Thermodynamics I - Enthalpy

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1 Thermodynamics I - Enthalpy Tinoco Chapter 2 Secondary Reference: J.B. Fenn, Engines, Energy and Entropy, Global View Publishing, Pittsburgh,

2 Thermodynamics CHEM Kinetics An essential foundation for understanding physical and biological sciences. Relationships and interconversions between various forms of energy (mechanical, electrical, chemical, heat, etc) An understanding of the maximum efficiency with which we can transform one form of energy into another Preferred direction by which a system evolves, i.e. will a conversion (reaction) go or not Understanding of equilibrium It is not based on the ideas of molecules or atoms. A linkage between these and thermo can be achieved using statistical methods. It does not tell us about the rate of a process (how fast). Domain of kinetics. 2

3 Surroundings, boundaries, system When considering energy relationships it is important to define your point of reference. 3

4 Types of Systems Open: both mass and energy may leave and enter Closed: energy can be exchanged no matter can enter or leave Isolated: neither mass nor energy can enter or leave. 4

5 Energy Transfer CHEM Kinetics Energy can be transferred between the system and the surroundings as heat (q) or work (w). This leads to a change in the internal energy (E or U) of the system. Heat the energy transfer that occurs when two bodies at different temperatures come in contact with each other - the hotter body tends to cool while the cooler one warms until thermal equilibrium is achieved (they are both at the same temperature) heat transfer depends on the heat capacity (C) of the -1 bodies involved - C (J K ) reflects how much energy is required to heat up an object or substance by 1 C -1-1 or 1 K (C = molar heat capacity, J K mol ) Note: the equation q = C T applies only when C is independent of T, which is usually true over a small T sign-convention: heat is positive if it flows into the system (from the surroundings) and negative if it 5

6 flows out of the system (into the surroundings) Work an energy transfer between system and surroundings that is not heat any energy transfer that has or could have as its sole effect the raising of a weight (Fenn, p.6) sign-convention: work is positive if it is done by the surroundings on the system and negative if it is done by the system on the surroundings can be mechanical, electrical, gravitational etc. common type discussed in thermodynamics is work of increasing or decreasing volume See Tinoco pp for more examples of work. 6

7 How do you tell if an energy transfer is heat or work? often depends on how your define the system and surroundings one method that can differentiate between the two types of energy transfer is considering what happens if a thermal insulator is placed between the system and the surroundings (Fenn, p.6) if the energy transfers as heat, then this will likely effect the energy transfer if there is no effect, then the transfer is likely work 7

8 Example: Consider a battery connected to a hotplate heating a water bath. (Fenn, p.7-8) If the system is defined as the battery alone and the hotplate and water bath are considered part of the surroundings, then the energy transfer is work surrounding the battery with a thermal insulator would not effect the heating of the water If the system is defined as the battery and hotplate and the water bath is part of the surroundings, then the energy transfer is heat placing a thermal insulator between the hotplate and the water bath would effect the heating of the water If the system is defined as the battery, hotplate and water bath, then no energy transfer takes place (neither work nor heat). There are however some exceptions to this method for distinguishing work and heat. e.g. Radiation a mirrored surface can reflect electromagnetic radiation, preventing any energy transfer, however this radiation can represent a heat (thermal radiation) or a work (radio waves) energy exchange 8

9 The First Law of Thermodynamics Energy is conserved Energy can be transferred between the system and surroundings and it can change form, but the total energy of the system plus surroundings remains constant if the only forms of energy exchanged are q and w if other forms of energy are involved then terms must be added to this equation for each form of energy for an isolated system U = 0 for any change in a system, U depends only on the initial and final state and not on the process by which the change occurred Sign Convention work done on the system by the surroundings + work done by the system on the surroundings - heat absorbed by the system from the surroundings + (endothermic) heat absorbed by the surroundings from the system - (exothermic) 9

10 State Variables and Functions state variables depend only on the state of the system and not how it arrived at that state e.g. P, V, T, n generally only a few state variables are required to completely describe a system - other variables can be determined from these few e.g. for a liquid if P, V, T and chemical composition are specified then density, surface tension, refractive index etc can be determined q, w, and C (among others) depend on the path taken and are not state variables state functions depend only on the state of the system and not how it arrived at that state e.g. U, H, S, G The enthalpy (H) of the system is defined as: 10

changes in state functions depend only on the initial and final states and not on the path taken from one state to another U (E) and is a state function, q and w are not in taking a system from an

11 changes in state functions depend only on the initial and final states and not on the path taken from one state to another U (E) and is a state function, q and w are not in taking a system from an initial state (A) to some new final state (B), a number of different paths are generally available. Although the internal energy change will be the same for all paths, the amounts of heat and work will generally be different for different paths. e.g. if glucose is combusted or metabolized, the same amount of energy is released if a system undergoes a long series of steps and returns to its initial state (a cyclic process), the change in U or any other state variable is zero 11

12 Intensive vs Extensive CHEM Kinetics intensive state properties are independent of the size of the system - they do not change when a system is divided e.g. T, P extensive state properties are proportional to the size of the system - they do change when a system is divided e.g. mass, V, n, U an extensive property can be converted to an intensive property by describing it per unit of material (m, n, or V) Extensive Intensive energy (J) -1 energy/mol (J mol ) -1 heat capacity (J K ) -1-1 molar heat capacity(j K mol ) mass (kg) -1 density (kg mol ) 3 and volume (m ) 12

13 Equations of State CHEM Kinetics link state variables, most frequently P, V and T Solids and Liquids solids and liquids don t change V much with P or T so a first approximation for an equation of state of a solid or liquid is V constant find density at one temperature and you can use that at any temperature to describe the system liquids and solids do change V slightly with P and T - Tinoco Fig 2.3 shows example for liquid water 13

$9 x 10-9 atm-1 is the isothermal compressibility of water at 293 K - fractional decrease in V of water for an increase in P of 1 atm the change in V with T at constant P (1 atm) is not linear, it$

14 CHEM Kinetics In both plots, the volume of water changes by ~3% over the temperature and pressure ranges shown the change in V with P at constant T (293 K) is linear 45.9 x 10-9 atm-1 is the isothermal compressibility of water at 293 K - fractional decrease in V of water for an increase in P of 1 atm the change in V with T at constant P (1 atm) is not linear, it has a minimum at 277 K 14

15 Gases V does change significantly with P and T first approximation is the ideal gas law PV = nrt this law assumes ideal gas behaviour gas molecules are perfectly elastic, hard spheres of negligible volume, with no attractive nor repulsive forces between them, moving in a completely chaotic manner with perfectly elastic collisions between molecules and with the walls of the container the ideal gas law applies to all gases at low pressures at higher pressures it is a good approximation for most gases, accurate to within ±10% at room temperature and atmospheric pressure other, more accurate equations of state for gases attempt to account for non-ideal behaviour and contain parameters relating to individual gases the van der Waals gas equation accounts for attractive forces between molecules (a) and the intrinsic volume of the molecules (b) 15

16 in this case, changing P from 0 to 1000 atm causes a 1000x decrease in V (compared to 3% decrease for liquid water) 16

17 Mixtures for mixtures, the amount (m or n) of each component must be specified for gases the situation is particularly simple as the ideal gas law applies to each component individually and the total pressure is equal to the sum of the partial pressures of each component liquids and solids are more complicated e.g. the volume of a mixture of liquids is not necessarily equal to the sum of the volumes of the components l 17

18 Paths CHEM Kinetics changes in state functions are independent of the path taken so it is often convenient to pick a path for which the energy change(s) are easy to calculate a change from P 1, V 1, T 1 to P 2, V 2, T 2 might be easiest to calculate considering a series of path steps where one state variable is held constant in each step first P, V, T to P, V, T then to P, V, T etc , there are several types of named paths: Constant P Isobaric P sys = 0 Constant T Isothermal T sys = 0 Constant V Isochoric V = 0 No heat transferred Adiabatic q = 0 final state = initial state Cyclic no change in any state variable or function in biological systems, the paths of most interest will usually be isobaric and isothermal rather than isochoric 18

19 Dependence of U and H on P, V and T Liquids and Solids considering changes in q and w for the types of paths discussed on the previous page: Isobaric q and C are not state variables, so C P must be used to calculate q P V at constant P will be negligible, so no work is done 19

20 Isochoric here, C must be used to calculate q V V V is constant, so w = 0 Isothermal again, V will be negligible, and w = 0 So for any change of state for a pure solid or pure liquid, U can be calculated using a convenient path and then summing the appropriate q and w values. e.g. a isobaric path might be combined with an isothermal path, or an isochoric path might be combined with an isothermal path 20

21 For solids and liquids, the volume does not change appreciably with pressure so C C and U U P V P V The enthalpy change for the same processes can also be calculated: H = U + (PV) but any volume change with a change in pressure or temperature in a solid or liquid is negligible and (PV) 0, thus for processes involving constant pressure or temperature H = U 21

22 Tinoco Example 2.9 Calculate U and H in joules for heating 1 mole of liquid water from 0 C and 1 atm to 100 C and 10 atm. The volume for the water is essentially independent of pressure; it can be calculated from the average density of -1 water, 0.98 g ml. The overall reaction is: H O (l) (0 C, 1 atm) H O (l) (100 C, 10 atm) 2 2 We can do this in two steps, one isobaric and one isothermal: H2O (l) (0 C, 1 atm) H2O (l) (100 C, 1 atm) H O (l) (100 C, 1 atm) H O (l) (100 C, 10 atm) 2 2 U = U(isobaric) + U(isothermal) U = q + w + q + w P P T T For a liquid, q T and w T are zero and since there is little change in volume, w is also zero. P 22

23 Because there is a change in pressure, H U. 23

24 Gases A similar analysis can be done to determine the energy changes for gases undergoing changes in P, V and T. The major difference is that changes in V and thus w are significant for gases. Isobaric w is not zero P Isochoric no change 24

25 Isothermal w is not zero, but for a reversible path where P = P T ex By comparing two different paths for the same overall process, one isothermal + isobaric, the other isothermal + isochoric: For any change in P, V and T for gases: Relationships between q, U and H For any system undergoing an isochoric process: q = U V For any system undergoing an isobaric process: q = H P 25

26 Phase Changes CHEM Kinetics Phase Change gas liquid or solid solid liquid liquid solid liquid gas solid gas Name Condensation Fusion, melting Freezing Vaporization Sublimation We are interested in phase changes at constant T and P. w P = -Pex V q P = H U = q + w = H - P V P P ex H values for phase changes for many substances have been tabulated For phase changes involving only solids and liquids, V is often negligible and the work term can be ignored. For any phase change involving the gas phase, the volume of the other phase (solid or liquid) is negligible compared to that of the gas and V = V. gas 26

27 If H or U are known at one temperature and need to be calculated at another, appropriate paths can be chosen. Example: We know H for the vaporization of water at 100 C and need to calculate it at the temperature of human skin 35 C. (See reaction scheme Tinoco p. 45) Follow the path: 1 H2O (l) (35 C, 1 atm) H2O (l) (100 C, 1 atm) 2 H2O (l) (100 C, 1 atm) H2O (g) (100 C, 1 atm) 3 H2O (g) (100 C, 1 atm) H2O (g) (35 C, 1 atm) H O (l) (35 C, 1 atm) H O (g) (35 C, 1 atm) 2 2 Step 1 is an isobaric heating of liquid water H = q = C (l) T 1 P P 1 Step 2 is the vaporization of water at 100 C which we can look up in our standard tables Step 3 is the isobaric cooling of water vapour H = q = C (g) T Note that T = - T 2 P P

28 Thus for the overall reaction H(35 C) = H(100 C) + C P(l) T 1 + C P(g) T2 = H(100 C) + (C (l) T - C (g)) T or more generalized: P 1 P This equation does not apply only to phase changes. It can be used to determine the enthalpy change for any reaction at another temperature. 28

29 Chemical Reactions CHEM Kinetics The energy derived from chemical reactions is of great importance in biological systems. Enthalpy ( H) Changes For a chemical reaction: n A + n B n C + n D the enthalpy change is: A B C D If a reaction gives off heat, an exothermic reaction: H products < H reagents and H is negative. If a reaction absorbs heat, an endothermic reaction: H products > H reagents and H is positive. The enthalpy changes for many reactions have been tabulated and these values can be used to determine enthalpy changes for other reactions by addition and subtraction - Hess Law. 29

30 Example Tinoco p. 48 CHEM Kinetics Given the following enthalpies of reaction at 25 C, calculate the heat of reaction for the oxidation of aqueous glycine to aqueous urea. Combustion (oxidation) of two moles of solid glycine kj mol -1 Hydrolysis of solid urea kj mol -1 Dissolution of solid glycine kj mol -1 Dissolution of solid urea kj mol -1 The corresponding reactions are: 1) 2NH2CH2COOH (s) + 3O 2 (g, 1 atm) kj mol -1 4CO (g, 1 atm) + 2H O (l) + 2NH (g, 1 atm) ) H2NCONH 2 (s) + H2O (l) kj mol -1 CO (g, 1 atm) + 2NH (g, 1 atm) 2 3 3) NH2CH2COOH (s) + H2O (l) kj mol -1 NH CH COOH (aq) 2 2 4) H2NCONH 2 (s) + H2O (l) kj mol -1 H NCONH (aq)

31 The overall reaction we are looking for is: 2NH2CH2COOH (aq) + 3O 2 (g, 1 atm) H NCONH (aq) + 3CO (g, 1 atm) + 3H O (l) So we must add, subtract and multiply the reactions we have been given to achieve our overall reaction. We need to reverse reaction 3 (changing the sign of H) and multiply it by 2 and leave reaction 4 as is to get the aqueous glycine and urea in the correct places -2x3) 2NH2CH2COOH (aq) kj mol -1 2NH CH COOH (s) + H O (l) ) H2NCONH 2 (s) + H2O (l) kj mol -1 H NCONH (aq) 2 2 we want the solid forms of glycine and urea to cancel, so reaction 1 stays as is and reaction 2 needs to be reversed. 1) 2NH2CH2COOH (s) + 3O 2 (g, 1 atm) kj mol -1 4CO (g, 1 atm) + 2H O (l) + 2NH (g, 1 atm) ) CO 2 (g, 1 atm) + 2NH 3 (g, 1 atm) kj mol -1 H NCONH (s) + H O (l)

32 2x-3) 2NH2CH2COOH (aq) kj mol -1 2NH CH COOH (s) + H O (l) ) H2NCONH 2 (s) + H2O (l) kj mol -1 H NCONH (aq) 2 2 1) 2NH2CH2COOH (s) + 3O 2 (g, 1 atm) kj mol -1 34CO (g, 1 atm) + 2H O (l) + 2NH (g, 1 atm) ) CO 2 (g, 1 atm) + 2NH 3 (g, 1 atm) kj mol -1 H NCONH (s) + H O (l) NH2CH2COOH (aq) + 3O 2 (g, 1 atm) H NCONH (aq) + 3CO (g, 1 atm) + 3H O (l) H = kj mol -1-1 H = kj mol 32

33 Energy ( U) Changes At constant pressure: CHEM Kinetics U = H + (PV) if only solids and liquids are involved, V 0 and U = H If gases are involved, then only the V of the gases is important and (PV) = nrt U = H + nrt where n = n(gaseous products) - n(gaseous reagents) 33

34 Standard Enthalpies of Formation A convention adopted for enthalpies of reaction is that the absolute enthalpy of all elements in their most stable state at 1 atm pressure is zero. These are called standard states and are designated by superscript zero. i.e. H The standard enthalpy of a compound is defined as the enthalpy of formation of 1 mole of the compound at 1 atm pressure from the elements in their standard states. This is the standard enthalpy of formation H H f values for thousands of compounds have been determined and tabulated. Note that standard state specifies 1 atm pressure but not a specific temperature, though most H values are determined at 25 C. f The H of a reaction can be determined using the tables of H f using H = H (products) - H (reagents) f f f 34

35 Bond Energies Bond dissociation energies can be used to approximate reaction enthalpies by determining a reaction path that involves break and forming individual bonds. Bond dissociation energies are the energies required to break bonds: A-B (g) A (g) + B (g) Note that these reaction always take place in the gas phase at 1 atm pressure and 25 C. To determine a change in enthalpy for a reaction, the difference between the total bond energies of the products and reactants must be determined H = BE(reagents) - BE(products) Notice that unlike similar equations already discussed, this is reagents - products not products - reagents. This is due to bond energies being defined as the energy to break bonds not the energy to form them. 35

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