Chemistry 124 Thermodynamics PDF

 matter is it uniform throughout ? NO YES · hetrogeneous...

 matter is it uniform throughout ? NO YES · hetrogeneous · homogeneous mixture mixture does it have variable composition ? YES No · homogeneous mixture (solution) · pure substance does it contain more than one kind of atom ? No YES · elemant · compound properties of matter - physical observed without changing a substance into another substance · eg. colour odor , , denisty melting/ boiling point and hardness , - chemical · only observed when a substance is changed into another substance eg. flammability ↑ intensive independent of of the substance that for identifying · the amount is present (important a substance eg. denisty boiling point , or colour · extensive · depend on amount of the substance present volume + eg. mass , energy is a chemical reaction in which one part of each two compounds interchange to form two new compounds. Chemistry 124 Dr K de Villiers Thermodynamics INTRODUCTION Hello, and welcome! I’m Dr K (as one of our postgrad students calls me), and I’ll be your lecturer for this section on Thermodynamics. Let’s get started by having a class discussion about a reaction that we ALL know about… H2O (s)  H2O (l) ©Stellenbosch University, 2023 1 Chemistry 124 Dr K de Villiers Thermodynamics Thermodynamics is the study of ENERGY, and in thermochemistry specifically, we study the energy changes that accompany chemical reactions. 14.1 THE NATURE OF ENERGY Energy is defined as the capacity to do work or to transfer heat. Work is the energy used to cause an object with mass to move; Heat is the energy that causes the temperature of an object to increase. Forms of energy Matter can possess different forms of energy, which are interconvertible. Kinetic energy 1 where m = mass of object and v = 𝐸 𝑚𝑣 2 velocity of object Potential energy 𝐸 𝑚𝑔ℎ where m = mass of object, h = height of object relative to reference point, g = gravitational constant (9.8 ms-2) Electrostatic potential energy 𝑘𝑞 𝑞 where k = constant (8.99  109 JmC-2); r 𝐸 𝑟 = distance between two charges (q1 and q2) Units of energy As much as I like the jewels in my wedding ring (diamond and sapphire), joules (J) are the SI unit of energy.  The hydrolysis of ATP in the body releases 7.3 kcal/mol free energy. Express this in J/mol. ©Stellenbosch University, 2023 2 Chemistry 124 Dr K de Villiers Thermodynamics Some important terms before we go further – system, surroundings and universe: A system can be further described as open, closed or isolated: Matter Energy ©Stellenbosch University, 2023 3 Chemistry 124 Dr K de Villiers Thermodynamics 14.2 THE FIRST LAW OF THERMODYNAMICS Also known as the Law of Conservation of Energy, the take home message of the First Law of Thermodynamics is that energy is conserved; that means that energy cannot be created or destroyed – rather there is a set amount in the universe which can be converted from one form to another, or transferred from one place to another. Internal Energy For a given system, the internal energy (U) is the sum of all kinetic and potential energy. It is not possible to measure the absolute value of U, but it is possible to measure the change in U (symbol = U (delta U)): U = Ufinal  Uinitial Internal energy is an example of a state function: the values of Uinitial and Ufinal are fixed and do not depend on how the change comes about. (More about this later!) What is important now is that for a chemical reaction, the initial state of the system refers to the reactants, while the final state refers to the products. ©Stellenbosch University, 2023 4 Chemistry 124 Dr K de Villiers Thermodynamics Relating U to heat and work When a system undergoes chemical or physical change, the change in internal energy can be determined from the extent of heat and work exchanged with the surroundings: U = q + w where q = heat transferred, and w = work performed With respect to the system therefore, the following table summarises the sign conventions: q w + Heat is + Work is done  Heat is  Work is done Endothermic and exothermic processes An endothermic process is one in which the system gains (absorbs) heat. Consequence for the surroundings? An exothermic process is one in which the system loses heat. Consequence for the surroundings? ©Stellenbosch University, 2023 5 Chemistry 124 Dr K de Villiers Thermodynamics 14.3 ENTHALPY ΔH = ΔU + PΔV Many of the reactions that take place around us, and in the lab, can be described as isobaric (i.e. they take place under the constant pressure of the atmosphere, Patm). The work (w) that accompanies such reactions is generally mechanical in nature, resulting in a change in volume of a system. It is thus referred to as pressure- volume (P-V) work. Patm ΔV If work is done by the system, the system expands (+ ΔV), but it also loses energy while doing so ( w). If work is done on/to the system, the system is compressed ( ΔV), and it gains energy in the process (+ w). FYI - No need to know derivations by heart Thus, there is an indirect relationship between work and volume: w =  PΔV Substituting this into the very first relationship for enthalpy: ΔH = (q + w) + (–w) H = qP Note – the subscript “P” is a reminder that the transfer of heat takes place under constant pressure. ©Stellenbosch University, 2023 6 Chemistry 124 Dr K de Villiers Thermodynamics For a given reaction, the following sign conventions apply: + ΔH + qP Heat ________ by system Process is __________  ΔH  qP Heat ________ by system Process is __________ PRACTICE Pieces of dry ice (CO2(s)) are placed in a zip-lock bag. The system absorbs 140 J of heat, which causes the CO2(s) to sublime. As a result, the system performs 85 J of work. Calculate the change in internal energy of the system, ΔU. 14.4 ENTHALPIES OF REACTION Enthalpy (H) is also a state function, and therefore the enthalpy of reaction (rxnH) depends only on the initial and final states (i.e. products and reactants), and not on how the reaction takes place (i.e. it’s mechanism): rxnH = Hproducts  Hreactants When information about rxnH is included with a balanced chemical equation, the result is called a thermochemical equation. For example: 2H2(g) + O2(g)  2H2O(g) H =  483.6 kJ ©Stellenbosch University, 2023 7 Chemistry 124 Dr K de Villiers Thermodynamics Let’s stick with this example to learn 3 ways in which we can use thermochemical equations: 2H2(g) + O2(g)  2H2O(g) H =  483.6 kJ ©Stellenbosch University, 2023 8 Chemistry 124 Dr K de Villiers Thermodynamics PRACTICE Exercise 14.28 Heating potassium chlorate used to be a common means of producing oxygen: 2KClO3(s)  2KCl(s) + 3O2(g) ΔH = – 89.4 kJ (a) Is heat absorbed or evolved during the course of this reaction? (b) Calculate ΔH for the formation of: (i) 0.855 mol of O2(g) (ii) 10.75 g of KCl(s) (c) 12.3 g of KClO3(s) (D) CALORIMETRY You will need: Molecular weight (gmol-1) O 16.0 Cl 35.5 K 39.0 ©Stellenbosch University, 2023 9 Chemistry 124 Dr K de Villiers Thermodynamics 14.5 CALORIMETRY The value of H can be determined experimentally, and the experimental device used to do so is called a calorimeter. The word literally means to “measure (meter) energy (calories)”. You will have a chance to build a simple calorimeter in Practical 4 using Styrofoam coffee cups, so look out for that! Heat capacity and specific heat capacity All substances change temperature when subjected to heat transfer, however the magnitude of this change differs from substance to substance. The temperature change experienced by an object is determined by its heat capacity, C – that is the amount of heat required to change the temperature (T) of the object by 1 K or 1°C. For pure substances, we can also consider the molar heat capacity, Cm (i.e. per 1 mole of a substance), and the specific heat capacity, C (i.e. per 1 gram of a substance). The following equation is used to determine the specific heat capacity, C: 𝑞 𝐶 𝑚 ∆𝑇 where q = quantity of heat (J); m = mass of substance (g); ΔT = Tfinal – Tinitial (°C or K) ©Stellenbosch University, 2023 10 Chemistry 124 Dr K de Villiers Thermodynamics PRACTICE Exercise 14.34 (a) The specific heat capacity of liquid water is _________________________. (b) Calculate the molar heat capacity, Cm, of liquid water. ©Stellenbosch University, 2023 11 Chemistry 124 Dr K de Villiers Thermodynamics Constant pressure calorimetry If pressure can be controlled (i.e. kept constant), then H can be readily measured (ΔH = qp). A simple coffee-cup calorimeter (CCC) can be used in the lab to obtain informative results. qrxn =  qsoln qsoln = Csoln  msoln  ΔT In this experimental set up, you may be wondering where the boundary is between the system (chemical reaction of interest) and the surroundings. This is a frequently-asked-question, so you’re not alone. Due to the insulating properties of the stacked cups, the contents = system + surroundings. Heat exchanges occur between the reaction (system) and the resultant solution (surroundings). ©Stellenbosch University, 2023 12 Chemistry 124 Dr K de Villiers Thermodynamics PRACTICE Exercise 14.36 When 3.88 g of ammonium nitrate (NH4NO3(s)) dissolve in 60.0 g of water in a coffee-cup calorimeter, the temperature drops from 23.0 °C to 18.4 °C. (a) Write a chemical equation for the process described above. NH4NO3 (s)  NH4+ (aq) + NO3- (aq) (c) Is the process endo- or exothermic? How do you know? (b) Calculate ΔH in kJ/mol NH4NO3(s), assuming that (i) the specific heat capacity of the dilute solution is the same as water, and (ii) the heat capacity of the calorimeter is negligible. ©Stellenbosch University, 2023 13 Chemistry 124 Dr K de Villiers Thermodynamics 14.6 HESS’S LAW Enthalpies of reactions for which experimental data is not available can be calculated using Hess’s law. For example, in the case below, ΔH1 = {ΔH2 + ΔH3+ ΔH4}. INITIAL STATE FINAL STATE Single step Final Reactants ΔH1 Products ΔH2 ΔH4 Multi step Intermediate Intermediate Products ΔH3 Products PRACTICE Exercise 14.40 Consider the following hypothetical reactions: A  B ΔH = + 30 kJ C  B ΔH =  60 kJ (a) Use Hess’s Law to calculate the enthalpy change for the reaction A  C. (b) Construct an enthalpy diagram for substances A, B and C, and show how Hess’s Law applies. ©Stellenbosch University, 2023 14 Chemistry 124 Dr K de Villiers Thermodynamics Class exercise Calculate the enthalpy change (ΔH) for the reaction ClF(g) + F2(g)  ClF3(l) given the following enthalpies of reaction: Reactants Products ΔH / kJ 2 ClF(g) + O2(g)  Cl2O(g) + OF2(g) 167.5 2 F2(g) + O2(g)  2OF2(g) -43.5 2 ClF3(l) + 2 O2(g)  Cl2O(g) + 3 OF2(g) 394.1 Exercise 14.41 Calculate the enthalpy change (ΔH) for the reaction P4O6(s) + 2 O2(g)  P4O10(s) given the following enthalpies of reaction: Reactants Products ΔH / kJ P4(s) + 3 O2(g)  P4O6(s)  1640.1 P4(s) + 5 O2(g)  P4O10(s)  2940.1 Ans: -1300 kJON ©Stellenbosch University, 2023 15 Chemistry 124 Dr K de Villiers Thermodynamics 14.7 ENTHALPIES OF FORMATION The standard enthalpy of formation of a compound, fH°, is defined as the enthalpy change for the formation of 1 mole of a substance from its constituent elements, with all species in their standard states. 2C (graphite) + 3H2 (g) + ½O2 (g)  C2H5OH (l) fH° =  277.7 kJmol-1 Important considerations: 1. For comparison purposes, the standard state is specified – this is the form of a substance when P = 1 bar and T = 298 K (25 °C). 2. If an element exists in more than one form under standard conditions, then the most stable form is selected; thus O2 is used instead of O or O3, and graphite instead of diamond. 3. By definition, formation reactions involve the formation of 1 mole of a substance; enthalpies of formation are therefore reported in units of kJ per mole (kJmol-1). ** PRACTICE Using the information in Table 4.3 (next page), write thermochemical equations that correspond to the standard enthalpy of formation of: (a) Ammonia, NH3 (g) (b) Sodium carbonate, Na2CO3 (s) ©Stellenbosch University, 2023 16 Chemistry 124 Dr K de Villiers Thermodynamics ©Stellenbosch University, 2023 17 Chemistry 124 Dr K de Villiers Thermodynamics Using enthalpies of formation to calculate enthalpies of reaction 1. We can apply Hess’s Law in order to determine rxnH°  to do this, we need to use the individual formation reactions for each species. 2. Alternatively, we can make use of the following relationship, which holds since any reaction can be broken down into a series of formation reactions: ∆ 𝐻° Σ𝑛∆ 𝐻° 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 Σ𝑚∆ 𝐻° 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 PRACTICE ** Practice exercise 14.10 Using the standard enthalpies of formation listed in Table 4.3, calculate ΔrxnH° for the combustion of 1 mol of ethanol: C2H5OH (l) + 3O2 (g)  2CO2 (g) + 3H2O (l) Let’s start by using the second approach (formula): ©Stellenbosch University, 2023 18 Chemistry 124 Dr K de Villiers Thermodynamics Using enthalpies of formation to calculate enthalpies of reaction 1. We can apply Hess’s Law in order to determine rxnH°  to do this, we need to use the individual formation reactions for each species. 2. Alternatively, we can make use of the following relationship, which holds since any reaction can be broken down into a series of formation reactions: ∆ 𝐻° Σ𝑛∆ 𝐻° 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 Σ𝑚∆ 𝐻° 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 PRACTICE ** Practice exercise 14.10 Using the standard enthalpies of formation listed in Table 4.3, calculate ΔrxnH° for the combustion of 1 mol of ethanol: C2H5OH (l) + 3O2 (g)  2CO2 (g) + 3H2O (l) Let’s start by using the second approach (formula): ©Stellenbosch University, 2023 18 Chemistry 124 Dr K de Villiers Thermodynamics In your own time, you should give the first approach a go (using Hess’s Law) – remember, practice helps to develop a growth mindset, which is an important factor underpinning success. Feel free to ask me for help if you’re struggling with this approach. Practice exercise 14.11 Given the standard enthalpy change below, and using the information provided in Table 4.3, calculate ΔfH° for CuO(s): CuO (s) + H2 (g)  Cu (s) + H2O (l) ΔH° =  129.7 kJ ©Stellenbosch University, 2023 19 Chemistry 124 Dr K de Villiers Thermodynamics So far, we have focused on enthalpy – it’s more tangible in a sense, and there are numerous approaches to determining enthalpy changes for different reactions. But remember, the BIG PICTURE is to be able to predict whether a reaction is likely to take place under a given set of conditions, and to WHAT EXTENT. Unfortunately, enthalpy alone is not a suitable predictor. So, shifting gear – we need to know how to determine whether a reaction is going to be spontaneous or not. What does this interesting word mean for us in chemistry? Spontaneous / adjective performed or occurring as a result of a sudden impulse or inclination and without premeditation or external stimulus. e.g. "the audience broke into spontaneous applause" 14.8 SPONTANEOUS PROCESSES Let’s consider a familiar example – dissolving table salt in water at 25C: 𝑵𝒂𝑪𝒍 𝒔  𝑵𝒂 𝒂𝒒 𝑪𝒍 𝒂𝒒 1. Change occurs _________________________________________________ 2. External intervention ____________________________________________ 3. The forward reaction is __________________________________________ 4. Stirring _______________________________________________________ 5. The reverse reaction is ___________________________________________ 6. The reverse reaction is ___________________________________________ ©Stellenbosch University, 2023 20 Chemistry 124 Dr K de Villiers Thermodynamics A note about reversible and irreversible processes: These terms are usually applied for a given set of conditions (temperature and pressure). Reversible Irreversible Restoring the system causes Restoring the system requires no work. requires work. There is no net change to the This causes a surroundings. change in the surroundings. If temperature and pressure are changed, the reverse reaction may become spontaneous. Think about the reaction we started off with: H2O (s)  H2O (l) Okay, but what if the reaction is not one that we are familiar with? How do we look at a balanced chemical reaction and know/decide/figure out if the forward or reverse reaction is spontaneous? Of course, we always hope the forward reaction will be spontaneous – this means that REACTANTS DO GET CONVERTED INTO PRODUCTS. Looking ahead… this cartoon gives us some clue as to what else needs to be considered – (and as a mom, I can easily relate)… https://i0.wp.com/comic.hmp.is.it/wp-content/uploads/2014/02/0018- Entropy.png?fit=1041%2C1200&ssl=1 ©Stellenbosch University, 2023 21 Chemistry 124 Dr K de Villiers Thermodynamics Entropy, S, is a tricky concept to define in a simple sentence. It is, however important for determining whether a reaction is spontaneous or not. For now, consider entropy as a characteristic of a system that gives us an indication of the extent to which energy is disordered. 14.10 THE MOLECULAR INTERPRETATION OF ENTROPY Molecular motions and energy Molecules can undergo three kinds of motion – translational (the entire molecule moves), vibrational (atoms and bonds move – think of two balls connected via a spring) and rotational (the molecule spins). When a molecule is heated, its motion (and therefore its kinetic energy) increases. Collectively, we refer to a molecule as having motional energy. Boltzmann’s equation and microstates A microstate is a single possible arrangement of the positions and kinetic energies of a given set of molecules (i.e. system) – consider it like a snapshot taken at a given instant. For any system, the entropy (S) is related to the total number of microstates (W): 𝑆 𝑘 ln 𝑊 and therefore 𝑾𝒇𝒊𝒏𝒂𝒍 ∆𝑺 𝒌 𝐥𝐧 𝑾𝒊𝒏𝒊𝒕𝒊𝒂𝒍 Ludwig Boltzmann ©Stellenbosch University, 2023 22 Chemistry 124 Dr K de Villiers Thermodynamics The number of microstates available to a system increases with an increase in volume, temperature and the number of independent particles. Making qualitative predictions about S 1. Changes of state 2. Dissolving a solid + 3. Chemical reactions PRACTICE 6.1 ** Practice exercise 14.14 Indicate whether each of the following processes produce an increase or decrease in the entropy of the system: CO2(s)  CO2(g) + S (s  g) CaO(s) + CO2(g)  CaCO3(s) - S (g  s) HCl(g) + NH3(g)  NH4Cl(s) - S (g  s) 2SO2(g) + O2(g)  2SO3(g) - S (3 mol g  2 mol g) ©Stellenbosch University, 2023 23 Chemistry 124 Dr K de Villiers Thermodynamics The third law of thermodynamics If the temperature of a system is decreased to such a point that all molecular motion is halted, then a single microstate (W = 1) is essentially achieved. The temperature at which this is achieved is called absolute zero (zero degrees Kelvin, 0 K). The entropy of a pure crystalline system at 0 K is zero (i.e. S = 0). When a pure crystalline solid is heated, the entropy increases from 0 as shown below: ©Stellenbosch University, 2023 24 Chemistry 124 Dr K de Villiers Thermodynamics. The second law of thermodynamics A spontaneous (irreversible) process is characterized by an overall increase in entropy of the universe: ΔunivS = [ΔsysS + ΔsurrS] > 0 Alternatively, if there is no change in entropy of the universe, then you are dealing with a reversible process: ΔunivS = [ΔsysS + ΔsurrS] = 0 Back to business: To determine whether a reaction is spontaneous or not, you need to calculate univS – in other words, you need to calculate sysS and surrS (and add them together). Entropy is also a state function (rxnS = Sfinal  Sinitial). Thus sysS can be readily determined using the equation below: ∆𝑆° Σ𝑛𝑆° 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 Σ𝑚𝑆° 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 ** For an isothermal (constant temperature) process at equilibrium, the following relationship (derived from G = 0) is a useful alternative to calculate sysS: 𝑞 ∆ 𝑆 𝑇 where qrev = heat transferred if the process were reversible, and T = the temperature (in K) at which the process occurs. Entropy changes in the surroundings The previous equation is useful to calculate surrS: 𝑞 ∆𝐻 ∆ 𝑆 𝑇 𝑇 ©Stellenbosch University, 2023 25 Chemistry 124 Dr K de Villiers Thermodynamics 14.11 ENTROPY CHANGES IN CHEMICAL REACTIONS Standard molar entropies, S°, for substances in their standard state are reported in units of Jmol-1K-1. Observations: 1. At standard temperature (298 K), S° for pure elements is not zero. 2. In general, S°(g) > S°(l) > S°(s). 3. The value of S° usually increases with increasing molar mass. 4. The value of S° usually increases with an increasing number of atoms. PRACTICE Consider the complete combustion of 1 mole of ethanol (C2H5OH). C2H5OH (l) + 3O2 (g)  2CO2 (g) + 3H2O (l) Is the reaction above spontaneous under standard conditions? (Use the information in Appendix C of the textbook). ©Stellenbosch University, 2023 26 Chemistry 124 Dr K de Villiers Thermodynamics 14.11 ENTROPY CHANGES IN CHEMICAL REACTIONS Standard molar entropies, S°, for substances in their standard state are reported in units of Jmol-1K-1. Observations: 1. At standard temperature (298 K), S° for pure elements is not zero. 2. In general, S°(g) > S°(l) > S°(s). 3. The value of S° usually increases with increasing molar mass. 4. The value of S° usually increases with an increasing number of atoms. PRACTICE Consider the complete combustion of 1 mole of ethanol (C2H5OH). C2H5OH (l) + 3O2 (g)  2CO2 (g) + 3H2O (l) Is the reaction above spontaneous under standard conditions? (Use the information in Appendix C of the textbook). ©Stellenbosch University, 2023 26 Chemistry 124 Dr K de Villiers Thermodynamics Practice exercise 14.15 a) Using the information in Appendix C of the textbook, calculate the standard entropy change for the following reaction: Al2O3(s) + 3H2(g)  2Al(s) + 3H2O(g) Ans: 180.4 JK-1 b) Does your answer (quantitative analysis) align with your qualitative prediction? Explain. ©Stellenbosch University, 2023 27 Chemistry 124 Dr K de Villiers Thermodynamics 14.12 GIBBS FREE ENERGY Reaction spontaneity depends on both enthalpy and entropy; their combined contributions are expressed in a new state function – Gibbs free energy, G: 𝐺 𝐻 – 𝑇𝑆 ∆𝐺 ∆𝐻 𝑇∆𝑆 (T is constant under isothermal conditions) From entropy considerations: ∆ 𝑆 ∆ 𝑆 ∆ 𝑆 Substitute so that right hand side of equation is only in terms of the system: ∆ 𝐻 ∆ 𝑆 ∆ 𝑆 𝑇 Cross multiply by -T: 𝑇∆ 𝑆 ∆ 𝐻 𝑇∆ 𝑆 Compare this to: ∆𝑮 ∆𝑯 𝑻∆𝑺 Therefore: ∆𝑮 𝑻∆𝒖𝒏𝒊𝒗 𝑺 ∆𝐆 ∆𝐆 𝟎 ∆𝐆 Thus, the change in Gibbs free energy for a reaction can be calculated if the changes in enthalpy and entropy for the same reaction are known. The change in Gibbs free energy for a reaction can also be calculated, as for enthalpy, using standard Gibbs free energies of formation: ∆𝐺° Σ𝑛∆ 𝐺° 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 Σ𝑚∆ 𝐺° 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 ©Stellenbosch University, 2023 28 Chemistry 124 Dr K de Villiers Thermodynamics PRACTICE Calculate the standard change in Gibbs free energy for the combustion of 1 mole of ethanol (C2H5OH) at 298 K: C2H5OH (l) + 3O2 (g)  2CO2 (g) + 3H2O (l) a) using ΔfG° data in Appendix C of the textbook b) using ΔH° and ΔS° information (calculated in previous lectures) ©Stellenbosch University, 2023 29 Chemistry 124 Dr K de Villiers Thermodynamics So, aren’t all exothermic reactions spontaneous, and endothermic ones non-spontaneous? If it were that simple, we needn’t have had the last few lectures. Remember, reaction spontaneity (direction) is a combination of enthalpy, entropy AND TEMPERATURE! ∆𝐺 ∆𝐻 – 𝑇∆𝑆 ∆𝑺 ∆𝑺 ∆𝑯 ∆𝑯 ©Stellenbosch University, 2023 30 Chemistry 124 Dr K de Villiers Thermodynamics PRACTICE Determine ΔH° and ΔS° for the reaction below, and hence determine whether or not the reaction is spontaneous at 400 K. 2 SO2(g) + O2(g)  2 SO3(g) For a particular reaction, ΔH = -32 kJ and ΔS = -98 J/K. Assume that these values do not change significantly with temperature. (a) At what temperature will the reaction have ΔG = 0? (b) If T is increased from the value determined in (a), will the reaction be spontaneous or non-spontaneous? ©Stellenbosch University, 2023 31 Chemistry 124 Dr K de Villiers Thermodynamics Okay, so BIG PICTURE complete – we can predict whether a reaction is spontaneous or not. But how does that really help in the real world (where PRODUCT YIELD is the big driver?) The final step for now… and looking forward to Chemistry 144 in the second semester where we will cover the topic in detail... is the EQUILIBRIUM CONSTANT, K. K is an expression of the relationship between products and reactants at equilibrium. For A  B, K = [B]/[A]. The ideal scenario is when [B] >>> [A], and K >>> 1. GIBBS (FREE) ENERGY AND THE EQUILIBRIUM CONSTANT Chapter 16.7 (p 635) ∆𝐺° 𝑅𝑇𝑙𝑛𝐾 ∆ °/ ∴𝐾 𝑒 ∆𝐆 𝐾 Species favoured 𝟎 ©Stellenbosch University, 2023 32 Chemistry 124 Dr K de Villiers Thermodynamics PRACTICE Calculate the equilibrium constant for the solubility (Ksp) of BaF2(s) in water at 25 C, given that G = 32.9 kJ/mol. 𝑩𝒂𝑭𝟐 𝒔 ⇌ 𝑩𝒂𝟐 𝒂𝒒 𝟐𝑭 𝒂𝒒 ©Stellenbosch University, 2023 33 Chemistry 124 Dr K de Villiers Chemical Kinetics Chemistry is concerned about change – the transformation of reactants to products. In thermodynamics Gibbs free energy determine whether or not a reaction would occur spontaneously (in a particular direction and A reaction can be classified as spontaneous ong it takes – it rs/days/years. The area of chemistry that deals with the speed (rate) of a reaction is chemical kinetics. 15.1 FACTORS THAT AFFECT REACTION RATES breaking and forming of bonds – the rate at which these take place depend on the nature of the reactants. The following factors influence the rate of a chemical reaction: Physical state of the reactants Concentration of reactants Reaction temperature Presence of a catalyst © Stellenbosch University, 2023 1 Chemistry 124 Dr K de Villiers Chemical Kinetics 15.7 CATALYSIS Homogenous catalysis is when the catalyst is in the same phase as the reacting molecules. For example, consider the decomposition of hydrogen peroxide: ( ) ( )+ ( ) In the absence of a catalyst, the reaction proceeds extremely slowly. The inclusion of ( )+ ( )+ ( ) ( )+ () ( )+ ( ) ( )+ ( )+ ( ) In this reaction, bromide ions + catalyse the reaction – they are present at the start of the reaction, and appear unchanged at the end. Bromine (Br2(aq is an intermediate as it is produced in reaction 1 and consumed in reaction 2. Complete the energy diagram for the catalysed reaction: Heterogenous catalysis is when the catalyst is in a different phase compared to the reacting molecules. gas or liquid. Reactant molecules first need to adsorb to the catalyst surface, which is highly reaction proceeds to yield product molecules, which later desorb from the surface. © Stellenbosch University, 2023 2 Chemistry 124 Dr K de Villiers Chemical Kinetics 15.2 REACTION RATES In general, speed is defined as the extent of change in a given time interval. In the case of a chemical reaction, speed is the same as the reaction rate. The measurable that changes with time is concentration. Therefore the units are expressed as M/s (Ms-1. Consider a hypothetical reaction: A B the appearance/formation of pr = © Stellenbosch University, 2023 3 Chemistry 124 Dr K de Villiers Chemical Kinetics = NB: By convention, rates are always expressed as positive quantities – when looking at the disappearance of a reactant, the minus sign in the expression ensures this. PRACTICE ** Exercise 15.15 Consider the following hypothetical aqueous reaction: A B. A flask is loaded with 0.065 mol of A in a total 3. The following data are collected: Time (min) 0 10 20 30 Mol (A) 0.065 0.051 0.036 0.031 Mol (B) 0 of disappearance of A for each 10. Between t = 10 min and t = of appearance of © Stellenbosch University, 2023 4 Chemistry 124 Dr K de Villiers Chemical Kinetics Change of Rate with Time Now let’s consider an actual chemical reaction: C4H9Cl (aq) + H2O (aq) C4H9OH (aq) + HCl (aq) The average rate decreases with increasing time. The instantaneous rate is the rate of reaction at a particular moment. [ ] = e.g. t = 600 s © Stellenbosch University, 2023 5 Chemistry 124 Dr K de Villiers Chemical Kinetics Reaction Rates and Stoichiometry reaction stoichiometry has been 1:1: A B For such reactions, rates of reactant disappearance and product formation are equal: [ ] [ ] = = Now let’s consider what happens in cases where the stoichiometry is not 1:1. 2A 3B , which spe experiencing the largest change in concentration ( [A] or [B ate of disappearance of [A] [B] rate of disappearance of A = = t t ate of appearance of B relate to the rate of disappearance of A [B] [A] rate of appearance of B = = t t 1 [A] 1 [B] rate = = t t If the rate of appearance of B is 0.15 M/s, determine the following: Rate of disappearance of A Rate © Stellenbosch University, 2023 6 Chemistry 124 Dr K de Villiers Chemical Kinetics PRACTICE ** Exercise 15.19(a) Consider the combustion of ethylene: C2H4 (g) + 3 O2 (g) 2 CO2 (g) + 2 H2O (g) If the concentration of ethylene is decreasing at a rate of 0.37 Ms-1, what are the rates of change for O2 2 15.3 CONCENTRATION AND RATE LAWS A rate law shows how the rate of a reaction depends on the concentration of all reactants. For the general reaction, aA+bB cC+dD the rate law has the form, = [ ] [ ] = [ ] [ ] … where k = rate constant/coefficient and the exponents (m and n) = reaction orders. © Stellenbosch University, 2023 7 Chemistry 124 Dr K de Villiers Chemical Kinetics Let’s consider the following reaction and experimental data, and see how this is done. NH4+ (aq) + NO2– (aq) N2 (g) + 2 H2O (l) Compare the data from experiments 1 and 2: Compare the data from experiments 5 and 6: © Stellenbosch University, 2023 8 Chemistry 124 Dr K de Villiers Chemical Kinetics m and n, you are next able to determine the order of reaction. Finally it is possible to determine rate constant, k, for this reaction: Reaction Orders: Exponents in the Rate Law the rate law is = [ ] [ ] … Exponents indicate how the reaction rate is affected by the concentration of each deduced f Exponent (m or n) Meaning Interpretation 0 Zero order 1 First order 2 Second order The overall reaction order is the sum of the orders reactants. © Stellenbosch University, 2023 9 Chemistry 124 Dr K de Villiers Chemical Kinetics Units of Rate Constants Unlike the rate of a reaction, the rate constant does not depend on concentration. It is affected by temperature and the presence of a catalyst. The units of the rate constant Overall Rate equation Units of k (t in seconds) reaction order = 0 Ms-1 = [ ] 1 = [ ][ ] 2 = [ ] = [ ][ ] 3 PRACTICE ** Exercise 15.27 2– – 2O8 at 25 C: S2O82– (aq) + 3 I– (aq) 2 SO42– (aq) + I3– (aq) The rate of disappearance of S2O82– following manner: Experiment [S2O82–] (M) [I–] (M) Initial rate (M s-1) no. 1 0.018 0.036 2.6 10–6 2 0.027 0.036 3.9 10–6 3 0.036 7.8 10–6 0.050 0.072 10–5 Determine the rate law for the reaction. © Stellenbosch University, 2023 10 Chemistry 124 Dr K de Villiers Chemical Kinetics Units of Rate Constants Unlike the rate of a reaction, the rate constant does not depend on concentration. It is affected by temperature and the presence of a catalyst. The units of the rate constant Overall Rate equation Units of k (t in seconds) reaction order = 0 Ms-1 = [ ] 1 = [ ][ ] 2 = [ ] = [ ][ ] 3 PRACTICE ** Exercise 15.27 2– – 2O8 at 25 C: S2O82– (aq) + 3 I– (aq) 2 SO42– (aq) + I3– (aq) The rate of disappearance of S2O82– following manner: Experiment [S2O82–] (M) [I–] (M) Initial rate (M s-1) no. 1 0.018 0.036 2.6 10–6 2 0.027 0.036 3.9 10–6 3 0.036 7.8 10–6 0.050 0.072 10–5 Determine the rate law for the reaction. © Stellenbosch University, 2023 10 Chemistry 124 Dr K de Villiers Chemical Kinetics of 2– 2O8 related to the rate of disappearance of I– What is the rate of disappearance of I– 2– 2O8 ] = 0.015 M and [I–] = © Stellenbosch University, 2023 11 Chemistry 124 Dr K de Villiers Chemical Kinetics of 2– 2O8 related to the rate of disappearance of I– What is the rate of disappearance of I– 2– 2O8 ] = 0.015 M and [I–] = © Stellenbosch University, 2023 11 Chemistry 124 Dr K de Villiers Chemical Kinetics Practice Exercise 15.6 The following data were measured for the reaction of nitric acid with hydrogen: 2 NO (g) + 2 H2 (g) N2 (g) + 2 H2O (g) Experiment [NO] (M) [H2] (M) Initial rate (M s-1) no. 1 0.100 0.100 1.23 10–3 2 0.100 0.200 10–3 3 0.200 0.100 10–3 Determine the rate law for this reaction. Calculate the rate constant. 2] = 0.150 M © Stellenbosch University, 2023 12 Chemistry 124 Dr K de Villiers Chemical Kinetics 15.4 THE CHANGE OF CONCENTRATION WITH TIME (INTEGRATED RATE EQUATIONS) how the rate law can be used to calculate the reaction rate from the rate constant and initial reactant concentrations This is possible to determine – we will use the integrated form of the rate equation as shown below. To illustrate the point, consider the reaction below, where A is the only reactant: A B Before we can perform integration, we need the differential form of the reaction rate : [ ] = Eeeek – there is some pretty crazy maths coming up! Please don’t stress – you will not be required to perform the integration yourselves (i.e. derive these integrated rate equations) – the most important skill is for you to be able to apply the relevant integrated rate equation to the correct scenario. Zero-Order Reactions The rate is independent of the concentration of A: [ ] = = [ ] = [ ]= [ ] [ ] = ( 0) [ ] = +[ ] © Stellenbosch University, 2023 13 Chemistry 124 Dr K de Villiers Chemical Kinetics First-Order Reactions The rate is directly proportional to the concentration of A: [ ] = = [ ] = [ ] 1 [ ]= [ ] [ ] [ ] = ( 0) [ ] = + [ ] [ ] =[ ] Second-Order Reactions The rate is directly proportional to the square of the concentration of A: [ ] = = [ ] 1 [ ]= [ ] 1 1 = ( 0) [ ] [ ] = + [ ] [ ] © Stellenbosch University, 2023 14 Chemistry 124 Dr K de Villiers Chemical Kinetics Half-Life The half-life of a reaction ( concentration of a reactant to reach one- (i.e. when [ ] = [ ]. [ ] Zero-order [ ] = +[ ] = First-order [ ] = + [ ] = Second-order = + = [ ] [ ] [ ] The half-life is constant for a first order process, but will change depending on the initial concentration ([ ] ) for both zero- and second-order processes. This can be really useful (i.e. save time) when trying to identify reaction order. © Stellenbosch University, 2023 15 Chemistry 124 Dr K de Villiers Chemical Kinetics PRACTICE ** Exercise 15.30 The gas-phase 2Cl2 SO2Cl2(g) SO2(g) + Cl2(g) At 600 K the half-life for this process is 2.3 105 s. What is the rate constant At 320 °C the rate constant is 2.2 10-6 s-1. What is the half-life at this ** Exercise 15.32 At 660 K 10-2 s-1. If the initial 2Cl2 is 375 Pa, what will the will it take for the pressure of 2Cl2 to reach one-tenth its initial © Stellenbosch University, 2023 16 Chemistry 124 Dr K de Villiers Chemical Kinetics The following data were determined for a reaction A B, at 298 K: t/s [A] / M 0 7 0.20 0.10 21 0.05 28 0.025 Determine the order of the reaction from the data. © Stellenbosch University, 2023 17 Chemistry 124 Dr K de Villiers Chemical Kinetics The graph below was obtained or the reaction: NO2 (g) NO (g) + ½ O2 (g),: 300 ] 2 200 1 /[N O 100 0 0 100 200 300 T im e / s From the graph, determine (i rate constant, k and -life for the reaction. If the experiment were repeated using an initial concentration of M, how long would it take for the concentration of NO2 to decrease to 0.30 © Stellenbosch University, 2023 18 Chemistry 124 Dr K de Villiers Chemical Kinetics 15.5 TEMPERATURE AND RATE The rates of most chemical reactions increase with increasing temperature. The faster rate at higher temperature is due to an increase in the rate constant with increasing temperature. Reaction rates are affected by both the concentration of reactants and by temperature. This brings us to the collision model – molecules must collide in order to react. (You can imagine that the greater the number of collisions, the greater the reaction o need to be orientated in a certain way for the collisions to result in a reaction – this is called the orientation factor. © Stellenbosch University, 2023 19 Chemistry 124 Dr K de Villiers Chemical Kinetics Activation Energy In addition to correct orientation and collision, molecules must possess a minimum amount of energy in order to react – this is called the activation energy (Ea Kinetic energy of colliding molecules is used to stretch, bend and ultimately break bonds – this is what leads to a chemical reaction. Potential energy Transition state Ea Ereactants E Eproducts Reaction pathway/progress Products may be lower in energy compared to reactants (– exothermic vice versa endothermic Rather, t a. What is the activation barrier of the reverse reaction? lar collisions lead to a reaction (i.e. the molecules gain distribution of energies – this is also affected by temperature. © Stellenbosch University, 2023 20 Chemistry 124 Dr K de Villiers Chemical Kinetics The Arrhenius Equation Interestingly, the increase in reaction rate with increasing temperature does not show a linear dependence. Rather, the exponential r Arrhenius equation: / = where k = the rate constant; Ea = “A” is called the frequency factor – it relates to - le The Arrhenius equation predicts that as the activation barrier increases, so the reaction rate decreases. Determining the Activation Energy The Arrhenius equation can be re-worked into a linear form, which is then easier to graph in order to determine Ea: = + © Stellenbosch University, 2023 21 Chemistry 124 Dr K de Villiers Chemical Kinetics stants at two different reaction temperatures are known: A is assumed to be constant. With rearrangement, : = = PRACTICE ** Exercise 15.43 A certain first order reaction has a rate constant of 2.75 10-2 s-1 at 20 °C. k at 60 °C if Ea = 75.7 kJmol-1 © Stellenbosch University, 2023 22 Chemistry 124 Dr K de Villiers Chemical Kinetics ** Exercise 15.41 ing reactions from fastest to slowest. Ea / kJ mol-1 -1 Reaction 1 25 2 35 10 3 55 10 Explain. © Stellenbosch University, 2023 23