Chemical Equilibrium
Chemical Equilibrium: Ch. 15 15-1 Dynamic Equilibrium 15-2 The Equilibrium Constant Expression 15-3 Relationships Involving Equilibrium Constants 15-4 The Magnitude of an Equilibrium Constant 15-5 The Reaction Quotient, Q: Predicting the Direction of a Net Change 15-6 Altering Equilibrium Conditions: Le Châtelier’s Principle 15-7 Equilibrium Calculations: Some Illustrative Examples
15-1 Dynamic Equilibrium • Equilibrium – two opposing processes taking place at equal rates. H2O(l) NaCl(s)
Prepared
Equilibrium
I2(H2O)
I2(H2O)
I2(CCl4)
I2(CCl4)
H2O(g) H2O
NaCl(aq)
CO(g) + 2 H2(g)
I2(H2O)
Dynamic Equilibrium
I2(CCl4)
CH3OH(g)
Reversible reactions: Both forward and reverse reactions take place
15-2 The Equilibrium Constant Expression • Methanol synthesis is a reversible reaction. CO(g) + 2 H2(g) CH3OH(g)
k-1
CO(g) + 2 H2(g)
k1
CH3OH(g)
Three Approaches to Equilibrium H2(g) + 2 ICl(g)
I2(g) + 2 HCl(g)
ICl
CO(g) + 2 H2(g) k1 k-1
CH3OH(g)
As is Ammonia production (Haber-Bosch process) N2(g) + 3H2(g)
2NH3(g)
H2 I2
The Equilibrium Constant: A kinetic perspective H2(g) + 2 ICl(g)
d[P]
I2(g) + 2 HCl(g)
dt
Mechanism:
= k[H2][ICl]
Briggs Raucher reaction:Oscillating reactions
At Equilibrium
Step 1
H2(g) + ICl(g)
HI(g) + HCl(g)
k1[H2][ICl] = k-1[HI][HCl]
Step 2
HI(g) + ICl(g)
I2(g) + HCl(g)
step1
[ HI ] =
k2[HI][ICl] = k-2[I2][HCl]
k1 [ H 2 ][ ICl ] step 2 k − 2 [ I 2 ][ HCl ] = k −1 [ HCl ] k 2 [ ICl ]
kk [ I 2 ][ HCl ]2 = 1 2 = Kc 2 k −1k − 2 [ H 2 ][ ICl ]
So,
The Equilibrium Constant and Activities; a
General Expressions
• Activity
a A + b B …. → g G + h H ….
– Thermodynamic concept introduced by Lewis. – Dimensionless ratio referred to a chosen reference state.
aB = γ B
[ B] cB0
coefficient of activity and cB0 is a standard reference state = 1 mol L-1
[G]g[H]h …. Equilibrium constant = Kc= [A]a[B]b …. Thermodynamic (a )g(a )h …. Equilibrium constant = Keq= G a H b …. (aA) (aB) =
[G]g[H]h …. ( γG)g(γH)h …. (a+b…)-(g+h…) × × [A]a[B]b …. (γA)a(γB)b …. (co)
•γB accounts for non-ideal behaviour in solutions and gases. •γB[B] can be considered an effective or active concentration.
15-3 Relationships Involving the Equilibrium Constant • When 2 chemical equations (with K1 and K2) are added, the overall equilibrium constant is K1×K2 • Reversing an equation causes inversion of K. • Multiplying all coefficients by a common factor raises the equilibrium constant to the corresponding power. • Dividing the coefficients by a common factor causes the equilibrium constant to be taken to that root.
≈1 under ideal conditions
= 1 mol L-1
Combining Equilibrium Constant Expressions N2O(g) + ½O2 N2(g) + ½O2 N2(g) + O2
2 NO(g) N2O(g)
2 NO(g)
Kc= ?
[N2O] Kc(2)= 2.7×10-18 =[N ][O ]½ 2 2 2 [NO] Kc(3)= 4.7×10-31 =[N ][O ] 2 2
[NO]2 1 [NO]2 [N2][O2]½ Kc= [N O][O ]½ = = Kc(3) K [N ][O ] [N O] 2 2 c(2) 2 2 2 = 1.7×10-13
Gases: The Equilibrium Constant, KP • Mixtures of gases are solutions just as liquids are. • Define KP, based on partial pressure (or activities) of gasses, rather than concentration 2 SO2(g) + O2(g)
aSO3 =
PSO3 P°
KP =
2 SO3(g)
aSO2 = KP =
PSO2
(PSO3 )2
2 SO2(g) + O2(g)
=
(aSO2 )2(aO 2) PO 2 P°
= P° RT
P°
2 SO2(g) + O2(g)
2 SO3(g)
PSO3 2 2 2 PSO3 ] [SO RT 3 = Kc = RT = 2 [SO2]2[O2] PSO 2 PO P P SO O 2 2 2 2 RT RT Kc = KP(RT)
;PX= n_xRT = [X] RT V
([SO3] RT)2
P°
([SO2] RT)2([O2] RT) [SO3]2
[SO2]2[O2]
=
KC RT
Where P° = 1 bar
(PSO2 )2(PO2)
Gases: The Equilibrium Constant, KC
2 SO3(g)
(PSO3)2 (a )2 KP = SO3 = P° 2 (PSO2)2(PO 2) (aSO2) (aO 2)
(aSO3 )2
aSO3 =
P°
Gases: The Equilibrium Constant, KC
Heterogeneous Reactions: Pure Liquids and Solids • Equilibrium constant expressions do not contain concentration terms for solid or liquid phases of a single component (that is, pure solids or liquids). C(s) + H2O(g)
CO(g) + H2(g)
KP = Kc(RT)-1
In general: KP = Kc(RT)Δn
15-4 The Significance of the Magnitude of the Equilibrium Constant.
Kc =
P P [CO][H2] = CO H2 (RT)1 [H2O] PH2O
15-4 The Significance of the Magnitude of the Equilibrium Constant. • Does a large equilibrium constant ensure that a reaction will take place if only reactants are present? – No!! The rate can be so low that no product is formed
Significant quantities of both reactants and products are likely to be present at equilibrium only if ~10-10 < K < ~1010
• Thermodynamic control: concentration of products and reactants depends on relative stability of the species. • Kinetic control: concentration of products and reactants depends rates of reaction (i.e. k and Ea)
15-5 The Reaction Quotient, Q: Predicting the Direction of Net Change. CO(g) + 2 H2(g)
k1 k-1
Three Approaches to Equilibrium CO(g) + 2 H2(g)
[G]tg[H]th… [A]ta[B]tb…
k-1
CH3OH(g)
CH3OH(g) H2
• Equilibrium can be approached various ways. • What changes in concentration do we expect as equilibrium is approached from the conditions prepared at some time, t? • The reaction quotient helps us predict:
Qc =
k1
CO
CH3OH
At equilibrium Qc = Kc
Reaction Quotient
15-6 Altering Equilibrium Conditions: Le Châtelier’s Principle • When an equilibrium system is subjected to a change in temperature, pressure, or the concentration of a reacting species, the system responds by attaining a new equilibrium that partially offsets the impact of the change.
What happens if we add some ‘extra’ SO3 to the reaction system at equilibrium?
Le Châtelier’s Principle 2 SO2(g) + O2(g)
k1 k-1
2 SO3(g)
Kc =
2.8×102
at 1000K
Effect of Changes in ‘P’ or ‘V’ • Adding a gaseous reactant or product changes Pgas. Equivalent to a change in concentration • Adding an inert gas changes the total pressure, but relative partial pressures are
unchanged.
[SO3]2 Q= [SO2]2[O2] = Kc
Q > Kc
• Changing the volume of the system causes a change in the equilibrium position. nSO32 2 V nSO3 Kc = [SO3] = = V [SO2]2[O2] nSO22 nO2 n 2n SO2 O2 V V
Effect of Change in Volume g
h g h Kc = [G] [H] = nG nH V(a+b)-(g+h) [A]a[B]b nAa nBa
2 SO2(g) + O2(g)
=
nAa
naB
V-Δn
• When the volume of an equilibrium mixture of gases is reduced, a net change occurs in the direction that produces fewer moles of gas. When the volume is increased, a net change occurs in the direction that produces more moles of gas.
Effect of a Catalyst on Equilibrium • A catalyst changes the reaction mechanism to one with a lower activation energy. • A catalyst has no effect on the condition of equilibrium: the relative thermodynamic stability of reactants and products in unchanged. A catalyst does affect the rate at which equilibrium is attained.
Synthesis of Ammonia Haber-Bosch process: N2(g) + 3H2(g)
k1 k-1
2 SO3(g)
ΔH° = -197.8 kJ/mol
• Raising the temperature of an equilibrium mixture shifts the equilibrium condition in the direction of the endothermic reaction.
g
nG nHh
Effect of Temperature on Equilibrium
2NH3(g)
The optimum conditions are only for the equilibrium position and do not take into account the rate at which equilibrium is attained.
•Lowering the temperature causes a shift in the direction of the exothermic reaction. What does this behaviour say about the temperature dependence of k1 compared to k-1?
15-7 Equilibrium Calculations: Some Illustrative Examples. • Five numerical examples are given in the text that illustrate ideas that have been presented in this chapter. • Refer to the “comments” which describe the methodology. These will help in subsequent chapters. • Exercise your understanding by working through the examples with a pencil and paper.
Problem Set: Ch15 7,9,13,21,23,27,36,47,51,55,56,69,79,83
15-1 Dynamic Equilibrium • Equilibrium – two opposing processes taking place at equal rates. H2O(l) NaCl(s)
Prepared
Equilibrium
I2(H2O)
I2(H2O)
I2(CCl4)
I2(CCl4)
H2O(g) H2O
NaCl(aq)
CO(g) + 2 H2(g)
I2(H2O)
Dynamic Equilibrium
I2(CCl4)
CH3OH(g)
Reversible reactions: Both forward and reverse reactions take place
15-2 The Equilibrium Constant Expression • Methanol synthesis is a reversible reaction. CO(g) + 2 H2(g) CH3OH(g)
k-1
CO(g) + 2 H2(g)
k1
CH3OH(g)
Three Approaches to Equilibrium H2(g) + 2 ICl(g)
I2(g) + 2 HCl(g)
ICl
CO(g) + 2 H2(g) k1 k-1
CH3OH(g)
As is Ammonia production (Haber-Bosch process) N2(g) + 3H2(g)
2NH3(g)
H2 I2
The Equilibrium Constant: A kinetic perspective H2(g) + 2 ICl(g)
d[P]
I2(g) + 2 HCl(g)
dt
Mechanism:
= k[H2][ICl]
Briggs Raucher reaction:Oscillating reactions
At Equilibrium
Step 1
H2(g) + ICl(g)
HI(g) + HCl(g)
k1[H2][ICl] = k-1[HI][HCl]
Step 2
HI(g) + ICl(g)
I2(g) + HCl(g)
step1
[ HI ] =
k2[HI][ICl] = k-2[I2][HCl]
k1 [ H 2 ][ ICl ] step 2 k − 2 [ I 2 ][ HCl ] = k −1 [ HCl ] k 2 [ ICl ]
kk [ I 2 ][ HCl ]2 = 1 2 = Kc 2 k −1k − 2 [ H 2 ][ ICl ]
So,
The Equilibrium Constant and Activities; a
General Expressions
• Activity
a A + b B …. → g G + h H ….
– Thermodynamic concept introduced by Lewis. – Dimensionless ratio referred to a chosen reference state.
aB = γ B
[ B] cB0
coefficient of activity and cB0 is a standard reference state = 1 mol L-1
[G]g[H]h …. Equilibrium constant = Kc= [A]a[B]b …. Thermodynamic (a )g(a )h …. Equilibrium constant = Keq= G a H b …. (aA) (aB) =
[G]g[H]h …. ( γG)g(γH)h …. (a+b…)-(g+h…) × × [A]a[B]b …. (γA)a(γB)b …. (co)
•γB accounts for non-ideal behaviour in solutions and gases. •γB[B] can be considered an effective or active concentration.
15-3 Relationships Involving the Equilibrium Constant • When 2 chemical equations (with K1 and K2) are added, the overall equilibrium constant is K1×K2 • Reversing an equation causes inversion of K. • Multiplying all coefficients by a common factor raises the equilibrium constant to the corresponding power. • Dividing the coefficients by a common factor causes the equilibrium constant to be taken to that root.
≈1 under ideal conditions
= 1 mol L-1
Combining Equilibrium Constant Expressions N2O(g) + ½O2 N2(g) + ½O2 N2(g) + O2
2 NO(g) N2O(g)
2 NO(g)
Kc= ?
[N2O] Kc(2)= 2.7×10-18 =[N ][O ]½ 2 2 2 [NO] Kc(3)= 4.7×10-31 =[N ][O ] 2 2
[NO]2 1 [NO]2 [N2][O2]½ Kc= [N O][O ]½ = = Kc(3) K [N ][O ] [N O] 2 2 c(2) 2 2 2 = 1.7×10-13
Gases: The Equilibrium Constant, KP • Mixtures of gases are solutions just as liquids are. • Define KP, based on partial pressure (or activities) of gasses, rather than concentration 2 SO2(g) + O2(g)
aSO3 =
PSO3 P°
KP =
2 SO3(g)
aSO2 = KP =
PSO2
(PSO3 )2
2 SO2(g) + O2(g)
=
(aSO2 )2(aO 2) PO 2 P°
= P° RT
P°
2 SO2(g) + O2(g)
2 SO3(g)
PSO3 2 2 2 PSO3 ] [SO RT 3 = Kc = RT = 2 [SO2]2[O2] PSO 2 PO P P SO O 2 2 2 2 RT RT Kc = KP(RT)
;PX= n_xRT = [X] RT V
([SO3] RT)2
P°
([SO2] RT)2([O2] RT) [SO3]2
[SO2]2[O2]
=
KC RT
Where P° = 1 bar
(PSO2 )2(PO2)
Gases: The Equilibrium Constant, KC
2 SO3(g)
(PSO3)2 (a )2 KP = SO3 = P° 2 (PSO2)2(PO 2) (aSO2) (aO 2)
(aSO3 )2
aSO3 =
P°
Gases: The Equilibrium Constant, KC
Heterogeneous Reactions: Pure Liquids and Solids • Equilibrium constant expressions do not contain concentration terms for solid or liquid phases of a single component (that is, pure solids or liquids). C(s) + H2O(g)
CO(g) + H2(g)
KP = Kc(RT)-1
In general: KP = Kc(RT)Δn
15-4 The Significance of the Magnitude of the Equilibrium Constant.
Kc =
P P [CO][H2] = CO H2 (RT)1 [H2O] PH2O
15-4 The Significance of the Magnitude of the Equilibrium Constant. • Does a large equilibrium constant ensure that a reaction will take place if only reactants are present? – No!! The rate can be so low that no product is formed
Significant quantities of both reactants and products are likely to be present at equilibrium only if ~10-10 < K < ~1010
• Thermodynamic control: concentration of products and reactants depends on relative stability of the species. • Kinetic control: concentration of products and reactants depends rates of reaction (i.e. k and Ea)
15-5 The Reaction Quotient, Q: Predicting the Direction of Net Change. CO(g) + 2 H2(g)
k1 k-1
Three Approaches to Equilibrium CO(g) + 2 H2(g)
[G]tg[H]th… [A]ta[B]tb…
k-1
CH3OH(g)
CH3OH(g) H2
• Equilibrium can be approached various ways. • What changes in concentration do we expect as equilibrium is approached from the conditions prepared at some time, t? • The reaction quotient helps us predict:
Qc =
k1
CO
CH3OH
At equilibrium Qc = Kc
Reaction Quotient
15-6 Altering Equilibrium Conditions: Le Châtelier’s Principle • When an equilibrium system is subjected to a change in temperature, pressure, or the concentration of a reacting species, the system responds by attaining a new equilibrium that partially offsets the impact of the change.
What happens if we add some ‘extra’ SO3 to the reaction system at equilibrium?
Le Châtelier’s Principle 2 SO2(g) + O2(g)
k1 k-1
2 SO3(g)
Kc =
2.8×102
at 1000K
Effect of Changes in ‘P’ or ‘V’ • Adding a gaseous reactant or product changes Pgas. Equivalent to a change in concentration • Adding an inert gas changes the total pressure, but relative partial pressures are
unchanged.
[SO3]2 Q= [SO2]2[O2] = Kc
Q > Kc
• Changing the volume of the system causes a change in the equilibrium position. nSO32 2 V nSO3 Kc = [SO3] = = V [SO2]2[O2] nSO22 nO2 n 2n SO2 O2 V V
Effect of Change in Volume g
h g h Kc = [G] [H] = nG nH V(a+b)-(g+h) [A]a[B]b nAa nBa
2 SO2(g) + O2(g)
=
nAa
naB
V-Δn
• When the volume of an equilibrium mixture of gases is reduced, a net change occurs in the direction that produces fewer moles of gas. When the volume is increased, a net change occurs in the direction that produces more moles of gas.
Effect of a Catalyst on Equilibrium • A catalyst changes the reaction mechanism to one with a lower activation energy. • A catalyst has no effect on the condition of equilibrium: the relative thermodynamic stability of reactants and products in unchanged. A catalyst does affect the rate at which equilibrium is attained.
Synthesis of Ammonia Haber-Bosch process: N2(g) + 3H2(g)
k1 k-1
2 SO3(g)
ΔH° = -197.8 kJ/mol
• Raising the temperature of an equilibrium mixture shifts the equilibrium condition in the direction of the endothermic reaction.
g
nG nHh
Effect of Temperature on Equilibrium
2NH3(g)
The optimum conditions are only for the equilibrium position and do not take into account the rate at which equilibrium is attained.
•Lowering the temperature causes a shift in the direction of the exothermic reaction. What does this behaviour say about the temperature dependence of k1 compared to k-1?
15-7 Equilibrium Calculations: Some Illustrative Examples. • Five numerical examples are given in the text that illustrate ideas that have been presented in this chapter. • Refer to the “comments” which describe the methodology. These will help in subsequent chapters. • Exercise your understanding by working through the examples with a pencil and paper.
Problem Set: Ch15 7,9,13,21,23,27,36,47,51,55,56,69,79,83
