Chemical reactions involve a transfer of energy between the system and the surroundings.

Transcription

1 THERMODYNAMICS science of the interconversion of different forms of ENERGY e.g. heat and work an essential foundation for understanding physical sciences. at the centre of any physical description Key Questions: 1. Interconversion of energy how, why, etc. 2. Is a particular reaction possible? 3. How much useful work can be obtained from a reaction? 4. Can we understand chemical equilibria? DEFINITIONS: Chemical reactions involve a transfer of energy between the system and the surroundings. SYSTEM the part of the universe under observation e.g., engine, electrical cell, reaction in a flask, etc. The system can be: Open both mass and energy may leave and enter. Closed energy can be exchanged but no mass can enter or leave. Isolated neither mass or energy may enter or leave. SURROUNDINGS the rest of the universe. Therefore, the UNIVERSE is made up of the SYSTEM and the SURROUNDINGS. The STATE of a system is characterised by a set of variables, e.g., P, V, T, n, etc. These are STATE VARIABLES also known as STATE FUNCTIONS. 1

2 P, V, T, n and mass can be classified as either: Intensive independent of size. e.g. P, T Extensive dependent on size. e.g. m,v, n ENERGY TRANSFER We can transfer energy in and out of a system, whereby the energy of the system may change. Internal Energy (U) is the sum of the kinetic and potential energy of system. Two ways U can change: 1. q heat flow 2. w work ENERGY TRANSFER SYSTEM HEAT (q) Thermal Surroundings WORK (w) Mechanical Surroundings INTERNAL ENERGY (U) We can ONLY measure changes in U, not absolute values, so for any reaction or process, i.e., U final U initial = ΔU We can change the internal energy, U, by any combination of q and w. ΔU = q + w This is a statement of the conservation of energy: 1 st LAW OF THERMODYNAMICS The Law of Conservation of Energy states energy may be converted from one form to another, but the total quantity of energy remains constant. 2

3 HEAT ENERGY, HEAT TRANSFER and TEMPERATURE ΔU = q + w tells us we cannot create or destroy energy, but it can be transferred in and out of the system! T 1 > T 2 T 1 T 2 Object 1 Object 2 TEMPERATURE an indication of thermal energy. objects at same T do not exchange heat energy. if T s are different, energy (heat) flows from high T (hot) to low T (cold). THERMOCHEMISTRY (Chapter 6) a subset of thermodynamics study of the heat given off or absorbed in a chemical reaction. CHEMICAL REACTIONS AND ENERGY CHANGES All chemical reactions are accompanied by an ENERGY CHANGE Chemical reactions may release heat to the surroundings, e.g., CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) (burning or combustion of natural gas) If heat is released (given off), the reaction is exothermic. This arises due to differences in the energies of reactants and products. ALL combustion reactions are EXOTHERMIC. If heat is absorbed (taken in), the reaction is endothermic. e.g., 6CO 2 (g) + 6H 2 O(l) C 6 H 12 O 6 (s) + 6O 2 (g) (photosynthesis) Note: The amount of heat depends on the size of the system. (This is an EXTENSIVE property, like mass, V and n.) 3

4 SIGN CONVENTION FOR HEAT PROCESSES PROCESS SIGN Heat absorbed by the system from the surroundings (endothermic) + Heat absorbed from the system by the surroundings (exothermic) Most reactions are run under constant pressure conditions. The heat given off (or absorbed) at constant pressure is known as the ENTHALPY change, and is denoted by the symbol, ΔH ENTHALPY - is a measure of the difference in heat content, H, between final and initial states (e.g., products and reactants). i.e., ΔH Rxn = Σ ΔH products Σ ΔH reactants = q p = heat flow EXOTHERMIC and ENDOTHERMIC PROCESSES Enthalpy Products Reactants ΔH Rxn < 0 (heat released) EXOTHERMIC Enthalpy Products Reactants ΔH Rxn > 0 (heat absorbed) ENDOTHERMIC Remember: Heat changes at constant P are equal to the change in ENTHALPY If heat is given off, the rxn is EXOTHERMIC (ΔH is negative) If heat is absorbed, the rxn is ENDOTHERMIC (ΔH is positive) We can measure this heat change through the science of calorimetry. 4

5 CALORIMETRY (Refer to page 238 in text ) measurement of the quantities of heat in bodies. Experimentally, we need to: measure temperature change during a reaction, then convert that change in temperature to change in energy. How? HEAT CAPACITY (C) energy required to raise the temperature of a substance by 1C or 1 K unit, i.e., C = q/δt YOU should distinguish between: A) Heat Capacity (C) energy required to raise a substance by 1C or 1 K unit. Units: J K 1 B) Molar Heat Capacity (c) energy needed to raise 1 mole through 1C or 1 K unit (or the heat capacity per mole). Units: J K 1 mol 1 C) Specific Heat (s, C SP or C S ) amount of energy to raise 1 g through 1C or 1 K unit (or the heat capacity per gram of substance). Units: J g 1 K 1 e.g., H 2 O(l) Specific Heat = J g 1 deg 1 Hence, q = C ΔT = n c ΔT = mass s ΔT In calorimetry the idea is to ISOLATE the SYSTEM from the SURROUNDINGS. e.g. "Coffee Cup" Calorimeter refer to Figure 6.12 in text To measure heat given off (or absorbed) by the chemical reaction: q = CΔT (calculate this usually) C know this for the vessel and water (sometimes given together, sometimes separately). ΔT measure this Note: Heat released in the reaction is equal to the heat gained by the surroundings, i.e., q System + q Surroundings = 0 therefore, q System = q Surroundings = q Rxn q Rxn = (q solution + q calorimeter ) 5

6 HEAT OF SOLUTION - enthalpy change associated with dissolution of 1 mole solid: i) take known mass of water at temperature, T 1 ii) add mass of solid, dissolve and measure temperature T 2 iii) calculate the change in temperature: ΔT = T 2 T 1 = T Final T Initial Example Dissolution of a solid (KF) in water: Exactly 1.23 g KF (s = 0.84 J g 1 deg 1 ) is dissolved in 10.0 g H 2 O (s = J g 1 deg 1 ). The H 2 O temp rises from 23.2 to 33.1 C. What is heat of solution per mole for KF? (MM KF = 58.1 g mol 1 ) Energy = KF dissolves in the water and liberates heat which warms the solution. Energy = Heat gained by water + Heat gained by KF Energy = q(h 2 O) + q(kf) Note the size of q(kf) versus q(h 2 O)! In many cases, we will just assume the specific hear of water is very close to that of the solution and simplify this calculation to Energy = q(h 2 O). MM for KF = 58.1 g/mol, therefore # moles of KF added = 1 mole of KF would therefore evolve of heat. 6

7 Bomb Calorimeter (refer to Figure 6.13) carries out combustion reactions at constant volume conditions the heat produced raises the temperature of the bomb and the surrounding liquid (the bomb is usually placed in a water bath). Sometimes we evaluate the heat capacity of the bomb and the water separately (i.e., q Total = q Bomb + q water bath ) and sometimes the heat capacity for the entire calorimeter (bomb + water bath) has been determined experimentally. You must read the question carefully to determine which case you have. Example: Exactly 1.00 g of (CH 3 ) 2 N 2 H 2 (MM = 60.1 g/mol) is burned in a bomb calorimeter with a heat capacity = 1840 J deg 1. A 5.00 kg water bath rises in temperature from to C. Calculate the energy released when 1 mole is burned. Total heat produced = warms bomb + warms water bath 1 mole of (CH 3 ) 2 N 2 H 2 evolves kj of heat energy. EXTRA PROBLEMS TO TRY Problem 1: 50.0 ml dilute AgNO 3 is added to a solution with OH ions in a coffee cup calorimeter. Ag 2 O precipitates and the temperature of the liquid goes from to C. Assume that the mixture has the same specific heat as water. Calculate "q surr " for a mass of 150. g. Is the reaction exothermic or endothermic? Answer: q Surroundings = 885 Joules Rxn exothermic (temp rose). 7

8 Problem 2: When ice at 0 C melts to liquid H 2 O, it absorbs 334 J per gram. The specific heat of water is J g 1 deg 1. Suppose the heat needed to melt a 35.0g ice cube is absorbed from water (0.210 kg) in a calorimeter at 21.0 C. What is the final water temperature? [This is the kind of calculation that all experienced party givers do before throwing a party!] Answer: T Final = 6.6 C WHAT ABOUT WORK? The first law of thermodynamics states that the change in the internal energy is dependent upon the flow of heat, but is also depend on WORK (refer to sec. 18.1) WORK the energy exchange that results when a force moves an object through a distance. EXAMPLES OF WORK Mechanical: w = force distance = F Δx = m a Δx Units: (mass) (acceleration) (distance) = kg m 2 s -2 = N m = Joule (J) Electrical: w = charge potential difference = ΔQ V Units: (coulombs) (volts) = Joule Note: We will look at this kind of work when we discuss electrochemistry later on in the course. COMPRESSION/EXPANSION or P-V WORK (see Figures 6.9 & 18.2 in the text) At a CONSTANT external pressure, P ext., the work (in Joules) done BY the surroundings ON a gas (i.e., system) is given by w = force distance = F Δx w = F/A A Δx = P ext. A Δx where the gas is compressed a distance Δx by a pressure, P ext.. But, A Δx can be represented by the volume decrease, or, w = P ext. ΔV where ΔV = V 2 V 1 Compression: work is done ON the system BY the surroundings Our definition says work is to be positive because for a gas to be compressed, energy is added to the system, i.e., the surroundings does work on the system. So our relationship must be: w = P ext. ΔV 8

9 Expansion work is negative since energy leaves the system as the system does work on the surroundings. Pressure-Volume (or P-V) work is the most common type of work encountered in chemical systems. UNITS FOR ENERGY 1 J = 1 kg m 2 s -2 (SI definition) i.e., mass velocity 2 Note: 1 Pa = 1 N m -2 = 1 kg m -1 s -2 1 J = Pa m 3 Hence PΔV Pa m 3 = Joule An older unit still in use is the calorie, where 1 cal = J (by definition) EXAMPLE: Calculate the work at 25 C and 1.00 atm for a) CO 2 (g) + NaOH(s) NaHCO 3 (s) b) 2O 3 (g) 3O 2 (g) Consider: only PV work, that the gases behave ideally (i.e., follow the ideal gas law) and the equations represent the amounts in moles. In a), we move from 1.00 mole of gaseous CO 2 on the reactant side to 0 moles of gas on the product side. Work is done the system the surroundings. This is an example of a. Work done is then (+/ ). First determine the volume of the CO 2 (g), then determine the work done. Note: The gas constant R = J K -1 mol -1 = L atm K 1 mol 1 In b), we move from 2 moles of gaseous O 3 on the reactant side to 3 moles of O 2 on the product side. Work is done the system the surroundings. This is an example of a. Work done is then (+/ ). 9

10 QUESTIONS: 1) A gas sample is heated in a cylinder using 550 kj of heat. A piston compresses the gas, doing 700 kj of work. What is the change in internal energy of the system? 2) Calculate the work done when 6300 L of an ideal gas (initially at 1.00 atm) is compressed (at constant T) by a constant pressure of 1.40 atm to a final volume of 45L. SIGN CONVENTION FOR WORK and HEAT PROCESSES PROCESS SIGN Work done the system the surroundings ( ) Work done the system the surroundings ( ) + Heat absorbed the system the surroundings ( ) + Heat absorbed the system the surroundings ( ) if w and q are positive, they tend to increase U of the system work done by the system is a reduction in U, hence negative. INTERNAL ENERGY & ISOTHERMAL EXPANSIONS Internal energy of an ideal gas depends only on its temperature. To change the temperature of a gas, the internal energy must be changed by doing work on it or transferring energy to it, i.e., ΔU = q + w. When a gas expands, the forces due to pressure push the container walls back, so the gas is doing work. Change in w is negative. For an isothermal (i.e., ΔT = 0) expansion of an ideal gas, ΔU = 0 because the kinetic energy of the gas is related to its temperature (which here is unchanged) and the potential energy is related to bonding and structure (which here is unchanged). Since ΔU = q + w = 0, w = q. The system does w joules of PV-work and absorbs an equivalent amount of heat q = w. 10

11 HEAT CHANGES IN REACTIONS a) Reactions at Constant Volume, i.e., a sealed container System cannot expand or compress (ΔV = 0), so the work is zero, i.e., w = 0 Using the First Law of Thermodynamics, ΔU = q + w; At constant volume, q V = ΔU (e.g., in a bomb calorimeter) i.e., at constant volume (say in a sealed, bomb calorimeter), the increase/decrease in internal energy corresponds to the heat supplied/absorbed by the surroundings. Therefore the molar heat capacity, c, becomes: c V = q V /ΔT = ΔU/ΔT b) Reactions at Constant Pressure, i.e., open to the atmosphere The heat will cause the volume to change, i.e., ΔV 0 Some work will be done, e.g., pushing back the atmosphere i.e., w = P ext.δv Hence, ΔU = q P ext.δv OR q P = ΔU + P ext. ΔV = ΔH At constant pressure, q P = ΔΗ Most reactions run under constant pressure (including biological processes). We can measure this heat change with a coffee-cup calorimeter (refer to page 239). When dealing with calorimetry measurements at constant pressure, the molar heat capacity becomes: c P = q P /ΔT = ΔH/ΔT Therefore, heat capacities give the rate of change of U and H with T. ΔU and ΔH are STATE functions. (q and w are not.) X Final Height X Initial ΔHEIGHT = HEIGHT Final HEIGHT Initial 11

12 A state function for a system depends on the differences between the final and initial state ONLY, NOT on the pathway. This will become important later when we discuss Hess Law. HOW DO ΔH and ΔU COMPARE FOR A REACTION? ΔH = ΔU + Δ(PV) ΔH and ΔU differ only in the PV product. For reactions at constant P, the above equation becomes: ΔH = ΔU + PΔV OR ΔH ΔU = w OR ΔU ΔH = w i.e., the difference between the change in internal energy and the enthalpy is the amount of work that needs to be done to make room for the products of the reaction. For reactions at constant pressure, the volume change of solids and liquids is very small. Hence, for these cases ΔH is similar to ΔU. For an ideal gas at constant T and P, the equation PV = nrt will let us calculate the volume change, i.e., Δ(PV) = Δ(nRT) PΔV = ΔnRT i.e., Δn = moles of gaseous products minus moles of gaseous reactants. ΔH = ΔU + ΔnRT Example: At 298 K, CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l), ΔH Rxn = kj mol 1 What is ΔU Rxn? Δn = ΔU = At constant pressure, q = ΔU Rxn = Could calculate the actual volume change from PV = nrt. Not necessary (in this case), since we can use PΔV = ΔnRT. 12

13 Net result is a shrinking volume (compression), therefore work done by on. At 25 C, RT = JK -1 mol 1 298K 2.5 kj mol -1, therefore, ΔH Rxn and ΔU Rxn are not that far apart in value. PROBLEM: When an ideal gas expands at constant temperature (i.e., isothermal expansion), ΔU is zero. Consider 1.00 L of an ideal gas initially at 9.00 atm and 15 C. (a) Calculate q and w if the gas expands isothermally against a pressure of 1.00 atm. (b) Calculate q and w if the gas expands isothermally first against a pressure of 3.00 atm. And then against 1.00 atm. (c) Calculate q and w if the gas expands isothermally first against a pressure of 3.00 atm, then 2.00 atm and then against 1.00 atm. (d) Comment on the results. Answers: a) q = w = kj b) w 1 + w 2 = kj q 1 + q 2 = kj c) w = w 1 + w 2 + w 3 = kj q = q 1 + q 2 + q 3 = kj d) The conclusion is that you would get the maximum amount of work done on the surroundings by doing the expansion in an infinite number of very small steps. 13