H2(g) 2NH3(g)
Kc is 2.4x10-3 at 1000K.
Find K for the following:
(a) 1/3 N2+ H2 2/3 NH3
(b) NH3 1/2 N2 + 3/2 H2
Multiples of K
N2(g) + 3H2(g) 2NH3(g), Kc = 2.4x10-3
(a) 1/3 N2+ H2 2/3 NH3
(b) NH3 1/2 N2 + 3/2 H2
13.0104.2 KK 31
331
cac,
.20104.2 KK 21
321
cbc,
MultiplesN2(g) + O2(g) 2 NO(g) Kc = 1 x 10-30
Write the expression for Q and determine its value for ½ N2(g) + ½ O2(g) 2 NO(g)
H2(g) + Cl2(g) 2 HCl(g) Kc = 7.6 x 108
Write the expression for Q and determine its value for 2/3 HCl(g) 1/3 H2(g) + 1/3 Cl2(g)
Heterogeneous Equilibrium
PURE solids and liquids do not appear in expression for K (or Q).
2CO K
Kc vs. Kp
For:For: mD lC kB jA n
p K(RT)K
molK
atmL 0.08206 R
gas moles # in change
)nn( n reactantsproducts
Kp and Kc
For the ammonia reaction,
N2(g) + 3H2(g) 2NH3(g),
Kc = 2.4x10-3
Find Kp at 1000 K.
Kp and Kc
N2(g) + 3H2(g) 2NH3(g), Kc = 2.4x10-3
Find Kp at 1000 K.
2)4()2( n
)nn( n reactantsproducts
7p
23p
2c
Δncp
103.56 K
1000) )(0.08206104.2( K
(RT)K(RT) K K
Kp and Kc
For the following reaction,
PCl3(g) + Cl2(g) PCl5(g),
Kc = 1.67 at 500 K
Find Kp at 500 K.
K vs. Q
For the reaction: CH4(g) + Cl2 CH3Cl(g) + HCl
Kp = 1.6x104 at 1500 K.
At a point during the reaction,
PCH4 = 0.13 atm, PCl2= 0.035 atm,
PCH3Cl = 0.24 atm, and PHCl = 0.47 atm.
(a) Find Q. Is the reaction at equilibrium?
(b) If not, in which direction is it progressing?
Problems
1.1. Given equilibrium concentrations or pressures, Given equilibrium concentrations or pressures, find K or Q.find K or Q.
2. Given K and initial conditions (conc’s or P’s), 2. Given K and initial conditions (conc’s or P’s), find equilibrium quantities (conc’s or P’s).find equilibrium quantities (conc’s or P’s).
Le Châtelier’s Principle
. . . if a change is imposed on a . . . if a change is imposed on a system at equilibrium, the system at equilibrium, the position of the equilibrium will position of the equilibrium will shift in a direction that tends to shift in a direction that tends to reduce that change.reduce that change.
Le Châtelier’s Principle
1.1. ConcentrationConcentration
2.2. TemperatureTemperature
3.3. PressurePressure
4.4. VolumeVolume
5.5. Catalysts*Catalysts*
13_315
H2
NH3N2
Time
Concentration
Equilibrium