Chemical Kinetics and Equilibrium

Reaction Rates How fast or slow the reaction occurs

Collision Theory 2 conditions must be satisfied for a chemical reaction to occur  Particles of reactants must collide with one another; and  Colliding particles must have sufficient energy

Activation Energy Minimum amount of energy needed for a reaction to occur F> Ea Ea

F < Ea Ea

Energy Diagram for Exothermic reaction increasing Average energy energy of reactants Energy difference bet. Average energy reactants and products of products

Energy Diagram for Endothermic reaction increasing energy energy of the products energy of reactants energy difference

Factors Affecting Rate of reaction Surface Area Concentration of reactants Temperature Presence of Catalyst

Surface area of reactants Consider the block of wood or wood shavings  The smaller the particle size of wood the greater the surface area exposed to oxygen thus the rate of reaction increases.

Concentration of Reactants An increase in concentration of the reactants causes an increase in the rate of chemical reaction

Temperature Higher temperature increases reaction rates Ex: ripe fruits are placed in the refrigerator to slow down the ripening process.

Presence of Catalyst Catalyst  Substance that speeds up chemical reaction without itself undergoing a chemical change Inhibitor --->substances that slow down reaction

Activation energy of uncatalized reaction catalyzed activation energy Uncatalyzed activation energy Note: Catalysts increases reaction rates by providing alternative reaction pathways with lower activation energies

Lower activation energy Means that a lower amount of energy is needed to surpass the energy barrier.

Two types of catalyst Homogeneous  Exists in the same phase as the reactant Heterogeneous  Exist as a separate phase from the reactant in a reaction mixture  Most are solids

Heterogeneous and Homogeneous Reactions Homogeneous Reaction  Reaction involving only one phase Heterogeneous reaction  Reaction involving reactants in two phases

Chemical Equibrium It is always assumed that chemical reaction go to completion However, only few chemical reactions proceed to completion Reversible reaction N 2 O 4 (g) 2NO

Do Now: Page 547 Focus Questions 1 and 2 for 10 pts

Reversible Reaction A chemical reaction in which the products can regenerate the original reactants N 2 O 4 (g) 2NO 2 colorless brown N 2 O 4 (g) > 2NO 2 (g) ( forward Reaction) 2NO 2 (g) > N 2 O 4 (g) (backward Reaction)

Chemical Equilibrium The rate of forward reaction and backward reactions are equal Is dynamic  The reactants constantly forms the products while the products constantly forms the reactants Note: In equilibrium, concentration of reactants and products do not change

Ex: dynamic Equilibrium Carbonated Drinks CO 2 + H > H 2 CO 3 carbonic acid When the bottle is opened the pressure is released, carbon dioxide evolved from the decomposition of carbonic acid H 2 CO > CO 2 + H 2 O CO 2 + H 2 0 H 2 CO 3

Equilibrium Constants(K)  Can be obtained in any equilibrium reaction  Equal to the Ratio of the equilibrium concentration of the products to the equilibrium concentration of the reactants  aA + bB ----->cC + dD  Where:  a,b,c & d -- are the coefficients A,B,C & D– are the chemical species

Equilibrium Constant(K) aA + bB ----->cC + dD K = [C] c [D] d = [ products ] [A] a [B] b [reactants] --> Experimentally at equilibrium, Concentration of the products raised to a certain power divided the concentration of the reactants also raised to a certain power is constant at( fixed temperature)  Is also called equilibrium expression

Note: K = [ products] [reactants] is the concentration of products over that of the reactants If the K value is small, it means that the equilibrium concentration of the products is small while the reactants are large. This indicates that at equilibrium, the system consist mostly of reactants If the K value is great, it means that the equilibrium system consist mostly of the products

Writing equilibrium expression If a pure liquid or solid is involved in a reaction, its concentration is omitted in the equilibrium expression because it has constant concentration in mol/L at constant Temperature

Ex: Write the equilibrium expression for the reaction below H 2(g) + F 2(g) 2HF (g) K = [ HF] 2 [H 2 ] [F 2 ]

Ex:2 Write the equilibrium expression for the reaction below N 2(g) + 3H 2(g) 2NH 3(g) K = [NH 3 ] 2 [N 2 ][H 2 ] 3

Ex:3 Write the equilibrium expression for the reaction below CaCO 3(s) CaO (s) +CO 2(g) K = [ CO 2 ]

Ex:3 Write the equilibrium expression for the reaction below 2Hg (l) + O2 (g) 2HgO (s) K = 1 since solid and liquid is omitted [O 2 ]

Ex:3 Write the expression of the following Reactions a) 2H 2 S (g) + 3O 2(g) 2H 2 O (g) + 2SO2(g) b) CO 2(g) + H 2(g) CO (g) + H 2 O (l) c) FeO (s) + H 2(g) Fe (s) + H 2 O (g)

Exercises Write the expression of the following Reactions a) 2H 2 S (g) + 3O 2(g) 2H 2 O (g) + 2SO2(g) b) CO2(g) + H2(g) ---> CO(g) + H 2 O(l) c) FeO(s) + H2(g) --->

Calculating Equilibrium Constant (K) Ex: For the reaction 2NBr (g) N 2(g) + 3Br 2(g) The system at equilibrium at a particular temperature is analyzed and the following concentrations are found: [NBr 3 ] = 2.07 x M [N 2 ] = 4.11x10 -2 M [Br 2 ] = 1.06x M Calculate the value of K for the reaction Solution: K = [4.11x10-2M][1.06 x 10-3M] 3 [2.07x 10-3M] 2 K = 1.14 x Note: the coefficients in the balanced equation are the exponents you see in the solution

What is K sp ? K sp is the sollubility product constant  It refers to a reaction in which a solid is dissolving in water Ex: AB(s)---> A + (aq) + B - (aq)

How are Ksp and K different? Similar? Solubility product constant(Ksp) is a specific type of chemical equilibrium (K ). It is used when the reactant is a solid and the products are ions. In both cases the rules for setting up the equation are the same and all terms have units of mol/L

Writing Solubility Product Expression When CaF2 dissolves in water it dissociates and forms ions: CaF 2(s) > Ca 2+ + F - unbalanced CaF 2(s) > Ca 2+ (aq) + 2 F - (aq) balanced In writing the Ksp expression in the above reaction, the solid is being omitted so Ksp = [Ca 2+ ] [F - ] 2

Write the balanced equation when PbCl2 solid is dissolving in water and write the Ksp expression PbCl2(s) ----> Pb 2+ (aq) + Cl - (aq) PbCl2(s) ----> Pb 2+ (aq) + 2Cl - (aq) balanced Ksp = [Pb 2+ ] [ Cl - ] 2

Exercises Write the balanced equation for the reaction describing the dissolving of each of the following solids in water. Also write the Ksp expression for each solid a) BaSO 4 (s) b) Fe(OH) 3 (s) c) Ag 3 PO 4 (s)

Le Chatelier’s Principle States that if a change in conditions is imposed on a system at equilibrium, the equilibrium position will shift in the direction that tends to reduce the effects of that change --> The effect of change in concentration  The effect of change in Volume  The effect of change in temperature

Effects of Change in Concentration on reaction Equilibrium Consider the equation below N 2 + 3H 2 2NH 3 (g) This equation shows equilibrium because of the two arrows of the same length that is rate of forward reaction is equal to the rate of backward reaction. What will happen if you increase the concentration of N 2 ?  increasing the concentration of the reactants means more collision between them that increases the rate of forward reaction as shown below N 2 + 3H 2 2NH 3 (g) Notice that the arrow pointing to the right has a greater length which indicates forward reaction increases meaning more production of NH 3

What happens if you increase the concentration of NH 3 in the previous example? N 2 + 3H 2 2NH 3 (g) Increasing the concentration of NH3 will shift the reaction to the left. Note:  If a reactant or product is added to the system, the system shifts away from the added component  If a reactant or product is removed, the system shifts toward the removed component

Exercise Suppose the reaction system 2SO 2 (g) + O 2 (g) 2SO 3 (g) Has reached equilibrium. Predict the effect of each of the following changes on the position of the equilibrium a) SO 2 (g) is added to the system b) The SO 3 (g) present is liquefied and removed from the system c) Some of the O 2 gas is removed from the system

The effect of Change in Volume and pressure Recall : Boyle’s law : Pressure is inversely proportional to Volume of gases The smaller the volume, the higher the pressure between molecules Avogadro’s Law : V is directly proportional to the # of moles at constant T Consider the equation below: suppose the gases below are mixed in a vessel at equilibrium 2 NOCl(g) 2NO(g) + Cl 2 (g) What will happen to the equilibrium position if we reduce the volume?  Reducing volume means increasing the pressure between the molecules of the gases, the system moves in the direction that lowers its pressure. In the equation above, the direction will shift to the left because the products has more molecules (3 molecules) than the reactant (2molecules).  In other words, the equilibrium position will shift toward the side of the reaction that involves the smaller number of gaseous molecules in the balanced equation 2 NOCl(g) 2NO(g) + Cl 2 (g)

What happens when volume is increased in the previous example? 2 NOCl(g) 2NO(g) + Cl 2 (g) When the volume is increased, it lowers the pressure of the system, thus the direction will shift to the right to increase the total number of gaseous molecules present

Predict the shift in equilibrium of the reaction below when volume is reduced a) P4(s) + 6Cl 2 (g) 4PCl 3 (l) 6 gaseous molecules 0 gaseous molecules The direction will shift toward the right since it has 0 gaseous molecules b) PCl 3 (g) + Cl 2 (g) PCl 5 (g) 2 gaseous molecules 1 gaseous molecule The direction will shift to the right since it has lesser number of gaseous molecules c) PCl 3 (g) + 3NH 3 (g) P(NH 2 ) 3 (g) + 3HCl(g) 4 molecules 4 molecules Since the # of gaseous molecules in both sides are equal, a change in volume will have no effect on the equilibrium position

Practice: For each of the following reaction predict the direction in which the equilibrium will shift when volume in the container is increased a) 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) b) FeO(s) + H 2 (g) Fe(s) + H 2 O(g)

The effect of change in Temperature The change in concentration and volume in the system alter the equilibrium position but not the equilibrium constant(K) The effect of temperature on equilibrium is different because the value of equilibrium constant (K) is change  If the reaction is exothermic that means it releases heat the direction will shift to the left since heat is is one of the products  If the reaction is endothermic, energy is one of the reactants so the direction will shift to the right

Ex: A) C(s) + O 2 (g) CO 2 (g) + heat B) N 2 (g) + O 2 (g) 2NO(g) – endothermic Answer: A. since heat is one of the products it is exothermic thus the direction of the shift is to the right B. since it is endothermic, heat is one of the reactants so the direction of the shift is toward the left

Exercise: Predict the shift would each of the changes below have on the reaction O 2 (g) + 2CO(g) 2CO 2 (g) + heat a)Increasing [CO] b)Increasing [CO 2 ] c) Increasing Temperature d) Decreasing volume