1 CHAPTER 7 – CHEMICAL SYSTEMS IN EQUILIBRIUM Dynamic Equilibrium Observations of some chemical systems in a closed system (where no matter is able to escape to the surroundings) will appear as though no change is occurring, i.e. an equilibrium has been achieved • however, this does not mean that there is no activity at the molecular level • in fact chemical systems in equilibrium will have a balance of “reactants moving forward towards “products” and the “products” moving in the reverse direction towards ‘reactants” this is considered dynamic equilibrium Examples of chemical systems in equilibrium: 1. Solubility Equilibrium • in solubility equilibrium the undissolved solute particles continuously dissolve into solution, while an equal number of dissolved solute particles in solution crystallize or precipitate out of solution 2. Phase Equilibrium • in phase equilibrium particles in both phases are gaining or losing kinetic energy, such that they are moving from one phase to another, while an equal number are moving in the reverse direction e.g. H2O(l) → H20(g) 3. Chemical Reaction Equilibrium Quantitative reactions are those reactions where all the limiting reagent is used up in the reaction e.g. CaCO3 (s) → CaO(s) + CO2(g) 2 where the CO2 (in an open system, like this, escapes so that the reverse reaction cannot take place) If this reaction takes place in a closed container, both reactants and products can be found This is not considered to be a quantitative reaction can be explained by considering the reverse reaction: CaCO3(s) ---- CaO(s) + CO2(g) The equilibrium equation is CaCO3(s) <==>CaO(s) + CO2(g) Using this knowledge leads to two types of manufacturing processes; 1) continuous processing where as in the limestone example - as limestone is used up, it is replaced with more to keep the reaction going 2) batch processing where the reaction is done in a closed container and the products are taken out when the reaction is complete. i.e. microwave popcorn N2O4(g) → 2NO2(g Another example is 3 reactions such as these have led to this generalization for reversible reactions in closed systems: 4 Percent Reaction at Chemical Equilibrium Using the reaction one can see how the reaction proceeds quickly causing the darkpurple colour of the iodine vapour fading and then becoming constant Table 3 shows how the system reaches a state of equilibrium in 2 ways-in terms of percentage reaction and in terms of an equilibrium constant Looking at system 1; theoretically, using the mole ratio we would expect a quantity of 2mmol/L of HI(g) but in fact the yield was 1.56mmol/L Percent reaction is the yield of product measured at equilibrium compared with the maximum possible yield of product If we did the same percent reaction calculations for the other 2 systems we would see this; 5 Table 4 summarizes that the percent reaction is constant at 78% and to communicate this the percent reaction is written above the equilibrium arrow All chemical reactions are considered to be reversible These reactions fall into 3 categories: 6 An ICE table can be useful to organize information needed to solve stoichiometric problems involving equilibrium systems. ICE stands for Initial, Change and Equilibrium. Look at the Sample Problems on pages 433-436. HW Nelson 12: Page 437 PP #’s 6 & 7 Page 437 SR #’s 1-9 odd numbers 7 Equilibrium Law in Chemical Reactions when considering chemical reactions that reach an equilibrium point both reactants and products will be in the mixture repetitive testing of various chemical reactions that reach equilibrium while beginning with different initial concentrations of reactants show that the concentrations of the reactants and products at equilibrium can be related through the equilibrium law: Consider the reaction: aA + bB → cC + dD The Equilibrium Law can be expressed as; K = [C]c[D]d [A]a[B]b where: A, B, C and D are chemical entities in gas or aqueous phases a, b, c and d are the coefficients of the balanced equation K is the equilibrium constant The law of mass action is a relation that states that the value of the equilibrium constant is constant for a given temperature — when the concentrations of the reactants and products at equilibrium are arranged according to the equilibrium law expression they will be equal to the equilibrium constant The equilibrium law describes the behaviour of almost all gaseous and aqueous chemical equilibria the concentrations of the chemical entities are expressed in moles per Litre (mol/L) 8 Kinetics Argument for the Equilibrium Law Consider the following example to demonstrate how analyzing rates of reaction can yield the equilibrium law: N204(g) ↔ 2NO2(g) • Experimental work has shown the rate law equation for the forward reaction is: rf= kf [N204] while the rate law equation for the reverse reaction is: r r= kr [N02]2 • at equilibrium the rate of reaction in the forward direction is equal to the rate of reaction for the reverse reaction rf =rr kr [N204]=kr [N02]2 manipulation of the above equation yields kf = [N02]2 kr = [N204] ……(1) the equilibrium law expression can be written as K = [N02]2 ……….(2) [N204] we can set equation (1) equal to equation (2), so that: K = kf Kr from the unit on rates of reaction we understood that the magnitude of the rate constant will determine the degree to which a reaction will proceed in the direction specified 9 Considering that the equilibrium constant is a ratio of the rate constant for the forward reaction to the rate constant for the reverse reaction, the magnitude of the equilibrium constant will allow us to determine whether products or reactants are favoured in an equilibrium reaction K >>1 . = <<1 Description Reaction proceeds to completion Products >> Reactants Products . Reactants = Products << Reactants Example 2O3 (g) ↔ 3O2 (g) K = 2.0 x 1057 CO (g) +H2O(g) ↔CO2 (g) +H2 (g) at 700º C K = 5.09 N2 (g) + O2 (g) ↔ 2NO2 (g) K = 1.0 x 10-25 Depending on the reaction the equilibrium constant may or may not vary significantly with temperature Eg. Production of ammonia from nitrogen and hydrogen gas @T=25°C K = 4.26x108 @T=300°C K = 1.02 x 10-5 @T=400ºC K = 8.00 x 10-7 Homogeneous and Heterogeneous Equilibria Homogeneous equilibrium involves reactants and products in a single phase, eg. all aqueous or all gas Heterogeneous equilibrium involves reactants and products in more than one phase, eg. liquid and gas Consider the following reaction with solki and gas phases: 3Fe(s) + 4H2O(g) ↔ Fe304 (s) + 4H2 (g) the equilibrium law expression is: K = [Fe304] [H2]4 [Fe]3 [H2O]4 But, the concentration of the pure solids Fe304 and Fe do NOT change during the reaction as their densities are fixed; they do not affect equilibrium 10 As such the quantities [Fe304] and [Fe] are constant and could be incorporated in the equilibrium constant K [Fe]3 = [Fe3O4] [H2]4 [Fe3O4] [Fe]3 [H2O]4 Or K' = [H2]4 where K’ is the new equilibrium constant [H2O]4 Similarly, in heterogeneous reactions that involve pure liquids and another phase, the concentration of the pure liquid is constant (fixed density) and can also be incorporated into the equilibrium constant The equilibrium constant reported in reference tables for heterogeneous equilibria will already incorporate the densities of pure solids or liquids In the case where a heterogeneous reaction involves ions and a pure solid, pure liquid, or gas, then a net ionic equation is developed to eliminate spectator ions and show only those ions altered or involved in the chemical reaction eg. see page 447 sample problem Determining Equilibrium Concentrations Simple calculations involving the equilibrium constant requires knowledge of a combination of the concentrations of the reactants and products and the equilibrium constant An example involves the following reaction: H2 (g) + CO2 (g) ↔ H2O(g) +CO(g) Where the concentrations are given as; [H2] = 0.24 mol/L [H20] = 0.88 mol/L [C02] = 1.80 mol/L [CO] = 0.88 mol/L the equilibrium law expression is: K = [H2O][ CO] [H2][ CO2] K = (0.88)(0.88) = 1.8 (0.24)(1.80) Variations of this question may involve: 11 a) Determining the concentration of one of the reactants or products (K given along with other reactant and product concentrations) b) Determining K where balanced chemical equation and other information makes it possible to calculate concentrations 12 Qualitative Changes In Equilibrium Systems Le Chatelier’s Principle: “When a chemical system at equilibrium is disturbed by a change in a property, the system adjusts in a way that opposes the change” Le Chatelier’s Principle & Concentration Changes this principle predicts if more reactant is added to a system at equilibrium, then the system will have a change in the equilibrium called an equilibrium shift E.g. to improve the yield more HF(g) can be added to the system, the added reactant “disturbs” the system and the system shifts to the right using up more of the CCl4 and producing more Freon-12 and making a new equilibrium state in chemical equilibrium shifts, the imposed change in concentration is usually only partly counteracted and the concentrations are usually different than the original concentrations (see Fig. 2 below) – adding CO2 13 by removing a product (decreasing concentration) will also shift an equilibrium forward, to the right (see Fig. 3) An example of the effects of both forward and reverse shifts can be seen in Fig. 4 (on page 452) 14 other examples of how Le Chatelier’s Principle is used are; as more NO(g) is removed by reacting it with O2 (g) the system shifts to the right and more HNO3 is produced as blood circulates to the lungs, the high concentration of O2 (g) shifts the equilibrium to the right and the haemoglobin gets oxygenated as the oxygenated blood travels through the body the reverse happens, the equilibrium shifts to the left and more oxygen is released Rate Theory & Concentration Changes kinetic theory explains equilibrium shifts when a reactant concentration increases by assuming that when more reactant is added, more collisions are occurring in the forward direction the reverse reaction, in this case, will also occur until a new equilibrium is reached where the reaction rates are equal again 15 the rates of the new equilibrium will be faster than the original rates because there are now more particles and therefore more collisions if a substance is removed causing an equilibrium shift, the initial effect is to decrease one of the rates additions or removal of substances in the solid or liquid state DO NOT change the concentration of that substance Le Chatelier’s Principle & Temperature Changes energy can be treated as though it were a reactant or product e.g. when the system is heated the equilibrium shifts to the right and more nitrogen dioxide is produced as seen by a darkening of the reddish-brown colour observed Rate Theory & Energy Changes kinetic theory can explain the shifts that occur when energy in a system is increased or decreased rate theory explains this by assuming that both forward & reverse rates are slower at lower temperatures, but that the reverse rate decreases more than the forward rate 16 even though the rates remain equal, the result observes is the production of more product & more energy this causes concentration changes that will increase the reverse rate and decrease the forward rate until they become equal at the new lower temperature (see Fig. 7) in endothermic reactions, the situation is reversed Le Chatelier’s Principle & Gas Volume Changes Boyle’s Law states that P & V are directly related when T is constant decreasing the volume of a container to half its original value will double the concentration of every gas in the container changing the volume may cause a shift in the equilibrium to predict whether this shift may occur we have to look at the total number of moles of gas reactants and gas products e.g. 17 if the volume is decreased the overall pressure is increased. Le Chatelier’s Principle suggests that the system will react in a way that resists the change i.e. in a way to reduce the pressure (see Fig. 9) systems with equal numbers of gas molecules on each side of the equation will not shift after a change in volume systems involving only solids or liquids are not affected by changes in pressure substances in these states are virtually incompressible and reactions involving them cannot counteract pressure changes Rate Theory & Gas Volume Changes kinetic theory explains the effect of decrease of volume by assuming that both the forward and reverse reaction rates increase because the concentrations of reactants and products both increase however, the forward rate increases more than the reverse because there are more particles involved in the forward reaction 18 which means the number of collisions is greater for the forward direction rates are about the same but more product is produced the shift causes concentration changes that will increase the reverse rate and decrease the forward rate until they become equal again Changes that do not affect the Position of Equilibrium Systems adding catalysts – decreases the time required to reach the equilibrium position but does not affect the final position of equilibrium lowers the activation energy for both the forward & reverse reactions by an equal amount equilibrium establishes itself faster but at the same position as if no catalyst were present 19 adding inert gases – the pressure of a system can be changed by adding a gas while keeping the volume constant if the gas is inert, the equilibrium position of the system will not change presence of the inert gas changes the probability of successful collisions for both reactants & products equally, which means there will be no shift in equilibrium 20 Quantitative Changes in Equilibrium Systems Remember a large value of K means the reaction favours the formation of products A small value of K means the reaction favours the formation of reactants The Reaction Quotient, Q In a reaction with both reactants and products present we don’t always know the direction of the reaction We can use the concentrations and the equilibrium law expression to produce a trial value called the reaction quotient or Q Q is similar to K BUT Q is not always calculated when the system is at equilibrium The value of Q can mean 3 things: 1. Q =K, and the system is at equilibrium 2. Q>K, system must shift left (toward reactants) to reach equilibrium, because the product to reactant ratio is too high 3. Q<K, system must shift to the right (toward products) to reach equilibrium, because the product to reactant ratio is too low Look at the Sample Problems in your textbook, pages 464-480. 21 The Solubility Product Constant Solubility Product special case of equilibrium involves cases where excess solute is in equilibrium with its aqueous solution can be done by either preparing a saturated solution with excess solute remaining or by mixing 2 solutions of salt that results with a product that precipitates out of solution Example: for any solute that forms ions in solution, the solubility equilibrium constant is found by multiplying the concentrations of the ions in solution raised to the power equal to the coefficient of each entity in the balanced equation this value of K is called the solubility product constant, Ksp in the above example, Ksp = [Cu+(aq)][Cl-(aq)] Ksp = 1.7 x 10-7 at 25°C 22 other examples include: 23 Calculating Solubility Using Ksp Values See Sample Problem on page 484. Predicting Precipitation from the Grade 11 course we used Solubility Tables to predict whether a precipitate would form or not from section 7.5, we used the reaction quotient, Q, to determine whether the reaction was at equilibrium we can use Q to determine whether a precipitate will form when we mix solutions of metal cations and non-metal anions we can calculate Q to see if the ions are present in a high concentration, in which case we will see a precipitate form or not the reaction quotient, Q, is sometimes called the trial ion product to predict whether a precipitate will form or not we compare the Ksp to the value of Q 24 Example: Using Trial Ion Product Ksp is the value of a system at equilibrium, Q however can be measured at any time this is important in predicting whether or not a precipitate will form 25 The Common Ion Effect when an equilibrium exists in a solution involving ions, the equilibrium can be shifted by dissolving into the solution another compound that adds a common ion, or reacts with one of the ions already in solution example: NaCl = Na+(aq) + Cl-(aq) add a few drops of HCl and additional crystals of NaCl will form the explanation for this from Le Chatelier’s Principle is that when we add HCl it releases large numbers of Cl ions into solution which will cause the reaction to shift to the left….producing more NaCl to precipitate out the lowering of the solubility of an ionic compound (in this case the NaCl) by adding a common ion is called the common ion effect See Sample Problem on p.491 26 SSC CH H44U UC CH HE EM MIISST TR RY YN NO OT TE ESS Spontaneous Reaction ◙ whether a reaction occurs at all- slow or fast. ◙ We’ll look at predicting whether reaction is spontaneous first (then rates of reactions). ◙ It can have exothermic and endothermic reactions which are spontaneous. ◙ Given the necessary Ea to begin with, reaction occurs without continued assistance. ◙ As well, see lab on heats of soln; make note that combution or exothermic reactions tend to be spontaneous while dissolution of NH4NO3 is endothermic. Entropy Spontaneity depends on ∆H° and change in entropy. Entropy is defined as the degree of randomness or disorder. Each chemical or physical reaction process has change in entropy. “Law of Disorder”: A spontaneous reaction in which an isolated system always proceeds in a direction of increased entropy. Eg. Solid –> Liquid Liquid –> Gas ∆S>0 ∆S>0 0 0 –> 0 0 0 ∆S>0 Increasing # of moles T1 –> T2 (T2 > T1) Two liquids –> 0 ∆S>0 Mixture of two liquids ∆S> 27 Solid Solute in –> Liquid solvent. soln ∆S>0 Perfect crystal of pure substance at absolute zero (0 k) • S = 0 Standard Entropy: Change from 0 k –> 298 k (23° C) and 100 kPa. There are values for standard entropy of elements. S = J -----------k · mol Standard Entropy Change o S increases with temperature. o Standard S Is change between 0 and 298 k for a chemical reaction: ∆ S° = ∑ S° products — ∑ S° reactants Eg. 4 Fe (s) + 3O 2 (g) –> 2 Fe 2O 3(s) Gibbs Free Energy A thermodynamic quantity that relates enthalpy and entropy. Predicts whether a reaction is spontaneous or not. ∆ G = ∆ H — T∆ S° ________________________________________________________ __________ At a standard state: ∆ G° = ∆ H — T∆ S° 28 ∆ G° < 0, reaction is spontaneous. ∆ G° > 0, reaction is not spontaneous. ∆ G° = 0, reaction is in equilibrium. Lets look at 4 cases that affect sign of ∆ G° 1. 2 H 2O 2 –> 2 H 2O + O 2 (g) ∆ H° = – 197 KJ o Exothermic. o ∆ S° is positive, more moles result. • Therefore, ∆ G° < 0 at all temperatures; spontaneous. 2. NH 3 (g) + HCl (g) –> NH 4CL (S) ∆ H° = – 176 KJ o Exothermic. o ∆ S° is positive, less moles are produced. o Occurs at low temperatures. ∆ S° = 94.6 – [ 192.5 + 186.9] = – 284.8 o At low T –> T∆ S is smaller than ∆ H°; therefore, it is spontaneous. o At high T –> T∆ S is larger than ∆ H°; therefore, it is nonspontaneous. 3. 2 H 2O(g) –> 2 H 2 (g) + O2 (g) ∆ H° = + 484 KJ 29 o Endothermic. o ∆ S° is positive, more moles are produced. ∆ S° = [ 2 (69.9) + 205.1] – 2 (188.8) = – 32.7 o At low T –> T∆ S is smaller than ∆ H°; therefore, it is nonspontaneous. o At high T –> T∆ S is larger than ∆ H°; therefore, it is spontaneous around 5000°C. o Reaction is entropy driven. 4. 3 O 2 (g) –> 2 O 3 (g) ∆ H° = + 285 KJ o Endothermic. o ∆ S° is positive, less moles are produced. • Therefore, ∆ G° > 0 is nonspontaneous or spontaneous in reverse. Calculating ∆ G ∆ G = ∆ H — T∆ S Use example. NH3 (g) + HCL (g) –> NH4CL(S) ∆ H = ∑n∆H°f products — ∑n∆H°f reactants 30 Standard Free Energy of Formation ∆ G° reaction = ∑n∆G°f products — ∑n∆G°f reactants o Note Error Within the text on Pg. 222. NH3 (g) + HCL (g) –> NH4CL(S) Predicting Changes in Spontaineity ∆ G° = ∆ H — T∆ S° = 0 T∆ S° = ∆ H° • Therefore, T = ∆ H° -----------∆ S° o For NH3 (g) + HCL (g) –> NH4CL(S) ∆ H° = – ∆ S° = – 176.0 KJ 0.2948 KJ • K – 1 31 • Therefore, T = – 176.0 -----------– 0.2948 = 618.0 K (345°C) • Therefore, above 618.0 K, the reaction is nonspontaneous. o Assumes ∆ S° and ∆ H° do not change substantially from a standard state.