Lecture 3 - Thermodynamics and chemical kinetics PDF
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Università degli Studi di Milano Bicocca
Andrew Smith, PhD
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This document is a lecture on thermodynamics and chemical kinetics. It covers topics such as the ideal gas equation, chemical thermodynamics, bond enthalpy, entropy, and Gibbs free energy. It also includes discussions of reaction rates and factors affecting them.
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Thermodynamics and chemical kinetics CHEMISTRY AND PROPEDEUTIC BIOCHEMISTRY I (H4102D001M) Thermodynamics and chemical kinetics Andrew Smith, PhD andrew.smith@u...
Thermodynamics and chemical kinetics CHEMISTRY AND PROPEDEUTIC BIOCHEMISTRY I (H4102D001M) Thermodynamics and chemical kinetics Andrew Smith, PhD [email protected] DIPARTIMENTO DI MEDICINA E CHIRURGIA Learning objectives In today’s lecture we should understand the following: Understand the ideal gas equation The basics of chemical thermodynamics Bond enthalpy Entropy Gibb’s free energy The basics of chemical kinetics Rate and order of reactions Factors that affect the rate of reactions DIPARTIMENTO DI MEDICINA E CHIRURGIA 3 KINETIC THEORY DIPARTIMENTO DI MEDICINA E CHIRURGIA Arrangement of particles in solids, liquids and gases ……… DIPARTIMENTO DI MEDICINA E CHIRURGIA Plasma, the fourth state of matter https://www.iasgyan.in/daily-current-affairs/plasma-the-fourth-state-of-matter DIPARTIMENTO DI MEDICINA E CHIRURGIA Plasma treatment in medicine https://doi.org/10.3390/biophysica1010005 DIPARTIMENTO DI MEDICINA E CHIRURGIA Changes in state Melting and freezing If energy is supplied by heating a solid, the heat energy causes stronger vibrations until the particles eventually have enough energy to break away from the solid arrangement to form a liquid. The heat energy required to convert 1 mole of solid into a liquid at its melting point is called the enthalpy of fusion. When a liquid freezes, the reverse happens. At some temperature, the motion of the particles is slow enough for the forces of attraction to be able to hold the particles as a solid. As the new bonds are formed, heat energy is evolved. DIPARTIMENTO DI MEDICINA E CHIRURGIA Changes in state Boiling and condensing If more heat energy is supplied, the particles eventually move fast enough to break all the attractions between them, and the liquid boils. The heat energy required to convert 1 mole of liquid into a gas at its boiling point is called the enthalpy of vaporisation. If the gas is cooled, at some temperature the gas particles will slow down enough for the attractions to become effective enough to condense it back into a liquid. Again, as those forces are re-established, heat energy is released. DIPARTIMENTO DI MEDICINA E CHIRURGIA Changes in state Evaporation of a liquid in a closed container As the gaseous particles bounce around, some of them will hit the surface of the liquid again, and be trapped there. There will rapidly be an equilibrium set up in which the number of particles leaving the surface is exactly balanced by the number rejoining it. DIPARTIMENTO DI MEDICINA E CHIRURGIA Changes in state Evaporation of a liquid in a closed container In this equilibrium, there will be a fixed number of the gaseous particles in the space above the liquid. When these particles hit the walls of the container, they exert a pressure. This pressure is called the saturated vapour pressure (also known as saturation vapour pressure) of the liquid. DIPARTIMENTO DI MEDICINA E CHIRURGIA Changes in state Sublimation Solids can also lose particles from their surface to form a vapour, except that in this case we call the effect sublimation rather than evaporation. Sublimation is the direct change from solid to vapour (or vice versa) without going through the liquid stage. DIPARTIMENTO DI MEDICINA E CHIRURGIA Changes in state Sublimation https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_1 DIPARTIMENTO DI MEDICINA E CHIRURGIA 13 THE IDEAL GAS EQUATION DIPARTIMENTO DI MEDICINA E CHIRURGIA Assumptions about ideal gases There is no such thing as an ideal gas, of course, but many gases behave approximately as if they were ideal at ordinary working temperatures and pressures. The assumptions are: Gases are made up of molecules which are in constant random motion in straight lines. The molecules behave as rigid spheres. Pressure is due to collisions between the molecules and the walls of the container. All collisions, both between the molecules themselves, and between the molecules and the walls of the container, are perfectly elastic. The temperature of the gas is proportional to the average kinetic energy of the molecules. DIPARTIMENTO DI MEDICINA E CHIRURGIA And then two absolutely key assumptions, because these are the two most important ways in which real gases differ from ideal gases: There are no (or entirely negligible) intermolecular forces between the gas molecules. The volume occupied by the molecules themselves is entirely negligible relative to the volume of the container. DIPARTIMENTO DI MEDICINA E CHIRURGIA Amadeo Avogadro 1776 – 1856 Native of Turin, Italy Hypothesized that equal volumes of gases at the same temperature and pressure contained equal numbers of molecules. (He was correct, too!) DIPARTIMENTO DI MEDICINA E CHIRURGIA Robert Boyle Natural philosopher, in particular within the field of chemistry 1627 – 1691 https://www.britannica.com/biography/Robert-Boyle DIPARTIMENTO DI MEDICINA E CHIRURGIA Boyle’s Law For a fixed mass of gas at constant temperature, the volume is inversely proportional to the pressure. You can express this mathematically as pV = constant This is called Boyle’s Law. DIPARTIMENTO DI MEDICINA E CHIRURGIA Charles’ Law For a fixed mass of gas at constant pressure, the volume is directly proportional to the kelvin temperature. V = constant x T https://en.wikipedia.org/wiki/Charles% 27s_law#/media/File:Charles_and_Ga y-Lussac's_Law_animated.gif This is called Charles’ Law. DIPARTIMENTO DI MEDICINA E CHIRURGIA Charles’ Law We have fulfilled what Charles' Law says. We have a fixed mass of gas. The pressure is the same before and after (in each case, the same as the external air pressure) whilst the volume increases when you increase the temperature of the gas. DIPARTIMENTO DI MEDICINA E CHIRURGIA The ideal gas equation pV = nRT DIPARTIMENTO DI MEDICINA E CHIRURGIA CONVERTING UNITS FOR IDEAL GAS EQUATION Before using pV = nRT, convert units to Pa, m3 and K DIPARTIMENTO DI MEDICINA E CHIRURGIA Calculating gas volumes DIPARTIMENTO DI MEDICINA E CHIRURGIA Calculating gas volumes DIPARTIMENTO DI MEDICINA E CHIRURGIA Calculating a relative molecular mass DIPARTIMENTO DI MEDICINA E CHIRURGIA Calculating a relative molecular mass DIPARTIMENTO DI MEDICINA E CHIRURGIA Ideal gas calculations (Q) Calculate the mass of ammonia in flask Q. Give your answer to the appropriate number of significant figures DIPARTIMENTO DI MEDICINA E CHIRURGIA Ideal gas calculations (A) DIPARTIMENTO DI MEDICINA E CHIRURGIA Ideal gas calculations (Q) Flask Q (volume = 1.00 x 10-3 cm3 is filled with ammonia (NH3) at 102 kPa and 300K. The tap is closed and there is a vacuum in flask P. (Gas constant R = 8.31 JK-1mol-1) When the tap is opened, ammonia passes into flask P. The temperature decreases by 5 °C. The final pressure in both flasks is 75.0 kPa. Calculate the volume, in cm3 , of flask P. DIPARTIMENTO DI MEDICINA E CHIRURGIA Ideal gas calculations (A) Flask Q (volume = 1.00 x 10-3 cm3 is filled with ammonia (NH3) at 102 kPa and 300K. The tap is closed and there is a vacuum in flask P. (Gas constant R = 8.31 JK -1mol-1) When the tap is opened, ammonia passes into flask P. The temperature decreases by 5 °C. The final pressure in both flasks is 75.0 kPa. Calculate the volume, in cm3 , of flask P. DIPARTIMENTO DI MEDICINA E CHIRURGIA 31 CHEMICAL THERMODYNAMICS DIPARTIMENTO DI MEDICINA E CHIRURGIA Thermodynamics Thermodynamics studies how changes in energy, entropy and temperature affect the spontaneity of a process or chemical reaction. Using thermodynamics we can predict the direction a reaction will go, and also the driving force of a reaction or system to go to equilibrium. DIPARTIMENTO DI MEDICINA E CHIRURGIA THERMODYNAMICS First Law Energy can be neither created nor destroyed but It can be converted from one form to another Energy changes all chemical reactions are accompanied by some form of energy change changes can be very obvious (e.g. coal burning) but can go unnoticed Exothermic Energy is given out Endothermic Energy is absorbed Examples Exothermic combustion of fuels respiration (oxidation of carbohydrates) Endothermic photosynthesis thermal decomposition of calcium carbonate DIPARTIMENTO DI MEDICINA E CHIRURGIA Simple energy diagrams A reaction in which heat energy is given off is said to be exothermic DIPARTIMENTO DI MEDICINA E CHIRURGIA Simple energy diagrams A reaction in which heat energy is absorbed is said to be endothermic DIPARTIMENTO DI MEDICINA E CHIRURGIA Simple energy diagrams A reaction in which heat energy is absorbed is said to be endothermic DIPARTIMENTO DI MEDICINA E CHIRURGIA Energetic stability? Kinetic stability! DIPARTIMENTO DI MEDICINA E CHIRURGIA Energetic stability? Kinetic stability! DIPARTIMENTO DI MEDICINA E CHIRURGIA THERMODYNAMICS - ENTHALPY CHANGES Enthalpy a measure of the heat content of a substance at constant pressure you cannot measure the actual enthalpy of a substance you can measure an enthalpy CHANGE written as the symbol DH , “delta H ” Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants DIPARTIMENTO DI MEDICINA E CHIRURGIA THERMODYNAMICS - ENTHALPY CHANGES Enthalpy a measure of the heat content of a substance at constant pressure you cannot measure the actual enthalpy of a substance you can measure an enthalpy CHANGE written as the symbol DH , “delta H ” Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants ENTHALPY REACTION CO-ORDINATE Enthalpy of reactants > products DH = - ive EXOTHERMIC Heat given out DIPARTIMENTO DI MEDICINA E CHIRURGIA THERMODYNAMICS - ENTHALPY CHANGES Enthalpy a measure of the heat content of a substance at constant pressure you cannot measure the actual enthalpy of a substance you can measure an enthalpy CHANGE written as the symbol DH , “delta H ” Enthalpy change (DH) = Enthalpy of products - Enthalpy of reactants ENTHALPY ENTHALPY REACTION CO-ORDINATE REACTION CO-ORDINATE Enthalpy of reactants > products Enthalpy of reactants < products DH = - ive DH = + ive EXOTHERMIC Heat given out ENDOTHERMIC Heat absorbed DIPARTIMENTO DI MEDICINA E CHIRURGIA STANDARD ENTHALPY CHANGES Why a standard? enthalpy values vary according to the conditions a substance under these conditions is said to be in its standard state... Pressure:- 100 kPa (1 atmosphere) A stated temperature usually 298K (25°C) as a guide, just think of how a substance would be under normal lab conditions assign the correct subscript [(g), (l) or (s) ] to indicate which state it is in any solutions are of concentration 1 mol dm-3 to tell if standard conditions are used we modify the symbol for DH. Enthalpy Change Standard Enthalpy Change (at 298K) DIPARTIMENTO DI MEDICINA E CHIRURGIA STANDARD ENTHALPY OF FORMATION Definition The enthalpy change when ONE MOLE of a compound is formed in its standard state from its elements in their standard states. Symbol DH°f Values Usually, but not exclusively, exothermic Example(s) C(graphite) + O2(g) ———> CO2(g) H2(g) + ½O2(g) ———> H2O(l) 2C(graphite) + ½O2(g) + 3H2(g) ———> C2H5OH(l) Notes Only ONE MOLE of product on the RHS of the equation Elements In their standard states have zero enthalpy of formation. Carbon is usually taken as the graphite allotrope. DIPARTIMENTO DI MEDICINA E CHIRURGIA STANDARD ENTHALPY OF COMBUSTION Definition The enthalpy change when ONE MOLE of a substance undergoes complete combustion under standard conditions. All reactants and products are in their standard states. Symbol DH°c Values Always exothermic Example(s) C(graphite) + O2(g) ———> CO2(g) H2(g) + ½O2(g) ———> H2O(l) C2H5OH(l) + 3O2(g) ———> 2CO2(g) + 3H2O(l) Notes Always only ONE MOLE of what you are burning on the LHS of the equation To aid balancing the equation, remember... you get one carbon dioxide molecule for every carbon atom in the original and one water molecule for every two hydrogen atoms When you have done this, go back and balance the oxygen. DIPARTIMENTO DI MEDICINA E CHIRURGIA ENTHALPY OF NEUTRALISATION Definition The enthalpy change when ONE MOLE of water is formed from its ions in dilute solution. Values Exothermic Equation H+(aq) + OH¯(aq) ———> H2O(l) Notes A value of -57kJ mol-1 is obtained when strong acids react with strong alkalis. DIPARTIMENTO DI MEDICINA E CHIRURGIA BOND DISSOCIATION ENTHALPY Definition Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms. Values Endothermic - Energy must be put in to break any chemical bond Examples Cl2(g) ———> 2Cl(g) O-H(g) ———> O(g) + H(g) Notes strength of bonds also depends on environment; MEAN values quoted making bonds is exothermic as it is the opposite of breaking a bond for diatomic gases, bond enthalpy = 2 x enthalpy of atomisation smaller bond enthalpy = weaker bond = easier to break DIPARTIMENTO DI MEDICINA E CHIRURGIA BOND DISSOCIATION ENTHALPY Definition Energy required to break ONE MOLE of gaseous bonds to form gaseous atoms. Values Endothermic - Energy must be put in to break any chemical bond Examples Cl2(g) ———> 2Cl(g) O-H(g) ———> O(g) + H(g) Notes strength of bonds also depends on environment; MEAN values quoted making bonds is exothermic as it is the opposite of breaking a bond for diatomic gases, bond enthalpy = 2 x enthalpy of atomisation smaller bond enthalpy = weaker bond = easier to break Mean Values H-H 436 H-F 562 N-N 163 C-C 346 H-Cl 431 N=N 409 Average (mean) C=C 611 H-Br 366 NN 944 values are quoted CC 837 H-I 299 P-P 172 because the actual C-O 360 H-N 388 F-F 158 value depends on the environment of C=O 743 H-O 463 Cl-Cl 242 the bond i.e. where C-H 413 H-S 338 Br-Br 193 it is in the molecule C-N 305 H-Si 318 I-I 151 C-F 484 P-H 322 S-S 264 C-Cl 338 O-O 146 Si-Si 176 UNITS = kJ mol-1 DIPARTIMENTO DI MEDICINA E CHIRURGIA HESS’S LAW “The enthalpy change is independent of the path taken” How The enthalpy change going from A to B can be found by adding the values of the enthalpy changes for the reactions A to X, X to Y and Y to B. DHr = DH1 + DH2 + DH3 DIPARTIMENTO DI MEDICINA E CHIRURGIA HESS’S LAW “The enthalpy change is independent of the path taken” How The enthalpy change going from A to B can be found by adding the values of the enthalpy changes for the reactions A to X, X to Y and Y to B. DHr = DH1 + DH2 + DH3 If you go in the opposite direction of an arrow, you subtract the value of the enthalpy change. e.g. DH2 = - DH1 + DH r - DH3 The values of DH1 and DH3 have been subtracted because the route involves going in the opposite direction to their definition. DIPARTIMENTO DI MEDICINA E CHIRURGIA HESS’S LAW “The enthalpy change is independent of the path taken” Use applying Hess’s Law enables one to calculate enthalpy changes from other data, such as... changes which cannot be measured directly e.g. Lattice Enthalpy enthalpy change of reaction from bond enthalpy enthalpy change of reaction from DH°c enthalpy change of formation from DH°f DIPARTIMENTO DI MEDICINA E CHIRURGIA Enthalpy of reaction from bond enthalpies Theory Imagine that, during a reaction, all the bonds of reacting species are broken and the individual atoms join up again but in the form of products. The overall energy change will depend on the difference between the energy required to break the bonds and that released as bonds are made. energy released making bonds > energy used to break bonds... EXOTHERMIC energy used to break bonds > energy released making bonds... ENDOTHERMIC Step 1 Energy is put in to break bonds to form separate, gaseous atoms Step 2 The gaseous atoms then combine to form bonds and energy is released its value will be equal and opposite to that of breaking the bonds Applying Hess’s Law DHr = Step 1 + Step 2 DIPARTIMENTO DI MEDICINA E CHIRURGIA Enthalpy of reaction from bond enthalpies Alternative view Step 1 Energy is put in to break bonds ATOMS to form separate, gaseous atoms. SUM OFTHE BOND Step 2 Gaseous atoms then combine ENTHALPIES OF SUM OFTHE to form bonds and energy is THE REACTANTS BOND released; its value will be equal ENTHALPIES OF and opposite to that of breaking THE PRODUCTS REACTANTS the bonds DH DHr = Step 1 - Step 2 Because, in Step 2 the route involves PRODUCTS going in the OPPOSITE DIRECTION to the defined change of bond enthalpy, it’s value is subtracted. DH = bond enthalpies – bond enthalpies of reactants of products DIPARTIMENTO DI MEDICINA E CHIRURGIA Enthalpy of reaction from bond enthalpies Calculate the enthalpy change for the hydrogenation of ethene DIPARTIMENTO DI MEDICINA E CHIRURGIA Enthalpy of reaction from bond enthalpies Calculate the enthalpy change for the hydrogenation of ethene DH2 1 x C=C bond @ 611 = 611 kJ 4 x C-H bonds @ 413 = 1652 kJ 1 x H-H bond @ 436 = 436 kJ Total energy to break bonds of reactants = 2699 kJ DIPARTIMENTO DI MEDICINA E CHIRURGIA Enthalpy of reaction from bond enthalpies Calculate the enthalpy change for the hydrogenation of ethene DH2 1 x C=C bond @ 611 = 611 kJ 4 x C-H bonds @ 413 = 1652 kJ 1 x H-H bond @ 436 = 436 kJ Total energy to break bonds of reactants = 2699 kJ DH3 1 x C-C bond @ 346 = 346 kJ 6 x C-H bonds @ 413 = 2478 kJ Total energy to break bonds of products = 2824 kJ Applying Hess’s Law DH1 = DH2 – DH3 = (2699 – 2824) = – 125 kJ DIPARTIMENTO DI MEDICINA E CHIRURGIA Enthalpy of reaction from enthalpies of formation If you formed the products from their elements you should need the same amounts of every substance as if you formed the reactants from their elements. Enthalpy of formation tends to be an exothermic process DIPARTIMENTO DI MEDICINA E CHIRURGIA Enthalpy of reaction from enthalpies of formation Step 1 Energy is released as reactants ELEMENTS are formed from their elements. SUM OFTHE Step 2 Energy is released as products ENTHALPIES OF are formed from their elements. FORMATION OF SUM OFTHE THE REACTANTS ENTHALPIES OF DHr = - Step 1 + Step 2 REACTANTS FORMATION OF THE PRODUCTS or Step 2 - Step 1 DH In Step 1 the route involves going in the OPPOSITE DIRECTION to the defined PRODUCTS enthalpy change, it’s value is subtracted. DH = DHf of products – DHf of reactants DIPARTIMENTO DI MEDICINA E CHIRURGIA Enthalpy of reaction from enthalpies of formation Sample calculation Calculate the standard enthalpy change for the following reaction, given that the standard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286, +33 and -173 kJ mol-1 respectively; the value for oxygen is ZERO as it is an element 2H2O(l) + 4NO2(g) + O2(g) ———> 4HNO3(l) DH = DHf of products – DHf of reactants By applying Hess’s Law... The Standard Enthalpy of Reaction DH°r will be... PRODUCTS REACTANTS [ 4 x DHf of HNO3 ] minus [ (2 x DHf of H2O) + (4 x DHf of NO2) + (1 x DHf of O2) ] DH°r = 4 x (-173) - 2 x (-286) + 4 x (+33) + 0 DIPARTIMENTO DI MEDICINA E CHIRURGIA ANSWER = - 252 kJ Enthalpy of reaction from enthalpies of combustion If you burned all the products you should get the same amounts of oxidation products such a CO2 and H2O as if you burned the reactants. Enthalpy of combustion is an exothermic process DIPARTIMENTO DI MEDICINA E CHIRURGIA Enthalpy of reaction from enthalpies of combustion Step 1 Energy is released as reactants REACTANTS undergo combustion. DH Step 2 Energy is released as products SUM OFTHE undergo combustion. ENTHALPIES OF PRODUCTS COMBUSTION DHr = Step 1 - Step 2 OF THE SUM OFTHE REACTANTS ENTHALPIES OF COMBUSTION OF THE PRODUCTS Because, in Step 2 the route involves going in the OPPOSITE DIRECTION to the OXIDATION defined change of Enthalpy of PRODUCTS Combustion, it’s value is subtracted. DH = DHc of reactants – DHc of products DIPARTIMENTO DI MEDICINA E CHIRURGIA Enthalpy of reaction from enthalpies of combustion Sample calculation Calculate the standard enthalpy of formation of methane; the standard enthalpies of combustion of carbon, hydrogen and methane are -394, -286 and -890 kJ mol-1. C(graphite) + 2H2(g) ———> CH4(g) DH = DHc of reactants – DHc of products By applying Hess’s Law... The Standard Enthalpy of Reaction DH°r will be... REACTANTS PRODUCTS [ (1 x DHc of C) + (2 x DHc of H2) ] minus [ 1 x DHc of CH4] DH°r = 1 x (-394) + 2 x (-286) - 1 x (-890) ANSWER = - 76 kJ mol-1 DIPARTIMENTO DI MEDICINA E CHIRURGIA MEASURING ENTHALPY CHANGES Calorimetry involves the practical determination of enthalpy changes usually involves heating (or cooling) known amounts of water water is heated up reaction is EXOTHERMIC water cools down reaction is ENDOTHERMIC Calculation The energy required to change the temperature of a substance can be calculated using... q = m x c x DT where q = heat energy kJ m = mass kg c = Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ] DT = change in temperature K DIPARTIMENTO DI MEDICINA E CHIRURGIA MEASURING ENTHALPY CHANGES Calorimetry involves the practical determination of enthalpy changes usually involves heating (or cooling) known amounts of water water is heated up reaction is EXOTHERMIC water cools down reaction is ENDOTHERMIC Calculation The energy required to change the temperature of a substance can be calculated using... q = m x c x DT where q = heat energy kJ m = mass kg c = Specific Heat Capacity kJ K -1 kg -1 [ water is 4.18 ] DT = change in temperature K Example On complete combustion, 0.18g of hexane raised the temperature of 100g water from 22°C to 47°C. Calculate its enthalpy of combustion. Heat absorbed by the water (q) = 0.1 x 4.18 x 25 = 10.45 kJ Moles of hexane burned = mass / Mr = 0.18 / 86 = 0.00209 Enthalpy change = heat energy / moles = 10.45 / 0.00209 ANS = 5000 kJ mol -1 DIPARTIMENTO DI MEDICINA E CHIRURGIA MEASURING ENTHALPY CHANGES Example 1 - graphical The temperature is taken every half minute before mixing the reactants. Reactants are mixed after three minutes. Further readings are taken every half minute as the reaction mixture cools. Extrapolate the lines as shown and calculate the value of DT. DIPARTIMENTO DI MEDICINA E CHIRURGIA MEASURING ENTHALPY CHANGES Example 1 - graphical The temperature is taken every half minute before mixing the reactants. Reactants are mixed after three minutes. Further readings are taken every half minute as the reaction mixture cools. Extrapolate the lines as shown and calculate the value of DT. Example calculation When 0.18g of hexane underwent complete combustion, it raised the temperature of 100g (0.1kg) water from 22°C to 47°C. Calculate its enthalpy of combustion. Heat absorbed by the water (q) = m C DT = 0.1 x 4.18 x 25 = 10.45 kJ Moles of hexane burned = mass / Mr = 0.18 / 86 = 0.00209 Enthalpy change = heat energy / moles = 10.45 / 0.00209 = 5000 kJ mol -1 DIPARTIMENTO DI MEDICINA E CHIRURGIA MEASURING ENTHALPY CHANGES Example 2 25cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial temperature of both solutions was 20°C. The highest temperature reached by the solution was 33°C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat capacity (c) of water is 4.18 kJ K -1 kg -1] NaOH + HCl ——> NaCl + H2O DIPARTIMENTO DI MEDICINA E CHIRURGIA MEASURING ENTHALPY CHANGES Example 2 25cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial temperature of both solutions was 20°C. The highest temperature reached by the solution was 33°C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat capacity (c) of water is 4.18 kJ K -1 kg -1] NaOH + HCl ——> NaCl + H2O Temperature rise (DT) = 306K – 293K = 13K Volume of resulting solution= 25 + 25 = 50cm3 = 0.05 dm3 Equivalent mass of water = 50g = 0.05 kg (density is 1g per cm3) Heat absorbed by the water (q) = m x c x DT = 0.05 x 4.18 x 13 = 2.717 kJ DIPARTIMENTO DI MEDICINA E CHIRURGIA MEASURING ENTHALPY CHANGES Example 2 25cm3 of 2.0M HCl was added to 25cm3 of 2.0M NaOH in an insulated beaker. The initial temperature of both solutions was 20°C. The highest temperature reached by the solution was 33°C. Calculate the Molar Enthalpy of Neutralisation. [The specific heat capacity (c) of water is 4.18 kJ K -1 kg -1] NaOH + HCl ——> NaCl + H2O Temperature rise (DT) = 306K – 293K = 13K Volume of resulting solution= 25 + 25 = 50cm3 = 0.05 dm3 Equivalent mass of water = 50g = 0.05 kg (density is 1g per cm3) Heat absorbed by the water (q) = m x c x DT = 0.05 x 4.18 x 13 = 2.717 kJ Moles of HCl reacting = 2 x 25/1000 = 0.05 mol Moles of NaOH reacting = 2 x 25/1000 = 0.05 mol Moles of water produced = 0.05 mol Enthalpy change per mol (DH) = heat energy / moles of water = 2.717 / 0.05 = 54.34 kJ mol -1 DIPARTIMENTO DI MEDICINA E CHIRURGIA Spontaneity Spontaneous Processes Processes that occur without outside intervention Spontaneous processes may be fast or slow – Many forms of combustion are fast – Conversion of graphite to diamond is slow – Kinetics is concerned with speed, thermodynamics with the initial and final state DIPARTIMENTO DI MEDICINA E CHIRURGIA Entropy (S) A measure of the randomness or disorder The driving force for a spontaneous process is an increase in the entropy of the universe (one of the laws of thermodynamics!) – The universe favors chaos!! Entropy is a thermodynamic function describing the number of arrangements that are available to a system Nature proceeds towards the states that have the highest probabilities of existing DIPARTIMENTO DI MEDICINA E CHIRURGIA Second law of thermodynamics ▪ "In any spontaneous process there is always an increase in the entropy of the universe" ▪ "The entropy of the universe is increasing" ▪ For a given change to be spontaneous, DSuniverse must be positive DSuniv = DSsys + DSsurr DIPARTIMENTO DI MEDICINA E CHIRURGIA Positional Entropy ▪ The probability of occurrence of a particular state depends on the number of ways (microstates) in which that arrangement can be achieved Ssolid < Sliquid 0) DIPARTIMENTO DI MEDICINA E CHIRURGIA What If? DG = 0? – The reaction is at equilibrium As a result DG < 0 The reaction is spontaneous. DG > 0 The reaction is non-spontaneous. DG = 0 The reaction mixture is at equilibrium. -It’s not moving forward or reverse overall – multiple processes, same speed DIPARTIMENTO DI MEDICINA E CHIRURGIA Dependence on Temperature DG = DH - TDS Notice there are two terms in the equation – Spontaneity can change when the temperature changes. – Temperature and spontaneity are not necessarily correlated. A reaction with a negative entropy change (loss of randomness) would be LESS spontaneous at higher temperatures – it doesn’t want to happen, but is pushed by the extra heat. DIPARTIMENTO DI MEDICINA E CHIRURGIA Reaction Rates Free energy (G) values tell us if reactions will occur They do not tell us how fast reactions will occur. During a reaction – Reactant particles must physically collide – They must collide with enough energy to break the bonds in the reactant Some reactions require the addition of heat energy. This gives the reactants the extra energy needed (more collisions, harder collisions) for this process to occur. DIPARTIMENTO DI MEDICINA E CHIRURGIA DH, DS, DG and Spontaneity DG = DH - TDS H is enthalpy, T is Kelvin temperature Value of DH Value of TDS Value of DG Spontaneity Negative Positive Negative Spontaneous Positive Negative Positive Nonspontaneous Negative Negative ??? Spontaneous if the absolute value of DH is greater than the absolute value of TDS (low temperature) Positive Positive ??? Spontaneous if the absolute value of TDS is greater than the absolute value of DH (high temperature) DIPARTIMENTO DI MEDICINA E CHIRURGIA Example of Gibbs Energy (I) For the reaction at 298 K, the values of DH and DS are 58.03 kJ and 176.6 J/K, respectively. What is the value of DG at 298 K? DIPARTIMENTO DI MEDICINA E CHIRURGIA Example of Gibbs Energy (I) For the reaction at 298 K, the values of DH and DS are 58.03 kJ and 176.6 J/K, respectively. What is the value of DG at 298 K? DG = DH - TDS DG = 58.03 kJ – (298 K)(176.6 J/K) DG = 5.40 kJ Not spontaneous at this temperature DIPARTIMENTO DI MEDICINA E CHIRURGIA Example of Gibbs Energy (II) Calculate the standard free-energy change at 25 oC for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes: N2(g) + 3H2(g) 2NH3(g) DHo = −92.2 kJ DSo = −198.7 J/K DIPARTIMENTO DI MEDICINA E CHIRURGIA Example of Gibbs Energy (II) Calculate the standard free-energy change at 25oC for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes: N2(g) + 3H2(g) 2NH3(g) DHo = −92.2 kJ DSo = −198.7 J/K DG = DH − TDS DG = −92.2 kJ − (298 K)(-198.7 J/K)(1 kJ/1000 J) = -33.0 kJ DIPARTIMENTO DI MEDICINA E CHIRURGIA Example of Gibbs Energy (III) Iron metal can be produced by reducing iron(III) oxide with hydrogen: Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g) DH = +98.8 kJ; DS = +141.5 J/K (a) Is this reaction spontaneous under standard- state conditions at 25 °C? (b) At what temperature will the reaction become spontaneous? DIPARTIMENTO DI MEDICINA E CHIRURGIA Example of Gibbs Energy (III) To determine whether the reaction is spontaneous at 25 °C, we need to determine the sign of DG = DH − TDS. At 25 C (298 K), DG for the reaction is DG = DH − TDS = (98.8 kJ) − (298 K)(0.1415 kJ/K) = (98.8 kJ) − (42.2 kJ) = 56.6 kJ Reaction is NOT spontaneous! DIPARTIMENTO DI MEDICINA E CHIRURGIA Example of Gibbs Energy (III) At what temperature does this become spontaneous? At temperatures above 698 K, the TDS term becomes larger than DH, making the Gibbs energy negative and the process spontaneous Why is this postive this time? – Remember, here we’re dealing with the system, not the surroundings! DIPARTIMENTO DI MEDICINA E CHIRURGIA Entropy Changes in Chemical Reactions Constant Temperature and Pressure – Reactions involving gaseous molecules The change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products Typically, more moles of gas, more entropy! DIPARTIMENTO DI MEDICINA E CHIRURGIA Third Law of Thermodynamics "The entropy of a perfect crystal at 0 K is zero" (NO disorder, since everything is in perfect position) – No movement = 0 K – No disorder = no entropy (DS = 0) DIPARTIMENTO DI MEDICINA E CHIRURGIA Standard State Conditions We’ve seen that quantities such as entropy (S), enthalpy (H), and free energy (G) are dependent upon the conditions present Standard State – One set of conditions at which quantities can be compared Pure solids/liquids/gases at 1 atm pressure Solutes at 1 M concentration Typically at 25 oC (298 K) Designated by o sign (similar to degree sign) DIPARTIMENTO DI MEDICINA E CHIRURGIA DS 0 reaction = np S 0 products − nr S o reactants ❖ Calculates standard entropy of a reaction, uses standard entropies of compounds ❖ Entropy is an extensive property (a function of the number of moles) ❖ Generally, the more complex the molecule, the higher the standard entropy value DIPARTIMENTO DI MEDICINA E CHIRURGIA Calculating Standard Entropy DIPARTIMENTO DI MEDICINA E CHIRURGIA Calculating Standard Entropy (I) DS 0 reaction = np S 0 products − nr S o reactants 2 (28 J/Kmol) + 3 (189 J/K.mol) – 51 J/K-mol – 3 (131 J/K.mol) = 179 J/K.mol System gained entropy – water more complex than hydrogen! DIPARTIMENTO DI MEDICINA E CHIRURGIA Evaluate the entropy change for the reaction: CO + 3 H2 -> CH4 + H2O in which all reactants and products are gaseous. So values: CO 198 J/K.mol H2 131 J/K.mol CH4 186 J/K.mol H2O 189 J/K.mol DIPARTIMENTO DI MEDICINA E CHIRURGIA Evaluate the entropy change for the reaction: CO + 3 H2 -> CH4 + H2O in which all reactants and products are gaseous. So values: CO 198 J/Kmol H2 131 J/Kmol CH4 186 J/Kmol H2O 189 J/Kmol Dsoreaction = [186 + 189] – [198 + 3(131)] = 216 J/Kmol DIPARTIMENTO DI MEDICINA E CHIRURGIA Standard Free Energy Change DG0 is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states – DG0 cannot be measured directly – The more negative the value for DG0, the farther to the right the reaction will proceed in order to achieve equilibrium – Equilibrium is the lowest possible free energy position for a reaction DIPARTIMENTO DI MEDICINA E CHIRURGIA Using standard free energy of formation (DGf0): DG = n p DG 0 0 f (products) − nr DG 0 f (reactants) DGf0 of an element in its standard state is zero! DIPARTIMENTO DI MEDICINA E CHIRURGIA DIPARTIMENTO DI MEDICINA E CHIRURGIA DG 0 = n p DG 0f (products) − nr DG 0f (reactants) 2 (-394kJ/mol) + 4 (-229 kJ/mol) – 2 (-163 kJ/mol) – 3 (0) = - 1378 kJ DIPARTIMENTO DI MEDICINA E CHIRURGIA Calculate the standard free energy for the reaction below and determine whether it is spontaneous at standard conditions. Fe2O3(s) + 3 CO (g) 2 Fe (s) + 3 CO2(g) DGo = -742.2 -137.2 0 -394.4 DIPARTIMENTO DI MEDICINA E CHIRURGIA Calculate the standard free energy for the reaction below and determine whether it is spontaneous at standard conditions. Fe2O3(s) + 3 CO (g) 2 Fe (s) + 3 CO2(g) DGo (kJ/mol)= -742.2 -137.2 0 -394.4 DGo = [0 + (3 x -394.4)] – [-742.2 + (3 x -137.2)] = -29.4 kJ/mol DIPARTIMENTO DI MEDICINA E CHIRURGIA 106 CHEMICAL KINETICS DIPARTIMENTO DI MEDICINA E CHIRURGIA Transformation that leads to a change of the nature of the particles interacting to each other, involves the formation and the breakage of chemical bonds. Most of the reactions proceed through multiple individual sub- steps in the transformation from reactants to products. These sub-steps involve the formation of “intermediate products” (sometimes difficult to detect because immediately used in the later stage). Reactants Products (initial stage) (final stage) Intermediate products (possible intermediate step) DIPARTIMENTO DI MEDICINA E CHIRURGIA Study of the speed of the reaction and of factors affecting the reaction rate A+B C+D v = variation of a certain property in time units DIPARTIMENTO DI MEDICINA E CHIRURGIA Rate of a chemical reaction Variation of the concentration of a reactant or of a product that happens in a defined period of time A+B=C+D V = D[C]/Dt quantity of product formed in a period of time V = -D[A]/Dt quantity of reactant consumed in a period of time Reaction rate usually has the units of mol/L·s DIPARTIMENTO DI MEDICINA E CHIRURGIA Collision theory For a reaction to happen, reactant particles must collide with enough energy for the collision to be successful ! They first have to collide, and then they may react. Why "may react"? It isn't enough for the two species to collide - they have to collide the right way around, and they have to collide with enough energy for bonds to break. DIPARTIMENTO DI MEDICINA E CHIRURGIA Collision theory DIPARTIMENTO DI MEDICINA E CHIRURGIA Collision theory DIPARTIMENTO DI MEDICINA E CHIRURGIA Collision orientation The molecules have to collide with an orientation that can allow the rearrangement of the atoms and the formation of the products. DIPARTIMENTO DI MEDICINA E CHIRURGIA Activation energy Even if the species are orientated properly, you still won't get a reaction unless the particles collide with a certain minimum energy called the activation energy of the reaction. Activation energy is the minimum energy required before a reaction can occur. DIPARTIMENTO DI MEDICINA E CHIRURGIA The Maxwell-Boltzmann Distribution In any system, the particles present will have a very wide range of energies. For gases, this can be shown on a graph called the Maxwell-Boltzmann Distribution which is a plot of the number of particles having each particular energy. DIPARTIMENTO DI MEDICINA E CHIRURGIA The Maxwell-Boltzmann Distribution Remember that for a reaction to happen, particles must collide with energies equal to or greater than the activation energy for the reaction. We can mark the activation energy on the Maxwell-Boltzmann distribution. DIPARTIMENTO DI MEDICINA E CHIRURGIA The effect of surface area on reaction rate The more finely divided the solid is, the faster the reaction happens. A powdered solid will normally produce a faster reaction than if the same mass is present as a single lump. The powdered solid has a greater surface area than the single lump. DIPARTIMENTO DI MEDICINA E CHIRURGIA The effect of surface area on reaction rate DIPARTIMENTO DI MEDICINA E CHIRURGIA The effect of surface area on reaction rate Increasing the number of collisions per second increases the rate of reaction. DIPARTIMENTO DI MEDICINA E CHIRURGIA The effect of temperature on reaction rate As you increase the temperature the rate of reaction increases. As a rough approximation, for many reactions happening at around room temperature, the rate of reaction doubles for every 10°C rise in temperature. DIPARTIMENTO DI MEDICINA E CHIRURGIA The effect of temperature on reaction rate Collision Frequency Particles can only react when they collide. If you heat a substance, the particles move faster and so collide more frequently. That will speed up the rate of reaction? How about a 10 K increase? Collision frequency plays only a minor part! DIPARTIMENTO DI MEDICINA E CHIRURGIA The effect of temperature on reaction rate Collision Frequency Particles can only react when they collide. If you heat a substance, the particles move faster and so collide more frequently. That will speed up the rate of reaction? How about a 10 K increase? Collision frequency plays only a minor part! DIPARTIMENTO DI MEDICINA E CHIRURGIA The effect of temperature on reaction rate Activation Energy DIPARTIMENTO DI MEDICINA E CHIRURGIA The effect of temperature on reaction rate Activation Energy DIPARTIMENTO DI MEDICINA E CHIRURGIA The effect of concentration on reaction rate Generally, the larger the concentration of reactant molecules, the faster the reaction. -This increases the frequency of reactant molecule contact. -Concentration of gases depends on the partial pressure of the gas. -Higher pressure = higher concentration Concentrations of solutions depend on the solute-to-solution ratio (molarity). DIPARTIMENTO DI MEDICINA E CHIRURGIA The effect of concentration on reaction rate Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) DIPARTIMENTO DI MEDICINA E CHIRURGIA The Rate Law The rate law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants and homogeneous catalysts as well. The rate law must be determined experimentally! The rate of a reaction is directly proportional to the concentration of each reactant raised to a power. For the reaction aA + bB → products the rate law would have the form given below. n is the order for each reactant k is called the rate constant DIPARTIMENTO DI MEDICINA E CHIRURGIA Reaction Order For the general reaction: aA + bB → cC + dD Rate = k [A] [B] x y Each concentration is expressed with an order (exponent). The rate constant converts the concentration expression into the correct units of rate (Ms−1). x and y are the reactant orders determined from experiment. DIPARTIMENTO DI MEDICINA E CHIRURGIA Reaction Order The exponent on each reactant in the rate law is called the order with respect to that reactant. The sum of the exponents on the reactants is called the order of the reaction. The rate law for the reaction 2 NO(g) + O2(g) → 2 NO2(g) is Rate = k[NO]2[O2]. The reaction is second order with respect to [NO], first order with respect to [O2], and third order overall. DIPARTIMENTO DI MEDICINA E CHIRURGIA Rate = k[A]n If a reaction is zero order, the rate of the reaction is always the same. – Doubling [A] will have no effect on the reaction rate. If a reaction is first order, the rate is directly proportional to the reactant concentration. – Doubling [A] will double the rate of the reaction. If a reaction is second order, the rate is directly proportional to the square of the reactant concentration. – Doubling [A] will quadruple the rate of the reaction. DIPARTIMENTO DI MEDICINA E CHIRURGIA Determining the Rate Law When There Are Multiple Reactants Changing each reactant will affect the overall rate of the reaction. By changing the initial concentration of one reactant at a time, the effect of each reactant’s concentration on the rate can be determined. In examining results, we compare differences in rate for reactions that only differ in the concentration of one reactant. DIPARTIMENTO DI MEDICINA E CHIRURGIA Determining the Order and Rate Constant of a Reaction Consider the reaction between nitrogen dioxide and carbon monoxide: The initial rate of the reaction is measured at several different concentrations of the reactants with the accompanied results. From the data, determine: a. the rate law for the reaction b. the rate constant (k) for the reaction Solution a. Begin by examining how the rate changes for each change in concentration. Between the first two experiments, the concentration of NO2 doubles, the concentration of CO stays constant, and the rate quadruples, suggesting that the reaction is second order in NO2. Between the second and third experiments, the concentration of NO2 stays constant, the concentration of CO doubles, and the rate remains constant (the small change in the least significant figure is simply experimental error), suggesting that the reaction is zero order in CO. Between the third and fourth experiments, the concentration of NO2 again doubles and the concentration of CO halves, yet the rate quadruples again, confirming that the reaction is second order in NO2 and zero order in CO. DIPARTIMENTO DI MEDICINA E CHIRURGIA Determining the Order and Rate Constant of a Reaction Write the overall rate expression. DIPARTIMENTO DI MEDICINA E CHIRURGIA Determining the Order and Rate Constant of a Reaction b. To determine the rate constant for the reaction, solve the rate law for k and substitute the concentration and the initial rate from any one of the four measurements. In this case, we use the first measurement. For Practice 13.2 Consider the equation: The initial rate of reaction is measured at several different concentrations of the reactants with the following results: From the data, determine: a. the rate law for the reaction b. the rate constant (k) for the reaction DIPARTIMENTO DI MEDICINA E CHIRURGIA RATE DETERMINING STEP In most mechanisms, one step occurs slower than the other steps. The result is that product production cannot occur any faster than the slowest step; the step determines the rate of the overall reaction. We call the slowest step in the mechanism the rate determining step. – The slowest step has the largest activation energy. The rate law of the rate determining step determines the rate law of the overall reaction. DIPARTIMENTO DI MEDICINA E CHIRURGIA In most mechanisms, one step occurs slower than the other steps. The result is that product production cannot occur any faster than the slowest step; the step determines the rate of the overall reaction. We call the slowest step in the mechanism the rate determining step. – The slowest step has the largest activation energy. The rate law of the rate determining step determines the rate law of the overall reaction. DIPARTIMENTO DI MEDICINA E CHIRURGIA RATE DETERMINING STEP In most mechanisms, one step occurs slower than the other steps. The result is that product production cannot occur any faster than the slowest step; the step determines the rate of the overall reaction. We call the slowest step in the mechanism the rate determining step. – The slowest step has the largest activation energy. The rate law of the rate determining step determines the rate law of the overall reaction. DIPARTIMENTO DI MEDICINA E CHIRURGIA In most mechanisms, one step occurs slower than the other steps. The result is that product production cannot occur any faster than the slowest step; the step determines the rate of the overall reaction. We call the slowest step in the mechanism the rate determining step. – The slowest step has the largest activation energy. The rate law of the rate determining step determines the rate law of the overall reaction. DIPARTIMENTO DI MEDICINA E CHIRURGIA In most mechanisms, one step occurs slower than the other steps. The result is that product production cannot occur any faster than the slowest step; the step determines the rate of the overall reaction. We call the slowest step in the mechanism the rate determining step. – The slowest step has the largest activation energy. The rate law of the rate determining step determines the rate law of the overall reaction. DIPARTIMENTO DI MEDICINA E CHIRURGIA Catalysts are substances that affect the rate of a reaction without being consumed. Catalysts work by providing an alternative mechanism for the reaction with a lower activation energy. Catalysts are consumed in an early mechanism step, and then made in a later step. Mechanism with catalyst: Mechanism without catalyst: O3(g) + O(g) → 2 O2(g) V. Slow Cl(g) + O3(g) O2(g) + ClO(g) Fast ClO(g) + O(g) → O2(g) + Cl(g) Slow DIPARTIMENTO DI MEDICINA E CHIRURGIA Polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals. DIPARTIMENTO DI MEDICINA E CHIRURGIA This diagram has been made by Finn Bjørklid, Norwegian Institute for Air Research (NILU). DIPARTIMENTO DI MEDICINA E CHIRURGIA This diagram has been made by Finn Bjørklid, Norwegian Institute for Air Research (NILU). DIPARTIMENTO DI MEDICINA E CHIRURGIA This diagram has been made by Finn Bjørklid, Norwegian Institute for Air Research (NILU). DIPARTIMENTO DI MEDICINA E CHIRURGIA Figure 13.19 pg 630 DIPARTIMENTO DI MEDICINA E CHIRURGIA Catalysts generally speed up a reaction. They give the reactant molecules a different path to follow with a lower activation energy. – Heterogeneous catalysts hold one reactant molecule in proper orientation for reaction to occur when the collision takes place. Sometimes they also help to start breaking bonds. – Homogeneous catalysts react with one of the reactant molecules to form a more stable activated complex with a lower activation energy. DIPARTIMENTO DI MEDICINA E CHIRURGIA Homogeneous catalysts are in the same phase as the reactant particles. – Cl(g) in the destruction of O3(g) Heterogeneous catalysts are in a different phase than the reactant particles. – Solid catalytic converter in a car’s exhaust system DIPARTIMENTO DI MEDICINA E CHIRURGIA DIPARTIMENTO DI MEDICINA E CHIRURGIA Because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate. Protein molecules that catalyze biological reactions are called enzymes. Enzymes work by adsorbing the substrate reactant onto an active site that orients the substrate for reaction. 1) Enzyme + Substrate Enzyme─Substrate Fast 2) Enzyme─Substrate → Enzyme + Product Slooooooow DIPARTIMENTO DI MEDICINA E CHIRURGIA Enzyme–Substrate Binding: The Lock and Key Mechanism DIPARTIMENTO DI MEDICINA E CHIRURGIA ENZYMATIC HYDROLYSIS OF SUCROSE DIPARTIMENTO DI MEDICINA E CHIRURGIA DIPARTIMENTO DI MEDICINA E CHIRURGIA References https://www.chemguide.co.uk/physical/ktmenu.html#top https://www.aqa.org.uk/subjects/science/as-and-a-level/chemistry-7404- 7405/assessment-resources http://www.knockhardy.org.uk/ppoints_htm_files/15deltahpp.ppt Chapter 16 Spontaneity, Entropy and Free Energy. Surry Schools. Dr. Walker. Available at: https://www.surryschools.net/cms/lib/VA02208192/Centricity/Domain/123/D E_-_Chapter_16_-_Free_Energy_and_Spontaneity.ppt https://www.chemguide.co.uk/physical/basicratesmenu.html#top Chemistry, The Central Science, Chapter 14, 10th editionTheodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten DIPARTIMENTO DI MEDICINA E CHIRURGIA