CHEM 112 - Physical Chemistry PDF

MUSLIM REFRESHER COURSE PROGRAMME OLD STUDENTS ASSOCIATION, AHMADU BELLO UNIVERSITY ZARIA CHAPTER (MUROSA ABU CHAPTER) CHEM112 ( INTRODUCTORY BASIC PHYSICAL CHEMISTRY) 2CU PRESENTED BY MUHAMMAD SHAMSUDDEEN MUHAMMAD CONTACT: 08134084713 08082310186 Email: [email protected] Naijmail: [email protected] Website: drmuhdshams.simdif.com COURSE OUTLINE * State of Matter * Kinetic Theory of Gases * Gas Laws * Real Gas and Ideal Gas * Thermochemistry * Chemical Equilibrium * Ionic Equilibrium * Electrochemistry CHEM112 ( INTRODUCTORY BASIC PHYSICAL CHEMISTRY) 2CU Physical chemistry is the branch of chemistry that deals with the physical properties of chemical substance. That means describing and explaining how specific chemical substances look and behave in particular situations, e.g. under certain temperatures and pressures. Physical Chemistry includes study of the physical properties of many different types of substances and on different scales (levels of physical detail). MATTER Matter generally means anything that has weight and occupy space. STATES OF MATTER Basically, there are three states of matter which are solid, liquid and gas. Solids, liquids and are three In solids, the particles are tightly packed together. In liquids, the particles have more movement, while in gases, they are spread out. Particles in chemistry can be atoms, or molecules. Below is a figure showing the features of the three states of matter. THE SOLID STATE Something is usually described as a solid if it can hold its own shape and is hard to compress (squash). The molecules in a solid are closely packed together – they have a high density. Right now, you are probably sitting on a chair, using a mouse or a keyboard that is resting on a desk – all those things are solids. THE LIQUID STATES In liquids, the molecules have the ability to move around and slide past each other. A liquid will take on the shape of the container it is being held in. While a liquid is easier to compress than a solid, it is still quite difficult – imagine trying to compress water in a confined container! Water is an example of a liquid, and so is milk, juice and the petrol you put in the car. THE GAS STATE In gases, the atoms are much more spread out than in solids or liquids, and the atoms randomly with one another. A gas will fill any container, but if the container is not sealed, the gas will escape. Gas can be compressed much more easily than a liquid or solid. Right now, you are breathing in air – a mixture of gases containing many such as oxygen, nitrogen and carbon. KINETIC THEORY OF GASES The (advanced) kinetic theory of gases is founded on the following six fundamental postulates: 1. Gases are composed of minute discrete particles (usually molecules). 2. The particles are in continuous chaotic motion moving in straight lines between very frequent collisions with each other and the sides of the container (approximately 10 9 /s). 3. The bombardment of the container walls by the particles causes the phenomenon we call pressure (i.e. force of impacts/unit area). The greater the force of collision and the more frequent the collisions the greater the gas pressure exerted on the container surface. 4. The collisions are perfectly elastic i.e. no energy loss on collision due to friction. 5. At relatively low pressures the average distance between particles is large compared to the diameter of the particles and therefore the inter–molecular forces between the particles is negligible. 6. The average kinetic energy of the particles is directly proportional to their absolute temperature on the Kelvin scale (K) i.e. KE(J) T (K) This means if you heat up a gas the average kinetic energy of the particles increases, therefore the average speed increases too. GAS LAWS Boyle's Law states that for given mass of gas at a constant temperature ( o C or K), the product of the pressure multiplied by the volume is a constant. p x V = constant Therefore, for initial values of p 1 and V 1 , which change to final values of p 2 and V 2 , the following equation applies... p 1 x V 1 = p 2 x V 2 (for fixed amount of gas at constant temperature) or p 2 = p 1 x V 1 /V 2 or V 2 = p 1 x V 1 /p2 Examples of Boyle's Law calculations (constant temperature assumed) Ex. Q 1. 240cm 3 of air at a pressure of 100kPa in a bicycle pump is compressed to a volume of 150cm 3. What is the pressure of the compressed air in the pump? p 1 x V 1 = p 2 x V 2 , rearranging to scale up for the new higher pressure p 2 = p 1 x V 1 /V 2 = 100 x 240/150 = 160 kPa Ex. Q 2. 10 m3 of butane gas at 1.2 atm was required to be stored at 6 atm pressure. To what volume must the gas be compressed to give the required storage pressure? p 1 x V 1 = p 2 x V 2 , rearranging to scale down for the new lower volume V 2 = p 1 x V 1 /p 2 = 1.2 x 10/6 = 2.0 m 3 Ex. Q 3. A 100 cm 3 gas syringe containing 80 cm 3 of gas that was compressed to 60 cm 3. If atmospheric pressure is 101325 Pa, and the temperature remains constant, what is the pressure of the gas in the syringe after compression. p x V = constant p1xV1=p2xV2 p 2 = p 1 x V 1 /V 2 p 2 = 101325 x 80/60 = 135100 Pa Ex. Q 4. In hospital the gas pressure in a 100 dm 3 cylinder of oxygen is 5.52 atm (5 x atmospheric pressure). What volume of gas can be released slowly to a patient on releasing it to an atmospheric pressure 1.01? p x V = constant V 2 = p 1 x V 1 /p 2 V 2 = p 1 x V 1 /p 2 = 5.52 x 100/1.01 = 546.5 dm 3 Charles's/Gay–Lussac's Law states that for a fixed mass of gas... (i) the volume of a gas is directly proportional to the absolute temperature (K) at constant pressure V = constant x T (left graph), or V/T = constant, or V 1 /V 2 = T 1 /T 2 for conditions changing from 1 (initial) to 2 (final), or V 1 /T 1 = V 2 /T 2 for constant pressure V1xT2=V2xT1 V 2 = V 1 x T 2 /T 1 or T 2 = T 1 x V 2 /V 1 OR (ii) the pressure of a gas is directly proportional to the absolute temperature (K) at constant volume , p = constant x T (right graph), or p/T = constant , or p 1 /p 2 = T 1 /T2 for conditions changing from 1 (initial) to 2 (final), or p 1 /T 1 = p 2 /T 2 for constant volume p1xT2=T1xp2 p 2 = p 1 x T 2 /T 1 or T 2 = T 1 x p 2 /p 1 therefore the three permutations for problem solving are... p 2 = (p 1 x V 1 x T 2 )/(V 2 x T 1 ) or V 2 = (p 1 x V 1 x T 2 )/(p 2 x T 1 ) or T 2 = (p 2 x V 2 x T 1 )/(V 1 x T 2) Note: If the temperature is constant you get Boyle's Law. If p or V is constant you get Charles's/Gay–Lussac's Law. You can use any volume or pressure units you like as long as both p's or both V's have the same unit. Examples of P–V–T calculations Ex. Q1 The pressure exerted by a gas in sealed container is 100kPa at 17o C. It was found that the container might leak if the internal pressure exceeds 120kPa. Assuming constant volume, at what temperature in o C will the container start to leak? 17o C + 273 = 290K p 1 /T 1 = p 2/T 2 rearranging to scale up to the higher temperature T 2 = T 1 x p 2/p 1 T 2 = 290 x 120/100 = 348 K or 348 – 273 = 75o C when the container might leaatm Ex. Q2 A cylinder of propane gas at 20 o C exerted a pressure of 8.5 atmospheres. When exposed to sunlight it warmed up to 28 o C. What pressure does the container side now experience? 20o C = 273 + 20 = 293K, 28 o C = 273 + 28 = 301K p 2 = p 1 x T 2 /T 1 p 2 = 8.5 x 301/293 = 8.73 atm Ex.Q 3. 25 cm 3 of a gas at 1.01 atm. at 25o C was compressed to 15 cm 3 at 35 o C. Calculate the final pressure of the gas. p 1 = 1.01 atm, p 2 = ?, V 1 = 25 cm 3 , V 2 = 15 cm 3 , T1 = 25 + 273 = 298 K, T2 = 35 + 273 = 308 K (p 1 x V 1 )/T 1 = (p 2 x V 2 )/T 2 p 2 = (p 1 x V 1 x T 2 )/(V 2 x T 1 ) p 2 = (1.01 x 25 x 308)/(15 x 298) = 1.74 atm Ex. Q4. 12.0 dm 3 of gas in a cylinder and piston system is heated from 290 K to 340 K. If the pressure remains constant, calculate the final volume of gas in the cylinder. V/T = constant V 1 /V 2 = T 1 /T 2 V1xT2=V2xT1 V 2 = V 1 x T 2 /T 1 V 2 = 12 x 340/290 = 14.1 dm 3 Ex. Q5. The fuel and air gases in the cylinders of a 1200 cm 3 car engine go from 25o C before combustion and rise to a peak temperature of 2100 o C after combustion. If normal atmospheric pressure is 101kPa, calculate the peak pressure reached after combustion. Although the movement of the piston changes the volume, (i) for the sake of argument assume the volume is constant. T 1 = 25 + 273 = 298 K, T 2 = 2100 + 273 = 2373 K, P1 = 101 KPa p/T = constant p 1 /p 2 = T 1 /T2 p 2 = p 1 x T 2 /T 1 p 2 = 101 x 2373/298 = 804 kPa (ii) To be more realistic, assume the initial volume of fuel vapour plus air was 400 cm3, now re- calculate the final pressure. You now need to use the full PVT expression. (p 1 x V 1 )/T 1 = (p 2 x V 2 )/T 2 p 2 = (p 1 x V 1 x T 2 )/(V 2 x T 1 ) p 2 = (101 x 400 x 2373)/(1200 x 298) = 268 kPa IDEAL GAS AND REAL GAS An ideal gas is a hypothetical gas that is assumed to have no intermolecular forces of attraction. Its molecules are also assumed to undergo elastic collisions with each other and with the walls of the container. An ideal gas exhibits proportionality between its temperature and its average kinetic energy. Further, the volume of a molecule of an ideal gas is assumed to be negligible compared to the volume of the entire gas itself. PV = nRT is the simplest form of the ideal gas law. There are many different models of the ideal gas law, trying to “compensate” for the non-ideal behavior. A real gas is a gas that does not obey these assumptions outlined above. However, at high temperatures and low pressures, because a real gas' molecules are quite far apart from each other, a real gas can behave like an ideal gas thought not completely. In summary; Ideal gases do not exist, real gases exist! Just as, ‘point’, ‘straight line’ etc are in geometry so is the ideal gas in physical sciences; none of them exists. You can do experiments in the lab using real gases, you cannot do experiments in the lab with ideal gases. There are a large number of real gases each with properties of its own, but there exists just one ideal gas with properties controlled by the equation PV = nRT. The symbols have their usual notation. Ex. Q 1. (a) What is the volume of 6g of chlorine at 27 o C and 101kPa (approx. 1 atm) ? pV = nRT, V = nRT/p T = 273 + 27 = 300K, n = 6/71 = 0.08451 mol chlorine, Mr(Cl 2) = 2 x 35.5 = 71 and p = 101 x 1000 = 101000 Pa. V = 0.08451 x 8.314 x 300/101000 = 0.002087 m 3 (b) What is the volume of the chlorine in dm 3 and cm 3 ? 1 m3 = 1000 dm 3 = 10 6 cm 3 V = 0.002087 x 1000 = 2.087 dm 3 V = 0.002087 x 10 6 = 2087 cm 3 Ex. Q 2. (a) A 5 litre container contained 0.5kg of butane gas (C 4H 10 ). Assuming ideal gas behaviour calculate the pressure of the gas if the cylinder is stored at 25o C. Mr(C 4 H 10 ) = (4 x 12) + 10 = 58, 0.5kg = 550g moles of gas n = 500/58 = 8.621, T = 273 + 25 = 298K R = 8.314 J mol –1 K –1 or 0.08206 atm litre mol –1 K –1 (i) SI units PV = nRT, P = nRT/V, 5 litre = 5 dm 3 = 5 x 10 –3 m3 P = 8.621 x 8.314 x 298/(5 x 10 –3 ) P = 4271830 Pa = 4272 kPa = 4.272 MPa (ii) 'old units' P = 8.621 x 0.08206 x 298/5 = 42.16 atm EXPRESSING PV = nRT INTERMS OF DENSITY Density is defined as mass per unit volume, which implies σ = m/V ______(1) For an ideal gas PV = nRT, V = nRT/P _________(2) Putting (2) in (1), we have; σ = Pm/nRT ________(3) But n= mass/molar mass = m/M ____(4) Putting (4) in (3), we have σ = PM/RT ______(5) Example: calculate the density of Ammonia (NH3) in gL–¹ at 0.989 ATM and 55°c. Solution: given P = 0.989atm, T = 55+ 273 = 328k, R = 0.0821 atmLmol–¹K–¹ The molar mass M of NH3 is 17.03 gmol-¹ from, σ = PM/RT = (0.989)(17.03)/(0.0821)(328); σ = 0.625gL-¹ Dalton's Law of partial pressures states that at constant temperature the total pressure exerted by a mixture of gases in a definite volume is equal to the sum of the individual pressures which each gas would exert if it alone occupied the same total volume. For a mixture of gases 1, 2, 3... p tot = p 1 + p 2 + p 3... where p 1 , p 2 etc. represent the partial pressures. Dalton's law of partial pressure was gotten by applicationon ideal gas equation PV = nRT separately to each gas of the mixture. Thus, we can write the partial pressure P1, P2, P3 of the three gases P1 = n1 (RT/V). P2 = n2 (RT/V). P3 =n3 (RT/V). Where n1, n2 and n3 are moles of gases 1, 2 and 3 respectively. The total pressure, Ptotal of the mixtures will be; Ptotal = (n1 + n2 + n3) RT/V______(2) or. Ptotal = ntotal RT/V __________(3) In order words, the total pressure of the mixture is determined by the total number of moles present, whether of just one gas or mixture of gases. Example: Q 1. What pressure is exerted by a mixture of 2.00g of H2 and 8.00g of N2 at 273k in a 10liter vessel? Solution: moles of H2 = 2.00/2.02 = 0.990moles, moles of N2 = 8.00/28 = 0.268moles Therefore, Ptotal = (nH2 + nNH2) RT/V Ptotal = (0.990 + 0.286) 0.0821 × 273/10 ==> Ptotal = 2.86atm The partial pressure ratio is the same as the % by volume of z and the same as the mole ratio of gases in the mixture. This means for a component gas z: p z = p tot x %z/100 or p z = p tot x mol z/total mol = p tot x mol fraction z Examples of partial pressure calculations Ex. Q 2. In the manufacture of ammonia a mixture of nitrogen : hydrogen in a 1 : 3 ratio is passed over an iron/iron oxide catalyst at high temperature and high pressure. N 2(g) + 3H 2(g) 2NH 3(g) What are the partial pressures of nitrogen and hydrogen if the total pressure of the gases is 200 atm prior to reaction? The 1 : 3, N 2 :H 2 ratio means that nitrogen forms 1 /4 of the mixture, therefore p N2 = 1 / 4 x 200 = 50 atm and p H 2 = p tot – p N 2 = 150 atm (or from 3 /4 x 200) Ex Q 3. 10g of N2 and 10g of He are placed together in a 10L container at 25°C, Calculate the partial pressure of each gas and the total pressure of the mixture of the gas. Solution: moles of N2 = 10/28 = 0.4moles. moles of He = 10/4 = 2.5moles For the total pressure; Ptotal = (nN2 + nHe) RT/V Ptotal = (0.4 + 2.5) 0.0821 × 298/10 ====> 7atm For the individual pressures; Partial pressure of N2 = n(N2)÷n(total) × total pressure = 0.4 ÷ 2.9 × 7atm = 0.9atm Partial pressure of He = n(He)÷n(total) × total pressure = 2.5 ÷ 2.9 × 7 = 6.1atm MOLE FRACTIONS By definition; nA + nB = ntotal_____________1 By dividing eqn. 1 through by ntotal; nA/ntotal + nB/ntotal = ntotal/ntotal = 1 xA + xB = 1, where xA = nA/ntotal, which is the mole fraction of A, and xB = nB/ntotal, the mole fraction of B. Then; the sum of the mole fractions of the components of a mixture is equal to 1. Example: Calculate the mole fraction of HCl in a solution of hydrochloric acid in water, containing 36 percent of HCl by weight. Solution: We are told that the solution contains 36g of HCl and 64g of H2O Number of moles of HCl = (36g HCl) (1 mole of HCl/36.5gmol-¹ HCl) = 0.99 Number of moles of H2O = (64g H2O) (1 mole of H2O/18gmol-¹) = 3.55 ==> X of HCl = moles of HCl/moles of HCl + moles of H2O = 0.99/0.99 + 3.55 = 0.218 Example 2: 0.100 mole of NaCl is dissolved into 100g of pure H2O. What is the mole fraction of NaCl? Solution: We calculate the moles of water (which is the solvent) ===>. 100g/18gmol-¹ = 5?56mole of H2O Add that to the 0.100 mole of NaCl = 5.56 + 0.100 = 5.66 (total mole) mole fraction of NaCl = 0.100/5.66 = 0.018 mole fraction of H2O = 5.56/5.66 = 0.982 RELATIONSHIP BETWEEN PARTIAL PRESSURE, MOLE FRACTION AND TOTAL PRESSURE Consider an ideal gas mixture (binary) of A and B from the ideal gas equation PV = nRT; PA = nART/V ________________________________i Ptotal = ntotalRT/V ___________________________i i By dividing eqn (i) by (i i ) PA/Ptotal = nA/ntotal = XA ____________________i i i ==> PA = XA Ptotal __________________________i v The partial pressure of a particular component is equal to its mole fraction multiplied by the total pressure. Example: A mixture of gases contains 2.1 moles of Argon (Ar) and 1.6 moles of Xenon (Xe). Calculate the partial pressures of the gases if the total pressure is 2.00 atm. Solution: We first calculate the mole fraction; X of Ar = nAr/nAr + nXe = 2.1/2.1 + 1.6 = 0.567 X of Xe = nXe/nXe + nAr = 1.6/2.1. + 1.6 = 0.432 The partial pressures will be P of Xe = xXePtotal = 0.432 × 2.00 atm = 0.864 atm P of Ar = xArPtotal = 0.567 × 2.00 atm = 1.134 atm FUNNDAMENTAL EQUATION FOR THE PRESSURE OF A GAS PV = ⅓mNu² This is the fundamental equation of the kinetic molecular theory of gases. It is called the Kinetic Gas Equation. The equation although was derived for a cubical vessel, is equally valid for a vessel of any shape. The available volume in the vessel could be considered as made up of a large number of infinitesimally small cubes for each of which the equation holds. u² is the mean of the square for the individual velocities of all the N molecules for the gas. KINETIC GAS EQUATION INTERMS OF KINETIC ENERGY Let N be the number of molecules in a given mass of gas, PV = ⅓mNu² __________________________1 = ⅔ × ½mNu² ________________________2 = ⅔N × e ____________________________3 Where e is the average kinetic energy of a single molecule. Therefore, PV = ⅔Ne = ⅔E ______________4 Where E is the total kinetic energy of all the N molecules. The equation (4) is called the kinetic gas equation in terms of kinetic energy. We know that the general ideal gas equation is PV = nRT _________5 From (4) and (5) above, ⅔E = nRT ________________________6 For one mole of a gas, the kinetic energy of N molecules is, E = 3/2 RT ______________________7 Example: Calculate the kinetic energy of two moles of N2 at 27°C. Take R = 8.314JK-¹ mol-¹ Solution: we know that; E = 3/2 RT ; where n = 2.0 moles, T = (27 + 273) K = 300K and R = 8.314JK-¹ mol-¹ On substitution, we have E = 3/2 × 2.0 × 8.314 × 300 = 7482.6J Therefore the kinetic energy of two moles of N2 is 7482.6J Graham's law of diffusion Diffusion, or the 'self–spreading' of molecules, naturally arises out of their constant chaotic movement in all directions, though on a time average basis, more molecules will move in the direction of a region of lower concentration if such a situation exists e.g. initially 'pouring bromine vapour into air' in gas jars. Molecules of differing molecular mass diffuse at different rates. It has been shown that, assuming ideal gas behaviour and constant temperature, the relative rate of diffusion of a gas through porous materials or a mixture of gases or a tiny hole (effusion ) is inversely proportional to the square root of its density. Since the density of an ideal gas is proportional to its molecular mass, the relative rate of diffusion of a gas is also inversely proportional to the square root of its molecular mass*. PV = nRT, PV = m/M rRT, Mr = mRT/PV, since d = m/V, then M r is proportional to density. r1 √d 2 √M 2 ––– = –––– = –––– r2 √d 1 √M 1 Which is the mathematical ratio representation of Graham's law of diffusion for comparing two gases of different molecular masses. Graham's Law arises from the fact that the average kinetic energy of gas particles is a constant for all gases at the same temperature. The formula for kinetic energy is KE = ½mu² , where m = mass of particle, u = velocity. This means the average mu² is a constant for constant kinetic energy, so u is proportional to 1/√m and the m can be shown via the Avogadro Constant to be proportional to M r, the molecular mass of the gas. ROOT MEAN SQUARE VELOCITY If V1, V2, V3..........................Vn are the velocities of n molecules in a has, U², the mean of the squares of all the velocities is: U² = V1 2 + V2 2 + V3 2......Vn 2 /n Taking the square root, we have, U = √V1 2 + V2 2 + V3 2......Vn 2 /n Where U is thus the Root Mean Square Velocity or RMS velocity. It is denoted by U. The value of RMS of velocity U, at a temperature can be calculated from the kinetic gas equation. PV = ⅓mNU² __________________________(1) U² = 3PV/mN __________________________(2) For one mole of gas, PV = RT _____________________(3) Therefore, U² = 3RT/mN And, U² = 3RT/M ________________________(4) Where M = molar mass then; ( U = √3RT/M ) ______________________(5) RMS velocity is superior to the average velocity considered earlier. With the help of U, the total kinetic energy of a gas sample can be calculated. REAL GAS Distinction Between Ideal and Real gases 1. An ideal gas is infinitely compressible, a real gas will condense this a liquid at some pressure. 2. The particles of an ideal gas lose no to its container. A real gas conduct and radiates heat, thereby losing energy. 3. There is no attraction between the molecules of an ideal gas. A real gas has particle attractions. 4. An ideal gas can follow the formula PV = nRT. A real gas does not follow this this formula. COMPRESSIBILITY FACTOR This is a quantity that measures the deviation of a real gas from ideal behaviors. It is defined by the equation below; z = PVm/RT Where Z = compression factor Vm = molar volume of the gas, given by ; Vm = V/n The alphabets P, R and T maintain their usual meanings. ==> Z= 1 at all conditions. For a real gas, Z depends on the value of PVm relative to RT At high pressure, PVm > RT so that Z > 1. The gas is said to be positively from ideality and is less compressible than an ideal gas. At low pressure, PVm < RT. Z is less than 1. The gas deviates negatively from ideality and more compressible than an ideal gas. EXPLANATION OF DEVIATION - VANER WAALS EQUATION Van dear Waals (1873) attributed the deviations of real gases from ideal behavior to two postulates. These are : a. The molecules in a gas are point masses and possess no volume. b. There are no intermolecular attraction in a gas. Therefore, the ideal gas equation PV = nRT derived from kinetic theory could not hold for real gases. Van der Waals pointed out that both the pressure (P) and volume (V) factors in the ideal gas equation need to be corrected in order to make it applicable to real gases. Volume Correction The volume of a gas is the free space in the container in which molecules move about. Volume V of an ideal gas is the same as the volume of the container. The dot molecules of ideal gas have zero-volume and the entire space in the container is available for their movement. However, Van der Waals assumed that the molecules of real gas are rigid spherical particles which possess define volume. The volume of a real gas is, therefore, ideal volume minus the volume occupied by gas molecules. If b is the effective volume of molecules per mole of a gas, the volume in the ideal gas equation is corrected as : (V-b) For n moles of the gas, the corrected volume is (V-nb) Were, b is termed the excluded volume which is constant and characteristics for each gas. Excluded volume is four times the actual volume of molecules. Therefore, excluded volume is note equal to the actual volume of the gas molecules. Pressure Correction A molecule in the interior of a gas is attracted by the other molecules on all sides. The attractive forces cancele out. But a molecule about to strike the wall of the vessel is attracted by molecules on one side only. Hence, it experiences an inward pull. ==> A molecule about to strike the walls has a net inward pull; ==> A molecule in the interior of a gas has balanced attractions, therefore , it strikes the wall with reduced velocity and actual pressure of the gas P, will be less than the ideal pressure. If the actual pressure P, is less than Pideal by a quality p, we have P = Pideal - p or Pideal = P + p. P is determined by the force of attraction between molecule (A ) striking the wall of container and the molecule (B) pulling them inward. The net force of attraction is, therefore, proportional to the concentration of (A) type molecules and also (B) type molecules. That is; P ∆ CA × CB or P : n/v × n/v or P = an²/v² Where n is the total number of gas molecules in volume, V and a is the proportionality constant. Thus the pressure P in the ideal has equation is corrected as (P + an²/v²) for n moles of gas. VAN DER WAALS EQUATION Substituting the values of corrected pressure and volume in the ideal gas equation, PV = nRT, we have ; (P + an²/v²) (V - nb) = nRT This is called the Vander Waals equation for n molecules of a gas. Constants a and b in Vander Waals equation are called the Vander Waals constants. The constant are characteristic of each gas. DETERMINATION OF a AND b From the expression P = an²/v², the value of a is given by the relation a = pn²/v² = atmlitre²mol-² if the pressure is expressed in atmosphere and volume in litres. Thus, a is expressed in atmlitre²mol-² units. Since nb is excluded, volume for n moles of gas, b = volume/n = litre/mol its volume is expressed in litres, b is in litre mol-¹ units. S I Units of a and b a = (pressure) (volume) ²/(mol)² = (Nm-²)(m³)²/(mol)² = Nm⁴mol-² b = Volume mol-¹ = m³mol-¹ Example 1 : One mole of diethyl ether occupies 15L at 227°C. Calculate the pressure if Van dear Waals constant for diethyl ether are a= 17.38 atmL²mol-² and b = 0.134 Lmol-¹ Solution: Using the formula (P + an²/v²) (V-nb) = nRT or in a term wise P = (nRT/V-nb) - (n²a/v²) Given ; a = 17.38 atmL²mol-², b= 0.134 Lmol-¹, n= 1 mol, V = 15L , R = 0.0821 atmLmol-¹K-¹ , T = 500K After substituting the quantities given, the pressure P = 2.6837 atm. Example 2: Calculate the pressure exerted by 1.00 mole of methane (CH4) in a 250mL container at 300k using Vander Waals equation. What pressure will be predicted by ideal gas equation? Take a = 2.253 atmL²mol-² , b = 0.0428Lmol-¹ and R = 0.021 atmLmol-¹K-¹ Solution : Using the equation (P + an²/v²) (V -nb) = nRT which is the Vander Waals, By making P the subject and substituting the values; P = 82.8atm. While using the ideal gas equation, the predicted pressure gives P = nRT/V = 98.5atm. LIMITATION OF VAN DER WAALS EQUATION It expains satisfactorily the general behavior of real gases. It is applicable over a wide range temperatures and pressures. However, it fails to give the exact agreement with experimental data at very high pressures and low temperatures. NB: The constant a varies considerably from gas to gas because of the wide variety of intermolecular forces e.g. very low for helium and non–polar hydrogen (2 e's each, just instantaneous dipole–induced dipole forces), to much higher a values for larger polar molecules like water or dichlorodifluoromethane (more electrons and extra permanent dipole–permanent dipole intermolecular forces). The constant b varies less, and not unexpectedly, just tends to rise with increase in molecule size. Critical values of gas behaviour. Critical temperature T c This is the maximum temperature at which a substance can exist as a liquid. Above T c , only the gaseous state can exist, however great the density or pressure! It might be truer to say that above T c , the gaseous and liquid state become indistinguishable as the meniscus just disappears! Critical pressure p c This is the pressure the gas exerts at the critical temperature. Generally speaking the critical values for a gas/liquid increase with increase in intermolecular forces e.g. due to increase in molecular mass or increasing polarity of molecule. THEMOCHEMISTRY Thermochemistry is the study of the heat energy associated with chemical reactions and/or physical transformations. A reaction may release or absorb energy, and a phase change may do the same, such as in melting and boiling. Thermochemistry focuses on these energy changes, particularly on the system's energy exchange with its surroundings. Thermochemistry is useful in predicting reactant and product quantities throughout the course of a given reaction. In combination with entropy determinations, it is also used to predict whether a reaction is spontaneous or non-spontaneous, favorable or unfavorable. Endothermic reactions absorb heat, while exothermic reactions release heat. Thermochemistry coalesces the concepts of thermodynamics with the concept of energy in the form of chemical bonds. The subject commonly includes calculations of such quantities as heat capacity , heat of combustion , heat of formation , enthalpy, entropy , free energy, and calories. Below are graphs for the exothermic and endothermic reactions NB: Activation energy: The energy needed to get the reaction started - in exothermic, ∆H = HP - HR , where HR , HP are enthalpies of reactants and products. ∆H is negative because HP is less than HR. - in endothermic, ∆H = HP - HR , ∆H is positive because HP is greater than HR. The bigger the difference in reactants/products, the more energy is given out/taken in The world’s first ice-calorimeter , used in the winter of 1782-83, by Antoine Lavoisier and Pierre-Simon Laplace , to determine the heat evolved in various chemical changes ; calculations which were based on Joseph Black ’s prior discovery of latent heat. These experiments mark the foundation of thermochemistry. History Thermochemistry rests on two generalizations. Stated in modern terms, they are as follows: 1. Lavoisier and Laplace’s law (1780): The energy change accompanying any transformation is equal and opposite to energy change accompanying the reverse process. 2. Hess' law (1840): The energy change accompanying any transformation is the same whether the process occurs in one step or many. These statements preceded the first law of thermodynamics (1845) and helped in its formulation. Lavoisier, Laplace and Hess also investigated specific heat and latent heat , although it was Joseph Black who made the most important contributions to the development of latent energy changes. Gustav Kirchhoff showed in 1858 that the variation of the heat of reaction is given by the difference in heat capacity between products and reactants: dΔH / dT = ΔC p. Integration of this equation permits the evaluation of the heat of reaction at one temperature from measurements at another temperature. Calorimetry The measurement of heat changes is performed using calorimetry , usually an enclosed chamber within which the change to be examined occurs. The temperature of the chamber is monitored either using a thermometer or thermocouple , and the temperature plotted against time to give a graph from which fundamental quantities can be calculated. Modern calorimeters are frequently supplied with automatic devices to provide a quick read-out of information, one example being the differential scanning calorimeter (DSC). Systems Several thermodynamic definitions are very useful in thermochemistry. A system is the specific portion of the universe that is being studied. Everything outside the system is considered the surroundings or environment. A system may be: ==> a (completely) isolated system which can exchange neither energy nor matter with the surroundings, such as an insulated bomb calorimeter a thermally isolated system which can exchange mechanical work but not heat or matter, such as an insulated closed piston or baloon ==> a mechanically isolated system which can exchange heat but not mechanical work or matter, such as an un insulated bomb calorimeter ==> a closed system which can exchange energy but not matter, such as an un insulated closed piston or balloon ==> an open system which it can exchange both matter and energy with the surroundings, such as a pot of boiling water Processes A system undergoes a process when one or more of its properties changes. A process relates to the change of state. An isothermal (same-temperature) process occurs when temperature of the system remains constant. An isobaric (same-pressure) process occurs when the pressure of the system remains constant. A process is adiabatic when no heat exchange occurs. DEFINITION OF SOME TERMS ENTHALPY (H) is the amount of heat absorbed or liberated during chemical reaction at constant pressure. ENTHALPY CHANGE (∆H) is the energy change that occur in a chemical reaction at constant pressure. It's the difference between the of enthalpies of products and that of reactants. ∆H = ∑H (products) - ∑H (reactants) Unit of ∆H The unit used in thermochemical measurements is Kilojoule (kJ). If the amount of substance is 1 mole, the unit of ∆H is Kilojoule per mole (kJmol-¹). The old unit of ∆H is kilocalorie (kcal) or kilocalorie per mole (kcal mol-¹). The two units are related below: Kcal mol-¹ = 4.18kJmol-¹ A thermochemical reaction is a balanced chemical equation, describing the physical states of all reactants and products and the enthalpy change of the reaction. Manipulating the thermochemical equations; (i ) if the balanced equation is reversed, the sign of ∆H is changed ( i.e. PLUS is changed to MINUS and vice-versa) For instance; H2(g) + ∂ O(g) ---> H2O(l) + ∆H = -286kJ mol-¹ Reversing the equation gives; H2O(l) ---> H2(g) + ∂ O2(g) - ∆H = +286kJ mol-¹ (ii) if the balanced equation is multiplied by a factor n, ∆H is also multiplied by the same factor. For Example: H2(g) + ½O2(g) ---> H2O(l). ∆H1 = -286kJmol-¹ Multiplying the equation by 3 gives; 3(H2(g) + ½O2(g) ----> H2O(l) , ∆H1 = -286kJ) 3H2(g) + 3/2O2(g) ----> 3H2O(l) , 3∆H1 = -858kJ NB: if ∆H is multiplied by a number, "mol-¹" is dropped in the unit. (iii) The physical states of all the reactants and the products must be indicated in thermochemical equations. STANDARD ENTHALPY CHANGE (∆H) When enthalpy changes are measured at a temperature of 25°C (298k), pressure of 1 atm, concentration of 1 moldm-³ and in the natural states of the reactants, the enthalpy change is called the standard enthalpy change denoted by (∆H°) Standard enthalpy change of formation (∆H°f ) This is the change in enthalpy that take place when one fole of compound is formed from its elements, at a standard states that's( 298k and 1 atm). * A thermochemical equation for the formation of water from its elements at standard state. At standard state, hydrogen, oxygen are gases, water is a liquid. So, the formation of water can be represented by the thermochemical equation; H2(g) + ½O2(g) ----> H2O(l) , ∆H°f =-286kJmol-¹ * Considere the two equations below. Determine whether the ∆H°f values represent standard enthalpy change of formation of glucose. 6CO2 + 6H2O ---> C6H12O6 + 6O2 ∆H°1 = -32KJmol-¹ _______ 6C(s) + 6H2(g) + 3O2(g) ---> C6H12O6(s) ∆H°2 = -276KJmol-¹ _______(2) Answer: Equation (2) represents standard enthalpy change of formation, whilequation (1) doesn't because: (a) In equation (2), every substance is shown in its standard state. That's physical state (solid, liquid or gas). That substance would be under standard conditions. (b) There is never a compound on the reactants side in equation (2), only elements. Standard enthalpy change of solution (∆H°s) This is the amount of energy evolved or absorbed when one of substance is dissolved. Standard enthalpy change of combustion (∆H°c) This is the amount of energy evolved or absorbed when one mole of substance is completely burnt in excess oxygen, in standard conditions. Lattice energy (∆H L ) Lattice energy is the energy released when one mole of an ionic solid is formed from gaseous ions. Lattice dissociation energy (∆H° lD ) This is the energy absorbed when one mole of an ionic solid is dissociated into its gaseous ions. ∆H is equal and opposite in sign to ∆H for any substance. Standard enthalpy change of sublimation (∆H sub ) This is the heat required to convert one mole of a solid substance to a gas. Usually at standard temperature and pressure. Standard enthalpy change of atomization (∆H° a or dissociation (∆H° diss ) This is the heat required to convert one mole of gaseous molecules to atoms at standard state. Electron affinity (EA) This is the released or absorbed when one mole of gaseous anion is formed from one mole of gaseous atoms. Group VII elements (F, Cl, Be etc ) gain only one electron. Gaining of electrons occur in stepwise. Ionisation potential (IP) This is the energy required to remove an electron completely from a gaseous metal atom. Group IA elements ( Li, Na, K etc.) lose only one electron. Bond Energy ( BE) This is the energy released when one mole of a chemical bond is formed from gaseous atoms. Bond Dissociation Energy (BDE) This is the energy absorbed when one mole of a chemical bond is dissociated into free gaseous atoms. BDE is equal and opposite in sign to BE for any substance. The BDE cycle of a compound is a mirror image of its BE cycle. So, in this case the first equation is the dissociation of cthe compound to the individual atoms. In the BE and BDE of a liquid substance, an additional step for its vaporization should be included in the cycle as the following example below; Example: Calculate the standard enthalpy of formation of ethanol (CH3CH2OH ) from the following data: CH3CH2OH(l) ====> CH3CH2OH(g) , ∆H° vap = + 39 KJ mol-¹ C(s) ====> C(g) , ∆H°1 = + 716 KJ mol-¹ H2(g) ====> 2H (g) , ∆H° 2 = + 433 KJ mol-¹ O2(g) ====> 2O(g) , ∆H° 3 = + 498 KJ mol-¹ BEC-H = - 413 KJ mol-¹, BE C-C = - 345 KJ mol-¹ BEC-O = - 351 KJ mol-¹ , BE O-H = - 463 KJ mol-¹ Solution : The Born-Haber cycle and the structure of ethanol ; From the structure above, ethanol has ; 5(C-H), 1(C-C), 1(C-O), and 1(O-H) bonds. ∆H°rxn = 5BEC-H + BE C-C + BEC-O + BE O-H = 5(-413) - 345 - 351 - 463 = -3224 KJ ∆H°rxn = - 2 ∆H°1 - 3 ∆H° 2 - ½ ∆H° 3 + ∆H°f + ∆H° vap - 3224 = - 2 (716) - 3 (433) - ½ (498) + 39 + ∆H°f - 3224 = - 2941 + ∆H°f ∆H°f = - 283 KJ mol-¹ Therefore, The standard enthalpy change of formation of ethanol is - 283 KJ mol-¹. Example 2 : Using the above values, calculate the strain energy of cyclopropane if its heat of formation is 53 KJ mol-¹. Solution: The structure of cyclopropane is; Born- Haber cycle By Hess's law, ∆H°rxn = - 3 ∆H°1 - 3 ∆H° 2 + ∆H°f..................(1) But, ∆H°rxn = 6BEC-H + 3BE C-C......................(2) By equating (1) and (2) , we have; 6BEC-H + 3BE C-C = - 3 ∆H°1 - 3 ∆H° 2 + ∆H°f 6(-413) + 3(-345) = - 3(716) - 3(433) + ∆H°f ∆H°f = - 66 KJ mol-¹ Strain energy = ∆H°f (actual) - ∆H°f (calculated) = 53 - (-66) = 169 KJ mol-¹ The strain energy of cyclopropane is + 169 KJ mol-¹. Standard enthalpy change of neutralization ( ∆H neut ) This is the heat evolved when one mole of hydrogen ion from an acid reactants with one of hydroxyl ion from a base which results in neutralization reaction. When a strong acid is neutralized by a strong base, the heat of neutralization his approximately a constant value of -57kJmol-¹. This is due to the complete neutralization of the strong acid and the strong base. However, when the acid or the base is weak, further heat change is involved in ionizing the weak acid or the weak base and so the enthalpy of neutralization is either greater or less than - 57kJmol-¹. Standard enthalpy change of solution (∆H°s) Integral heat of solution and heat of solution at infinite dilution Integral heat of solution is the total heat change when one mole of a solute is dissolved in a specified amount of solvent. Heat of solution of ionic solids The enthalpy change of ionic solid depends on the energy absorbed in breaking up the crystal lattice (∆H L ) and the energy released in hydrating the ions ( ∆Hhdydration ). ∆Hsolution. = ∆HLD + ∆Hhydration  HESS'S LAW OF CONSTANT HEAT SUMMATION  The law states that; When reactants are converted to products, the enthalpy change is the same regardless of the path followed, provided the initial and final conditions are the same. As a consequence of this law, thermochemical problems can be solved either from : (a) a thermochemical cycle or (b) a combination of thermochemical equations THERMOCHEMICAL ( OR BORN-HARBER ) CYCLE Considere the following reactions. A + B _____∆H1 ======> P A ______∆H2 ======> C B _______∆H3 ======> D P ______∆H4 =======> C + D The reactions can be represented in a cycle as follows ; The conversion of P can be achieved either along path 1 or path 2. By Hess's law, the energy changes along the two paths are the same. Energy change along path 1 = ∆H1 Energy change along path 2 = ∆H2 + ∆H3 - ∆H4 NB: The negative sign attached to ∆H4 shows that path 2 is opposite to the direction of the flow of the arrow. Born-Haber cycle involving formation and combustion reactions. The general cycle involving formation and combustion reactions is given below; If A and B react to give a product P, then will give the same combustion products as P. This is the basis of the above cycle. Example: The heats of combustion of H2(g) and C(s) are -286 and -393kJmol-¹ respectively. The heat of formation of benzene is +48.1kJmol-¹. (a) write the thermochemical equations for the energy changes. (b) construct a Born-Haber cycle and used it to calculate the heat of combustion of benzene. Solution: (a) The thermochemical equations for the changes are: Combustion of H2(g)  H2(g) + ½O2(g) ====> H2 O(l) , ∆H 1 = -286kJmol-¹ ______(1) Combustion of C(s) C(s) + O2(g). =====> CO2(g) , ∆H2 = -393kJmol-¹ ______(2) Formation of C6H6(l) 6C(s) + 3H2(g) ====> C6H6(l) , ∆H3 = +48kJmol-¹ ________(3) combustion of C6H6.(l)  C6H6(l) + 15/2 O2(g) ===> 6CO2(g) + 3H2O(l) , ∆H4 = ? _______(4) Next, use the thernochemical equations to construct a B-H cycle STEPS Write down the formation equation; 6C(s) + 3H2(g) ====> C6H6(l) Draw arrow to indicate the complete combustion of 6C(s) and 3H2(g) C6H6(l) will give the same combustion products as 6C(s) + 3H2(g). So, draw an arrow from C6H6(l) to 6CO2(g) + 3H2O(l) The amount of O2(g) required to completely burn C6H6(l) is ;  6O2(g) + 3/2 O2(g) + = 15/2 O2(g) After writing the sketch and the correct enthalpy change on each arrow, ∆H1 is the enthalpy change for the combustion of H2(g). So, 3∆H2 is enthalpy change for combustion of 3H2(g) Similarly, 6∆H2 is the enthalpy change for 6C(s). ∆H3 and ∆H4 are the enthalpy changes for the formation of C6H6(l) respectively. The reactants 6C(s) + 3H2(g) can be converted to the product, C6H6(l) either along path 1 or path 2. Enthalpy change along path 1 = ∆H3 Enthalpy change along path 2 = 6 ∆H2 + 3 ∆H1 - ∆H4 NB: the negative sign attached to ∆H4 shows that the path taken is opposite to the direction of the arrow. By Hess's law, the enthalpy changes along the two paths are equal. ∆H3 = 6 ∆H2 + 3 ∆H1 - ∆H4 48 = -2358 + 858 - ∆H4 ∆H4 = -2358 + 858 - 48 = -3268kJ. Therefore, the heat of combustion of C6H6(l) is -3268kJ. NB: Read on the following B-H cycles ; * B-H cycle for polymerisation reaction * B-H cycle for bond energy calculations * B-H cycle for bond dissociation energy calculations USES OF BOND ENERGIES (a) Understanding the mechanism of chemical reactions (b) Comparing the strength of bonds (c) Understanding bonding and structure (d) Estimating enthalpy changes of reactions COMBINATION OF THERMOCHEMICAL EQUATIONS It is not always necessary to draw a Born-Haber cycle when solving a thermochemical problem. An alternative method of solution is by combination of thermochemical equations. Example: Calculate the enthalpy change of the reaction. C2H2(g) + 2H2(g) ====> C2H6(g) , by combination of the following thermochemical equations; 2C2H2(g) + 5O2(g) ====> 4CO2(g) + 2H2O(l) , ∆H1 = -2602kJ ______(1) 2 C2H6(g) + 7O2(g) ====> 4CO2(g) + 6H2O(l) , ∆H2 = -3132kJ ______(2) H2(g) + ½O2(g) =====> H2O(l) , ∆H3 = -286kJ _____________(3) Solution: The target equation is C2H2(g) + 2H2(g) ∆H = ? =====> C2H6(g) The following manipulations are performed using equation (1), (2) and (3) until the target equation is obtained. The target equation has 1 C2H2(g) on the left, So, equation (1) is divided by 2. C2H2(g). + 5/2 O2(g) =====> 2CO2(g) + H2O(l) , ∆H1 /2 __________(4) There are 2H2(g) on the left side of the target equation. Therefore, equation (3) is multiplied by 2. 2H2(g) + O2(g) =====> 2H2O(l) , 2∆H3 ________________(5) But, the target equation doesn't contain O2 , CO2 , and H2O(l). So we have to find a way to remove them. To remove 2CO2 on the right of equation (4), divide equation (2) by 2 and reverse. 2CO2(g) + 3H2O(l) ====> C2H6(g) + 7/2 O2(g) , -∆H2 /2 _________(6) Adding equation (4), (5) and (6) ; C2H2(g) + 2H2O(l) + 7/2 O2(g) + 2CO2(g) + 3H2O(l) ====> 2CO2(g) + 3H2O(l) + C2H6(g) + 7/2 O2(g) ______________(7) Cancelling elements and compounds that are the same on both sides of (7) gives the target equation. C2H2(g) + 2H2(g + C2H6(g) ______________(8) ∆H = ∆H1 /2 + 2∆H3 - ∆H2 /2 = (-2602/2) + 2(-286) - (-3123/2) = -311.5kJ Therefore; the enthalpy change of the reaction is = -311.5kJ. CHEMICAL EQUILIBRIUM (PLURAL EQUILIBRIA) It is a common and well established fact that many reactions do not reach to completion. They proceed to a certain point and then stop apparently, often leaving considerable amount of unaffected reactants. Under any given conditions of temperature, pressure, and concentration, the point at which any reaction seems to stop is always the same, i.e. there exist at this point among the concentrations of the various reactants and products of any a reaction a relationship which is definitely fixed. When a reaction reaches this stage at its course, then it's said to be in equilibrium. These types of reactions can generally be represented as; aA + bB cC + dD Chemical equilibrium may be defined as a state of a reversible reaction when the two opposing reactions occur at the same rate and the concentrations of the reactants and the products do not change with time. A + B C + D However, the true equilibrium of a reaction can be attained from both sides. Thus, the equilibrium concentrations of the reactants and the products are the same whether they start with A and B or C and D. Characteristics of Chemical Equilibrium * Value of Equilibrium Constant does not depend upon the initial concentration of the reactants. It has been found that equilibrium constant must be the same when the concentrations of the reacting species veried over a wide range. * A catalyst cannot change the equilibrium point When a catalyst is added to a system in equilibrium, it speed up the rate of both forward and reverse reaction to an equal extent. Therefore, a catalyst cannot change the equilibrium point except that it's earlier achieved. * Equilibrium cannot be attained in an Open Vessel The equilibrium can be established only if the reaction vessel is closed and no part of the reactants or products is allowed to escape out. In an open vessel, the gaseous reactants and/or products may escape into the atmosphere leaving behind no possibility of attaining equilibrium. * Equilibrium can be initiated from either side The state of equilibrium of a reversible reaction can be approached whether we start with reactants or products. * Constancy of concentrations When a chemical equilibrium is established in a closed vessel at constant temperature, concentrations of the various species in the reaction mixture becomes constant. EQUILIBRIUM CONSTANT Let consider a general reaction; A + B C + D And let [A], [B] , [C] and [D] represent the molar concentrations of A, B, C, and D at equilibrium point. According to the law of Mass Action, Rate of forward reaction is ∆ [A] [B] = K1 [A] [B] Rate of reverse reaction is ∆ [C] [D] = K2 [C] [D] Where K1 and K2 are rate constants for the forward and reverse reactions. At equilibrium, the rate of forward reaction = rate of reverse reaction. Therefore; K1 [A] [B] = K2 [C] [D] or K1 / K2 = [C] [D] / [A] [B] At any specified temperature, both K1 and K2 are constant so K1 / K2 is also constant called the equilibrium constant, Kc or K. Kc = [C] [D] / [A] [B] This equation is know as the equilibrium constant expression or Equilibrium law. Generally for a reaction; aA + bB cC + dD The equilibrium constant is; Kc = [D]d [C]c / [A]a [B]b RATES OF REACTIONS (REVRSIBLE AND IRREVERSIBLE REACTIONS) Rates of chemical reaction is defined as the speed with which reactants disappear or the speed with which the products are formed. It is the rate of change of reactants or products with time. Different types of reactions proceed at different rates. Normally the ordinary (simple) laboratory reactions such as precipitation, neutralisation etc. are fast reactions. But reactions such as radioactive decays, rusting, and some biochemical reactions are relatively slow and can take years to reach completion. REVERSIBLE REACTIONS Those reactions which proceeds in both forward and backward direction and never reaches completion are called reversible reaction. These reactions can proceed in either direction. For instance; PCl5(g) ∆ PCl3(g) + Cl2(g) (Thermal dissociation) N2(g) + 3H2(g) 2NH3(g) (Synthesis reaction) CH3COOH + NaOH CH3COONa + H2O (Neutralisation reactions) IRREVERSIBLE REACTIONS Those reactions that proceeds in forward direction and almost reaches to completion are called irreversible reactions. For instance; 2KClO3(s) MnO2 catalyst ======> 2KCl(s) + 3O2(g) (Thermal decomposition) NaOH(aq) + HCl(aq) ====> NaCl(aq) + H2O ( Strong-acid base neutralisation reaction) 2AgNO3 + BaCl2 ====> 2AgCl + Ba(NO3)2 (Precipitation reaction) STATE OF EQUILIBRIUM It has generally been observed that many changes (both physical and chemical) do not proceed to completion when the are carried out in a container. Consider for instance, vaporisation of water; Water Vapour Vaporisation of water occur at any temperature. Initially, the concentration of water is much greater than the concentration of vapour, but with the progress of time, concentration of vapour increase whereas that of water remains constant and after a certain interval of time, there is no change in the concentration of vapour, this state is known as the physical state of equilibrium. In a similar way, this has also been found for chemical reactions. For instance, when PCl5(g) is heated in a closed container, its dissociation starts with the formation of PCl3(g) and Cl2(g). Initially, only PCl5(g) was taken, but with the progress of the reaction, PCl3(g) and Cl2(g) are formed due to dissociation of PCl5(g). After a certain interval of time, the concentration of PCl5(g) , PCl3(g) and Cl2(g) each becomes constant. It doesn't mean that this point of time, dissociation of PCl5(g) and its formation from PCl3(g) and Cl2(g) has been stopped. Actually, the rate of dissociation of PCl5(g) and the rate of formation of PCl5(g) becomes equal. This state is called the state of chemical equilibrium. So, the state of chemical equilibrium is dynamic. PCl5(g) PCl3(g) + Cl2(g) This can be shown graphically; In general, "state of equilibrium" can be defined as the state when the rate of forward reaction becomes equal to the rate of reverse reaction and the concentration of all the species becomes constant. LAW OF MASS ACTION Gulberg and Waage in 1807 gave this law and it states that; " At constant temperature, the rate at which a substance react is directly proportional to its active mass, and the rate at which a chemical reaction proceeds is directly proportional to the product of the active masses of the reacting species. The term "active mass" of a reacting species is the effective concentration or its activity which is related to the molar concentration. Example: Some nitrogen and hydrogen are placed in an empty 5.0 litre container at 500°C. When equilibrium is established, 3.01 mol of N2 , 2.10 mol of H2 and 0.565 mol of NH3 are present. Evaluate Kc for the following reaction at 500°C. N2(g) + 3H2(g) 2NH3(g) The equilibrium concentrations are : [N2 ] = 3.01/5.00L = 0.602M, [ H2 ] = 2.10/5.00L = 0.420M, [NH3 ] = 0.565/5.00L = 0.113M Kc = [NH3 ]² / [N2 ] [ H2 ]³ = (0.113)²/(0.602)(0.420)³ =0.206 Variation of Kc With the Form of The Balanced Equation The value of Kc depends on how we write the balanced equation for the reaction. Example: Considere the following chemical equation and its equilibrium constant at a given temperature 2HBr(g) + Cl2(g) 2HCl(aq) + Br2(g) Kc = 4.0×10⁴ Write the the expression for and calculate the numerical value of the equilibrium constant for the equation at given temperature. 4HBr(g) + 2Cl2(g) 4HCl(aq) + 2Br2(g) Solution: Kc = [HCl]² [Br2 ] / [HBr]² [Cl2 ] = 4.0×10⁴ (From the original equation) Since the original equation has been multiplied by 2, so Kc must be squared Kc = [HCl] [Br2 ]² / [HBr]⁴ [Cl2 ]² = (4.0×10⁴)² = 1.6×(10³)³ The Reaction Quotient The reaction quotient Q for the general reaction is given as; aA + bB cC + dD Q = [C]c [D]d / [A]a [B]d (not necessarily equilibrium concentrations) The reaction quotient, Q is a measure of the progress of a reaction Q < Kc = forward reaction predominates until equilibrium is established Q = Kc = The system is at equilibrium Q > Kc = reverse reaction predominates until equilibrium is established Example: Some nitrogen and hydrogen gases are pumped into an empty five-litre glass bulb at 500°C. When equilibrium is established, 3.00 moles of N2 , 2.10 moles of H2 and 0.298 mole of NH3 are found to be present. Find the value of Kc for the reaction. N2(g) + 3H2(g) 2NH3(g) Solution: The equilibrium concentrations are obtained by dividing the number of moles of each reactant and product by the volume, 5.00L, thus, [N2 ] = 3.00moles/5.00L = 0.600M [ H2 ] = 2.10/5.00L = 0.420M [NH3 ] = 0.298/5.00L = 0.0596M By substituting the above concentrations (not number of moles) in the equilibrium constant expression, we get the value of Kc Kc = [NH3 ]² / [N2 ] [ H2 ]³ = (0.0596)²/(0.600)(0.420)³ = 65.0 For the reaction below; 2HI(g) H2(g) + I2(g) The following concentrations were detected in a mixture. Is the system at equilibrium? If not, in which direction must the reaction proceed for equilibrium to be established? [HI] = 0.500M, [H2] = 2.80M, [I2 ] = 3.40M Solution: Q = [H2] [I2 ]/ [HI]² = (2.80)(3.40)/(0.500)² = 38.1 But Kc = 65.0, so Q < Kc the system is not at equilibKc um. For equilibrium, the value of Q must be equal to that of Kc. This can only be achieved when the numerator increases and the denominators decreases. Thus, the forward (left to right) reaction must occur to a greater extent than the reverse, that's some cHI must react to form more H2 and I2 to attain equilibrium. Uses of the equilibrium constant, Kc Once the value of Kc has been calculated from the equilibrium concentrations, the process can be turned around to calculate the equilibrium concentrations from the equilibrium constant. Example: The equation for the following reaction and the value of Kc at a given temperature are given. Am equilibrium mixture in a 1.00L container contains 0.25 mol of PCl5 and 0.16 mol of PCl3. What equilibrium concentrations of Cl2 must be present? PCl3(g) + Cl2(g) PCl5(g) Kc = 1.9 Solution: The volume of the container is 1.00L, the molar concentration (mol/L) of each substance is numerically equal to the number of moles. Kc = [ PCl5 ] / [ PCl3 ] [ Cl2 ] = 1.9 Then for Cl2 will be; [ Cl2 ] = [ PCl5 ] / Kc [ PCl3 ] = 0.25 / (1.90)(0.16) = 0.82 In as much asxwe know the starting concentrations and wish to know how much of each reactant and product would be present at equilibrium is possible. Example 2: At a certain temperature, 0.100mole of H2 and 0.100mole of I2 were placed in a one-litre flask. The purple colour of iodine vapour was used to monitor the reaction. After a time, the equilibrium was established and it was found that the concentration of I2 decreased to 0.020mole/litre. Calculate the value of Kc for the reaction at a given temperature. Solution: Let see the equilibrium constant expression for the reaction; H2 + I2 2HI Kc = [HI]² / [ H2 ] [ I2 ] Also, let us proceed to find the equilibrium concentrations of the various species. The initial moles of H2 , I2 , and HI present at equilibrium are ; H2 + I2 2HI Moles at equilibrium: (0.100-x) (0.100-x) 2x Since the volume of the reaction vessel is one litre, the moles at equilibrium also represent the equilibrium concentrations. Thus; [ I2 ] = 0.100-x = 0.020 (as given) x = 0.100-0.020 x = 0.080 Knowing the value of x, we can find the equilibrium concentrations of H2 , I2 and HI [ H2 ] = 0.100-x = 0.100-0.080 = 0.020 [ I2 ] = 0.020 (from above) [2HI] = 2x = 2x0.080 = 0.160 By substituting the values; Kc = [HI]² / [ H2 ] [ I2 ] = (0.160)² / (0.02)(0.02) = 64 Partial Pressures and The Equilibrium Constant It is often more convenient to measure pressures rather than concentration of gases. In general, for reaction involving gases; aA(g) + bB(g) cC(g) + dD(g) Kp = (Pc)c (PD) d / (PA)a (PB)b N2(g) + 3H2(g) 2NH3(g) Kp= (PNH3)² / (PN2) (PH2)³ Example : In an equilibrium mixture at 500°C we find, PNH3 = 0.147atm , PN2 = 6.00atm, PH2 = 3.70atm. Evaluate Kp at 500°C for the following reaction. N2(g) + 3H2(g) 2NH3(g) Solution: Kp = (PNH3)² / (PN2) (PH2)³ = (0.147)² / (6.00) (3.70)³ = 0.0711×10-³ Relationship Between Kp and Kc If the ideal gas equation is rearranged, the molar concentration of a gas is M = P/RT or n/V = P/RT. By substituting P/RT for n/V in the Kc expression for the N2 - H2 - NH3 in ammonia production reaction, equilibrium gives the relationship between Kc and Kp for the reaction. Kc = [NH3 ]² / [N2 ] [ H2 ]³ = Kp = (PNH3 /RT)² / (PN2 /RT) (PH2 /RT)³ = Kp = (PNH3)² / (PN2) (PH2)³ × (1/RT)² /(1/RT)⁴ Kc = Kp (RT)² and Kp = Kc (RT)-² In a general, the relationship between Kc and Kp is Kp = Kc(RT)∆n or Kc = Kp(RT)-∆n Where ∆n = (ngas product(s)) - (ngas reactant(s)) For reactions in which equal number of moles of gases appear on both sides of the equation, ∆n = 0 and Kp = Kc. ∆n < 0 , Kp < Kc. ∆n > 0, Kp > Kc. Where ∆n = (c + d) - (a + b) Example: Calculate the Kc and Kp for the following reactants; (i) 2SO2(g) + O2(g) 2SO3(g). (ii) 1/2N2(g) + 2/3H2 NH3(g) Solution: (i) 2SO2(g) + O2(g) 2SO3(g). By applying law of mass action; Kc = [SO3 ]² / [SO2 ]²[ O2 ] ____(i) Kp = (PSO3 )² / (PSO2 )² (PO2 ) _____(i i ) We know that; Kp = Kc (RT)∆n But; ∆n = 2 - (2 + 1 ) = -1 therefore; Kp = Kc /RT (ii) 1/2N2(g) + 2/3H2 NH3(g) Kc = [NH3 ] / [N2 ]½ [H2 ]^3/3 _____(i) Kp = (PNH3 ) / (PN2 )½ ( PH2 )^2/3 _____(ii) Knowing that; Kp = Kc (RT)∆n ∆n = 1 - ( 3/2 + 1/2 ) = -1 therefore; Kp = Kc /RT Example 2: At 400k for the gaseous reaction, 2A + 3B 3C + 4D The value of Kp = 0.05. Calculate the value of Kc (R = 0.082 dm³ at K-¹ mol-¹ ) Solution: From the given data; From the reaction; 2A + 3B 3C + 4D ∆n = (3 + 4 ) - ( 2 + 3 ) = 2 Since R is given in dm³ atm mol-¹ K-¹. We shall take P = 1atm By substituting the values in the below equation; Kp = Kc (CRT/P)∆n we get; 0.05 = Kc (1 × 0.082 × 400/1)² Kc = 0.05 / 0.08 × 0.08 × 400 × 400 Kc = 4.88×10^-5 Heterogeneous Equilibrium Heterogeneous equilibrium is the equilibrium that involves reactants and products in different physical state or an equilibrium involving species in more than one phase. Consider the following reversible reaction at 25°C. 2HgO(s) 2Hg(l) + O2(g) For any pure solid of pure liquid, the activity is taken as 1, so terms of pure liquids or pure solids do not appear in the K expressions for heterogeneous equilibrium, ie for above reaction Kc = [O2] or Kp = PO2 Example : Write both Kc and Kp of the following reactions; (i) N2(g) + 3H2(g) 2NH3(g) (ii) S(s) + H2 SO3(aq) H2S2O3(aq) Solution: (i) N2(g) + 3H2(g) 2NH3(g) Kc = [NH3 ]²/[H2 ]³[N2 ] Kp = (PNH3 /RT)² / (PN2 /RT) (PH2 /RT)³ (ii) S(s) + H2 SO3(aq) H2S2O3(aq) Kc = [H2S2O3 ]/[H2 SO3 ][S] Kp = is undefined because no gases are involved. Relationship Between ∆G°rxn and K The relationship between free energies both at standard and non standard conditions is given as; ∆G rxn = ∆G°rxn + RTlnQ Where R is the universal gas constant, T is the absolute temperature, and Q is the reaction quotient. When a system is at equilibrium, ∆G rxn = 0 and Q = K, upon substitution, 0 = ∆G°rxn + RTlnK, ∆G°rxn = -RTlnK When the relationship ∆G°rxn = -RTlnK is used with ; a. All gaseous reactants and products, K represents Kp b. All solution reactants and products, K represents Kc c. A mixture of solution and gaseous reactants, K represents the thermodynamic equilibrium constant and there is distinction between Kp and Kc. The thermodynamic equilibrium constant is defined in terms of activities of the species involved. aA(g) + bB(g) cC(g) + dD(g) K = (aC )c (aD )d / (aA )a (aB )b Where a is the activity of the substances. Example: Calculate Kp for the following reaction at 25°C. 2N2O(g) 2 N2(g) + O2(g) where, ∆G°f (N2O) = 104.2KJ/mol Solution: Remember, for all gaseous reaction; K = Kp and ∆G°rxn = -RTlnKp ∆G°rxn = [2∆G°f (N2(g)) + ∆G°f ( O2(g)) ] - [2∆G°f (N2O(g)) ] = [2(0) + 0 ] - [2(104.2) ] = -208.4KJ/mol or -2.084×10^-5 J/mol But, ∆G°rxn = -RTlnKp lnKp = ∆G°rxn /- RT = -2.084×10^-5/-8.314×298) = 84.1 Kp = e84.1 = 3.3×10^36 Evaluation of Equilibrium Constant at Different Temperatures This is achieved by the use of van't Hoff equation. ln(KT2 / KT1 ) = ∆H° /R (1/T1 - 1/T2 ) Where KT1 is the equilibrium constant at T1 KT2 is the equilibrium constant at T2 Thus, knowing the value of ∆H° for the reaction and R at a given temperature, van't Hoff equation can be used to calculate the value of K at any other temperature. Example: N2(g) + O2(g) 2NO(g) If Kp for the above reaction is 4.4×10-³¹ at 25°C (298K) and ∆H° is 180.5KJ/mol. Evaluate Kp at 2400K. Solution: We know that; ln(KT2 / KT1 ) = ∆H° /R (1/T1 - 1/T2 ) ln(KT2 / KT1 ) = 1.805×10^-5 /8.314 ( 1/298 - 1/2400) = 63.8 KT2 / KT1 = e 63.8 = 5.1×10^-27 KT2 = 5.1×10^-27 × 4.4×10-³¹ = 2.2×10-³ Displacement of Equilibrium : Factors Affecting Chemical Equilibrium The establishment of chemical equilibrium is the result of a balance between the forward and reverse reactions of a reversible reaction. Therefore, the equilibrium position is altered by the factors which affects the rate of forward and reverse reactions unequally and their effects were summarised by H. leChatelier in 1885. Le Chatelier's Principle sates that; If a chemical system is in equilibrium and one of the factors involved in the equilibrium is altered, the equilibrium will shift so as to annul the effect of change. Displacement caused by concentration changes Consider the general reaction; aA + bB cC + dD = [C]c aq [D]d aq / [A]a aq [B]d aq Provided the total volume and other factors such as temperature are kept constant, an increase in the concentration of (say) A will produce an increase in the concentration of C and D, since Kc must be kept constant. Example: The equilibrium constant for the formation of ethyl ethanoate is 4.0 at 70°C. What concentration of ester would be produced at equilibrium starting from one mole of ethanoic acid (i) 2 moles of alcahol (ii) 10 moles of alcahol Solution: C2H5OH + CH3COOH CH3COOC2H5 + H2O Initial concentration 2mol 1mol 0M 0M ∆ in conc. due to reaction -xmol -xmol +xmol +xmol Concentration at equilibrium (2-x)mol (1-x)mol xmol xmol (i) Kc = [CH3COOC2H5 ][H2O] / [CH3COOH][C2H5OH] = 4 = x²/(1-x)(2-x) = 4 3x² - 12x + 8 = 0 , x = 0.84 mol (ii) Kc = x²/(1-x)(10-x) = 4 3x² - 44x + 40 = 0, x = 0.97 mol Thus, a greater yield of ester follows an increase in the initial concentrations of alcohol (or acid) and the equilibrium is displaced to the right, but if one of the products is removed from the reaction medium, as it is formed, then the equilibrium is displaced so far that the reaction may go to completion. Displacement caused by changes in pressure Pressure affects only those equilibrium which shows a change in volume between the reactants and products. The effect of pressure changes on reactions in which gases involved are particularly important for the following reasons. a. The composition of equilibrium mixture (but not the Kc ) is altered in those cases where there is a volume change. b. In any gaseous reaction, an increase in pressure brings the reaction molecules closer together and the reaction speeds up, that affects both forward and reverse reactions. Example: In the reaction; 2SO2(g) + O2(g) 2SO3(g). ∆n = (ngas product(s)) - (ngas reactant(s)) = 2-(2+1) = -1 Increase in pressure favours smaller volume ie more SO3 will be formed, while decrease in pressure favours higher volume ie more SO2 and O2 will be formed. Displacement caused by changes in temperature Generally, in all reactions, an increase in temperature causes an increase in the rate at which a reaction takes place. Ie approximately double for every 10°C rise in temperature. This means that, at higher temperatures, the equilibrium position is attained more rapidly. Temperature has an effect on the value of the equilibrium constant. It may be shown that log K α -∆H/T For an endothermic reactions, ∆H is positive, K increases and at any temperature, log K is negative. Thus, log K α -∆H/T or log 1/K.... ∆H/T Hence, as the temperature increases, K increases and the product concentration will rise. Example: N2(g) + 3H2(g) 2NH3(g) = -92KJ The forward reaction is favoured by a decrease in temperature been exothermic, while reverse reaction is favoured by an increase in temperature. Effect of catalyst A catalyst is a substance that speed up the rate of a chemical reaction. In equilibrium reactions, both forward and reverse reactions are equally affected so that the position of equilibrium is not changed, although the equilibrium state is arrived at more quickly. IONIC EQUILIBRIUM Introduction: We have learnt previously about non-ionic equilibrium whose name implies those that do not form any ions. In this section, we are going to continue with ionic equilibrium, which is concerned with dissociation of of substances in solution into ions and the reverse process. Ionic equilibrium is in two forms, namely; 1. Dissociation equilibrium 2. Solubility equilibrium Therefore, the basic reaction of ionic equilibrium is ionisation and the members include the acids, bases and salts. While dissociation equilibrium is concerned with the ionisation of substances. The solubility equilibrium is primarily concerned with the solubility properties of salts in solution. As we shall see both types of equilibria are going to be discussed strictly as chemical changes. An acid or a base can be defined either in terms of dissociation or proton transfer. For example, an acid can be defined as; a. A substance which dissociates to give hydrogen ion or hydroxonium ion as the only positive ion (Arrhenius definition) b. A proton donor (Bronsted definition) A base may therefore be defined as a proton acceptor (Bronsted definition) or as a substance which dissociates to give hydroxyl ions the only negative ion (Arrhenius definition). We are going to consider the Bronsted acid and base in many of our acid-base equilibrium (also called Bronsted equilibrium. Acid-base equilibrium occur when at least one component (either the acid or base) is weak (ie partially dissociates). Let's take some examples of Bronsted equilibrium. NH3 + HCl ====> NH4 + + Cl- (Base) (Acid) (Acid) (Base) It would be seen from the chemical equation that an a I'd reacts with a base to form the corresponding 'conjugate acid' and 'conjugate base'. Dissociation equilibrium Generally, any weak acid HA will dissociates in water according to the following equation: e.g. HA + H2O ====> H3O+ + Cl- CH3COOH + H2O ====> H 3 O+ + CH3COO- Strong acid very weak base strong acid strong base HCl + H2O ====> H 3 O+ + Cl- Strong acid very weak base strong acid extremely weak base The equilibrium constant for the dissociation of the acid can be given in terms of the concentration of the reactants and products as follows: K = [H3O+ ][A] / [HA][H2O] So, K[H2O] = [H3O+ ][A] / [HA] The concentration of water in a solution containing 1 moldm-³ solute is only 2% lower than that of water (which is 55.5 moldm-³ if 1000g of water is considered). Therefore, [H2O] = constant. Because K[H2O] = Ka , we can write the following equation: Ka = [H3O+ ][A] / [HA] Where Ka is the acid dissociation constant (acidity constant or acid ionisation constant). The Bronsted equilibrium of a base in water is given by: B + H 2O ====> HB+ + OH- e.g. NH3 + H2O ====> NH4 + + OH- By analogous approach as in the case of acid Kb = [HB+ ][OH- ] / [B] Where Kb is the base dissociation constant (base ionisation constant). It is important to note that the numerical values of Ka and Kb are characteristics of particular acid or base, and is a measure of their strength. Like all equilibrium constants. Example: If 0.1 mole of ethanoic acid is dissolved in 1 dm³ of water and the concentration of hydroxonium ions is 6.64×10^-7 moldm-³, calculate Ka for the ethanoic acid. Solution: Write the equation for the ionisation of CH3COOH CH3COOH + H2O ====> H3O+ + CH3COO- Ka = [ H3O+ ][CH3COO- ] / [CH3COOH ] But after dissociation, partial ionisation of the acid occurs, [ H3O+ ] = [CH3COO- ] = 6.64×10^-7 moldm-³, The resulting concentration of CH3COOH after this ionisation is given by; [ CH3COOH ] =(0.1-6.64×10^-7 moldm-³) approx. = 0.1moldm-³ ====> Ka = [ H3O+ ][CH3COO- ] / [CH3COOH ] = (6.64×10^-7)² / 0.1 Therefore; Ka = 4.41×10-¹² moldm-³ Amphiprotic behaviour of water Water is amphoteric, it can even react with itself by a process called autoionisation (or autoprotolysis of water). The equation for autoionisation of water is given as: H2O + H2O ====> H3O+ + OH- (Acid) (Base) (Acid) (Base) K = [H3O+ ][OH- ] / [H2O]² K[H2O]² = [H3O+ ][OH- ] But K[H2O]² = Kw ==> Kw = [H3O+ ][OH- ] At 25°C, pure water [H3O+ ] = 1×10^-7 moldm-³ = [OH- ] So Kw = 1×10-¹⁴ mol²dm^-6 It is also true that Kw = Kb × Ka Concept of pH PH scale represents the widely different hydronium ion concentrations conveniently on graph. By definition, " pH is the negative log to base ten of the molar hydrogen ion concentration" Mathematically, pH = -log[H+ ] In analogous manner, pOH = -log[OH- ] The above notation has been extended to equilibrium constant. For instance, pKa = -logKa If pH = 7 (e.g pure water), pOH = 7. Therefore, pH + pOH = 14 (In support, observation has shown that as pH = 0, pOH = 14. Similarly, when pH = 5, pOH = 9 ). But Kw = [H3O+ ][OH- ] pKw = -log[H+ ] + -log[OH- ] = pH + pOH pH + pOH = 14 = pKw pH of strong acid and strong base In strong acids, dissociation is virtually complete and stable salts are formed. Hence, the formulae to use for the calculation of pH and pOH do not require any modification. pH = - log[H3O+ ] pOH = - log[OH- ] pH = pOH = 7 pKa for strong acids is negative (for instance pKa (HNO3 ) = -1 and pKa (HCl) = -3 ). This implies that Ka < 1. In contrast, pKb for strong base is positive (for example pKb (NO3 - ) = 15 and pKb (Cl- ) = 17 ). This implies that Kb > 1. Example: Calculate the pH of (a) 0.05 moldm-³ KOH (b) 0.1 moldm-³ HCl Solution: (a) Write the equation for the ionisation of KOH KOH + H2O ====> K+ (aq) + OH- (aq) Since KOH ionises completely in water; [OH- ] = [KOH] = 0.05 moldm-³ For strong base, pOH = -log[OH- ] Or pOH = -log 0.05 = 1.3 But pH + pOH = 14 ==> pH = 14 - pOH pH = 14 - 1.3 = 12.7 (b) Write the equation for ionisation of HCl HCl + H2O ====> H3O+ (aq) + Cl- (aq) Since HCl ionises completely in water, [ Cl- ] = [HCl] = 0.01 moldm-³ For strong acid, pH = - log[H3O+ ] or pH = -log 0.1 = 1.0 Degree of ionisation of weak acids and bases The degree of ionisation (α) of a weak acid or base is the extent to which it ionises in aqueous solution. If C is the concentration of a weak acid HA then; α = [A] /C or [H+ ] /C Similarly, for a weak base B α = [B]/C or [OH- ] /C Aso, percentage ionisation is given by; = α × 100 NB: the Ka of weak acid or Kb for a weak base usually reduces to a quadratic equation. Solution of the equation is either by approximation or by quadratic formula. 1. To predict the method of solution of quadratic equation, calculate C/Ka of the weak acid or C/Kb of the weak base. If this value is equal to or greater than 1000, solution of the equation by approximation MAY be valid. 2. Apply the approximation to determine the [H+ ] or [OH- ] 3. Calculate [H+ ]/C or [OH- ]/C if the value is less or equal to 5%, the solution of the equation by approximation is VALID. 4. If [H+ ]/C or [OH- ]/C is greater than 5%, the equation is solved by quadratic formula. Example: Calculate (a) [H+ ] (b) [CH3COO-] (c) [CH3COOH] (d) α (e) pH of a 0.049 mol/L acetic acid ( Ka =1.8×10^-5 ) Solution: CH3COOH(aq) + H2O ====> CH3COO- (aq) + H+ (aq) Initial conc 0.049 mol/L 0 0 Equilibrium conc 0.049 - x 0+x 0+x The equilibrium concs are [CH3COOH] = 0.049 - x , [CH3COO-] = x, [H+ ]= x Ka = [CH3COO-][H+ ] /[CH3COOH] 1.8×10^-5 = (x)(x) /(0.049 - x) We now predict the method of solution of the above equation C/Ka = 0.049/1.8×10^-5 = 2722 Since C/Ka is greater than 1000, we can apply the approximation method 0.049 - x is approx = 0.05 1.8×10^-5 = x²/0.049 x = √(0.049 × 1.8×10^-5) = 9.4×10-⁴ mol/L [H+ ] = 9.4×10-⁴ mol/L We check the validity of the approximation [H+ ]/C = 9.4×10-⁴ / 0.049 = 0.019 or 1.9% Since [H+ ]/C is less than 5%, the approximation is valid Then proceed to determine the other parameters (a) [H+ ] = 9.4×10-⁴ mol/L (b) [CH3COO-] = x = 9.4×10-⁴ mol/L (c) [CH3COOH] = 0.049 - x = 0.049 - 9.4×10-⁴ = 0.048 mol/L (d) α = [H+ ]/C = 9.4×10-⁴ / 0.049 = 0.019 or 1.9% (e) pH = -log [H+ ] = - log(9.4×10-⁴) = 3.03 Example 2: Calculate the pH of a 0.036 mol/L nitrous acid (HNO2 ) solution (Ka = 4.5×10-⁴) Solution: Let x be the concentration of HNO2 which dissociate at quilibrium HNO2 (aq) + H2O (l) + ====> H+ (aq) + NO2 - (aq) Equilibrium conc 0.036 - x x x Ka = x²/(0.036 - x) = 4.5×10-⁴ Predicting the method of solution; C/Ka = 0.036/4.5×10-⁴ = 80 Since this value is less than 1000, the equation cannot be solved by approximation, the quadratic formula has to be used. x²/(0.036-x) = 4.5×10-⁴ The equation expands to x² + 4.5×10-⁴ x - (4.5×10-⁴)(0.036)=0 This equation is comparable to x² + Kax - KaC = 0 which has a solution of x= -Ka ± √(Ka ² + 4KaC)/2 Therefore, Ka = 4.5×10-⁴ and C = 0.036 The positive root is x = -4.5×10-⁴ + √(4.5×10-⁴)² + 4(4.5×10- ⁴)(0.036) /2 = 3.8×10-³ mol/L [H+ ] = 3.8×10-³ mol/L and pH = -log(3.8×10-³) = 2.42 Buffer Solutions A buffer solution is a solution containing a mixture of a weak acid and its salt (i.e. the conjugate base) or a weak base and its salt (i.e. the conjugate acid). A good example of a buffer solution is consists of a mixture of ethanoic acid ( i.e. a weak acid) and its esther ammonium ethanoate (i.e. the conjugate base). CH3COOH + H2O ====> H 3 O+ + CH3COO- Amount: Relatively large Huge Relatively small Relatively large Because the corresponding salt is present (e.g. CH3COONa), common ion effect is observed owing to the low concentration of hydroxonium ions formed by the weak ethanoic acid. (Refer to the common ionic effect under solubility equilibrium). CH3COONa ====> CH3COO- + Na+ Buffer solution resist changes in pH on either addition of acid or base. For instance, in the above reaction, any addition of strong acid will cause the added H3O+ to reactant with excess acetate to form the union ionised ethanoic acid. On addition of strong base, the added OH- reacts with H3O+ to form water. In the preparation of buffers, some factors need to be considered which include buffer range and buffer capacity. Buffer range refers to the pH range over which the buffer is effective, while buffer capacity is the amount of strong base (or strong acid) required to alter the pH by 1 unit. The buffer capacity determines the amount of strong acid or strong base that may be added without producing a significant change in pH. The empherical equation that computes the pH of a buffer is the "Henderson-Hasselbaleh equation". This equation can be derived as follows; For the dissociation of weak acid HA to give conjugate base A- Ka = [ H3O+ ][A- ]/[HA] ==> Ka [HA]/[A-] = [ H3O+ ] If we take negative logarithm to base 10 of both sides of the above equation, we obtain; pKa -log [HA]/[A- ] = pH Where [HA] is the concentration of the acid and [A-] i the conjugate base which in disguise, stands for the concentration of the salt. pH = -log [Acid]/[Salt] (Henderson-Hasselbaleh equation) Similarly, for weak base and its salt: [OH- ] = [B]/[BH+] × Kb Therefore, pOH = pKb -log [Base]/[Salt] For ionisation of a weak base: B + H 2O + BH+ (aq) + OH- (aq) Example:  Calculate the pH of an aqueous buffer solution which contains 0.1 moldm- ³ NH3 and 0.2 moldm-³ NH4Cl. (pKa ( NH4 + ) = 9.3 )  Solution:  Step 1: write the equations  NH4Cl ====> NH4 + + Cl-  0.2 moldm-³  NH3 + H2O ====> NH4 + + OH-  Step 2: select a suitable Henderson-Hasselbaleh equation  pH = pKa -log [Acid]/[Base]  Step 3: Identity the acid HA and the conjugate A  pH = 9.3 -log (0.2)/(0.1)  pH = 9.3 - 0.3 = 9.0    Acid-base titration and indicators Basically, acid-base titration is a neutralisation reaction in which the titrant is either an acid or a base and the solution is being acidic or basic respectively. Usually pH indicator is added in order to show the endpoint. An acid-base indicator is a weak acid or base with conjugate forms having different colours. The following equations represent the change that takes place in an indicator which is a weak acid;  HInd + H2O H3O+ + Ind- Weak acid conjugate base The indicator constant Kind is given in the following equation: Kind = [Ind-][H3O+ ]/[HInd] By taking the -log of both sides and rearrangement, we obtain the following equation; pKind -pH = log [HInd]/[Ind-] pH = pKind at the endpoint pH < pKind when the indicator and corresponding colour are predominantly in acidic form pH > pKind when the indicator and the corresponding colour are predominantly in basic form Endpoint is indicated by a sharp colour change resulting from the addition of only a slight excess of the titrant. The choice of appropriate indicator becomes necessary in acid-base titration so as to achieve rapid colour change as endpoint is reached. The table below gives a guide to the choice of indicator, depending on the titrant and titrant used. Acid-base titrations and suitable indicators for use Acid-base Titration Rapid pah change occurs Suitable indicator Strong acid against 3.5 - 9.5 Any indicator strong base Strong base against weak 7.0 - 9.5 Phenolphthalein acid Strong acid against weak 3.5 - 7.0 Methyl orange base Weak acid against weak No sharp change No suitable indicator base The curve or titration curve is graph of pH used. The pH curve displays the progress of titration. At this junction, it is important to distinguish between stoichiometric point (or equivalence point) and endpoint. The equivalence point is a specific point on the graph at exact which neutralisation is deemed complete. The pH of the indicator must change at the equivalence point of the reaction. In order words, end point must coincide with equivalence point. The equivalent point can be seen by inspection, but can also be calculated precisely through integral calculus by computing the points of inflection on the pH curve. Case I : in case of titration of strong acid (e.g. HCl) versus strong base (e.g. NaOH) where the pH of the resulting solution is ultimately 7, the endpoint could be called the neutral point (even though the use of endpoint has been preferred by analysis). The pH curve begins at high pH typical of strong base and ends at low pH typical of strong acid. There is a rapid change in pH near the equivalence point (pH = 7) from the fig. Below ; pH = pOH = 7 Strong acid versus strong base pH curve Hydrolysis: An acid and base react to form salt and water. The reverse process (in which the salt forms the acid and base in the presence of water) is called hydrolysis. The salt of strong acid and strong base does not undergo hydrolysis. Case II : For titration of weak acid (e.g. CH3COOH ) against strong base (e.g. NaOH) HA + BOH ====> BA + H 2O Where B can be metal ion such Na+ + NH4 + e.g. CH3COOH + NaOH ====> CH3COONa + H 2O Recall: Kw = [H3O+ ][OH- ] Because Ka = [H3O+ ][A- ] / [HA] we can have the following expression for Kw ; Kw = [OH- ] × Ka [HA] /[A- ] The majority of the of the [OH- ] in solution is due to Bronsted equilibrium above and the levels have outnumbered that of autoionisation. Hence [OH- ] = [HA] ==> Kw = [OH- ]² × Ka [HA] /[A- ] If we make [OH- ] the subject of the formula and take -log of both sides of the equation, we have; pOH = ½ (pKw - pKa - log [A- ] ) The titration curve begins at a higher acidic pH and ends at high basic pH (see fig. Below). The pH change at the equivalence point (pH>7) is not so great and after the equivalence point a buffer solution is formed. Because pKw = pH - pOH, the pH at that equivalence point is given by; pH = ½ (pKw + pKa + log [A- ] ) Where [A- ] = [Salt] Weak against strong base pH curve Example: Calculate the value of pH at stoichiometric point of titration 20cm³ of 0.1 moldm-³ HClO(aq) with 0.1 moldm-³ NaOH. pKa ( NaClO ) = 7.4 Solution : The first thing to know is that the titration is between weak acid and strong base. HClO + NaOH ====> NaClO + H 2O Therefore, pH = ½pKw + ½pKa + ½log [A- ] At the stiochiometric point the volume is twice 20 cm³ Recall: MV = constant So, 0.1 moldm-³ × 20cm³ = 40cm³ × [NaClO] The concentration of the salt = [NaClO] = 0.05 moldm-³ pH = ½(7.4) + ½(14) + ½log(0.05) pH = 9.4 Hydrolysis: For the hydrolysis of the salt of weak acid and strong base in water (which is the reverse reaction to case II, the same pH formula (above) holds. However, we can include hydrolytic constant Kb in the equation; For the hydrolysis of salt BA (of weak acid and strong base) A + H2O ====> HA + OH- (the spectator of B is not shown) The hydrolytic constant Kh can be expressed by; Kh = [HA][OH- ]/[A- ] Kh = [HA][OH- ]/[A- ] × [H+ ]/[H+ ] ==> K h = Kw / Ka If use this relationship, the pH equation above turns to the following; pH = ½pKw + ½pKa + ½log [A- ] Where [A- ]= [Salt BA] The pH of the solution at equivalence point ( and so is the pKind ) as well as on hydrolysis is greater than 7. Case III : For strong acid (e.g. HCl ) against weak base (e.g. NH3 ) the curve begins at low pH and ends at a less high alkaline pH. The pH change at the equivalence point (pH < 7) is not great. HA + BOH ====> BA + H 2O e.g. CH3COOH + NaOH ====> CH3COONa + H2O Because Ka = [H3O+ ][A- ] / [HA] = [H3O+ ]² /[A- ] We can have the following expression for [H3O+ ] : [H3O+ ] = ([A- ] × Ka )½ If we take the -log of both sides of the equation, we arrive at; -log[H3O+ ] = -½logKa - ½log[A- ] Therefore the pH at stiochiometric point of strong acid weak base titration is given by; pH =½logKa - ½log[A- ] Where [A- ] = [Salt] Example: If 10 ml of 0.3 moldm-³ NH3(aq) is reacted with H2SO4. What will be the pH at stiochiometric point? pKa (NH+ ) = 9.3 Solution: The first thing to know is that the titration is between strong acid and weak base. Therefore, pH = ½pKa - ½log [A- ] Now, the concentration of salt ( [A- ] ) will depend on the concentration of the weak base. At the stiochiometric point, the volume will double and the concentration of the salt can be obtain from the relationship MV = constant: 0.3 moldm-³ × 10 cm³ = 20 cm³ × [A- ] The concentration of the salt = [A- ] = 0.15 moldm-³ pH = ½(9.3) - ⅓log 0.15 = 4.65 - 0.41 pH = 4.24 Hydrolysis: For hydrolysis of the salt of strong acid and weak base in water ( which is the reverse reaction of Case III), the same pH formula (above) holds. However, we may need to include the hydrolytic constant Kh in the equation. For hydrolysis of salt BA (of strong acid and weak base) BA + H2O ====> HA + BOH The hydrolytic constant Kh can be expressed by; Kh = [B][H3O+ ] / [BA] Kh = [B][H3O+ ] / [BA] × [OH- ] / [OH- ] ==> Kh = Kw / Kb which means pKh = pKw - pKb If use this relationship, the pH equation above turns to the following; pH = ½pKw - ½pKb + ½log [A- ] Where [A- ] = [Salt BA] The pH of this solution is on hydrolysis is Kh = Kw / Ka Kb SOLUBILITY EQUILIBRIUM On addition of water to a sparingly soluble ionic salt, a saturated solution is formed. Eventually, the ions so formed will be in equilibrium with the excess undissolved solute. MxXY(s) ====> xM+ (aq) + YX- (aq) e.g. BaSO4 ====> Ba2+ + SO4 2- K = [Ba2+ ][SO4 2- ] / [BaSO4 ] ==> [Ba2+ ][SO4 2- ] = K[BaSO4 ] The product [Ba2+ ][SO4 2- ] at any instant is known as the ion activity product (IAP). The IAP at saturation is known as the solubility product (Ksp ). ==> Ksp = [Ba2+ ][SO4 2- ] The following points should be noted about IAP. 1. In saturated solution, there is a state of equilibrium between salt and the solution and IAP = Ksp 2. In undersaturated solution, IAP < Ksp 3. In supersaturated solution, IAP > Ksp Solubility product must not be confused with solubility. Solubility of a salt refers to the total, maximum of a salt dissolvable. Example: Find the solubility product of BaSO4 , if the saturated solution of the salt contains 2.39×10-³ g/dm³ at 20°C Solution: The solubility (total concentration) of BaSO4 = 2.39×10-³ g/dm³ at 20°C We need this concentration to that in molar unit. Since the molar mass of BaSO4 = 233.2 g/mol, the concentration in moldm-³ is given by; Concentration (moldm-³) = (2.39×10-³ g/dm³) / 233.3 = 1 ×10^-5 BaSO4 ====> Ba2+ + SO4 2- Ksp = [Ba2+ ][SO4 2- ] = 1×10^-5 moldm-³ × 1×10^-5 moldm-³ Ksp = 1×10^-10 mol²dm^-6 NB: dm³ = L² Example 2: Find the solution of Ca(OH)2 in water if Ksp = 1.32×10-¹¹ mol³/dm^9 Solution: Ca(OH)2 ====> Ca2+ + 2OH- Knowing that Ksp = [Ca2+ ][OH- ]² Let [Ca2+ ] = x, then [OH- ] = 2x Ksp = 1.32×10-¹¹ = x × (2x)² 4x³ = 1.32×10-¹¹ x = 1.5×10-⁴ Application of solubility product 1. Prediction of precipitation If IAP is greater than Ksp value, it means precipitation can occur. Example: Determine whether Ca precipitation will form when equal volumes of 0.2 moldm-³ Pb2+ and 0.3 moldm-³ SO4 2- solution, Ksp (PbSO4 ) = 2×10^-8 Solution: PbSO4 ====> Pb2+ + SO4 2- Immediately after mixing, the concentration of Pb2+ and SO4 2- ions gets halved and ion activity product becomes; IAP = [Pb2+ ][SO4 2- ] = (0.1) × (0.15) = 1.5×10-² mol²dm^-6 Because IAP is greater (>>) than Ksp precipitation will occur (formed). 2. Common effect The solubility of an ionic salt will reduce if mixed with the solution of another salt containing the same ion. Generally, the solubility of salt say, A+ B- will reduce in the presence of A+ or B- from the second source. Example: Calculate the solubility of AgCl (a) in water (b) in 0.1 moldm-³ NaCl Ksp (AgCl) = 2×10^-10 mol²dm^-6 Solution: (a) in water AgCl(s) ====> Ag+ (aq) + Cl- (aq) Let the solubility of AgCl be x Ksp (AgCl) = [Ag+ ][Cl- ] = (x) × (x) Or 2×10^-10 mol²dm^-6 = x² ==> x = 1.14×10^-5 moldm-³ (b) in 0.1 moldm-³ NaCl AgCl(s) ====> Ag+ (aq) + Cl- (aq) AgCl(s) ====> Ag+ (aq) + Cl- (aq) 0.1 moldm-³ 0.1 moldm-³ 0.1 moldm-³ [Ag+ ] = x moldm-³ [Cl- ] = (x + 0.1) moldm-³ Ksp (AgCl) = [Ag+ ][Cl- ] = (x) × (x + 0.1) 2×10^-10 mol²dm^-6 = (x) × (x + 0.1) But x (x + 0.1) approx. = 0.1 2×10^-10 mol²dm^-6 = (x) × (0.1) x = 2×10^-9 3. Prediction of concentration of ions in solution If an excess solution of a given cation (e.g. Ag+ ) is mixed with a solution of a suitable anion (e.g. Cl- ), the concentration of the cation remaining in solution can be calculated using the Ksp of AgCl. ELECTROCHEMISTRY Electrochemistry is the branch of physical chemistry that is concerned with the behaviour of ions in solutions, their interaction with one another and with a metallic conductor (Electrode) and the subsequent relationship between chemical energy and electrical energy. Uses of Electrochemistry 1. Power source a. Batteries b. Fuel cell (space craft) 2. Biomedical research 3. Pollution analysis 4. Corrosion prevention ELECTROLYTIC AND ELECTROCHEMICAL CELLS Electrolytic Cell Electrolytic cell is a device in which electrical energy is converted into chemical energy, e.g. any set-up used in electrolysis; which comprises of ; 1. An electrolyte 2. An electrodes a. Anode which is the positive electrode b. Cathode which is the negative electrode 3. Both the electrode and the electrolyte are existing in the same container, with the two electrodes connected externally by an electric circuit which serves as a source of direct current (DC). Electrochemical Cell Electrochemical cell is a device that converts chemical energy into electrical energy e.g. batteries. In this tyre of cell, Anode is the negative while cathode is the positive. Each electrode is dipped into its electrolyte or paste containing ions. The two separate electrolyte are connected by a salt bridge, which is a paste of KCl or KNO3. CELL POTENTIAL When a metal is in contact with water or ionic solution, it dissociates to give ions. The solution leaving behind electrons on the remanning surface of metal e.g. M(s) H2O====> Mn+ (aq) + ne- __________________(a) The force with which these electrons are transported from the negative electrode to the positive electrode is called electromotive force (EMF) or cell potential measured in volts ; is given by; Volt = joule/coulomb = J or Nm/C or A sec mol-¹ _____________(b) Reduction Potential Generally, when a metal electrode is dipped in its aqueous solution, the ions in the solution try to withdraw electrons from the metal electrode in order to get reduced. The extent with which this process takes place is referred to as reduction potential. STANDARD HYDROGEN ELECTRODE (SHE) The SHE consists of hydrogen gas (H2(g) ) at 1 atm bubbling over a platinum plate (electrode) immersed in a solution of 1 M HCl at 25°C. The platinised electrode has the following uses in the SHE ; 1. It acts as an inert metal connection to the H2 /H+ system. 2. It allows H2(g) to be adsorbed on its surface. 3. Being platinised, it has increased surface area so that it can establish an equilibrium between H2(g) and H+ (aq) as rapidly as possible. The SHE is assumed to have a standard electrode potential E°, of zero (0.00V); i.e. for the reaction ½H2(g) ====> H+ (aq) + e-, E° = 0.00V. Standard Electrode (Half-Cell) Reduction Potential To measure the standard reduction potential of a given hal-cell; 1. All the ionic species must have a concentration of 1 M. 2. All gases should be at 1 atm pressure. 3. Working temperature should be 25°C, and 4. Platinum should be used as the electrode when the half-cell does not include a metal. Generally, the standard reduction potential (E°) of any half-cell is obtained by reference t that of the SHE. For instance, the standard reduction potential of a Zn(s) /Zn2+ (aq) half-cell, is the potential difference between the SHE and the zinc half-cell. Generally, E°cell = E°RHS - E° LHS = E° Zn/Zn2+ - E° H2/H + = E° Zn/Zn2+ - 0 Therefore, E° Zn/Zn2+ = -0.76V VOLTAIC CELL This is formed by combining two types of half-cells , e.g. Daniel cell is formed by combining zinc and copper half-cells. In this cell, the reaction taking place is Zn(s) + Cu2+ (aq) ====> Zn2+(aq) + Cu(s).The standard reduction potential of this type of cell is defined as "the reduction potential for the substance that actually undergoes reduction minus the reduction potential for the substance that is forced to undergo oxidation". Thus; E°cell = E°sub.Red. - E° sub.Oxd. E°cell = E°Cu²+/Cu - E° Zn/Zn²+ = E° LHS - E° RHS = +1.10V CELL DIAGRAM The following conventions or notations are applied for writing the cell diagram in accordance with IUPAC recommendations. The Daniel cell is represented as follows : Zn(s) | Zn2+ (C1) || Cu2+ (C2) | Cu(s) (a) Anode half cell is written on the left hand side while cathode half cell on right hand side. (b) A single vertical line separates the metal from aqueous solution of its own ions. Zn(s) |

CHEM 112 - Physical Chemistry PDF

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