Chapter 6 Chemistry PDF

C H E M I S T RY I (FCH0216) Semester I CHAPTER 6: CHEMICAL EQUILIBRIUM  Objectives At the end of this chapter, students should be able to:  Explain basic ideas of chemical equilibrium, equilibrium constant and reaction quotient.  Use equilibrium constants t...

C H E M I S T RY I (FCH0216) Semester I CHAPTER 6: CHEMICAL EQUILIBRIUM  Objectives At the end of this chapter, students should be able to:  Explain basic ideas of chemical equilibrium, equilibrium constant and reaction quotient.  Use equilibrium constants to describe systems at equilibrium.  Identify the factors that affect equilibrium and predict the resulting effects.  Apply the relationship between Kp and Kc.  Describe heterogeneous equilibrium and write their equilibrium constant. CHAPTER 6: CHEMICAL EQUILIBRIUM 2  Outline 6.1 Chemical equilibrium 6.2 Dynamic equilibrium 6.3 Equilibrium constant 6.4 Reaction Quotient 6.5 Le Chatelier’s Principle 6.6 Haber Process CHAPTER 6: CHEMICAL EQUILIBRIUM 3  6.1 Chemical equilibrium  Chemical reaction takes place when concentration of reactant decreases, and concentration of products increases with time. Irreversible reaction Reversible reaction Chemical reactions: entire Chemical reaction that can amounts of reactants are proceed in both the forward converted into products and reverse direction  Double-headed arrow is used to indicate a reversible reaction.  [ ] indicates concentrations.  Chemical equilibrium is achieved when the [reactants] and [products] are remains constant and both forward and reverse reactions are occurring at the SAME RATE. Equilibrium which all the reactant and product are in the same phase is called homogeneous equilibrium. If different phase are involved, it is CHAPTER 6: CHEMICAL EQUILIBRIUM referred as heterogeneous equilibrium. 4  Example Equilibrium [HI] H2 (g) + I2 (g) 2HI (g) [H2] + [I2] Graph concentrations vs time for reaction between iodine (I2) and hydrogen (H2). As reaction proceed, the rate for forward reaction decreases because the concentration of reactant decreases. In contrast, the rate of reverse reaction increases because the concentration of product increases. When rate forward and reverse become equal, the concentration of reactant and product remain constant, the system is said to be in equilibrium. CHAPTER 6: CHEMICAL EQUILIBRIUM 5  Characteristic of Chemical Equilibrium Rate of forward = Rate of reverse reaction [reactants] and [products] remain constant The reaction is continuous (dynamic process) but composition of the mixture constant. A state in which there is no observation changes as time goes by CHAPTER 6: CHEMICAL EQUILIBRIUM 6  Closed System. Rate forwards = Rate reverse Properties such as concentration, Characteristic of a pressure and color are constant. system in equilibrium Equilibrium can be approached from forward and reverse reaction and the final equilibrium position is same. CHAPTER 6: CHEMICAL EQUILIBRIUM 7  6.2 Dynamic equilibrium A state of equilibrium is reached when the [reactants] and [products] become Rate forward = rate reverse constant. However, this does not mean that the reaction stopped at equilibrium. In this Rate stage, the reaction proceed in such a way that the rate of formation of product from Equilibrium region reactant is equal to rate of formation of reactant from product. This is called dynamic equilibrium. Time [Product] > [Reactant] [Reactants] > [Product] Concentration Product Reactants Concentration Equilibrium region Reactant Equilibrium region Product Time CHAPTER 6: CHEMICAL EQUILIBRIUM 8 Time  6.3 Equilibrium constant A relationship that relates the concentration of reactant and product at equilibrium is referred as the Law of Mass Action or Equilibrium Law. The relationship is constant, known as the equilibrium constant (K). For a chemical equilibrium, when the concentration of reactant and product are expressed in concentrations. K is written as Kc. If the reaction is a gas phase, the equilibrium expression can be written in term of partial pressure, Kp. Unit of concentration: mol/L Unit of partial pressure: atm CHAPTER 6: CHEMICAL EQUILIBRIUM 9  Homogeneous Equilibrium The term homogeneous equilibrium applies to reaction in which all reactants and products are in the same phase. pA (g) + qB (g) ⇌ rC (g) + sD (g) Using units of concentration, however in thermodynamic definition, concentration = activity. ➔ K has NO unit ❖ Equilibrium constant 𝐾= 𝐩𝐫𝐨𝐝𝐮𝐜𝐭 K values 𝐫𝐞𝐚𝐜𝐭𝐚𝐧𝐭 Constant at a given [𝑪]𝒓 [𝑫]𝒔 𝐾𝑐 = temperature. 𝑨 𝒑 [𝑩]𝒒 𝒓 𝒔 Change if the temperature 𝑷𝑪 𝑷𝑫 Kp = 𝒑 𝒒 changes. 𝑷𝑨 𝑷𝑩 Does not depend on the CHAPTER 6: CHEMICAL EQUILIBRIUM initial concentration 10  Equilibrium expression in term of Kc and Kp  Consider the below reaction N2 O4 𝑔 ⇌ 2NO2 (𝑔) [𝑝𝑟𝑜𝑑𝑢𝑐𝑡] (𝑃𝑝𝑟𝑜𝑑𝑢𝑐𝑡 ) 𝐾𝑐 = 𝐾𝑝 = [𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡] (𝑃𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡) [𝑁𝑂2 ]2 (𝑃𝑁𝑂2 )2 𝐾𝑐 = 𝐾𝑝 = [𝑁2 𝑂4 ] (𝑃 ) 𝑁2 𝑂4 Example Write the expression for Kc and Kp for this reaction: N2(g) + 3H2(g) ⇌ 2NH3 (g) Solution 𝑃𝑁𝐻3 2 NH3 2 𝐾𝑐 = 𝐾𝑝 = N2 H2 3 𝑃𝑁2 𝑃𝐻 3 2 CHAPTER 6: CHEMICAL EQUILIBRIUM 11  Example Write the expressions for Kc and Kp for the following reaction: (a) 2SO g + O2 g ⇌ 2SO2(g) (b) CO g + H2O g ⇌ CO2 g + H2(g) Solution SO2 2 𝑃SO2 2 (a) 𝐾𝑐 = 2 𝐾𝑝 = 2 O2 SO 𝑃O2 𝑃SO CO2 [H2] 𝑃 (𝑃𝐻2 ) (b) 𝐾𝑐 = 𝐾𝑝 = 𝐶𝑂2 CO H2O 𝑃CO 𝑃H2O CHAPTER 6: CHEMICAL EQUILIBRIUM 12 Heterogeneous equilibrium For heterogeneous equilibrium (different phases), the concentration term of pure liquid and pure solids are not included in the expression of equilibrium constant. Reason:  Concentration of liquids and solids are related by: 𝑔ൗ 3 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑐𝑚 𝑚𝑜𝑙 = 𝑔 = 𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 ൗ𝑚𝑜𝑙 𝑐𝑚3  The density of any given liquid and solid will not / will changes slightly with temperature (constant).  Therefore, the concentration of pure liquids and pure solid can be neglected. CHAPTER 6: CHEMICAL EQUILIBRIUM 13  Example Consider the following reaction: CaCO3 𝑠 ⇌ CaO 𝑠 + CO2 (𝑔) CaO [CO2 ] Normally, 𝐾𝑐 = , but CaCO3 and CaO are solids. [CaCO3 ] Therefore, [CaCO3] and [CaO] are constant and will not be included in the equilibrium expression. 𝐾𝑐 = CO2 and 𝐾𝑝 = 𝑃CO2 CHAPTER 6: CHEMICAL EQUILIBRIUM 14  Exercise 1 For reactions below, express Kc and Kp. 1. 2KClO3(s) ⇌ 2KCl(s) + 3O2 (g) 2. CO2 (aq) + H2O (l) ⇌ H2CO3 (aq) CHAPTER 6: CHEMICAL EQUILIBRIUM 15 Example (a) Calculate the value of equilibrium constant for the reaction: N2 O4 𝑔 ⇌ 2 NO2 (𝑔) Given that, at equilibrium: [N2O4] = 0.0292 mol/L, and [NO2] = 0.0116 mol/L. (b) The equilibrium pressure for the reaction between carbon monoxide and chlorine gases to form COCl2 (g) at 80C are 𝑃𝐶𝑂 = 0.1 atm, 𝑃𝐶𝑙2 = 1.0 atm and 𝑃𝐶𝑂𝐶𝑙2 = 1.6 atm. Calculate the equilibrium constants, Kp. CO(g) + Cl2 (g) ⇌ COCl2 (g) Solution [NO2 ]2 (𝑃𝐶𝑂𝐶𝑙 ) (a) 𝐾𝑐 = (b) 𝐾𝑝 = 2 [N2 O4 ] 𝑃CO 𝑃𝐶𝑙 2 (0.0116)2 1.6 = = (0.0292) (0.1)(1.0) = 𝟒. 𝟔𝟏 × 𝟏𝟎−𝟑 = 16 CHAPTER 6: CHEMICAL EQUILIBRIUM 16  Exercise 2 For Haber process, N2 g + 3H2 (g) ⇌ 2NH3 (g) Kp = 1.45 x 10-5 at 500 oC At equilibrium, the partial pressure of H2 is 0.793 atm and N2 is 0.357 atm. What is the partial pressure of NH3? CHAPTER 6: CHEMICAL EQUILIBRIUM 17  Calculation of Kc and Kp  From Kc and Kp expression, the concentration of product and reactant can be determined and vice versa.  If the concentration or partial pressure given at equilibrium, value of Kc and Kp can be determined.  If the value of Kc and Kp given, the concentration or partial pressure need to be determined using ICE table.  ICE means Initial, Change, Equilibrium.  Since reaction is reversible, the reactant will not fully converted to product. CHAPTER 6: CHEMICAL EQUILIBRIUM 18 Example Calculate the equilibrium concentrations of N2, O2 and NO gases for below reaction: N2 𝑔 + O2 𝑔 ⇌ 2NO 𝑔 𝐾𝑐 = 4.8 × 10−31 Nitrogen and oxygen react to form nitrogen monoxide. In air at 25 oC and 1 atm, the [N2] and [O2] are initially 0.033 M and 0.0081 M, respectively. Solution Step 1: ICE Table As reaction As reaction executed, N2 𝑔 + O2 𝑔 ⇌ 2NO 𝑔 executed, product reactant will I (M) 0.033 0.0081 0 will increase reduce according C(M) -𝑥 -𝑥 + 2𝑥 according to to number of mol number of mol in in BALANCED BALANCED equation E(M) 0.033 – 𝑥 0.0081 – 𝑥 2𝑥 equation CHAPTER 6: CHEMICAL EQUILIBRIUM 19  Assuming that 𝑥 is too small when Kc ≤ 10-4 Step 3 N2 𝑔 + O2 𝑔 ⇌ 2 NO 𝑔 Substitute 𝑥 into I (M) 0.033 0.0081 0 equilibrium equation C (M) - 𝑥 -𝑥 + 2𝑥 E (M) 0.033 – 𝑥 0.0081 – 𝑥 2𝑥 At equilibrium; [N2] = 0.033 – 𝑥 = 0.033 M Don’t forget to put [O2] = 0.0081 – 𝑥 = 0.0081 M concentration unit, M [NO] = 2 𝑥 = 1.132 × 10-17 M CHAPTER 6: CHEMICAL EQUILIBRIUM 20  Example A mixture of 0.003 mol of H2 and 0.017 mol of I2 is placed in 5 L container at 419 K and allowed to become equilibrium. Calculate concentrations of H2, I2 and HI at the equilibrium. H2 g + I2 (g) ⇌ 2HI(g) Kc = 9.72 x 10-4 Solution H2 g + I2 g ⇌ 2 HI 𝑔 I (M) 6 x 10-4 3.4 x 10-3 0 C (M) -𝑥 -𝑥 + 2𝑥 E (M) 6 x 10-4 - 𝑥 3.4 x 10-3 – 𝑥 2𝑥 [HI]2 (2𝑥)2 𝐾𝑐 = = −4 −3 = 9.72 x 10−4 H2 [I2 ] 6 x 10 − 𝑥 3.4 x 10 − 𝑥 Assume 𝑥 is too small, thus, (6 x 10−4 − 𝑥) ≈ 6 x 10−4 and (3.4 x 10-3−𝑥) ≈3.4 x 10-3 4𝑥 2 6 x 10−4 3.4 x 10−3 = 9.72 x 10 -4 Therefore at equilibrium; [H2] = 6x10 – 𝑥 = 5.777 × 10-4 M -4 𝑥 = 2.23 x 10-5 [I2] = 3.4x10-3 – 𝑥 = 3.3777 × 10-3 M CHAPTER 6: CHEMICAL EQUILIBRIUM [HI] = 2 𝑥 = 4.46 × 10-5 M 21  Exercise 3 The equilibrium constant, Kc for the reaction is 55.3. Br2 (g) + F2 (g) ⇌ 2BrF (g) What are the equilibrium concentrations of all gases if the initial concentrations of bromine and fluorine were both 0.220 mol/L? CHAPTER 6: CHEMICAL EQUILIBRIUM 22  Significance of Magnitude K Value of K can be used to interpret the direction of reaction. The equilibrium can be approached from either direction, the direction in which we write the equilibrium is arbitrary. K >> 1 K Kc Net reverse reaction will occur (shift to the left) CHAPTER 6: CHEMICAL EQUILIBRIUM 32  Q vs K 𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭 ⇌ 𝐏𝐫𝐨𝐝𝐮𝐜𝐭 Q If Q = K,The system is at equilibrium EQUILIBRIUM K K Q K If Q > K, There is too much product and the equilibrium shifts to the LEFT to form more reactants. Q If Q < K, There is too much reactant, and the equilibrium shifts to the RIGHT to form more products. Reaction Reaction forms forms products reactants CHAPTER 6: CHEMICAL EQUILIBRIUM 33  Example At 400oC, the Kc for the reaction of H2 (g) + I2 (g) ⇌ 2HI (g) is 49. Predict how the reaction will proceed to reach equilibrium if starting with 0.5 x10-2 M of HI, 3.0 x 10-3 M of H2 and 1.0 x 10-2 M of I2. K Reaction Solution forms products [HI]2 𝑄𝑐 = H2 [I2] Q (0.5 x 10−2 )2 𝑄𝑐 = (3.0 x 10 −3 ) (1.0 x 10 −2 ) 𝑄𝑐 = 𝟎. 𝟖𝟑𝟑 Qc < Kc, therefore [HI] will need to increase and [H2] and [I2] decrease to reach equilibrium. Thus the reaction will proceed to FORWARD reaction to form more product. CHAPTER 6: CHEMICAL EQUILIBRIUM 34  6.5 Le Chatelier’s Principle The equilibrium in a system depends upon factors such as temperature, pressure and concentration of the system. Any change in any of these factors may affect the position of equilibrium. System in Equilibrium Le Chatelier suggested that, if an external stress is applied to an equilibrium system, the system undergoes a change in the direction that counteracts the external stress and, if possible, returns the system to equilibrium. External stress subjected to the system CHAPTER 6: CHEMICAL EQUILIBRIUM 35 Factors that affect chemical equilibrium Changes Shift the Change in K Equilibrium Concentration YES NO Pressure YES NO Volume & Pressure Volume YES NO Temperature YES YES Concentration Catalyst NO NO Inert gas NO NO Inert Gas Catalyst Henri –Louis Le Chatelier (1850-1936) CHAPTER 6: CHEMICAL EQUILIBRIUM Studied mining engineering, Interested in glass & ceramic 36  1. Effect of changes in concentration of a reactant and product Adding or removing a product or reactant The equilibrium shifts to remove reactants or products that have been added. The equilibrium shifts to replace reactants or products that have been removed. For example: Fe3+ (aq) + SCN- (aq) ⇌ FeSCN2+ (aq) Pale yellow Colorless Red If increase [Fe3+] or [SCN-]: ✓ Equilibrium shift to the right, producing more FeSCN2+. ✓ Solution become a darker red color. CHAPTER 6: CHEMICAL EQUILIBRIUM 37  Fe3+ (aq) + SCN- (aq) ⇌ FeSCN2+ (aq) Pale yellow Colorless Red If increase [FeSCN2+]: ✓ Equilibrium shifts to the left, producing more Fe3+ and SCN-. ✓ Intensity of red color reduces. Removing some of the Fe3+ or SCN- ✓ Equilibrium shifts to the left, to produce more Fe3+ and SCN-. ✓ Intensity of red color reduces as FeSCN2+ is consumed. CHAPTER 6: CHEMICAL EQUILIBRIUM 38  2. Changing the volume (V) and pressure (P) ▪ If the volume of a closed system at equilibrium is decreased, the pressure will be increase. The system will respond to decrease the number of moles of gaseous molecules, to offset the increase in pressure. ▪ Reaction will move towards less mol of gas. ▪ If number of mole gas in product and reactant are the same, changing the volume and pressure does not affect equilibrium. V ,P (shift to lower number of moles) V ,P (shift to higher number of moles) ▪ Gas pressure – related to number of gas molecules in the system. ( gas molecules, gas pressure) CHAPTER 6: CHEMICAL EQUILIBRIUM 39 Example N2O4 (g) ⇌ 2NO2 (g) 1 mole gas 2 moles gas If volume pressure Equilibrium shift to the left, it will favor reverse reaction. H2 (g) + CO2 (g) ⇌ H2O (g) + CO (g) 2 moles gases 2 moles gases Changes in P ➔ has no effect in equilibrium CHAPTER 6: CHEMICAL EQUILIBRIUM 40  3. Changing the temperature In an endothermic reaction (ΔH is +ve), energy can be considered as a reactant of the reaction (written on the left). System in H Equilibrium In an exothermic reaction (ΔH is -ve), energy can be considered as a product of the reaction (written on the right). System in Equilibrium H CHAPTER 6: CHEMICAL EQUILIBRIUM 41  i. Endothermic reaction: H2 (g) + I2 (g) ⇌ 2 HI (g) ΔH = +52 kJ/mol Can be written as H2 (g) + I2 (g) + 52 kJ ⇌ 2HI (g) Temperature , Adding additional heat. Since the forward reaction is an endothermic reaction, hence equilibrium shifts to the right. The increased heat will be consumed to produce more HI product. Kc of an endothermic reaction increases with increasing T. Temperature - Opposite effect CHAPTER 6: CHEMICAL EQUILIBRIUM 42  ii. Exothermic reaction: Ag+ (aq) + Cl- (aq) ⇌ AgCl (s) ΔH = -112 kJ/mol Can be written as: Ag+ (aq) + Cl- (aq) ⇌ AgCl (s) + 112 kJ Temperature , Adding additional heat. Since the forward reaction is an exothermic reaction, hence equilibrium shift to the left. The increased heat will be consumed to produce more reactants. Kc of an exothermic reaction decreases with increasing T. Temperature - Opposite effect CHAPTER 6: CHEMICAL EQUILIBRIUM 43  Experiment (a) (b) (c) N2O4 (g) ⇌ 2 NO2 (g) ΔH = +ve Colorless Red- brown A tube containing a mixture of N2O4 and NO2 is immersed at room temperature (Figure b). By immersing the tube in ice water causes the mixture to become lighter in color due to a shift in the equilibrium to reverse reaction to form N2O4 (Figure a). By immersing the same tube in boiling water causes the mixture to become darker due to a shift in the equilibrium to forward reaction to form NO2 (Figure c). CHAPTER 6: CHEMICAL EQUILIBRIUM 44  4. Effect of catalyst Have no effect on the position of equilibrium. Catalysts increase the rate of reaction to achieve equilibrium, not the relative distribution of reactants and products. Cause the reaction to proceed by an alternative route. 5. Adding an inert gas at constant volume If the added gas cannot react with any reactants or products, it is inert towards the substances in the equilibrium. NO changes in concentration occur, so no shift in equilibrium. CHAPTER 6: CHEMICAL EQUILIBRIUM 45  Le Chatelier’s Principle Rate forward & backward increase equally CHAPTER 6: CHEMICAL EQUILIBRIUM 46  Example Consider the following reaction at equilibrium H2 (g) + CO2 (g) ⇌ H2O (g) + CO (g) ∆H = +ve How will the amount of H2 change if :- 1. CO2 is added 2. CO is added 3. a catalyst (Osmium) is added 4. temperature decreased 5. the pressure increased Solution 1. Reaction favour forward reaction. Amount of H2 decreases. 2. Reaction favour reverse reaction. Amount of H2 increases. 3. Catalyst alter alternative path and speed up both forward and reverse reaction. No change in the amount of H2. 4. Since this is endothermic reaction, reaction will favour reverse reaction when the temperature decreased. Amount of H2 increases. 5. Same number of mole, change in pressure will not effect the equilibrium. No change in the amount of H2. 47 CHAPTER 6: CHEMICAL EQUILIBRIUM  Exercise 5 Consider the following reaction. PCl3 (g) + Cl2 (g) ⇌ PCl5 (g) ∆H = -ve What will be the effect on the equilibrium position for each of the following changes? 1. Increase pressure 2. Doubling the volume 3. Increase of temperature 4. Addition of Cl2 CHAPTER 6: CHEMICAL EQUILIBRIUM 48  6.6 The Haber Process Most of the industrial processes are reversible reactions. In industry, optimum condition are used to get maximum product profitability. http://upload.wikimedia.org/wikipedia/commons/3/37/Bundesarchiv_Bild_183-S13651%2C_Fritz_Haber.jpg The Haber process is named after its developer, German chemist, Fritz Haber (1868- 1934). The apparatus used by Haber to first synthesis ammonia Haber Process are widely used in manufacturing ammonia. Reaction equation as per below: N2(g) + 3H2(g) ⇌ 2NH3(g) ∆𝐻 = −𝑣𝑒 CHAPTER 6: CHEMICAL EQUILIBRIUM 49  Equilibrium constant of this reaction can be written as: 𝑃𝑁𝐻3 2 [𝑁𝐻3 ]2 𝐾𝑝 = 𝑃 3 𝐾𝑐 = 𝐻2 𝑃 𝑁2 [𝐻2 ]3 [𝑁2 ] A higher yield of NH3 can be obtained by carrying out reaction under high pressures and at the lowest temperature possible. However the rate of production is low at a lower temperature. Commercially, faster production is preferable. So combination of high-pressure, high temperature and proper catalyst is the most efficient way to produce ammonia on a large scale. Preferable condition: Temperature used = 500ºC Pressure = 400 – 1000 atm Catalyst = Iron, Osmium and Ruthenium CHAPTER 6: CHEMICAL EQUILIBRIUM 50  Reaction mechanism in the production of ammonia using Haber Process CHAPTER 6: CHEMICAL EQUILIBRIUM 51  Exercise 6 You are in charge of ammonia manufacturing plant. Describe the process to be used in your manufacturing plant. Your plan of action needs to include: 1. Name of reaction 2. All equations including whether exothermic or endothermic reaction 3. Factor(s) affecting the yield 4. Give suggestion on the preferable condition 5. Explain why suggested factor is preferred in term of economy. CHAPTER 6: CHEMICAL EQUILIBRIUM 52  References Books. Seng, C.E., Lim, B.T. and Lau, K. P., Chemistry for Matriculation 1, Revised 6th Ed., Oriental Academic Publication, Kuala Lumpur, 2013. Tio Mei Ling, William L. Masterton, William H. Brown, Chemistry for Matriculation, Cengage Learning, 2013. Whitten, K.W., Davis, R.E., Peck M.L., and Stanley G.G., Chemistry 9th Ed., Brooks/Cole, Cengage Learning, Canada, 2010. Brady, J.E., Senese, F.A., Jespersen, N.D., Chemistry: The Study of Matter and Its Changes, 6th Ed, John Wiley & Sons (Asia), International Student Version, 2011. Chang, R., Goldsby, K.A., Chemistry, 11th Ed., Mc GrawHill Higher Education, International ed., USA, 2013. Web. http://faculty.ccri.edu/aahughes/GenChemII/.../Chemical%20Equilibrium.ppt http://catalog.flatworldknowledge.com/bookhub/4309?e=averill_1.0- ch15_s05 CHAPTER 6: CHEMICAL EQUILIBRIUM 53