Br
UNIVERSITY OF ALBERTA INTRODUCTORY UNIVERSITY CHEMISTRY II (CHEM 102) Chemical Equilibrium Introduction • •
Many physical / chemical processes are reversible. If the reversible process is in a closed vessel (neither the reactant nor the product can escape), a state of equilibrium will develop.
For example: I2 in water
I2 in dichloromethane
Water
Saturated salt vsolution NaCl
H2O (l)
H2O (g)
NaCl (s)
Na+ (aq) + Cl- (aq)
I2 (aq)
I2 (CH2Cl2)
Chemical equilibrium In the following reversible reaction: N2O4 (g) Colorless gas
2NO2 (g) Brown color gas
Reaction starts from NO2
Reaction starts from N2O4 Equilibrium is reached
1
Characteristics of equilibrium state: 1. The equilibrium is dynamic (i.e. both the forward and backward reactions are still reacting but there is no net change) 2. Rate of forward reaction = rate of reversed reaction 3. The concentrations of the reactants and products are constant The Equilibrium Constant Expression For a chemical reaction: aA
+
bB
cC
+
dD
Both the forward and reversed reactions are elementary steps and so, Rate (forward) = k (forward) [A]a x [B]b k (forward) = rate constant of the forward reaction. Rate (reversed) = k (reversed) [C]c x [D]d k (reversed) = rate constant of the reversed reaction. Since at equilibrium, rate (forward) = rate (reversed) k (forward) [A]a x [B]b = k (reversed) [C]c x [D]d [C]c x [D]d k (forward) / k (reversed) = • • •
[A]a x [B]b
= Kc
Kc is called the equilibrium constant and it is dependent only on the temperature. The ratio is called the equilibrium constant expression. In most cases, the values of K are shown as dimensionless numbers. This is because in thermodynamics, the concentration is expressed as the ‘activity’ which is without unit.
For the reaction: N2O4 (g)
2NO2 (g) [NO2]2
Kc = [N2O4]
2
Significance of K K is an indication of how far a reaction proceeds toward product at a given temperature. •
Small K:
2NO (g) K = 1 x 10-30
N2 (g) + O2 (g)
Very little product is formed. There is actually no reaction. •
Large K:
2CO2 (g) K = 2.2 x 1022
2CO (g) + O2 (g)
High concentration of product is formed. The reaction goes to completion •
Intermediate K: 2BrCl (g)
Br2 (g) + Cl2 (g)
K=5
The reaction mixture contains a considerate amount of both the reactant and product.
Equilibrium position Equilibrium position is the concentrations of the reactants and the products at equilibrium. For the chemical reaction: N2(g) + 3H2(g)
2NH3(g)
The equilibrium constant expression is: [NH3]2 Kc =
[N2][ H2] 3
The following are two different sets of equilibrium positions at 127oC: [0.203]2 Kc =
= 0.0602 [0.399][1.197] 3 [1.82]2
Kc =
= 0.0602 [2.59][2.77]
3
Thus, there is only one equilibrium constant for a particular system at a particular temperature but you can have an infinite number of equilibrium positions.
3
Equilibria Involving Gases The ideal gas law:
PV = nRT n P=
RT V
Concentration Thus partial pressure of a gas can be used to express concentration if T is kept constant. 2SO2 (g)
O2 (g) 2SO3 (g)
+
pSO32 Kp =
pSO22 x pO2
pSO3 , pSO2, pO2 are the partial pressures of SO3, SO2 and O2 respectively. Reaction Quotient (Q) – Direction of a chemical reaction How does a reaction behave when it is not at equilibrium? Determine the reaction quotient, Q of the reaction. aA
Qc =
[C]c [D]d [A ]a [B]b
+
bB cC
Kc =
+
dD
[C]c eq [D]deq [A ]aeq [B]beq
If Q < K System is not at equilibrium, it needs to form more products. Reaction system is shifting to the right (forward reaction favored) If Q > K System is not at equilibrium, it needs to form more reactants. Reaction system is shifting to the left (reverse reaction favored) If Q = K at a given T, the system is at equilibrium.
4
For reaction: N2O4 (g) Q>K
2NO2 (g) Q=K
Q 4.0 x 10-5 0.0486 Therefore, equilibrium will shift to form more reactant What is going to happen if some PCl3 is removed?
7
**In heterogeneous equilibria, no change would occur if solid or a pure liquid is added or removed provided there is some in the equilibrium mixture. For example: CaCO3 (s) CaO (s) + CO2 (g) No effect for any change in the amount of CaCO3 or CaO.
Pressure Change Only applies to equilibrium system involving gases. PCl5 (g) PCl3 (g) + Cl2 (g) If the pressure of the system is increased (by decreasing the volume), the equilibrium is shifted to the side with lower pressure (the side with less number of gas molecules) Equilibrium therefore shifts to the left. More PCl5 is formed. What happens if pressure is decreased? Which direction does the equilibrium shift? For the equilibrium: H2 (g) + I2 (g) 2HI (g) Does pressure change affect this equilibrium? Why?
Adding an inert gas to an equilibrium system N2 (g) + 3H2 (g) 2NH3 (g) If an inert gas such as argon is added, the total pressure would be increased. However, the partial pressure of the gases is still the same. When the partial pressure remains the same, their concentration remains the same and so there is no change in the equilibrium. Therefore, adding an inert gas has no effect on the equilibrium position. Temperature Change Remember: Only temperature change affects the value of K ∆Ho ln = K1 R K2
1
1 -
T1
T2
Equation is very similar to the Arrhenius equation
The Van’t Hoff equation (we will see in the section of Thermodynamics) The direction depends on the sign of ∆H for the reaction: 8
For endothermic reactions (∆H = +): PCl5 (g) PCl3 (g) + Cl2 (g) ∆H = +111 kJ What is going to happen if we apply heat to an endothermic reaction? According to the Le Chatelier Principle, it would shift to the direction where heat is removed (i.e. it moves towards the endothermic side of the reaction) How will an increase in temperature influence the reactions below? PCl3 (g) + Cl2 (g) PCl5 (g) ΔH = -111 kJ ΔH = +151 kJ
CH3CHO (g) C2H2 (g) + H2O (g)
Effect of a catalyst A catalyst speeds up both forward and reverse reactions. It does not affect the position of equilibrium. It simply speeds up the equilibrium. N2 (g) + 3H2 (g) 2NH3 (g) The catalyst used in this process (Haber Process) is iron. It does not increase the yield of ammonia but it shortens the time to reach equilibrium.
How to solve equilibrium problems (Quantitative aspects): Calculation of Equilibrium Concentrations from given initial concentrations When equilibrium concentrations are required from given initial concentrations, we need to use a reaction table (ICE table) which relates them: Example problems 1 For the following reaction: H2 (g) + F2 (g) 2HF (g) The equilibrium constant Kc at 100oC is 1.15 x 102 If 3.000 mol of H2 and 6.000 mol of F2 are mixed in a 3.000 L flask, find the equilibrium concentrations of each component. The equilibrium constant expression: [HF]2 Kc = [H2] [F2] 9
The initial concentrations: [H2] = 3.000 mol / 3.000 L = 1.000 M [F2] = 6.000 mol / 3.000 L = 2.000 M Construct ICE [H2]
[F2]
[HF]
I C E Substitute into equilibrium constant expression:
2
Phosphorus pentachloride (PCl5) decomposes to phosphorus trichloride (PCl3) and Cl2 according to the following equation: PCl5 (g) PCl3 (g) + Cl2 (g) Find the equilibrium concentration of each component if 0.10 mol of PCl5 was introduced into a 2.0 L flask at 35oC. The equilibrium constant Kc for the reaction at 35oC is 4.2 x 10-5
10
The initial concentration:
Construct ICE [PCl5]
[PCl3]
I C E
Substitute into equilibrium constant expression:
11
[Cl2]
3
For the following reaction:
H2S (g) + I2 (s) 2HI (g) + S (s) Kp = 1.34 x 10-5 at 60oC If H2S (g) at 747.6 mm Hg pressure is added to excess I2 (s) in a flask at 60oC. What will be the total pressure of the flask at equilibrium?
P H2S
P HI
I C E
12
Many physical / chemical processes are reversible. If the reversible process is in a closed vessel (neither the reactant nor the product can escape), a state of equilibrium will develop.
For example: I2 in water
I2 in dichloromethane
Water
Saturated salt vsolution NaCl
H2O (l)
H2O (g)
NaCl (s)
Na+ (aq) + Cl- (aq)
I2 (aq)
I2 (CH2Cl2)
Chemical equilibrium In the following reversible reaction: N2O4 (g) Colorless gas
2NO2 (g) Brown color gas
Reaction starts from NO2
Reaction starts from N2O4 Equilibrium is reached
1
Characteristics of equilibrium state: 1. The equilibrium is dynamic (i.e. both the forward and backward reactions are still reacting but there is no net change) 2. Rate of forward reaction = rate of reversed reaction 3. The concentrations of the reactants and products are constant The Equilibrium Constant Expression For a chemical reaction: aA
+
bB
cC
+
dD
Both the forward and reversed reactions are elementary steps and so, Rate (forward) = k (forward) [A]a x [B]b k (forward) = rate constant of the forward reaction. Rate (reversed) = k (reversed) [C]c x [D]d k (reversed) = rate constant of the reversed reaction. Since at equilibrium, rate (forward) = rate (reversed) k (forward) [A]a x [B]b = k (reversed) [C]c x [D]d [C]c x [D]d k (forward) / k (reversed) = • • •
[A]a x [B]b
= Kc
Kc is called the equilibrium constant and it is dependent only on the temperature. The ratio is called the equilibrium constant expression. In most cases, the values of K are shown as dimensionless numbers. This is because in thermodynamics, the concentration is expressed as the ‘activity’ which is without unit.
For the reaction: N2O4 (g)
2NO2 (g) [NO2]2
Kc = [N2O4]
2
Significance of K K is an indication of how far a reaction proceeds toward product at a given temperature. •
Small K:
2NO (g) K = 1 x 10-30
N2 (g) + O2 (g)
Very little product is formed. There is actually no reaction. •
Large K:
2CO2 (g) K = 2.2 x 1022
2CO (g) + O2 (g)
High concentration of product is formed. The reaction goes to completion •
Intermediate K: 2BrCl (g)
Br2 (g) + Cl2 (g)
K=5
The reaction mixture contains a considerate amount of both the reactant and product.
Equilibrium position Equilibrium position is the concentrations of the reactants and the products at equilibrium. For the chemical reaction: N2(g) + 3H2(g)
2NH3(g)
The equilibrium constant expression is: [NH3]2 Kc =
[N2][ H2] 3
The following are two different sets of equilibrium positions at 127oC: [0.203]2 Kc =
= 0.0602 [0.399][1.197] 3 [1.82]2
Kc =
= 0.0602 [2.59][2.77]
3
Thus, there is only one equilibrium constant for a particular system at a particular temperature but you can have an infinite number of equilibrium positions.
3
Equilibria Involving Gases The ideal gas law:
PV = nRT n P=
RT V
Concentration Thus partial pressure of a gas can be used to express concentration if T is kept constant. 2SO2 (g)
O2 (g) 2SO3 (g)
+
pSO32 Kp =
pSO22 x pO2
pSO3 , pSO2, pO2 are the partial pressures of SO3, SO2 and O2 respectively. Reaction Quotient (Q) – Direction of a chemical reaction How does a reaction behave when it is not at equilibrium? Determine the reaction quotient, Q of the reaction. aA
Qc =
[C]c [D]d [A ]a [B]b
+
bB cC
Kc =
+
dD
[C]c eq [D]deq [A ]aeq [B]beq
If Q < K System is not at equilibrium, it needs to form more products. Reaction system is shifting to the right (forward reaction favored) If Q > K System is not at equilibrium, it needs to form more reactants. Reaction system is shifting to the left (reverse reaction favored) If Q = K at a given T, the system is at equilibrium.
4
For reaction: N2O4 (g) Q>K
2NO2 (g) Q=K
Q 4.0 x 10-5 0.0486 Therefore, equilibrium will shift to form more reactant What is going to happen if some PCl3 is removed?
7
**In heterogeneous equilibria, no change would occur if solid or a pure liquid is added or removed provided there is some in the equilibrium mixture. For example: CaCO3 (s) CaO (s) + CO2 (g) No effect for any change in the amount of CaCO3 or CaO.
Pressure Change Only applies to equilibrium system involving gases. PCl5 (g) PCl3 (g) + Cl2 (g) If the pressure of the system is increased (by decreasing the volume), the equilibrium is shifted to the side with lower pressure (the side with less number of gas molecules) Equilibrium therefore shifts to the left. More PCl5 is formed. What happens if pressure is decreased? Which direction does the equilibrium shift? For the equilibrium: H2 (g) + I2 (g) 2HI (g) Does pressure change affect this equilibrium? Why?
Adding an inert gas to an equilibrium system N2 (g) + 3H2 (g) 2NH3 (g) If an inert gas such as argon is added, the total pressure would be increased. However, the partial pressure of the gases is still the same. When the partial pressure remains the same, their concentration remains the same and so there is no change in the equilibrium. Therefore, adding an inert gas has no effect on the equilibrium position. Temperature Change Remember: Only temperature change affects the value of K ∆Ho ln = K1 R K2
1
1 -
T1
T2
Equation is very similar to the Arrhenius equation
The Van’t Hoff equation (we will see in the section of Thermodynamics) The direction depends on the sign of ∆H for the reaction: 8
For endothermic reactions (∆H = +): PCl5 (g) PCl3 (g) + Cl2 (g) ∆H = +111 kJ What is going to happen if we apply heat to an endothermic reaction? According to the Le Chatelier Principle, it would shift to the direction where heat is removed (i.e. it moves towards the endothermic side of the reaction) How will an increase in temperature influence the reactions below? PCl3 (g) + Cl2 (g) PCl5 (g) ΔH = -111 kJ ΔH = +151 kJ
CH3CHO (g) C2H2 (g) + H2O (g)
Effect of a catalyst A catalyst speeds up both forward and reverse reactions. It does not affect the position of equilibrium. It simply speeds up the equilibrium. N2 (g) + 3H2 (g) 2NH3 (g) The catalyst used in this process (Haber Process) is iron. It does not increase the yield of ammonia but it shortens the time to reach equilibrium.
How to solve equilibrium problems (Quantitative aspects): Calculation of Equilibrium Concentrations from given initial concentrations When equilibrium concentrations are required from given initial concentrations, we need to use a reaction table (ICE table) which relates them: Example problems 1 For the following reaction: H2 (g) + F2 (g) 2HF (g) The equilibrium constant Kc at 100oC is 1.15 x 102 If 3.000 mol of H2 and 6.000 mol of F2 are mixed in a 3.000 L flask, find the equilibrium concentrations of each component. The equilibrium constant expression: [HF]2 Kc = [H2] [F2] 9
The initial concentrations: [H2] = 3.000 mol / 3.000 L = 1.000 M [F2] = 6.000 mol / 3.000 L = 2.000 M Construct ICE [H2]
[F2]
[HF]
I C E Substitute into equilibrium constant expression:
2
Phosphorus pentachloride (PCl5) decomposes to phosphorus trichloride (PCl3) and Cl2 according to the following equation: PCl5 (g) PCl3 (g) + Cl2 (g) Find the equilibrium concentration of each component if 0.10 mol of PCl5 was introduced into a 2.0 L flask at 35oC. The equilibrium constant Kc for the reaction at 35oC is 4.2 x 10-5
10
The initial concentration:
Construct ICE [PCl5]
[PCl3]
I C E
Substitute into equilibrium constant expression:
11
[Cl2]
3
For the following reaction:
H2S (g) + I2 (s) 2HI (g) + S (s) Kp = 1.34 x 10-5 at 60oC If H2S (g) at 747.6 mm Hg pressure is added to excess I2 (s) in a flask at 60oC. What will be the total pressure of the flask at equilibrium?
P H2S
P HI
I C E
12
