chapter 17

Post on 20-Jan-2016

31 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

DESCRIPTION

Chapter 17. Equilibrium. 17.1 Reaction Rates and Equilibrium. To understand the collision model of chemical reactions To understand activation energy To understand how a catalyst speeds up a chemical reaction To explore reactions with reactants or products in different phases - PowerPoint PPT Presentation

TRANSCRIPT

Chapter 17

Equilibrium

17.1 Reaction Rates and Equilibrium

1. To understand the collision model of chemical reactions

2. To understand activation energy 3. To understand how a catalyst speeds up a chemical

reaction 4. To explore reactions with reactants or products in

different phases 5. To learn how equilibrium is established 6. To learn about the characteristics of chemical

equilibrium

• Rates of chemical reactions are determined by the details of the molecular collisions

How Chemical Reactions Occur

• How do atoms in reactants rearrange to form products?

• Collision model – molecules react by colliding with each other. – Some collisions are violent enough to break

bonds, allowing reactants to rearrange to form products

• Many reactions proceed via a series of elementary reactions

• Elementary reactions are mediated by collisions between molecules. Only collisions having sufficient energy and relative orientation of reactants lead to products

Conditions that Affect Reaction Rates

• Concentration of reactants –

• Temperature -

• Activation energy – the minimum amount of energy needed for a reaction to occur.– Collision > Ea:

– Collision < Ea:

• Catalyst - substance that speeds up a chemical reaction without being used in the reaction– Enzymes – biological catalyst– Work by lowering activation energy

• Catalysts function by lowering the activation energy of an elementary step in a reaction mechanism, and by providing a new and faster reaction mechanism

• Important classes in catalyst include acid-base catalysis, surface catalysis, and enzyme catalysis

Heterogeneous Reactions

• Homogeneous reactions – all reactants and products are in same phase

• Heterogeneous reactions – reactants are in different phases

• Heterogeneous reaction rates are affected by the surface area of the reactants– Lower surface area (smaller pieces) = faster

reaction

The Equilibrium Condition

• Equilibrium – exact balancing of two 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

Chemical Equilibrium: A Dynamic Condition

Equal numbers of moles of H2O and CO are mixed in a closed container.

The reaction begins to occur, and some products (H2 and CO2) are formed.

The reaction continues as time passes and more reactants are changed to products.

Although time continues to pass, the numbers of reactant and product molecules are the same as in (c). No further changes are seen as time continues to pass. The system has reached equilibrium.

• Why does equilibrium occur?– As concentration of reactants decreases, forward

reaction slows down– As concentration of products increases, reverse

reaction speeds up

17.2 Characteristics of Equilibrium

1. To understand the law of chemical equilibrium

2. To learn to calculate values for the equilibrium constant

3. To understand how the presence of solids or liquids affects the equilibrium expression

• Any bond or intermolecular attraction that can be formed can be broken. These two processes are in a dynamic competition, sensitive to initial conditions and external perturbations.

The Equilibrium Constant

• Law of chemical equilibrium– For a reaction of the type

• Each set of equilibrium concentrations is called an equilibrium position.

aA + bB ↔ cC + dD – Equilibrium expression

• Write the equilibrium expression for the following:– H2(g) + F2(g) ↔2HF(g)

– N2(g) + 3H2(g) ↔ 2NH3(g)

Initial Concentrations Equilibrium Concentrations

Exp. [N2] [H2] [NH3] [N2] [H2] [NH3] K1 1.00M 1.00M 0 0.921M 0.763M 0.157M 0.06022 0 0 1.00M 0.399M 1.197M 0.203M 0.06023 2.00M 1.00M 3.00M 2.59M 2.77M 1.82M 0.0602

• Value of K is always the same regardless of initial amounts of reactants and products

• Equilibrium concentrations are not always the same• Equilibrium position – a particular set of equilibrium

concentrations of all reactants and products in a chemical system

• Suppose that for the reaction: 2SO2(g) + O2(g)↔2SO3(g) it is determined that at a particular temperature the equilibrium concentrations are: [SO2] = 1.50M, [O2] = 1.25M, [SO3] = 3.50M. Calculate the value of K for this reaction.

Heterogeneous Equilibria

• An equilibrium system where all the products and reactants are not in the same state.

• Position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present because their concentrations cannot change

• Write the equilibrium expression for the decomposition of water:

2H2O(l) ↔ 2H2(g) + O2(g)

17.3 Application of Equilibria

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

2. To learn to calculate equilibrium concentrations

3. To learn to calculate the solubility product of a salt

4. To learn to calculate solubility from the solubility product

Le Chatelier’s Principle

• when 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

• Effect of a Change in ConcentrationN2(g) + 3H2(g) ↔ 2NH3(g)

• When a reactant or product is added to a system at equilibrium, the system shifts away from the added component

• If a reactant or product is removed from a system at equilibrium, the system shifts toward the removed component

• For the reaction: As4O6(s) + 6C(s) ↔ As4(g) + 6CO(g) Predict the direction of the shift in the equilibrium position in response to each of the following:– Addition of carbon monoxide– Addition or removal of C(s) or As4O6(s)

– Removal of As4(g)

• Effect of a Change in Volume – Decreasing the volume of a gas increases the pressure

The system is initially at equilibrium.

The piston is pushed in, decreasing the volume and increasing the pressure. The system shifts in the direction that consumes CO2 molecules, lowering the pressure again.

• Decreasing the volume – system shifts in direction that gives fewest number of gas molecules

• Increasing the volume – system shifts in the direction to increase the number of gas molecules (increase the pressure)

• Predict the direction of the equilibrium shift for each of the following when the volume of the container is decreased:– CO(g) + 2H2(g) ↔ CH3OH(g)

– H2(g) + F2(g) ↔ 2HF(g)

• Effect of a Change in Temperature– Value of K changes with temperature– Use Le Chatelier’s principle to predict the direct of

the change in K

• Exothermic reaction – heat is a product– Adding energy shifts equilibrium away from products

• Endothermic reaction – heat is a reactant– Adding energy shifts equilibrium away from reactants

• Predict the direction of the shift in equilibrium position the same way as if a product or reactant were added or removed

• For the exothermic reaction: 2SO2(g) + O2(g) ↔ 2SO3 (g) Predict the equilibrium shift caused by each of the following changes:– SO2 is added– SO3 is removed– Volume is decreased– Temperature is decreased

Applications Involving the Equilibrium Constant

• Inherent tendency of a reaction to occur• The meaning of K:

– K > 1 equilibrium position is far to the left– K < 1 equilibrium position is far to the right

• Calculation of equilibrium concentration of reactants and products

• Gaseous phosphorus pentachloride decomposes to chlorine gas and gaseous phosphorus trichloride. In a certain experiment, at a temperature where K = 8.96 x 10-2, the equilibrium concentrations of PCl5 = 6.70 x 10-3 M and PCl3 = 0.300 M. What was the equilibrium concentration of Cl2?

Solubility Equilibria

• CaF2(s) ↔ Ca2+ + 2F-

• At equilibrium, solution is saturated

• Ksp = [Ca2+][F-]2

• Ksp = solubility product constant

• Write the balanced equation describing the reaction for dissolving each of the following in water, also write the Ksp expression for each solid:– PbCl2(s)

– Ag2CrO4(s)

• The solubility of CuBr is 2.0 x 10-4 mol/L. Calculate the Ksp for copper (I) bromide.

• The Ksp value for solid AgI is 1.5 x 10-16. Calculate the solubility of AgI in water.

top related