+ Equilibrium Topic 7

+ 7.1 Equilibrium Reversible rxn  a rxn that can go either way A double arrow shows that it is reversible CaCO 3 (s)  CaO(s) + CO 2 (g)

+ Equilibrium Decomposition of CaCO 3 is reversible, but what if we influence the system such that it is kept in a container at a constant high temp? Goes to CaO and not back to CaCO 3 Why? All the CO 2 escapes and the reaction will not go in reverse

+ Con’t If the same experiment is run, but in a closed container, what will the outcome be? After the same amount of times as the first exp. There is still CaCO 3 Why? The rxn reaches equilibrium and amounts of product and reactant remain constant

+ Dynamic equilibrium Macroscopic properties are constant (concentrations of all reactants and products remain constant) and the rate of the forward rxn is equal to the rate of the reverse rxn Forward  reactants going to products Reverse  products going to reactants

+ Equilibrium and rate of rxn Consider the reaction: H 2 (g) + I 2 (g)  2HI(g) If we begin with just hydrogen and iodine in a closed container, how will their concentrations change?

+ Reaching Equilibrium

+ Con’t What must the rates of the forward and reverse rxns of a reversible rxn equal to be at equilibrium? Each other

+ Characteristics of the equilibrium rate 1. Macroscopic properties are constant at equilibrium [products] and [reactants] remain constant 2. At equi. The rate of forward rxn = rate of reverse rxn 3. Equi. can only be obtained in a closed system If in an open system, products can escape and have no opportunity to recombine

+ Con’t 4. All species in the equation are present in the equi. Rxn mixtures All reactants and all products 5. Equi. can be obtained from either direction Does not matter if beginning with reactants or with products

+ 7.2 Position of equilibrium What does the position of equi. refer to? relative amounts of reactants and products present at equi. Equi. does not imply 50% reactants and 50% products Some reactions favor one over the other

+ Ex. 2NO(g)  N 2 (g) + O 2 (g) Δ H= +10 kJ mol -1 At 700 K the position of the equi lies mostly to the right There is a large amount of products and not very much of the reactant There is about 1.5 million times as many product

+ The effect of changing conditions on the position of equilibrium Le Chatelier’s princliple: If a system at equilibrium is subjected to some change, the position of equilibrium will shift in order to minimize the effect of the change This allows us to predict which direction equi will shift

+ Effect of temperature 2HI(g)  H 2 (g) + I 2 (g) At room temp, there is about 28 times more of the reactant than products (96.6% and 3.4%) At 700 K there is only about 7 times more of the reactants than products (88% and 12%) Which way did the equilibrium shift? To the right

+ Use Le Chatelier’s with temp. Looking at the first rxn with HI, the enthalpy is positive, meaning what? The reaction is endothermic When the temp is increased the rxn is going to shift to minimize the effects For this rxn, the equi shifts more toward the endothermic direction to take in the added heat, thus the equi shifted to the right

+ Summed up Heat reaction mixture  position of equi is shifted in the endothermic direction Cool reaction mixture  position of equi is shifted in the exothermic direction

+ Effects of pressure 2NO 2 (g)  N 2 O 4 (g) Brown colorless If this rxn is performed in a sealed syringe, as the pressure increases the color initially gets darker and then becomes colorless as a new equi position is established Which way did equi shift? To the right

+ Con’t At higher pressure the equi shifted from 2 gaseous molecules to the side where there is just 1 One molecule takes up less space than two Which way would the equi for this rxn shift? 2SO 3 (g)  2SO 2 (g) + O 2 (g) left Which direction will this equi shift? 2HI(g)  H 2 (g) + I 2 (g) neither

+ Effects of concentration 2[CrO 4 ] 2- (aq) + 2H + (aq)  [Cr 2 O 7 ] 2- (aq) + H 2 0(l) yelloworange

+ Con’t Looking as the first yellow solution, which direction does the equi lie? To the left When acid is added, the color changes to orange, meaning the equi is shifted which direction? To the right

+ What this means Adding the acid increases the H + ions, thus shifting the equi to the right to minimize the effects If OH - is added, it does what to the H + ? Neutralizes it to form water and the equi shifts back to the left

+ In general If the concentration of one of the species in an equilibrium mixture is increased, the position of the equi shifts to the opposite side ot reduce the concentration of this species

+ 7.3 The equilibrium constant CH 3 COOH(l) + C 2 H 5 OH(l)  CH 3 COOCH 2 CH 3 (l) + H 2 O(l) Ethanoic acidethanolethyl ethanoate Various amounts of ethanoic acid and ethanol are reacted and allowed to come to equi at the same temp. The equilibrium concentrations of each is determined and a ration is found (this is the constant): [CH 3 COOCH 2 CH 3 (l)][H 2 O(l)] [CH 3 COOH(l)][C 2 H 5 OH(l)]

+ Constants All concentrations are measured at equilibrium Equilibrium law  for the rxn aA + bB  cC + dD Where all reactants are in the same phase, the following ratio is constant at a particular temp K c = [C] c [D] d c=in terms of concentration [A] a [B] b

+ Purpose of equilibrium constant Provides info about how far a rxn proceeds at a particular temp K c >> 1the rxn proceeds almost totally towards the products K c << 1the rxn hardly proceeds toward the products Allows for preduction about which direction the equi will lie for a rxn

+ How changed conditions affects the value of the constant Effect of pressure  NOT affected by pressure Effect of concentration  NOT affected by concentration If either of these two are changed, we can often conclude that the system is not at equilibrium

+ Effect of temperature The value of the constant for a particular rxn is affected only by a change in temp For exo rxns, the value of the equi constant decreases as the temp is increased For an endo rxn, the value of the equi constant increases as the temp is raised

+ Catalysts and the equi constant The presence of a catalyst does NOT affect the position of equi or the value of the constant, it simply reduces the time required to reach equi Increases the rate of the forward and reverse rxns equally

+ Constant and rate The equi constant give info about how far a rxn goes towards completion (about the extent of the rxn) It gives us NO info about how quickly the rxn occurs Kinetic data, such as rate constant, indicates how quickly equi is attained by privies no info about the position of equi and how far the rxn proceeds

+ HL 2

+ 7.4 Calculations with the equilibrium constants Calculating units: CH 3 COOH(l) + C 2 H 5 OH(l)  CH 3 COOCH 2 CH 3 (l) +H 2 O(l) K c = [CH 3 COOCH 2 CH 3 (l)][H 2 O(l)] [CH 3 COOH(l)][C 2 H 5 OH(l)] All units are mol dm -3

+ Con’t All units will cancel out Thus, K c has no units Always the case when you have the same # of species on the reactant and product side

+ Another Ex N 2 (g) + 3H 2 (g)  2NH 3 (g) K c = [NH 3 ] 2 [N 2 ][H 2 ] 3 Δ n= total power on top- total power on bottom What do the units equal? K c = (mol dm -3 ) - Δ n K c = (mol dm -3 ) -2 = mol -2 dm 6

+ Calculating equi constants Generally as simple as plugging in knows to solve for an unknown H 2 (g) + I 2 (g)  2HI(g) SubstanceH2H2 I2I2 HI Equi concentration / mol dm

+ Con’t K c = [HI] 2 [H 2 ][I 2 ] Plug in values: = *0.39 K c has no units for this rxn

+ Two different values of the constant for the same rxn under the same conditions For the rxn:2NO 2 (g)  N 2 O 4 (g) Set up the constant equation K c = 0.69 mol dm -3 at 400 K However, the rxn could also be written as: N 2 O 4 (g)  2NO 2 (g) The constant (K c ’ ) for this is 1.46 mol dm -3 at 400 K K c ’ = 1/ K c

+ Another Ex The rxn:H 2 (g) + I 2 (g)  2HI(g) Can also be written:½ H 2 (g) + ½ I 2 (g)  HI(g) If both are set up in the equi constant equation, the answers will be different Thus the constant is dependent upon how the equation is written

+ SL & HL 1/2

+ 7.5 Industrial Processes Applying these ideas to industrial processes considering these two factors: 1. How much product is obtained? 2. How quickly is the product obtained?

+ Haber process What is the product of this process? N 2 (g) + 3H 2 (g)  2NH 3 The N is obtained from the air and one of the sources on H is methane Yield of ammonia is about % per pass of the process

+

+ Pressure and Temperature

+ Contact process What is the product of this process? Sulfuric acid It involves the following rxns: S(s) + O 2 (g)  SO 2 (g) Δ H=-297 kJ mol -1 2SO 2 (g) + O 2 (g)  2SO 3 (g) Δ H=-197 kJ mol -1 SO 3 (g) + H 2 O(l)  H 2 SO 4 (l) Δ H=-130kJ mol -1

+ Temperature considerations This process is run at a relatively low temperature that causes a lower rate of rxn Why would industries run this process at a lower temp rather than a higher temp? The higher temp is not cost-affective due to the higher costs of fuel and the quicker corrosion of the rxn vessels

+ Increasing pressure Increasing the pressure would increase the yield of SO 3 and would increase costs because a plant that is designed for high pressures would have to be built

+ 7.6 Phase equilibria Physical equilibria  equilibria involving change of state; specifically that between liquid and its vapor Evaporation  when a liquid is left in an open container, it evaporates On the molecular level, the particles overcome the forces that hold them together and escape into the gaseous phase

+

+ Liquid- vapor equilibrium Consider a volatile liquid in a closed container

+ Rate of vaporization When rate of vaporization and condensation are equal, the system is at dynamic equilibrium Dynamic  two opposing processes occurring at the same rate

+ HL 2

+ HL 2  Vapor pressure The pressure exerted by the vapor in equilibrium with a liquid (or a solid) Depends on nature of liquid The higher the temp, the more energy the molecules have to break free from intermolecular forces Thus, depends ONLY on the nature of liquid and temp

+ Con’t Independent of: Volume of container Volume of the liquid Volume of the vapor Shape of the container Surface area of liquid

+ Variation of vapor pressure and temp

+ Boiling point of a liquid A liquid boils when its vapor pressure equals the external pressure Thus the normal boiling point of a liquid is the temp at which the vapor pressure of the liquid is 1 atm

+ Bp and pressure Water boils at 70°C on top of Mount Everest Pressure decreases as the altitude increases Boiling point is dependent on the external pressure

+ Vapor pressure and intermolecular forces At all temps 1-butanol has lower vapor pressure than ethoxyethane This is due to the stronger intermolecular forces of 1- butanol They have about the same masses and Van der Waal’s forces, but the molecules are bonded together differently 1-butanol has hydrogen bonding and ethoxyethane has mainly dipole-dipole forces

+ Enthalpy of vaporization The standard enthalpy change of vaporization ( Δ H vap ) is the energy needed to convert 1 mol of a liquid to vapor under standard conditions Intermolecular forces Vapor pressure Boiling point Enthalpy of vaporization WeakHighLow StrongLowHighhigh