Topic 11:Thermodynamics (22/10/20) • Kimia Dasar Pertanian (General Chemistry) Basic Chemistry of Thermodynamics 1. Gas Laws and the behaviour of gases 2. Enthalpy - Exothermic and endothermic reactions - Hess’s Law - Calorimetry - Bond enthalpy 2. Entropy - Second and Third Laws of Thermodynamics 3. Spontaneity and Gibbs Free Energy Laws of Thermodynamics - relationships Charles’ Law At constant pressure, the volume occupied by a fixed amount of gas is directly proportional to the temperature P and n fixed V T Boyle’s Law At constant temperature, the volume occupied by a fixed amount of gas is inversely proportional to the applied (external) pressure V 1/P T and n fixed Avogadro’s Law At a fixed temperature and pressure, equal volumes of any gas contain equal numbers of particles (or moles) T and P fixed V n Ideal Gas Law Combination of these laws… PV = nRT R is the universal gas constant Gas Law Problem A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O 2 at 21oC. PV = nRT Don’t forget to change the units!! V=438L R = 0.0821 atmLmol-1K-1 n = number of moles T = 21oC = 294K Given mass = 0.885 kg n = mass in g/molar mass of O2 n = 885g/32gmol-1 = 27.7 mol of O2 P = nRT/V P = (27.7mol)(0.0821 atmLmol-1K-1)(294K) (438L) P = 1.53 atm Behaviour of gases Dalton’s Law of Partial Pressure •Gases mix homogeneously in any proportions •Each gas in a mixture behaves as if it were the only gas present (as long as there is no chemical reaction) Dalton’s Law: In a mixture of unreacting gases, the total pressure is the sum of the partial pressures of the individual gases Ptotal = P1 + P2 + P3 + …. Each component in a mixture contributes a fraction of the total number of moles in the mixture….the mole fraction (X) of that component P =X xP a a Xa = n a n total total Where a is the one component you are interested in Question A mixture of noble gases consisting of 5.5 g of He, 15.0 g of Ne and 35.0 g of Kr is place in an tank at STP. Calculate the partial pressure of each gas. P =X xP a a Xa = n a n total total The gases are noble gases, therefore they are stable (inert) and will not react with each other. Need to work out mole fraction (X) of each gas. Molar mass of He = 4.003 gmol1 Mass of He = 5.5 g nHe = 1.37 mol Molar mass of Ne = 20.18 gmol 1 Mass of Ne = 15.0 g nNe = 0.74 mol Molar mass of Kr = 83.8 gmol1 Mass of Kr = 35.0 g nKr = 0.42 mol n total - =n +n +n He Ne X =n /n He total = 1.37 / 2.53 = 0.54 P =X xP He He total =0.54 x 1 atm = 0.54 atm X =n Ne Ne /n total = 0.74 / 2.53 = 0.29 P =X xP Ne He total = 0.29 x 1 atm = 0.29 atm X =n / n Kr Kr total = 0.42 / 2.53 = 0.17 Kr = 1.37 mol + 0.74 mol + 0.42 mol = 2.53 mol Ptotal = 1 atm He P =X xP Kr Kr total = 0.17 x 1 atm = 0.17 atm Behaviour of gases Kinetic-Molecular Theory An observation of the behaviour of gases at the molecular level….looking at their motion and their speed; also helps us to make sense of the gas laws. The theory is based on 3 assumptions. 1. Particle volume. Any gas consists of a large amount of individual particles. The volume of an individual particle is extremely small compared with the volume of the container – therefore the theory says that the gas particles have mass but no volume. 2. Particle motion. Gas particles are in contant, random, straight line motion, except when they collide with the container walls or each other. 3. Particle collisions. Collisions are elastic. This means that the colliding molecules exchange energy (but do not lose energy through friction). Therefore their total kinetic energy (Ek) is contant. In between collisions, the molecules do not influence each other at all. Behaviour of gases The average kinetic evergy of the molecules is proportional to the absolute temperature. At a given temperature, the molecules of all gases have the same average kinetic energy. Ek T Therefore, if two different gases are at the same temperature, their molecules have the same average kinetic energy. If the temperature of a gas is doubled, the average kinetic energy of its molecules is doubled. Behaviour of gases Although the molecules in a sample of gas have an average kinetic energy and therefore an average speed, the individual molecules move at various speeds, i.e. they exhibit a distribution of speeds – some move fast, others relatively slowly. Collisions can change individual molecular speeds but the distribution of speeds remains the same. Ek T Behaviour of gases At the same temperature, lighter gases move on average faster than heavier gases. Behaviour of gases Mean Free Path The mean free path (λ) of any molecule is the average distance travelled by a molecule between two successive collisions. Every time we carry out a reaction in the lab, we get a change in energy taking place. We study thermodynamics in order to keep track of these energy changes. Our first task is to identify the system in the universe we are interested in observing Open system -can exchange mass or energy with surroundings Surroundings System ΔEsys + ΔEsurr = 0 = ΔE Closed system univ -can exchange only energy with × surroundings Isolated system - No transfer of either energy or mass Law of conservation of energy states that energy cannot be made or destroyed, but can be coverted from one form to another Conclusion: The total energy of the universe remains constant The First Law of Thermodynamics Enthalpy: method for measuring energy changes during chemical reactions At constant pressure, the change in enthalpy of a system is the amount of energy released or absorbed Standard reaction enthalpy, ΔH°rxn: change in enthalpy when reactants in their standard states are changed to products in their standard states Standard enthalpy of formation, ΔH°form: change in enthalpy when one mole of product is formed from reactants in their standard states Standard enthalpy of combustion, ΔH°comb: change in enthalpy when one mole of reactant is burned completely in oxygen Exothermic reactions • release heat to their surroundings • negative enthalpy change, ΔH < 0 Endothermic reactions • take in heat from their surroundings • positive enthalpy change, ΔH > 0 reactants Heat given out products reactants Heat taken in products Hess’s Law • The overall reaction enthalpy is equal to the sum of the individual enthalpies for the reactions which make it up Example Given that for the combustion of glucose C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) ΔH = -2816 kJ and for the combustion of ethanol, C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) ΔH = -1372 kJ Calculate ΔH (in kJ) for the fermentation of glucose: CH O 6 12 6(s) CH O 6 12 4CO 6(s) 2(g) CH O 6 12 2C H OH + 2CO 2 + 6O 2(g) + 6H O 6(s) 2 (l) 5 (l) 6CO ΔH=? 2(g) 2(g) + 6H O 2 (g) 2C H OH + 6O 2 5 (l) 2C2H5OH(l) + 2CO2(g) 2(g) ΔH = -2816 kJ ΔH = +2744 kJ ΔH = -72 kJ Exothermic reaction Example Given that ΔH°for formation (ΔH°f) of Pb3O4 is -175.6 kJ mol-1 and ΔH° for the reaction 3PbO2 Pb3O4 + O2 is 22.8 kJ mol-1. What is the ΔHf for PbO2 (in kJ mol-1)? Step 1: Write equation for reactions given Enthalpy of formation: 3Pb + 2O (s) 3PbO Pb O 2(g) 3 Pb O 2(s) 3 4(s) 4(s) +O 2(g) Step 2: Write down what you are looking for Enthalpy of formation of PbO2: Pb + O (s) 2(g) PbO 2 ΔH°f = -175.6 kJ mol1 ΔH° = 22.8 kJ mol-1 ΔH°f = ? Step 3: Combine the equations we know to get the answer we require 3Pb (s) + 2O Pb O ΔH° = -175.6 kJ mol-1 Pb O O 3PbO (s) ΔH° = -22.8 kJ mol-1 2(g) 3 4(s) + 3Pb 1 Pb 1 2(g) + 3O (s) (s) 3 4(s) 2 3PbO 2(g) +O 2(g) 2(s) PbO 2(s) f ΔH° = -198.4 kJ molf ΔH° = -66.13 kJ molf Question 1. Given the following information: 2C + 3H (s) 2(g) C H 2 C +O CO H + ½O H O ΔH = -84.68 kJ mol-1 6(g) ΔH = -393.51 kJ mol-1 ΔH = -285.83 kJ mol-1 Calculate the standard enthalpy of combustion of ethane: 2(g) 2(g) 2 (l) C2H6(g) + 3½O2 2CO2(g) + 3H2O(l) Answer: -1559.8 kJ mol-1 2. Given the following information: S (s) + O SO 2(g) 2(g) C (s) + O CO 2(g) ΔH = -296.1 kJ mol-1 ΔH = -393.5 kJ mol-1 2(g) CS 2(l) + 3O 2(g) CO 2(g) + 2SO 2(g) ΔH = -1072 kJ mol-1 Calculate the enthalpy of formation of carbon disulfide, CS2: C + 2S CS (s) (s) 2(l) Answer: +86.3 kJ mol-1 By using a bomb calorimeter in the lab, we can determine the reaction enthalpy Mechanical stirrer To electrical source Thermometer Material combusted in oxygen The equation to calculate the heat change is q = m s ΔT where q is the heat change m is the mass of the sample s is the specific heat of the sample ΔT is the temperature change during reaction The specific heat of a substance is the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius Example A sample of 350g of water is heated from 10.5°C to 15.0°C. The specific heat of water is 4.184 J g-1 °C-1. Calculate the heat change. m = 350g 4.5°C q = m s ΔT s = 4.184 J g-1 °C-1 ΔT = (15.0 – 10.5)°C = q = (350 g)(4.184 J g-1 °C-1)(4.5 °C) = 6589.8 J = 6.59 kJ Question A 560g sample of mercury is heated from 40°C to 78°C. The specific heat of mercury is 0.139 J g-1 °C-1. What is the heat change for the reaction? Answer: 2.96 kJ A 782g sample of water is cooled from 25°C to 1°C. The specific heat of water is 4.184 J g-1 °C-1. What is the heat change for the reaction? Answer: -78.5 kJ Bond Enthalpy • Measure of stability of molecule • Enthalpy change required to break a given bond in 1 mole of gaseous molecules • Bond formation: exothermic process, negative sign enthalpy • Bond breakage: endothermic process, positive sign enthalpy • Average values since energy of given bond varies from molecule to molecule, e.g. due to electronegative atoms Example Given the following data: H +I 2HI 2(g) 2(s) H 2(g) 2H(g) I 2I 2(g) ΔH°= +54 kJ mol-1 (g) ΔH°= +436 kJ mol-1 ΔH°= +214 kJ mol-1 What is the bond dissociation energy for HI? (g) Step 1: Draw out structures for each molecule and decide what bonds are broken The equation we are interested in is: 1. H H I hydrogen I H iodine I H I hydrogen iodide hydrogen iodide ΔH°= +54 kJ mol- The information we are given is: 2. H H 1 2H ΔH°= +436 kJ mol-1 2I ΔH°= +214 kJ mol-1 hydrogen 3. I I iodine Bonds broken in equation 1: H HΔH°= +436 kJ mol-1 I I ΔH°= +214 kJ mol-1 Bonds broken, endothermic, ΔH°= +650 kJ mol-1 Step 2: Draw out structure and decide what bonds are formed H H hydrogen I I H iodine I H hydrogen iodide I hydrogen iodide ΔH°rxn = +54 kJ mol-1 Bonds formed, exothermic, ΔH°= -2 (H ΔH° = ΔH° rxn bonds broken + ΔH° bonds formed +54 kJ mol-1 = 650 kJ mol-1 – 2 ΔH° (H 2 ΔH°(H 2 ΔH°(H ΔH° (H I) I) = (650 – 54) kJ mol-1 I) = 596 kJ mol-1 I) = 298 kJ mol-1 I) Question Given the following data: HCl + F HF + ClF 1(g) 2(g) (g) ClF Cl + F (g) (g) (g) HCl H + Cl (g) (g) (g) F 2F 2(g) (g) ΔH°= -232 kJ mol(g) ΔH°= 256 kJ mol-1 ΔH°= 427 kJ mol-1 ΔH°= 158 kJ mol-1 What is the bond enthalpy for HF? Answer: +561 kJ mol-1 Entropy • Measure of disorder of a system • May be increased by increasing number of ways of arranging components. Explained by Boltzmann equation: S = k lnW where S = entropy of system k = Boltzmann constant W = number of possible arrangements • Has relationship with spontaneous change. Second Law of Thermodynamics: spontaneous processes (those which occur naturally without any external influence) are accompanied by an increase in entropy of the universe • Absolute entropies may be determined from Third Law of Thermodynamics: At zero degrees Kelvin, the entropy of a perfect crystal is zero • Because this starting point exists, can measure standard molar entropies: entropy change for 1 mol of a pure substance at 1 atm pressure (usually 25°C) Predicting Entropy Changes • An increase in temperature leads to greater kinetic energy of moving particles, more motion and hence greater S° • Going from solid to liquid to gas (i.e. to less ordered systems) leads to an increase in S° • For spontaneous change, ΔS must be greater than zero. For negative ΔS values, the process is spontaneous in the reverse direction Example Predict whether the entropy change for the following reaction will be positive or negative: CH 3 8(g) + 5O 2(g) 3CO 2(g) + 4H O 2 (l) 6 gas molecules 3 gas molecules + 4 liquid molecules A decrease of the more disordered gas system indicates the entropy change for the reaction should be negative Given the following information, calculate ΔS° for the reaction S° (J / mol K) ΔS° = ΣS° rxn CO2 213.7 H2O 69.9 C3H8 269.9 O2 205.0 products - ΣS° reactants = [(3 mol CO2)(S° of CO2) + (4 mol H2O)(S° of H2O)] – [(1 mol C3H8)(S° of C3H8) + (5 mol O2)(S° of O2)] = [(3 mol)(213.7 J/molK) + (4 mol)(69.9 J/molK)] – [(1 mol)(269.9 J/molK) + (5 mol)(205.0 J/molK)] = 641.1 J/K + 279.6 J/K – (269.9 J/K + 1025 J/K) = 920.7 J/K – 1294.9 J/K = -374.2 J/K Question Predict the sign of ΔS° and calculate its value from the following: 2NO + O (g) 2(g) 2NO 2(g) S° (J / mol K) NO 210.65 O2 205.0 NO2 239.9 Answer: should be negative S° = -146.5 J/K Spontaneity and Gibbs Free Energy • Gibbs Free energy is a measure of the spontaneity of a process • ΔG is the free energy change for a reaction under standard state conditions • At constant temperature and pressure: ΔG=ΔH–TΔS – an increase in ΔS leads to a decrease in ΔG – if ΔG < 0, the forward reaction is spontaneous – if ΔG > 0, the forward reaction is nonspontaneous – if ΔG = 0, the process is in equilibrium – if ΔH and ΔS are positive, then ΔG will be negative only at high T, i.e. forward reaction spontaneous – if ΔH is positive and ΔS is negative, ΔG will always be positive, i.e. forward reaction nonspontaneous – if ΔH is negative and ΔS is positive, ΔG will always be negative, i.e. forward reaction spontaneous – if ΔH and ΔS are negative, ΔG is negative only at low T, i.e. forward reaction spontaneous Example At 27°C, a reaction has ΔH = +10 kJ mol-1 and ΔS = +30 J K-1 mol1 . What is the value of ΔG? ΔG=ΔH–TΔS T=300K ΔG = (+10 kJ mol-1) – (300 K)(+30 J K-1 mol-1) ΔG = +10 kJ mol-1 – 9000 J mol-1 ΔG = +10 kJ mol-1 – 9 kJ mol-1 ΔG = + 1.0 kJ mol-1 Question For the reaction 4KClO3(s) 3KClO4(s) + KCl(s) Calculate ΔG for the process at 298 K if ΔH = -144.3 kJ and ΔS = -36.8 J K-1 Answer: -133.3 kJ Kinetics Refers to reaction rates – the speed of the reaction, i.e. the change in concentration of reactants (or products) with respect to time Equilibrium Refers to the extent of the reation – when no further change occurs, i.e. the (final) concentration of the product given unlimited time An equilibrium reaction: Rate forward Reactants Rate Products backward At equilibrium, the rate of the forward reaction = the rate of the backwards reaction ∆G0 = -RTlnK ∆G0 = Standard Gibb’s free energy change R = gas contant T= temperature K = equilibrium constant Le Chatelier’s Principle When a chemical system at equilibrium is disturbed, the system shifts to (22/10/20) • Kimia Dasar Pertanian (General Chemistry) Topic 12 Solubility Equilibria 10/18/2020 30 Why Study Solubility Equilibria? • Many natural processes involve precipitation or dissolution of salts. A few examples: – Dissolving of underground limestone deposits (CaCO3) forms caves • Note: Limestone is water “insoluble” (How can this be?) – Precipitation of limestone (CaCO3) forms stalactites and stalagmites in underground caverns – Precipitation of insoluble Ca3(PO4)2 and/or CaC2O4 in the kidneys forms kidney stones – Dissolving of tooth enamel, Ca5(PO4)3OH, leads to tooth decay (ouch!) – Precipitation of sodium urate, Na2C5H2N4O2, in joints results in gouty arthritis. 10/18/2020 31 Why Study Solubility Equilibria? • Many chemical and industrial processes involve precipitation or dissolution of salts. A few examples: – Production/synthesis of many inorganic compounds involves their precipitation reactions from aqueous solution – Separation of metals from their ores often involves dissolution – Qualitative analysis, i.e. identification of chemical species in solution, involves characteristic precipitation and dissolution reactions of salts – Water treatment/purification often involves precipitation of metals as insoluble inorganic salts • Toxic Pb2+, Hg2+, Cd2+ removed as their insoluble sulfide (S2-) salts • PO43- removed as insoluble calcium salts • Precipitation of gelatinous insoluble Al(OH)3 removes suspended matter in water 10/18/2020 32 Why Study Solubility Equilibria? • To understand precipitation/dissolution processes in nature, and how to exploit precipitation/dissolution processes for useful purposes, we need to look at the quantitative aspects of solubility and solubility equilibria. 10/18/2020 33 Solubility of Ionic Compounds • Solubility Rules – general rules for predicting the solubility of ionic compounds – strictly qualitative 10/18/2020 34 Solubility of Ionic Compounds • Solubility Rule Examples – All alkali metal compounds are soluble – The nitrates of all metals are soluble in water. – Most hydroxide compounds are insoluble. The 2+ exceptions are the alkali metals, Ba , and Ca 2+ – Most compounds containing Cl- are soluble. The exceptions are those with Ag+, Pb2+, and Hg22+ – All chromates are insoluble, except those of the + alkali metals and the NH4 ion 10/18/2020 35 Solubility of Ionic Compounds large excess added + NaOH Fe(OH)3 Cr(OH)3 Fe3+ Cr3+ 10/18/2020 Precipitation of both Cr3+ and Fe3+ occurs 36 Solubility of Ionic Compounds small excess added slowly+ NaOH Cr3+ Fe(OH)3 Fe3+ Cr3+ 10/18/2020 less soluble salt precipitates only 37 Solubility of Ionic Compounds • Solubility Rules – general rules for predicting the solubility of ionic compounds – strictly qualitative • Do not tell “how” soluble • Not quantitative 10/18/2020 Chapter 17 38 Solubility Equilibrium saturated solution solid 10/18/2020 M A y+ xM y+ yAx- My+ Ax- x- MA x y 39 Solubility of Ionic Compounds Solubility Equilibrium MxAy(s) <=> xMy+(aq) + yAx-(aq) The equilibrium constant for this reaction is the solubility product, Ksp: y+ x x- y Ksp = [M ] [A ] 10/18/2020 40 Solubility Product, Ksp • Ksp used to compare relative solubilities – smaller Ksp = less soluble – larger Ksp= more soluble • Ksp is related to molar solubility – qualitative comparisons – quantitative calculations 10/18/2020 41 Calculations with Ksp • Basic steps for solving solubility equilibrium problems – Write the balanced chemical equation for the solubility equilibrium and the expression for Ksp – Derive the mathematical relationship between Ksp and molar solubility (x) • Make an ICE table • Substitute equilibrium concentrations of ions into Ksp expression – Using Ksp, solve for x or visa versa, depending on what is wanted and the information provided 10/18/2020 42 Example 1 • Calculate the Ksp for MgF2 if the molar solubility of this salt is 2.7 x 10-3 M. (ans.: 7.9 x 10-8) 10/18/2020 Chapter 17 43 Example 2 (2 on Example Problems Handout) • Calculate the Ksp for Ca3(PO4)2 (FW = 310.2) -4 if the solubility of this salt is 8.1 x 10 g/L. (ans.: 1.3 x 10-26) 10/18/2020 Chapter 17 44 Example 3 • The Ksp for CaF2 (FW = 78 g/mol) is 4.0 x 10-11. What is the molar solubility of CaF2 in water? What is the solubility of CaF2 in water in g/L? (ans.: 2.2 x 10-4 M, 0.017 g/L) 10/18/2020 Chapter 17 45 Precipitation • Precipitation reaction – exchange reaction • one product is insoluble • Example Overall: CaCl2(aq) + Na2CO3(aq) --> CaCO3(s) + 2NaCl(aq) Na+ and Ca2+ “exchange” anions Net Ionic: Ca2+(aq) + CO32-(aq) <=> CaCO3(s) 10/18/2020 46 Precipitation • Compare precipitation to solubility equilibrium 2+ 2- Ca (aq) + CO3 (aq) <=> CaCO3(s) prec. vs CaCO3(s) <=> Ca2+(aq) + CO32-(aq) sol. Equil. saturated solution Precipitation occurs until solubility equilibrium is established. Key to forming ionic precipitates: Mix ions so concentrations exceed those in saturated solution (supersaturated solution) 10/18/2020 47 Predicting Precipitation • To determine if solution is supersaturated: – Compare ion product (Q or IP) to Ksp • For MxAy(s) <=> xMy+(aq) + yAx-(aq) – Q = [My+]x[Ax-]y – Q calculated for initial conditions • Q > Ksp supersaturated solution, precipitation occurs, solubility equilibrium established (Q = Ksp) – Q = Ksp – Q < Ksp 10/18/2020 saturated solution, no precipitation unsaturated solution, no precipitation 48 Basic Steps for Predicting Precipitation – Consult solubility rules (if necessary) to determine what ionic compound might precipitate – Write the solubility equilibrium for this substance • Pay close attention to the stoichiometry – Calculate the moles of each ion involved before mixing • moles = M x L or moles = mass/FW – Calculate the concentration of each ion involved after mixing assuming no reaction – Calculate Q and compare to Ksp 10/18/2020 49 Example 4 • Will a precipitate form if (a) 500.0 mL of 0.0030 M lead nitrate, Pb(NO3)2, and 800.0 mL of 0.0040 M sodium fluoride, NaF, are mixed, and (b) 500.0 mL of 0.0030 M Pb(NO3)2 and 800.0 mL of 0.040 M NaF are mixed? (ans.: (a) No, Q = 7.5 x 10-9; (b) Yes, Q = 7.5 x 10-7) 10/18/2020 50 Solubility of Ionic Compounds • Solubility Rules – All alkali metal compounds are soluble – The nitrates of all metals are soluble in water. – Most compounds containing chloride are soluble. The exceptions are those with Ag+, Pb2+, and Hg22+ – Most compounds containing fluoride are soluble. The 2+ 2+ 2+ 2+ exceptions are those with Mg , Ca , Sr , Ba , and Pb2+ • Ex. 4: Possible precipitate = PbF2 (Ksp = 4.1 x 10-8) 10/18/2020 Chapter 17 51 Example 5 • A student carefully adds solid silver nitrate, AgNO3, to a 0.0030 M solution of sodium sulfate, Na2SO4. What [Ag+] in the solution is needed to just initiate precipitation of -5 silver sulfate, Ag2SO4 (Ksp = 1.4 x 10 )? (ans.: 0.068 M) 10/18/2020 Chapter 17 52 Problem Solving Strategy • Precipitation does not occur until Q exceeds Ksp. (Q > Ksp) – We need to add enough Ag+ to make the solution supersaturated • Use the saturated solution (Q = Ksp) as a reference point + – Calculate the [Ag ] needed to give a saturated solution. – Add more Ag+ than this to give a precipitate 10/18/2020 53 Factors that Affect Solubility • Common Ion Effect • pH • Complex-Ion Formation 10/18/2020 Chapter 17 These sure sound familiar. Where have I seen them before? 54 1. Common Ion Effect and Solubility • Consider the solubility equilibrium of AgCl. AgCl(s) <=> Ag+(aq) + Cl-(aq) • How does adding excess NaCl affect the solubility equilibrium? + NaCl(s) Na (aq) + Cl (aq) 2 sources of ClCl- is common ion 10/18/2020 55 Example 6 • What is the molar solubility of AgCl (Ksp = -10 1.8 x 10 ) in a 0.020 M NaCl solution? What is the molar solubility of AgCl in pure water? (ans.: 8.5 x 10-9, 1.3 x 10-5) 10/18/2020 Chapter 17 56 1. Common Ion Effect and Solubility • How does adding excess NaCl affect the solubility equilibrium of AgCl? 1.3 x 10-5 M 10/18/2020 + 0.020 M NaCl Molar solubility Molar solubility AgCl in H2O AgCl in 0.020 M NaCl 8.5 x 10-9 M 57 1. Common Ion Effect and Solubility • Why does the molar solubility of AgCl decrease after adding NaCl? – Understood in terms of LeChatelier’s principle: NaCl(s) --> Na+ + ClAgCl(s) <=> Ag+ + ClCl - Reverse reaction removes some excess 10/18/2020 Chapter 17 58 1. Common Ion Effect and Solubility • Why does the molar solubility of AgCl decrease after adding NaCl? – Understood in terms of LeChatelier’s principle: NaCl(s) --> Na+ + ClAgCl(s) <=> Ag+ + ClCommon-Ion Effect Shifts towards reactants Equilibrium reestablished More AgCl present = less dissolved = lower solubility 10/18/2020 59 2. pH and Solubility • How can pH influence solubility? – Solubility of “insoluble” salts will be affected by pH changes if the anion of the salt is at least moderately basic • Solubility increases as pH decreases • Solubility decreases as pH increases 10/18/2020 Chapter 17 60 2. pH and Solubility • Salts contain either basic or neutral anions: – basic anions • Strong bases: OH-, O2• Weak bases (conjugate bases of weak molecular acids): F-, S2-, CH3COO-, CO32-, PO43-, C2O42-, CrO42-, etc. • Solubility affected by pH changes – neutral anions (conjugate bases of strong monoprotic acids) • Cl-, Br-, I-, NO3-, ClO4• Solubility not affected by pH changes 10/18/2020 Chapter 17 61 2. pH and Solubility • Example: – Fe(OH)2-Add acid s t r e s =s Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq) Decr ease More Fe(OH)2 dissolves in response Solubility increases Stress relief = increase [OH-] 2H3O+(aq) + 2OH-(aq) 10/18/2020 4H2O 62 2. pH and Solubility • Example: – Fe(OH)2 Fe(OH)2(s) <=> Fe2+(aq) + 2OH-(aq) 2H3O+(aq) + 2OH-(aq) 4H2O(l) + 2+ overall Fe(OH)2(s) + 2H3O (aq) <=> Fe (aq) + 4H2O(l) decrease pH solubility increases increase pH solubility decreases 10/18/2020 63 2. pH, Solubility, and Tooth Decay Enamel (hydroxyapatite) = Ca10(PO4)6(OH)2 (insoluble ionic compound) weak base strong base Ca10(PO4)6(OH)2 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq) 10/18/2020 Chapter 17 64 2. pH, Solubility, and Tooth Decay metabolism + food organic acids (Yummy) (H3O+) bacteria in mouth 10/18/2020 Chapter 17 65 2. pH, Solubility, and Tooth Decay Solubility increases Leads to tooth decay OH-(aq) + H3O+(aq) Decrease = stress More Ca10(PO4)6(OH)2 dissolves in response Decreas e=s tress Ca10(PO4)6(OH)2(s) 10Ca2+(aq) + 6PO43-(aq) + 2OH-(aq) 2H2O(l) PO43-(aq) + H3O+(aq) HPO43-(aq) + H2O(l) 10/18/2020 Chapter 17 66 2. pH, Solubility, and Tooth Decay • Why fluoridation? – F- replaces OH- in enamel Ca10(PO4)6(F)2(s) 10Ca2+(aq) + 6PO43-(aq) + 2F-(aq) weaker base than OHmore resistant to acid attack Less soluble (has lower Ksp) than Ca10(PO4)6(OH)2 Factors together fight tooth decay! 10/18/2020 Chapter 17 67 2. pH, Solubility, and Tooth Decay • Why fluoridation? – F- replaces OH- in enamel Ca10(PO4)6(F)2(s) 10Ca2+(aq) + 6PO43-(aq) + 2F-(aq) – F- added to drinking water as NaF or Na2SiF6 • 1 ppm = 1 mg/L – F- added to toothpastes as SnF2, NaF, or Na2PO3F • 0.1 - 0.15 % w/w 10/18/2020 Chapter 17 68 3. Complex Ion Formation and Solubility • Metals act as Lewis acids (see Chapter 15) – Example Fe3+(aq) + 6H2O(l) Fe(H2O)63+(aq) Complex ion Complex ion/complex contains central metal ion bonded to one or more molecules or anions called ligands Lewis acid = metal Lewis base = ligand 10/18/2020 69 3. Complex Ion Formation and Solubility • Metals act as Lewis acids (see Chapter 15) – Example Fe3+(aq) + 6H2O(l) Fe(H2O)63+(aq) Complex ion Complex ions are often water soluble Ligands often bond strongly with metals Kf >> 1: Equilibrium lies very far to right. 10/18/2020 Chapter 17 70 3. Complex Ion Formation and Solubility • Metals act as Lewis acids (see Chapter 15) – Other Lewis bases react with metals also • Examples 3+ - 3- Fe (aq) + 6CN (aq) Fe(CN)6 (aq) Lewis acid Lewis base Complex ion 2+ 2+ Ni (aq) + 6NH3(aq) Ni(NH3)6 (aq) Lewis acid Lewis base Complex ion Ag+(aq) + 2S2O32-(aq) Ag(S2O3)23-(aq) Lewis acid 10/18/2020 Lewis base Chapter 17 Complex ion 71 3. Complex-Ion Formation and Solubility • How does complex ion formation influence solubility? – Solubility of “insoluble” salts increases with addition of Lewis bases if the metal ion forms a complex with the base. 10/18/2020 Chapter 17 72 Complex-Ion Formation and Solubility • Example – AgCl AgCl(s) Ag+(aq) + Cl-(aq) overall Ag+(aq) + 2NH (aq) Ag(NH ) +(aq) 3 32 (aq) Ag(NH ) (aq) + Cl-(aq) AgCl(s) + 2NH 3 + 32 Addition of ligand solubility increases 10/18/2020 Chapter 17 73 Summary: Factors that Influence Solubility • Common Ion Effect – Decreases solubility • pH – pH decreases • Increases solubility – pH increases • Decreases solubility – Salt must have basic anion • Complex-Ion Formation – Increases solubility 10/18/2020 Chapter 17 74 • Kimia Dasar Pertanian (General Chemistry) Topic 13 Reversible Reactions and Equilibrium (22/10/20) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium CHEMISTRY & YOU •How did chemists help farmers produce more food? Fertilizers can increase the amount of a crop per unit of land. Most fertilizers contain ammonia or nitrogen compounds made from ammonia. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium •What happens at the molecular level in a chemical system at equilibrium? • You may have inferred that chemical reactions always progress in one direction. • This inference is not true. Some reactions are reversible. • A reversible reaction is one in which the conversion of reactants to products and the conversion of products to reactants occur at the same time. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium •Here is an example of a reversible reaction. 2SO2(g) + O2(g) 2SO3(g) 2SO2(g) + O2(g) 2SO3(g) • The first reaction is called the forward reaction. • The second reaction is called the reverse reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium 2SO2(g) + O2(g) 2SO3(g) 2SO2(g) + O2(g) 2SO3(g) •The two equations can be combined into one using a double arrow. 2SO2(g) + O2(g) 2SO3(g) The double arrow tells you that the reaction is reversible. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Establishing Equilibrium •When the rates of the forward and reverse reactions are equal, the reaction has reached a state of balance called chemical equilibrium. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Conditions at Equilibrium •At chemical equilibrium, both the forward and reverse reactions continue, but because their rates are equal, no net change occurs in the concentrations of the reaction components. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Concentrations at Equilibrium •Although the rates of the forward and reverse reactions are equal at equilibrium, the concentrations of the components usually are not. • The relative concentrations of the reactants and products at equilibrium mark the equilibrium position of a reaction. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Concentrations at Equilibrium •The equilibrium position tells you whether the forward or reverse reaction is more likely to happen. • Suppose a single reactant, A, forms a single product, B. • If the equilibrium mixture contains 1% A and 99% B, then the formation of B is said to be favored. A B 1% 99% Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium •Why is equilibrium considered to be a dynamic state? Both the forward and reverse reactions are constantly taking place, but their rates are equal, so no net change occurs in the concentrations of the products or reactants. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Factors Affecting Equilibrium: Le Châtelier’s Principle •What 3 stresses can cause a change in the equilibrium position of a chemical system? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium •Stresses that upset the equilibrium of a chemical system include (1) changes in the concentration of reactants or products, (2) changes in temperature, and (3) changes in pressure. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium 1. Concentration •Changing the amount, or concentration, of any reactant or product in a system at equilibrium disturbs the equilibrium. • The system will adjust to minimize the effects of the change. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Suppose carbon dioxide is added to the system. Add CO2 Direction of shift H2CO3(aq) CO2(aq) + H2O(l) • This increase in the concentration of CO2 causes the rate of the reverse reaction to increase. • Adding a product to a reaction at equilibrium pushes a reversible reaction in the direction of the reactants. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Suppose carbon dioxide is removed. Add CO2 H CO (aq) 2 Direction of shift 3 CO (aq) + H O(l) 2 2 Remove CO2 Direction of shift • This decrease in the concentration of CO2 causes the rate of the reverse reaction to decrease. • Removing a product always pulls a reversible reaction in the direction of the products. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium • (2) Temperature Increasing the temperature causes the equilibrium position of a reaction to shift in the direction that absorbs heat. • In other words, it will shift in the direction that reduces the stress. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium • Temperature Add heat N2(g) + 3H2(g) Direction of shift 2NH3(g) + heat Remove heat (cool) Direction of shift Heat can be considered to be a product, just like NH3. • Heating the reaction mixture at equilibrium pushes the equilibrium position to the left, which favors the reactants. • Cooling, or removing heat, pulls the equilibrium position to the right, and the product yield increases. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium 3. Pressure Equilibrium systems in which some reactants and products are gases can be affected by a change in pressure. • A shift will occur only if there are an unequal number of moles of gas on each side of the equation. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Pressure •When the plunger is pushed down, the volume decreases and the pressure increases. Initial equilibrium Equilibrium is disturbed by an increase in Copyrightpressure©Peason.Education, Inc., or its affiliates. All Rights Reserved. A new equilibrium position is established with fewer molecules. Reversible Reactions and Equilibrium Pressure •You can predict which way the equilibrium position will shift by comparing the number of molecules of reactants and products. Add pressure N2(g) + 3H2(g) Direction of shift 2NH3(g) Reduce pressure Direction of shift • When two molecules of ammonia form, four molecules of reactants are used up. • A shift toward ammonia (the product) will reduce the number of molecules. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Sample Problem 18.2 Applying Le Châtelier’s Principle •What effect will each of the following changes have on the equilibrium position for this reversible reaction? PCl5(g) + heat PCl3(g) + Cl2(g) a. Cl2 is added. b. Pressure is increased. c. Heat is removed. d. PCl3 is removed as it forms. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium •The equilibrium constant (Keq) is the ratio of product concentrations to reactant concentrations at equilibrium. • aA + bB cC + dD • From the general equation, each concentration is raised to a power equal to the number of moles of that substance in the balanced chemical equation. K = [C] c x [D] eq a b [A] x [B] d Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium •The value of Keq depends on the temperature of the reaction. • The flask on the left is in a dish of hot water. • The flask on the right is in ice. Dinitrogen tetroxide is a colorless gas. Nitrogen dioxide is a brown gas. N2O4(g) 2NO2(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Sample Problem 18.3 1 Analyze List the knowns and the unknowns. Modify the general expression for the equilibrium constant and substitute the known concentrations to calculate Keq. KNOWNS UNKNOWN [N2O4] = 0.0045 mol/L Keq (algebraic expression) = ? [NO2] = 0.030 mol/L Keq (numerical value) = ? Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 18.3 Reversible Reactions and Equilibrium 2 Calculate Solve for the unknowns. • Start with the general expression for the equilibrium constant. Place the concentration of c d the product in the [C] x [D] K = eq [A]a x [B]b • Write the equilibrium constant expression for this reaction. numerator and the concentration of the reactant in the denominator. Raise each concentration to the power equal to its coefficient in the chemical equation. [NO2]2 K = eq [N2O2] Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 18.3 Reversible Reactions and Equilibrium 2 Calculate Solve for the unknowns. Substitute the concentrations that are known and calculate Keq. (0.030 mol/L)2 Keq = (0.0045 mol/L) = (0.030 mol/L x 0.030 mol/L) (0.0045 mol/L) Keq = 0.20 mol/L = 0.20 You can ignore the unit mol/L; chemists report equilibrium constants without a stated unit. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Sample Problem 18.3 3 Evaluate Does the result make sense? • Each concentration is raised to the correct power. • The numerical value of the constant is correctly expressed to two significant figures. • The value for Keq is appropriate for an equilibrium mixture that contains significant amounts of both gases. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Finding the Equilibrium Constant •One mole of colorless hydrogen gas and one mole of violet iodine vapor are sealed in a 1-L flask and allowed to react at 450oC. At equilibrium, 1.56 mol of colorless hydrogen iodide is present, together with some of the reactant gases. Calculate Keq for the reaction. H2(g) + I2(g) 2HI(g) Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 18.4 Sample Problem 18.4 Reversible Reactions and Equilibrium 1 Analyze List the knowns and the unknown. Find the concentrations of the reactants at equilibrium. Then substitute the equilibrium concentrations in the expression for the equilibrium constant for this reaction. KNOWNS UNKNOWN [H2] (initial) = 1.00 mol/L Keq = ? [I2] (initial) = 1.00 mol/L [HI] (equilibrium) = 1.56 mol/L Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Sample Problem 18.4 2 Calculate Solve for the unknown. First find out how much and I2 are consumed H2 in the reaction. x + x = 1.56 mol 2x = 1.56 mol x = 0.780 mol Let mol H2 used = mol I2 used = x. The number of mol H2 and mol I2 used must equal the number of mol HI formed (1.56 mol). Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Sample Problem 18.4 Reversible Reactions and Equilibrium 2 Calculate Solve for the unknown. • Calculate how much H2 and I2 remain in the flask at equilibrium. mol H2 = mol I2 = (1.00 mol – 0.780 mol) = 0.22 mol • Write the expression Use the general expression for Keq as a for Keq. K = eq guide: [HI]2 [H2] x [I2] = K eq Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. [C]c x [D]d [A]a x [B]b Reversible Reactions and Equilibrium Sample Problem 18.4 2 Calculate Solve for the unknown. Substitute the equilibrium concentrations of the reactants and products into the equation and solve for Keq. Keq = K = (1.56 mol/L)2 0.22 mol/L x 0.22 mol/L 1.56 mol/L x 1.56 mol/L 0.22 mol/L x 0.22 mol/L Keq = 5.0 x 101 eq Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Sample Problem 18.4 3 Evaluate Does the result make sense? • Each concentration is raised to the correct power. • The value of the constant reflects the presence of significant amounts of the reactions and product in the equilibrium mixture. Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Glossary Terms • reversible reaction: a reaction in which the conversion of reactants into products and the conversion of products into reactants occur simultaneously • chemical equilibrium: a state of balance in which the rates of the forward and reverse reactions are equal; no net change in the amount of reactants and products occurs in the chemical system Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium Glossary Terms • equilibrium position: the relative concentrations of reactants and products of a reaction that has reached equilibrium; indicates whether the reactants or products are favored in the reversible reaction • Le Châtelier’s principle: when a stress is applied to a system in dynamic equilibrium, the system changes in a way that relieves the stress Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium aA + bB Keq Glossary Terms cC + dD Keq = (C)c(D)d (A)a(B)b • equilibrium constant (Keq ): the ratio of product concentrations to reactant concentrations at equilibrium, with each concentration raised to a power equal to the number of moles of that substance in the balanced chemical equation Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved. Reversible Reactions and Equilibrium End Copyright © Pearson Education, Inc., or its affiliates. All Rights Reserved.
