Part 2 Objectives – Explain and describe equilibrium in terms of molecular motion (when forward and reverse reaction rates are equal) – Be able to write an equilibrium constant expression for a reaction – Use Le Chatelier’s Principle to predict effects of changes on equilibrium mixtures

Chemical Equilibrium Ch Equilibrium: A State of Dynamic Balance When a reaction results in almost complete conversion of reactants to products, chemists say the reaction goes to completion. Example: Combustion reaction (products removed quickly, escape) C 6 H 12 O 6 + 6O 2  6CO 2 + 6H 2 O

Section 18.1 Equilibrium: A State of Dynamic Balance Reversible reactions, can occur in both the forward and the reverse direction. Example: lots of reactions in aqueous solution and in gaseous phase N 2 O 4 (g)  2NO 2(g)

Section 18.1 Equilibrium: A State of Dynamic Balance Chemical equilibrium is a state in which the forward and reverse reactions take place at equal rates. U6A

Section 18.1 Equilibrium: A State of Dynamic Balance In a sealed bottle half full of water, equilibrium will be attained when water molecules H 2 O (l) H 2 O (g)

Section 18.1 Equilibrium: A State of Dynamic Balance In a sealed bottle half full of water, equilibrium will be attained when water molecules evaporate and condense at equal rates. H 2 O (l) H 2 O (g) Note that this is not a chemical reaction, but there are molecules moving and changing phase (physical state)

Section 18.1 Equilibrium: A State of Dynamic Balance Chemical equilibrium is a state in which the forward and reverse reactions take place at equal rates. At equilibrium, the concentrations of reactants and products are constant, but that does not mean that the amounts or concentrations are equal. N 2 O 4 (g)  2NO 2(g)

Section 18.1 Equilibrium: A State of Dynamic Balance Chemical equilibrium is a state in which the forward and reverse reactions take place at equal rates. At equilibrium, the concentrations of reactants and products are constant, but that does not mean that the amounts or concentrations are equal. N 2 O 4 (g)  2NO 2(g) as fast as NO 2 forms it also reacts to form N 2 O 4 NO observable change in [reactants] or [products]

Section 18.1 Equilibrium: A State of Dynamic Balance Equilibrium is a state of action not inaction. N 2 O 4 (g)  2NO 2(g)

N 2(g) + 3H 2(g) 2NH 3(g)

The Law of Chemical Equilibrium The law of chemical equilibrium states that at a given TEMPERATURE, a chemical system may reach a state in which a particular ratio of reactant to product concentrations has a constant value. aA + bBcC + dD

The Law of Chemical Equilibrium The law of chemical equilibrium states that at a given TEMPERATURE, a chemical system may reach a state in which a particular ratio of reactant to product concentrations has a constant value. aA + bBcC + dD K eq = [C] c [D] d [A] a [B] b

The Law of Chemical Equilibrium aA + bBcC + dD K eq = [C] c [D] d [A] a [B] b This ratio is called the equilibrium constant expression. Keq is the equilibrium constant and it is the numerical value of the ratio of product concentrations to reactant concentrations raised to the power corresponding to the coefficient in the balanced equation.

The Law of Chemical Equilibrium aA + bBcC + dD K eq = [C] c [D] d [A] a [B] b The value of Keq is constant only at a specified Temperature. (If change Temperature, equilibrium shifts until a new equilibrium is established and there will be a new Keq for that particular temperature.)

If an equilibrium constant has a value less than one, the reactants are favored in the equilibrium. K eq = [products] x = 1 [reactants] y >1 If an equilibrium constant has a value greater than one, the products are favored in the equilibrium. K eq = [products] x = >1 [reactants] y 1

Heterogeneous equilibrium reactants and products are present in more then one state. Example a gas and a solid or a liquid and a gas. Homogeneous equilibrium H 2(g) + I 2(g) 2HI (g) all the reactants and products in the same phase Keq =

Heterogeneous equilibrium reactants and products are present in more then one physical state. Example a gas and a solid or a liquid and a gas. Homogeneous equilibrium H 2(g) + I 2(g) 2HI (g) all the reactants and products in the same phase Keq = [HI] 2 [H 2 ] [I 2 ]

§18.2 Factors Affecting Chemical Equilibrium Le Châtelier’s Principle If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. Stress? – Any kind of change in a system at equilibrium that UPSETS the equilibrium.

§18.2 Factors Affecting Chemical Equilibrium Le Châtelier’s Principle If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. Stress? – Any kind of change in a system at equilibrium that UPSETS the equilibrium.

Use Le Châtelier’s Principle to predict how systems at equilibrium will react to changes. Changes in concentration CO (g) + 3 H 2(g) CH 4(g) + H 2 O (g) carbon monoxide (CO) is injected into the reaction vessel System will try to reduce concentration of CO again forward reaction speeds up (more collisions possible)

CO (g) + 3 H 2(g) CH 4(g) + H 2 O (g) Higher concentration of CO – speeds up forward reaction right away – also a stress on the equilibrium – equilibrium system tries to reduce [CO] to relieve the stress equilibrium shifts to right (produces more products) K eq = [CH 4 ] [H 2 O] [CO][H 2 ] 3

Changes in volume What will happen to the equilibrium system in the equation below if a piston is used to decrease the volume of the vessel in which the equilibrium has been established? (Pushing particles closer together, effectively increasing concentration) CO (g) + 3 H 2(g) CH 4(g) + H 2 O (g)

Stress: Decrease volume = increase pressure Le Chatelier’s Principle: Equilibrium will adjust to relieve the stress. System tries to reduce pressure How? (Notice all gases; 1 mole of any gas occupies same volume if they are at same T and P) K eq = [CH 4 ] [H 2 O] [CO][H 2 ] 3

CO (g) + 3 H 2(g) CH 4(g) + H 2 O (g) Will adjust to side with fewest moles of gas particles LHS: 4 moles of gas/ 4 volumes of gas RHS: 2 moles gas / 2 volumes

CO (g) + 3 H 2(g) CH 4(g) + H 2 O (g) Will adjust to side with fewest moles of gas particles LHS: 4 moles of gas/ 4 volumes of gas RHS: 2 moles gas / 2 volumes Answer: If the volume of the container for the above equilibrium is decreased then the equilibrium will shift to right and form more products to relieve the stress of being in a smaller volume.

What effect will increasing pressure have on this system at equilibrium? What about decreasing the pressure? H 2(g) + I 2(g) 2HI (g) 2 volumes of gas on each side Pressure won’t affect the system N.B. Solids also not affected by changes in P

Changes in temperature CO (g) + 3 H 2(g) CH 4(g) + H 2 O (g) kJ Stress: Increase Temperature Le Chatelier’s Principle: System will try to relieve stress by …

CO (g) + 3 H 2(g) CH 4(g) + H 2 O (g) kJ Stress: Increase Temp System will try to lower temp will shift away from the side with heat – Result: equilibrium shifts to the left (carbon monoxide and hydrogen side) (Note changing temperature will also give a totally new value for Keq.)

CO (g) + 3 H 2(g) CH 4(g) + H 2 O (g) kJ Stress: Lower Temperature Le Chatelier’s Principle: System will…. Result:

CO (g) + 3 H 2(g) CH 4(g) + H 2 O (g) kJ Stress: Lower Temperature Le Chatelier’s Principle: System will try to raise the temperature (by favoring the exothermic reaction direction, side with heat) Result: equilibrium will shift to the right, forming more products, methane and water

Homework Study Guide- Equilibrium (Ch 18) Chemical Equilibrium Practice Problems for Mastery Chapter 18 Equilibrium – take home quiz. Review Atomic Structure and Nuclear Processes notes from semester 1

Part 2 Objectives – Explain and describe equilibrium in terms of molecular motion (when forward and reverse reaction rates are equal) – Use Le Chatelier’s Principle to predict effects of changes on equilibrium mixtures – Be able to write an equilibrium constant expression for a reaction