Chapter 17: Equilibrium

17.1: How chemical reactions occur 17.2: Conditions that affect Reaction Rates 17.3 Heterogeneous Reaction

17.1 How Chemical Reactions Occur Objectives: To understand the collision model of how reactions occur.

17.1 How Chemical Reactions Occur How does a chemical reaction occur????

Atoms bump into each other hard enough to break bonds allowing the reactants to rearrange to form the products.

Collision model : says that reactions occur during molecular collisions.

A reaction proceeds faster -higher concentrations -higher temperatures

Figure 17.2: (a)Two BrNO molecules approach each other at high speeds.

(b) The collision occurs. (c) The energy of the collision causes Br-N bonds to break and Br-Br bonds to form.

(d) The products: one Br 2 molecules.

and two NO

17.2 Conditions that Affect Reaction Rates Objectives: To understand activation energy.

To understand how a catalyst speeds up a reaction.

17.2 Conditions that Affect Reaction Rates WHY DO REACTION RATES increase w/ Increasing Temperature?

Not all collisions have enough energy to break bonds Activation Energy : the minimum amount of energy necessary to break bonds.

Higher Temp  Higher speeds  More high-energy  More broken bonds  FASTER collisions

17.2 Conditions that Affect Reaction Rates Is it possible to speed up a reaction w/o changing the temperature or the reactant concentration?

YES!

Catalyst : a substance that speeds up a reaction without being consumed.

Enzymes : Biological catalysts Catalysts work by lowering the activation energy of a reaction.

17.2 Conditions that Affect Reaction Rates IMPT. EXAMPLES CO 2 (g) + H 2 O(l) H + (aq) + HCO 3 (aq) Catalyzed by carbonic anhydrase: prevents the accumulation of CO 2 in your blood Cl 2 +O 3 ClO + O 2 O + ClO Cl + O 2 O + O 3 2O 2 Freons such as CF 2 Cl 2 ; Banned in 1996.

17.3 Heterogeneous Reactions

Objectives: To consider reactions with reactants or products in different phases.

17.3 Heterogeneous Reactions

Homogeneous Reaction : reactions involving only one phase. (gas, liquid or solid) Heterogeneous Reaction : reactants in 2 phases.

17.3 Heterogeneous Reactions

Factors that Affect Reaction Rates Nature of reactants: substances vary greatly in their tendency to react depending on their bond strengths and structures.

Concentration(Pressure): Increase/Increase Temperature: Increase/Increase Surface Area : Heterogeneous rxns increase w/ increased surface area.

17.4 The Equilibrium Condition

Objectives: To learn how equilibrium is established.

REMEMBER THIS?????

Figure 17.8: (a) Net transfer of molecules from the liquid state to the vapor state.(b) The amount of the substance in the vapor state becomes constant (c) The equilibrium state.

17.4 The Equilibrium Condition

Equilibrium : exact balancing of 2 processes, one of which is the opposite of the other.

Chemical equilibrium : a dynamic state where the concentrations of all reactants and products remain constant.

Forward 2NO 2 (g) N 2 O 4 (g) Reverse

17.5 Chemical Equilibrium: A Dynamic Condition Objective: To learn about the characteristics of chemical equilibrium.

17.5 Chemical Equilibrium: A Dynamic Condition Objective: To learn about the characteristics of chemical equilibrium.

Figure 17.9: (a) Equal numbers of moles of H 2 O and CO are mixed in a closed container.

(b) The reaction begins to occur.

(c) The reaction continues, and more reactants are changed to products.

(d) No further changes are seen as time continues to pass.

RATE of the FORWARD REACTION EQUALS the RATE of the REVERSE REACTION

17.6 The Equilibrium Constant: An Introduction Objective: To understand the law of chemical equilibrium and to learn how to calculate values for the equilibrium constant.

17.6 The Equilibrium Constant: An Introduction 1864: Cato Maximilian Guldberg and Peter Waage LAW of Chemical Equilibrium (Law of Mass Action) aA + bB cC + dD A, B, C, D: reactants and products a, b, c, d: coefficients Equilibrium Expression: equilibrium constant K=[C] c [D] d [A] a [B] b [ ] mol/L

17.6 The Equilibrium Constant: An Introduction Equilibrium Expression: equilibrium constant K=[C] c [D] d [A] a [B] b 2O 3(g) 3O 2(g) K=[O 2 ] 3 [O 3 ] 2 [ ] mol/L The equilibrium constant means that for a given reaction at a given temperature the ratio of the concentrations of the products to reactants defined by the equilibrium expression will always be equal to the same number.

17.6 The Equilibrium Constant: An Introduction Write the equilibrium constant for the following CH 3 OH(g) CH 2 O(g) + H 2(g) 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6 H 2 O(g) [ ] mol/L

17.6 The Equilibrium Constant: An Introduction N 2 (g) + 3 H 2 (g) 2NH 3 K=6.02 x 10 -2 If [N 2 ]=0.921 M, [H 2 ]=0.763M and [NH 3 ]=0.157M

Each set of equilibrium concentrations is called an equilibrium position THERE IS only ONE EQUILIBRIUM CONSTANT, but INFINITE EQUILIBRIUM POSITIONS

17.7 Heterogeneous Equilibria

Homogeneous equilibria : all substances are in the same state.

Heterogenous equilibria : more than one state CaCO 3 (s) CaO(s) + CO 2 (g) Lime Does not depend on the amounts of pure solids or liquids present…... SO…… K=[CO 2 ]

17.7 Heterogeneous Equilibria

Write the expression for K for the following processes.

CO 2 (g) + H 2 O(l) H 2 CO 3 (l) ZrI 4 (s) Zr(s) + 2I 2 (g)

17.8 Le Chatelier’s Principle

Objective: To learn to predict the changes that occur when a system at equilibrium is disturbed.

17.8 Le Chatelier’s Principle

How do we understand the factors that control the position of a chemical equilibrium?

LeChatelier’s Principle : a change is imposed on a system at equilibrium, the position of the equilibrium, shifts in a direction that tends to reduce the effect of that change.

17.8 Le Chatelier’s Principle

The Effect of a Change in Concentration N 2 (g) + 3 H 2 (g) 2NH 3 Add 1 mol of N 2 What do you think will happen to equilibrium?

2) What about the equilibrium constant?

17.8 Le Chatelier’s Principle

The Effect of a Change in Concentration N 2 (g) + 3 H 2 (g) 2NH 3 Equilibrium Position I Equilibrium Position II [N 2 ]= 0.399 M [H 2 ]= 1.197 M [NH 3 ]= 0.203 M [N 2 ]= 1.348M

[H 2 ]= 1.044 M [NH 3 ]=0.304 M Position I: [NH 3 ] 2 [N 2 ][H 2 ] 3 = (0.203) 2 (0.399)(1.197) 3 Position II: ?

LeChatelier’s says: when a reactant or product is added to a system at equilibrium, the system shifts away from the added component.

17.8 Le Chatelier’s Principle

A real-life example: In the mountains, if you feel dizzy why?

Lower air pressure, lower supply of oxygen Hb(aq) + 4O 2 (g) Hb(O 2 ) 4 (aq)

17.8 Le Chatelier’s Principle

2SO 2 (g) + O 2 (g) 2SO 3 (g) What would happen if SO 2 (g) is added to the system?

The SO 3 (g) present is liquefied and removed from the system?

Some of the O 2 (g) is removed from the system.

17.8 Le Chatelier’s Principle

The Effect of a Change in Volume CaCO 3 (s) CaO(s) + CO 2( g) Shift to the left When the volume of a gas is DECREASED, the PRESSURE INCREASES.

By Le Chatelier’s principle, the system will shift in the direction that reduces the pressure.

When the volume of a gas is decreased, the system shifts in the direction that gives the smaller number of gas molecules.

17.8 Le Chatelier’s Principle

N 2 (g) + 3 H 2 (g) 2NH 3 4 molecules 2 gaseous molecuels If we reduce the volume, what happens?

Equilibrium shifts to the right

17.8 Le Chatelier’s Principle

Predict the shift in equilibrium when volume is increased 4NH 3 (g) + 5 O 2 (g) 4NO(g) + 6H 2 O(g) Shift to the right NH 4 NO 3 (s) 4N 2 O(g) + 2H 2 O(g) Shift to the right

17.8 LeChatelier’s Principle

The Effect of a Change in Temperature This affects the equilibrium constant In an endothermic reaction, heat can be considered a reactant.

In an exothermic reaction, heat can be considered a product.

N 2 (g) + 3 H 2 (g) 2NH 3 + 92 kJ CaCO 3 (s) +556kJ CaO(s) + CO 2( g) In summary to use LeChatelier’s principle: treat heat as a reactant or product.

17.8 Le Chatelier’s Principle

The reaction C 2 H 2 (g) + 2Br 2 (g) C 2 H 2 Br 4 (g) This is exothermic…an increase in temperature shifts the equilibrium Shifts to the left ZrI 4 (s) Zr(s) + 2 I 2 (g) This is endothermic…an increase in temperature will shift equilibrium to… Shifts to the right

17.9 Applications Involving the Equilibrium Constant • Objective: To learn to calculate equilibrium concentrations from equilibrium constants.

17.9 Applications Involving the Equilibrium Constant K tells us: >1 that the equilibrium favors products <1 that the equilibrium favors reactants Also, we can calculate the equilibrium constant.

17.9 Applications Involving the Equilibrium Constant Assume the equilibrium constant for the reaction H 2 (g) + F 2 (g) 2HF(g) Has the value 2.1x10

3 at a particular temperature. When the system is analyzed at equilibrium at this temperature, the concentrations of both H 2 (g) and F 2 (g) are 0.0021 M. What is the concentration of HF(g) in the equilibrium system under these conditions?

[HF(g)]= 9.6 x 10 -2 M

17.9 Applications Involving the Equilibrium Constant Assume the equilibrium constant for the reaction 2H 2 O(g) 2H 2 (g) + O 2 (g) Has the value 2.4x10

-3 at a particular temperature. When the system is analyzed at equilibrium at this temperature, it is found that [H2O(g)]=1.1 x 10-1 M and [H 2 (g)]=1.9x10-2M. What is the concentration of O 2 (g) in the equilibrium system under these conditions?

[O2(g)]= 8.0 x 10 -2 M

17.10 Solubility Equilibria

Objective: To learn to calculate the solubility product of a salt given its solubility and vice versa.

17.10 Solubility Equilibria

Consider the equilibria associated with dissolving solids in water to form aqueous solutions.

CaF 2 (s)  Ca 2+ (aq) + 2F (aq) K sp =[Ca 2+ ][F-] 2 K sp =solubility product constant or solubility product So where’s the solid….dissolving happens at the surface so if you increase the surface increase the rates in both directions.

17.10 Solubility Equilibria

Write the balanced equation for dissolving each of the following solids in water. Also write the Ksp expression for each solid.

Cr(OH) 3 Ca 3 (PO 4 ) 2

17.10 Solubility Equilibria

Zinc carbonate, ZnCO 3 (s), dissolves in water to give a solution that is 1.7 x 10 -5 M at 22 o C. Calculate K sp for ZnCO 3 (s) at this temperature (ignoring the fact that CO 3 2 reacts with water).

Ksp=2.9x10-10

17.10 Solubility Equilibria

Copper(II) chromate, CuCrO 4 (s) dissolves in water to give a solution that contains 1.1 x 10 -5 g CuCrO 4 per liter at 23 o C. Calculate K sp for CuCrO 4 (s) at this temperature.

Ksp=3.8x 10-15

17.10 Solubility Equilibria

PbCrO 4 (s) Pb 2+ + CrO 4 2 Calculate the solubility at 25 o C if the Ksp value is 2.0 x 10 -16