Dr. Donald VanDerveer Office: Boggs 2-5 Office Hours: 8:00-9:30 MWF e-mail: don.vanderveer@ chemistry.gatech.edu Phone/voice-mail: 404-894-8517
Location of Boggs 2-5 B6A
Equilibrium
Equilibrium In order to have equilibrium for a chemical reaction, the process must be able to move both forwards and backwards.
2 H2 + O2 2 H2O spark
2 H2 + O2 2 H2O spark 2 H2O 2 H2 + O2
Equilibrium processes include both reactions which have chemically different reactants and products and phase changes.
Equilibrium H2O(l) H2O(g) 25oC
H2O(l) H2O(g)
After a given period of time, the pressure does not increase nor does it decrease.
H2O(l) H2O(g)
H2O(g) H2O(l) H2O(g) H2O(l)
As long as there is liquid water present, there will be a liquid-gas equilibrium.
As long as there is liquid water present, there will be a liquid-gas equilibrium. At any given temperature, the equilibrium pressure will be the same.
As long as there is liquid water present, there will be a liquid-gas equilibrium. At any given temperature, the equilibrium pressure will be the same. This pressure is independent of the volume of the container.
Equilibrium constant:
Equilibrium constant: Equilibrium constants are calculated based on the phases of the reactants and products.
Equilibrium constant: If reactants and products are gasses, the equilibrium constant is calculated from partial pressures.
Equilibrium constant: If reactants and products in solution, the equilibrium constant is calculated from concentrations.
Equilibrium constant: How the value of an equilibrium constant is determined varies, depending on the type of reaction.
Equilibrium constant: For a liquid - gas equilibrium, the constant, K, is equal to the vapor pressure of the gas at equilibrium.
Equilibrium constant: For a liquid - gas equilibrium, the constant, K, is equal to the vapor pressure of the gas at equilibrium. Px K = 1 atm
H2O(l) H2O(g) K = 0.03126 25o C H2O(g) H2O(l) K = 0.04187 30o C
Chemical equilibrium N2O4(g) 2 NO2(g)
2 17 e-
Place different amounts of pure NO2 in a container, equilibrate at 25oC and measure the partial pressures of both gasses.
N2O4(g) 2 NO2(g) PNO 2
. Know PN O and PNO 2 4 2
. Know PN O and PNO 2 4 2 Make equilibrium constant from an expression of partial pressures.
Partial pressures P1 + P2 + … Pn = Ptotal
PN O = K(PNO )2 2 4 2 PNO 2 N2O4(g) 2 NO2(g)
PN O = K(PNO )2 2 4 2 PNO 2 N2O4(g) 2 NO2(g)
PN O = K(PNO )2 2 4 2 PN O K = 2 4 (PNO )2 2 PNO 2 N2O4(g) 2 NO2(g)
PN O = K(PNO )2 2 4 2 PN O K = 2 4 (PNO )2 2 K = 8.8 @ 25oC PNO 2
Characteristics of equilibrium states 1. No changes in macroscopic states (pressure, volume, etc.)
Characteristics of equilibrium states 1. No changes in macroscopic states (pressure, volume, etc.) 2. Equilibrium is reached through spontaneous processes.
Characteristics of equilibrium states 1. No changes in macroscopic states (pressure, volume, etc.) 2. Equilibrium is reached through spontaneous processes. 3. Dynamic balance of forward and reverse processes.
Characteristics of equilibrium states 1. No changes in macroscopic states (pressure, volume, etc.) 2. Equilibrium is reached through spontaneous processes. 3. Dynamic balance of forward and reverse processes. 4. Are the same regardless of direction of approach.
General form of equilibrium constant when all species are low-pressure gasses
General form of equilibrium constant when all species are low-pressure gasses Balanced chemical equation: aA + bB cC + dD c d PCPD = K a b PAPB
c d PCPD = K a b PAPB 2 NOCl 2 NO + Cl2 PNOPCl 2 = K PNOCl 2
Calculating equilibrium constants
Calculating equilibrium constants Data PNO 2
Calculating equilibrium constants Data PN O K = 2 4 (PNO )2 2 PNO 2
Calculating equilibrium constants PN O K = 2 4 (PNO )2 2 PNO 2
Calculating equilibrium constants PN O K = 2 4 (PNO )2 2 = 5 atm PN O 2 4 PNO 2
Calculating equilibrium constants PN O K = 2 2 (PNO )2 2 = 5 atm PN O 2 4 = 0.75 atm PNO 2 PNO 2
Calculating equilibrium constants 5 atm K = (0.75 atm)2 = 5 atm PN O 2 4 = 0.75 atm PNO 2 PNO 2
Calculating equilibrium constants 5 atm K = = (0.75 atm)2 K = 8.9 PNO 2
Calculating equilibrium constants 5 atm K = = (0.75 atm)2 K = 8.9 PNO K = 8.8 @ 25oC 2
Calculating equilibrium constants without all partial pressures.
Calculating equilibrium constants without all partial pressures. 4 NO2 2 N2O + 3 O2
Calculating equilibrium constants without all partial pressures. 4 NO2 2 N2O + 3 O2 Given: starting partial pressures, equilibrium partial pressure for NO2.
4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) ? ? ? Equilibrium 2.4 ? ?
4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) -1.2 ? ? Equilibrium 2.4 ? ?
4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) -1.2 ? ? Equilibrium 2.4 ? ? For every 4 molecules of NO2 that react, 2 molecules of N2O and 3 molecules of O2 are formed.
Volume changes are coefficients 4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) -4x +2x +3x Equilibrium 2.4 ? ?
Volume changes are coefficients 4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) -4x +2x +3x Equilibrium 2.4 5.1 + 2x 8.0 + 3x
Volume changes are coefficients 4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) -4x +2x +3x Equilibrium 2.4 5.1 + 2x 8.0 + 3x 3.6 - 4x = 2.4
Volume changes are coefficients 4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) -4x +2x +3x Equilibrium 2.4 5.1 + 2x 8.0 + 3x 2.4 - 3.6 3.6 - 4x = 2.4 x = -4
Volume changes are coefficients 4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) -4x +2x +3x Equilibrium 2.4 5.1 + 2x 8.0 + 3x 2.4 - 3.6 3.6 - 4x = 2.4 x = = 0.3 atm -4
Volume changes are coefficients 4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) -1.2 +0.6 +0.9 Equilibrium 2.4 5.7 8.9 2.4 - 3.6 3.6 - 4x = 2.4 x = = 0.3 atm -4
Volume changes are coefficients 4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) -1.2 +0.6 +0.9 Equilibrium 2.4 5.7 8.9 (PN O)2(PO)3 2 2 (PNO )4 2
Volume changes are coefficients 4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) -1.2 +0.6 +0.9 Equilibrium 2.4 5.7 8.9 (PN O)2(PO)3 (5.7)2(8.9)3 2 2 = (2.4)4 (PNO )4 2
Volume changes are coefficients 4 NO2 2 N2O + 3 O2 Initial (atm) 3.6 5.1 8.0 change (atm) -1.2 +0.6 +0.9 Equilibrium 2.4 5.7 8.9 (PN O)2(PO)3 (5.7)2(8.9)3 = 6.9 x 102 2 2 = (PNO )4 (2.4)4 2
Relationships among equilibrium expressions.
Relationships among equilibrium expressions. 1. The equilibrium constant of a reverse reaction is the reciprocal of the forward reaction.
Relationships among equilibrium expressions. 1. The equilibrium constant of a reverse reaction is the reciprocal of the forward reaction. 2 NO2 N2O4 PN O K = 2 4 (PNO )2 2
Relationships among equilibrium expressions. 1. The equilibrium constant of a reverse reaction is the reciprocal of the forward reaction. 2 NO2 N2O4 N2O4 2 NO2 (PNO)2 PN O K1 = 2 4 K2 = 2 PN O (PNO )2 2 2 4
2 NO2 N2O4 N2O4 2 NO2 (PNO)2 PN O K1 = 2 4 K2 = 2 PN O (PNO )2 2 2 4 K1K2 =1
2 NO2 N2O4 N2O4 2 NO2 (PNO)2 PN O K1 = 2 4 K2 = 2 PN O (PNO )2 2 2 4 K1K2 =1 1 = 1 K1 K1
2. If coefficients in a balanced equation are all multiplied by a single factor, the equilibrium constant is raised to a power equal to that factor.
2. If coefficients in a balanced equation are all multiplied by a single factor, the equilibrium constant is raised to a power equal to that factor. 2 NO2 N2O4 4 NO2 2 N2O4 PN O (PN O )2 K1 = 2 4 K2 = 2 4 (PNO )2 (PNO )4 2 2
2. If coefficients in a balanced equation are all multiplied by a single factor, the equilibrium constant is raised to a power equal to that factor. 2 NO2 N2O2 4 NO2 2 N2O2 PN O (PN O )2 K1 = 2 4 K2 = 2 4 (PNO )2 (PNO )4 2 2 K2 = K12
3. Operations of addition and subtraction, applied to chemical equations, lead to operations of multiplication and division of their corresponding equilibrium expressions and constants.
3. Operations of addition and subtraction, applied to chemical equations, lead to operations of multiplication and division of their corresponding equilibrium expressions and constants. Exercise page 290 CF4 + 2 H2O 2 CO2 + 4 HF K1 = 5.9 x 1023
3. Operations of addition and subtraction, applied to chemical equations, lead to operations of multiplication and division of their corresponding equilibrium expressions and constants. Exercise page 290 CF4 + 2 H2O CO2 + 4 HF K1 = 5.9 x 1023 CO + 1/2 O2 CO2 K2 = 1.3 x 1035
CF4 + 2 H2O CO2 + 4 HF K1 = 5.9 x 1023 CO + 1/2 O2 CO2 K2 = 1.3 x 1035 2 CF4 + 4 H2O 2 CO + 8 HF + O2
CF4 + 2 H2O CO2 + 4 HF K1 = 5.9 x 1023 CO + 1/2 O2 CO2 K2 = 1.3 x 1035 2 CF4 + 4 H2O 2 CO + 8 HF + O2
CF4 + 2 H2O CO2 + 4 HF K1 = 5.9 x 1023 CO + 1/2 O2 CO2 K2 = 1.3 x 1035 2 CF4 + 4 H2O 2 CO + 8 HF + O2
CF4 + 2 H2O CO2 + 4 HF K1 = 5.9 x 1023 2 CO + O2 2 CO2 K2 = 1.7 x 1070 2 CF4 + 4 H2O 2 CO + 8 HF + O2
2 CF4 + 4 H2O 2 CO2 + 8 HF K1 = 34.8 x 1046 2 CO + O2 2 CO2 K2 = 1.7 x 1070 2 CF4 + 4 H2O 2 CO + 8 HF + O2
2 CF4 + 4 H2O 2 CO2 + 8 HF 2 CO + O2 2 CF4 + 4 H2O 2 CO + 8 HF + O2
2 CF4 + 4 H2O 2 CO2 + 8 HF K1 = 34.8 x 1046 2 CO + O2 2 CO2 K2 = 1.7 x 1070 2 CF4 + 4 H2O 2 CO + 8 HF + O2 K3 = K1 x K2-1
2 CF4 + 4 H2O 2 CO2 + 8 HF K1 = 34.8 x 1046 2 CO + O2 2 CO2 K2 = 1.7 x 1070 2 CF4 + 4 H2O 2 CO + 8 HF + O2 K3 = K1 x K2-1 34.8 x 1046 = 20.5 x 10-24 K3 = 1.7 x 1070
Reaction quotient c d PCPD Q = a b PAPB
Reaction quotient c d PCPD Q = a b PAPB PA , PB, etc. not at equilibrium
Reaction quotient c d PCPD Q = a b PAPB Q will go to the value of K as the partial pressures go to equilibrium
Q vs K can predict where the reaction is in respect to equilibrium.
c d K = Q < K PCPD a b PAPB aA + bB cC + dD c d PCPD Q = a b PAPB
c d K = Q < K PCPD a b PAPB aA + bB cC + dD c d PCPD Too much reactant, not enough product. Q = a b PAPB
c d K = Q < K PCPD a b PAPB aA + bB cC + dD c d PCPD Q > K Q = a b PAPB aA + bB cC + dD
c d K = Q < K PCPD a b PAPB aA + bB cC + dD c d PCPD Q > K Q = a b PAPB aA + bB cC + dD Too much product, not enough reactant.
c d K = Q < K PCPD a b PAPB aA + bB cC + dD c d PCPD Q > K Q = a b PAPB aA + bB cC + dD c Q < K a b d PAPB Large vs PCPD
c d PCPD Q = a b PAPB
Exercise page 291 P4 2 P2 K = 1.39 @ 400oC 1.40 mol P4 1.25 mol P2 Volume = 25.0 L
P4 2 P2 (PP )2 Q = 2 PP 4
P4 2 P2 (PP )2 Q = 2 PP 4 T = 673 K nRT P = V = 25.0 L V R = 0.0820578 L atm mol-1 K-1
P4 2 P2 (PP )2 Q = 2 PP 4 (1.4)(0.0820578)(673) nRT = PP = = 3.09 atm 4 V 25.0
P4 2 P2 (PP )2 Q = 2 PP 4 (1.4)(0.0820578)(673) nRT = PP = = 3.09 atm 4 V 25.0 (1.25)(0.0820578)(673) nRT = PP = = 2.76 atm 2 V 25.0
P4 2 P2 (PP )2 (2.76)2 Q = 2 = = 3.09 PP PP = 2.76 atm 4 2 PP = 3.09 atm 4
P4 2 P2 (PP )2 (2.76)2 Q = 2 = = 3.09 PP PP = 2.76 atm 4 2 PP = 3.09 atm Q = 2.46 4 K = 1.39
P4 2 P2 (PP )2 (2.76)2 Q = 2 = = 3.09 PP PP = 2.76 atm 4 2 PP = 3.09 atm Q = 2.46 4 K = 1.39 P4 2 P2
Exercise page 292 CO + H2O CO2 + H2
Exercise page 292 CO + H2O CO2 + H2 PCO = 2.00 atm PCO = 0.80 atm 2 = 0.48 atm PH 2 K = 0.64 @ 900 K
Exercise page 292 CO + H2O CO2 + H2 PCO = 2.00 atm PH PCO K = 2 2 PCO PH O PCO = 0.80 atm 2 2 = 0.48 atm PH 2 K = 0.64 @ 900 K
Exercise page 292 CO + H2O CO2 + H2 PCO = 2.00 atm PH PCO K = 2 2 PCO PH O PCO = 0.80 atm 2 2 PH PCO = 0.48 atm PH PH O 2 2 = 2 2 PCO K K = 0.64 @ 900 K
Exercise page 292 CO + H2O CO2 + H2 PCO = 2.00 atm PH PCO PH O 2 2 = = PCO = 0.80 atm 2 PCO K 2 (0.80)(0.48) = 0.48 atm PH = 0.30 atm 2 (2.00) (0.64) K = 0.64 @ 900 K
Converting between partial pressures and concentrations.
Converting between partial pressures and concentrations. moles Concentration = V
Converting between partial pressures and concentrations. moles Concentration = V PV = nRT
Converting between partial pressures and concentrations. moles Concentration = V nA For gas ‘A’ [A] = V PV = nRT
Converting between partial pressures and concentrations. moles Concentration = V PA nA For gas ‘A’ [A] = = RT V PV = nRT
Converting between partial pressures and concentrations. moles Concentration = V PA nA For gas ‘A’ [A] = = RT V PV = nRT PA = RT[A]
PA = RT[A] N2O4(g) 2 NO2(g)
PA = RT[A] N2O4(g) 2 NO2(g) Pref = 1 atm
PA = RT[A] N2O4(g) 2 NO2(g) Pref = 1 atm PN O /Pref 2 4 = K /Pref)2 (PNO 2
PA = RT[A] N2O4(g) 2 NO2(g) PN O = RT[N2O4] Pref = 1 atm 2 4 PN O /Pref 2 4 = K /Pref)2 (PNO 2
PA = RT[A] N2O4(g) 2 NO2(g) PN O = RT[N2O4] Pref = 1 atm 2 4 PNO = RT[NO2] PN O /Pref 2 2 4 = K /Pref)2 (PNO 2
( ) N2O4(g) 2 NO2(g) PN O = RT[N2O4] PN O /Pref 2 4 2 4 = K /Pref)2 -1 [N2O4] ( RT ) [N2O4](RT/Pref) x = K = Pref [NO2](RT/Pref) [NO2]
( ) ( ) N2O4(g) 2 NO2(g) -1 [N2O4] RT [N2O4](RT/Pref) x = K = Pref
( ) ( ) [N2O4] RT = K [NO2] Pref aA + bB cC + dD c d a+b-c-d PCPD PAPB [A]a[B]b
( ) a+b-c-d [C]c[D]d RT = K Pref [A]a[B]b Exercise page 297 CH4 + H2O CO + 3 H2 [H2]=[CO]=[H2O]= 0.00642 mol L-1 K = 0.172
( ) ( ) a+b-c-d [C]c[D]d RT = K Pref [A]a[B]b CH4 + H2O CO + 3 H2 [H2]=[CO]=[H2O]= 0.00642 mol L-1 K = 0.172 ( RT ) -2 (0.00642)4 = 0.172 Pref [CH4](0.00642)
( ) CH4 + H2O CO + 3 H2 RT -2 (0.00642)3 = 0.172 Pref [CH4] K(RT)-2 = 3.153 x 10-5 (0.00642)3 [CH4] = = 8.39 x 10-2 mol L-1 3.153 x 10-5