Chapter 16 Equilibrium Collision Theory Collision theory 1

Chapter 16 Equilibrium

Collision Theory • Collision theory: 1. atoms, ions, and molecules must collide in order to react. – Only a small number of collisions produce reactions

2. Reacting substances must collide in the correct orientation. CO + NO 2 CO 2 + NO

3. Reacting substances must collide with sufficient energy to form an activated complex. – Activated complex (transition state) – temporary, unstable arrangement of atoms in which old bonds are breaking and new bonds are forming.

• Activation energy (Ea) – the minimum amount of energy needed to form the activated complex and lead to a reaction – Collision > Ea – Collision < Ea

Exothermic Reaction

Endothermic Reaction

Factors Affecting Reaction Rates • Nature of reactants – Some things are more reactive than others (activity series) • Concentration of reactants • Temperature • Catalysts

• Lowers activation energy required for a reaction to take place

Chemical Equilibrium • Equilibrium – the exact balancing of two processes, one of which is the exact opposite of the other

Reversible reactions and chemical equilibrium • Not all reactions go to completion • Reversible reaction – a reaction that occurs in both the forward and reverse direction – Forward: – Reverse: – Both occurring:

• Chemical equilibrium – a dynamic state in which the forward and reverse reactions balance each other out because they take place at equal rates – Rate forward reaction = Rate reverse reaction – Concentration of all reactants and products remains constant

Equal numbers of moles of H 2 O and CO are mixed in a closed container. The reaction begins to occur, and some products (H 2 and CO 2) 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

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Equilibrium expressions • Law of chemical equilibrium – at a given temperature a chemical system reaches a state in which the ratio of reactant and product concentration has a constant value

The Equilibrium Constant • Law of chemical equilibrium – For a reaction of the type a. A + b. B ↔ c. C + d. D – Equilibrium expression • Each set of equilibrium concentrations is called an equilibrium position.

• Interpreting Keq – Keq > 1: – Keq < 1:

Homogeneous equilibria • All reactants and products are in the same physical state • Write the equilibrium expression for the following: – H 2(g) + F 2(g) ↔ 2 HF(g) – N 2(g) + 3 H 2(g) ↔ 2 NH 3(g)

• N 2 O 4(g) ↔ 2 NO 2(g) • CO(g) + 3 H 2(g) ↔ CH 4(g) + H 20(g)

• Suppose that for the reaction: 2 SO 2(g) + O 2(g)↔ 2 SO 3(g) it is determined that at a particular temperature the equilibrium concentrations are: [SO 2] = 1. 50 M, [O 2] = 1. 25 M, [SO 3] = 3. 50 M. Calculate the value of K for this reaction.

Heterogeneous equilibria • Reactants and products are present in more than one state

• Keq does not depend on amounts of solids or liquids because their concentration cannot change.

• Write the equilibrium expression for the following: • 2 Na. HCO 3(s) ↔ Na 2 CO 3(s) + CO 2(g) + H 2 O(g) • 2 H 2 O(l) ↔ 2 H 2(g) + O 2(g)

Using Equilibrium Constants • Equilibrium constant expressions can be used to calculate concentrations and solubilities

Calculating equilibrium concentrations • At a temperature of 1405 K, hydrogen sulfide decomposes to form hydrogen and diatomic sulfur according to the reaction: 2 H 2 S(g) ↔ 2 H 2(g) + S 2(g). The equilibrium constant for the reaction is 2. 27 x 10 -3. What is the concentration of hydrogen gas if [S 2] = 0. 0540 M and [H 2 S] = 0. 184 M

• PCl 5 ↔ PCl 3 + Cl 2. In a certain experiment, at a temperature where K = 8. 96 x 10 -2, the equilibrium concentrations of PCl 5 = 6. 70 x 10 -3 M and PCl 3 = 0. 300 M. What was the equilibrium concentration of Cl 2?

Equilibrium Constants • For a reaction at a constant temperature Keq will always be the same regardless of initial concentration of reactants and products N 2 + H 2 ↔ 2 NH 3 Initial Concentrations Equilibrium Concentrations Exp. [N 2] [H 2] [NH 3] [N 2] [H 2] 1 1. 00 M 0 0. 921 M 0. 763 M 0. 157 M 0. 0602 2 0 0 1. 00 M 0. 399 M 1. 197 M 0. 203 M 0. 0602 3 2. 00 M 1. 00 M 3. 00 M 2. 59 M 2. 77 M [NH 3] 1. 82 M K 0. 0602

Equilibrium Characteristics 1. Reaction must take place in a closed system 2. Temperature must remain constant 3. All reactants and products are in constant dynamic motion

Factors Affecting Chemical Equilibrium • When changes are made to a system at equilibrium the system shifts to a new equilibrium position

• Le Chatelier’s principle: if a stress is applied to a system at equilibrium, the system shifts in the direction that relieves that stress.

Applying Le Chatelier’s Principle • Change in Concentration N 2(g) + 3 H 2(g) ↔ 2 NH 3(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: As 4 O 6(s) + 6 C(s) ↔ As 4(g) + 6 CO(g) Predict the direction of the shift in the equilibrium position in response to each of the following: – Addition of carbon monoxide – Addition of As 4 O 6(s) – removal of C(s) – Removal of As 4(g)

• Changes in volume and pressure (gas only) – Remember: for gasses there is an inverse relationship between volume and 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 CO 2 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) + 2 H 2(g) ↔ CH 3 OH(g) – H 2(g) + F 2(g) ↔ 2 HF(g)

• Change in temperature: • 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

• Decide whether higher or lower temperatures will produce more CH 3 CHO in the reaction: C 2 H 2(g) + H 2 O(g) ↔ CH 3 CHO ΔH = -151 k. J

• For the exothermic reaction: 2 SO 2(g) + O 2(g) ↔ 2 SO 3 (g) Predict the equilibrium shift caused by each of the following changes: – SO 2 is added – SO 3 is removed – Volume is decreased – Temperature is decreased – Ne(g) is added

• Catalysts – speeds up a reaction equally in both directions therefore does not affect equilibrium

The Solubility Product Constant • Upon dissolving, all ionic compounds dissociate into ions • Na. Cl(s) • Ba. SO 4(s) • Mg(OH)2(s)

Solubility Equilibria • Ca. F 2(s) ↔ Ca 2+(aq) + 2 F-(aq) • At equilibrium, solution is saturated • Ksp = [Ca 2+][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: – Pb. Cl 2(s) – Ag 2 Cr. O 4(s)

• Small Ksp = not very soluble compound

• The Ksp value for solid Ag. I is 1. 5 x 10 -16. Calculate the solubility of Ag. I in water.

• The Ksp value for solid Cu. CO 3 is 2. 5 x 10 -10. Calculate the solubility of Ag. I in water.

• The solubility of Cu. Br is 2. 0 x 10 -4 mol/L. Calculate the Ksp for copper (I) bromide.