Thermodynamics

Introduction

Thermodynamics is a fundamental branch of chemistry that deals with the study of energy changes, especially heat and work, during physical and chemical processes. It provides a quantitative description of energy transformations and helps us understand how energy is conserved and transferred. The study of thermodynamics is essential because all chemical reactions and physical changes involve energy changes. For example, when a fuel like methane burns, chemical energy stored in the molecules is released as heat and can be used to do mechanical work or generate electricity. Thermodynamics allows us to analyze these energy changes systematically and predict the direction of chemical reactions and physical processes. The laws of thermodynamics govern these energy changes and provide a framework to calculate quantities such as internal energy, enthalpy, entropy, and free energy. This chapter introduces the basic concepts and laws of thermodynamics, focusing on their application to chemical systems. **Table on page 1 (1×2)** | It is the only physical theory of universal content concernin which I am convinced that, within the framework of th applicability of its basic concepts, it will never be overthrown After studying this Unit, you will be Albert Einstei able to • explain the terms : system and surroundings; • discriminate between close, open and isolated systems; Chemical energy stored by molecules can be released a • explain internal energy, work and heat during chemical reactions when a fuel like methane heat; cooking gas or coal burns in air. The chemical energy ma • state first law of thermodynamics and express it mathematically; also be used to do mechanical work when a fuel burn • calculate energy changes as in an engine or to provide electrical energy through work and heat contributions in galvanic cell like dry cell. Thus, various forms of energ chemical systems; are interrelated and under certain conditions, these ma • explain state functions: U, H. be transformed from one form into another. The stud • correlate ∆U and ∆H; of these energy transformations forms the subject matte • measure experimentally ∆U and of thermodynamics. The laws of thermodynamics dea ∆H; with energy changes of macroscopic systems involvin • define standard states for ∆H; a large number of molecules rather than microscopi • calculate enthalpy changes for various types of reactions; systems containing a few molecules. Thermodynamics i • state and apply Hess’s law of not concerned about how and at what rate these energ constant heat summation; transformations are carried out, but is based on initial an • differentiate between extensive final states of a system undergoing the change. Laws o and intensive properties; thermodynamics apply only when a system is in equilibrium • define spontaneous and non- or moves from one equilibrium state to another equilibrium spontaneous processes; state. Macroscopic properties like pressure and temperatur • e x p l a i n e n t r o p y a s a do not change with time for a system in equilibrium state thermodynamic state function and apply it for spontaneity; In this unit, we would like to answer some of the importan • explain Gibbs energy change (∆G); questions through thermodynamics, like: and How do we determine the energy changes involved in | | | --- | --- | | | g e . n s , y s a y y y r l g c s y d f e . t a | **Table on page 9 (1×3)** | | 1 mol of water is vapourised at 1 bar pressure and 100°C. Solution (i) The change H O (l) → H O (g) 2 2 ∆H = ∆U + ∆ngRT | Fig. 5.6(a) A gas at volume V and temperature T | | --- | --- | --- | | or ∆U = ∆H – ∆n RT, substituting the g values, we get ∆U = 41.00 kJ mol–1 – 1 × 8.3 J mol–1 K–1 × 373 K = 41.00 kJ mol-1 – 3.096 kJ mol-1 = 37.904 kJ mol–1 Fig. 5.6 (b) Partition, each part having half th volume of the gas (b) Extensive and Intensive Properties (c) Heat Capacity In thermodynamics, a distinction is made In this sub-section, let us see how to measur between extensive properties and intensive heat transferred to a system. This hea appears as a rise in temperature of the system properties. An extensive property is a in case of heat absorbed by the system. property whose value depends on the quantity or size of matter present in the system. For The increase of temperature is proportiona example, mass, volume, internal energy, to the heat transferred enthalpy, heat capacity, etc. are extensive q  coeff T properties. The magnitude of the coefficient depend Those properties which do not depend on the size, composition and nature of th on the quantity or size of matter present system. We can also write it as q = C ∆T are known as intensive properties. For The coefficient, C is called the hea example temperature, density, pressure etc. capacity. are intensive properties. A molar property, Thus, we can measure the heat supplie χ , is the value of an extensive property χ of m by monitoring the temperature rise, provide the system for 1 mol of the substance. If n is  we know the heat capacity.   the amount of matter, m n is independent When C is large, a given amount of hea of the amount of matter. Other examples are results in only a small temperature rise. Wate molar volume, V and molar heat capacity, m has a large heat capacity i.e., a lot of energ C . Let us understand the distinction m is needed to raise its temperature. between extensive and intensive properties by C is directly proportional to amount o considering a gas enclosed in a container of substance. The molar heat capacity of volume V and at temperature T [Fig. 5.6(a)]. L ise t au ls em d,a k ee ca p pa ar rt ti ti [o Fn .s u 5c .6h (t bh )]a t ov wol u hm ae substance, C m=  C n  , is the heat capacit h v a h ig n s V for one mole of the substance and i one half of the original volume, , but the | or ∆U = ∆H – ∆n RT, substituting the g values, we get ∆U = 41.00 kJ mol–1 – 1 × 8.3 J mol–1 K–1 × 373 K = 41.00 kJ mol-1 – 3.096 kJ mol-1 = 37.904 kJ mol–1 | | **Table on page 18 (1×3)** | | 6C graphite6O g6CO g; 2 2 ∆ H  = – 2361 kJ mol–1 f 3 3H g O g3H O1; 2 2 2 2 ∆ H  = – 857.49 kJ mol–1 f | In this case, the enthalpy of atomization i same as the enthalpy of sublimation. (c) Bond Enthalpy (symbol: ∆ H ) bond Chemical reactions involve the breaking an making of chemical bonds. Energy is require to break a bond and energy is released when bond is formed. It is possible to relate heat o | | --- | --- | --- | | Summing up the above two equations : reaction to changes in energy associated wit 15 6C graphite3H g O g6CO g breaking and making of chemical bonds. Wit 2 2 2 2 reference to the enthalpy changes associate 3H O l; with chemical bonds, two different terms ar 2 used in thermodynamics. ∆ H  = – 3218.49 kJ mol–1... (v) f (i) Bond dissociation enthalpy Reversing equation (ii); (ii) Mean bond enthalpy 15 6CO g3H O lC H l O ; Let us discuss these terms with referenc 2 2 6 6 2 2 to diatomic and polyatomic molecules. ∆ H  = – 3267.0 kJ mol–1... (vi) Diatomic Molecules: Consider the followin f Adding equations (v) and (vi), we get process in which the bonds in one mole o dihydrogen gas (H ) are broken: 6C graphite3H gC H l; 2 2 6 6 H (g) → 2H(g); ∆ H  = 435.0 kJ mol–1 2 H–H ∆ H  = – 48.51 kJ mol–1... (iv) f The enthalpy change involved in this proces is the bond dissociation enthalpy of H– bond. The bond dissociation enthalpy is th (b) Enthalpy of Atomization change in enthalpy when one mole of covalen (symbol: ∆ H ) bonds of a gaseous covalent compound i a Consider the following example of atomization broken to form products in the gas phase. of dihydrogen Note that it is the same as the enthalpy o H (g) → 2H(g); ∆ H  = 435.0 kJ mol–1 atomization of dihydrogen. This is true for a 2 a diatomic molecules. For example: You can see that H atoms are formed by breaking H–H bonds in dihydrogen. The Cl (g) → 2Cl(g); ∆ H  = 242 kJ mol–1 2 Cl–Cl enthalpy change in this process is known O (g) → 2O(g); ∆ H = 428 kJ mol–1 as enthalpy of atomization, ∆ H . It is the 2 O=O a enthalpy change on breaking one mole of In the case of polyatomic molecules, bon bonds completely to obtain atoms in the gas dissociation enthalpy is different for differen bonds within the same molecule. phase. Polyatomic Molecules: Let us now conside In case of diatomic molecules, like a polyatomic molecule like methane, CH dihydrogen (given above), the enthalpy of The overall thermochemical equation for it atomization is also the bond dissociation atomization reaction is given below: enthalpy. The other examples of enthalpy of atomization can be CH (g)→C(g)+4H(g); 4 CH 4(g) → C(g) + 4H(g); ∆ aH  = 1665 kJ mol–1 ∆ H  = 1665 kJ mol a Note that the products are only atoms of C | Summing up the above two equations : 15 6C graphite3H g O g6CO g 2 2 2 2 3H O l; 2 ∆ H  = – 3218.49 kJ mol–1... (v) f Reversing equation (ii); 15 6CO g3H O lC H l O ; 2 2 6 6 2 2 ∆ H  = – 3267.0 kJ mol–1... (vi) f Adding equations (v) and (vi), we get 6C graphite3H gC H l; 2 6 6 ∆ H  = – 48.51 kJ mol–1... (iv) f | | **Table on page 19 (3×14)** | In such cases we use mean bond enthalpy and products in gas phase reactions as: of C – H bond. ∆ H   bond enthalpies For example in CH , ∆ H  is calculated as: r reactants 4 C–H ∆ C–HH = ¼ (∆ a H) = ¼ (1665 kJ mol–1) bond enthalpies products (5.17)* = 416 kJ mol–1 This relationship is particularly mor We find that mean C–H bond enthalpy useful when the required values of ∆ H ar f in methane is 416 kJ/mol. It has been not available. The net enthalpy change of found that mean C–H bond enthalpies differ reaction is the amount of energy require slightly from compound to compound, as to break all the bonds in the reactan in CH CH Cl, CH NO , etc., but it does not molecules minus the amount of energ 3 2 3 2 differ in a great deal*. Using Hess’s law, bond required to break all the bonds in the produc enthalpies can be calculated. Bond enthalpy molecules. Remember that this relationship i values of some single and multiple bonds are approximate and is valid when all substance Table 5.3(a) Some Mean Single Bond Enthalpies in kJ mol–1 at 298 K H C N O F Si P S Cl Br I 435.8 414 389 464 569 293 318 339 431 368 297 H 347 293 351 439 289 264 259 330 276 238 C 159 201 272 - 209 - 201 243 - N 138 184 368 351 - 205 - 201 O 155 540 490 327 255 197 - F 176 213 226 360 289 213 Si 213 230 331 272 213 P 213 251 213 - S 243 218 209 CI 192 180 Br 151 I Table 5.3(b) Some Mean Multiple Bond Enthalpies in kJ mol–1 at 298 K N = N 418 C = C 611 O = O 498 N ≡ N 946 C ≡ C 837 C = N 615 C = O 741 C ≡ N 891 C ≡ O 1070 * Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same. | | | | | | | | | | | | | * e e a d t y t s s | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | | H 435.8 | C 414 347 | N 389 293 159 | O 464 351 201 138 | F 569 439 272 184 155 | Si 293 289 - 368 540 176 | P 318 264 209 351 490 213 213 | S 339 259 - - 327 226 230 213 | Cl 431 330 201 205 255 360 331 251 243 | Br 368 276 243 - 197 289 272 213 218 192 | I 297 238 - 201 - 213 213 - 209 180 151 | H C N O F Si P S CI Br I | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | **Table on page 21 (4×3)** | g is endothermic. Therefore the solubility o = 2) and is equal to + 783 kJ mol–1. most salts in water increases with rise o Now we use the value of lattice enthalpy temperature. If the lattice enthalpy is ver to calculate enthalpy of solution from the high, the dissolution of the compound ma expression: not take place at all. Why do many fluoride tend to be less soluble than the correspondin ∆ H = ∆ H + ∆ H sol lattice hyd chlorides? Estimates of the magnitudes o For one mole of NaCl(s), enthalpy changes may be made by usin lattice enthalpy = + 788 kJ mol–1 tables of bond energies (enthalpies) and lattic and ∆ H = – 784 kJ mol–1( from the hyd energies (enthalpies). literature) ∆ H = + 788 kJ mol–1 – 784 kJ mol–1 (f) Enthalpy of Dilution sol = + 4 kJ mol–1 It is known that enthalpy of solution is th The dissolution of NaCl(s) is accompanied enthalpy change associated with the additio by very little heat change. of a specifei d amount of solute to the specifei (e) Enthalpy of Solution (symbol : ∆ H ) amount of solvent at a constant temperatur sol and pressure. This argument can be applie Enthalpy of solution of a substance is the to any solvent with slight modification enthalpy change when one mole of it dissolves Enthalpy change for dissolving one mole o in a specifei d amount of solvent. The enthalpy gaseous hydrogen chloride in 10 mol of wate of solution at infinite dilution is the enthalpy can be represented by the following equation change observed on dissolving the substance For convenience we will use the symbol aq in an infinite amount of solvent when the for water interactions between the ions (or solute molecules) are negligible. HCl(g) + 10 aq. → HCl.10 aq. When an ionic compound dissolves in a ∆H = –69.01 kJ / mo solvent, the ions leave their ordered positions Let us consider the following set o on the crystal lattice. These are now more free in enthalpy changes: solution. But solvation of these ions (hydration (S-1) HCl(g) + 25 aq. → HCl.25 aq. in case solvent is water) also occurs at the ∆H = –72.03 kJ / mo same time. This is shown diagrammatically, for an ionic compound, AB (s) (S-2) HCl(g) + 40 aq. → HCl.40 aq. ∆H = –72.79 kJ / mo (S-3) HCl(g) + ∞ aq. → HCl. ∞ aq. ∆H = –74.85 kJ / mo The values of ∆H show general dependenc of the enthalpy of solution on amount o solvent. As more and more solvent is used | | | | --- | --- | --- | | | | | | | | | | | | the enthalpy of solution approaches a limitin value, i.e, the value in infinitely dilut solution. For hydrochloric acid this value o ∆H is given above in equation (S-3). | | | | | **Table on page 23 (11×13)** | 158 | | | | | | | | | | | | tr | | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | | | | | | | | | chemIs | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Fig. 5.10 (a) Enthalpy diagram for exothermic reactions These reactions though endothermic, are spontaneous. The increase in enthalpy may be represented on an enthalpy diagram as shown in Fig. 5.10(b). Fig. 5.11 Diffusion of two gases respectively and separated by a movabl partition [Fig. 5.11 (a)]. When the partition i withdrawn [Fig. 5.11(b)], the gases begin t diffuse into each other and after a period o time, diffusion will be complete. Let us examine the process. Befor partition, if we were to pick up the ga molecules from left container, we would b Fig. 5.10 (b) Enthalpy diagram for endothermic sure that these will be molecules of gas reactions and similarly if we were to pick up the ga molecules from right container, we woul Therefore, it becomes obvious that while be sure that these will be molecules of ga decrease in enthalpy may be a contributory B. But, if we were to pick up molecules from factor for spontaneity, but it is not true for container when partition is removed, we ar all cases. not sure whether the molecules picked are o (b) Entropy and Spontaneity gas A or gas B. We say that the system ha Then, what drives the spontaneous process become less predictable or more chaotic. in a given direction ? Let us examine such a We may now formulate another postulate case in which ∆H = 0 i.e., there is no change in in an isolated system, there is always enthalpy, but still the process is spontaneous. tendency for the systems’ energy to becom Let us consider diffusion of two gases more disordered or chaotic and this could b into each other in a closed container which a criterion for spontaneous change ! is isolated from the surroundings as shown At this point, we introduce anothe | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |

• Thermodynamics studies energy changes during physical and chemical processes.
• It focuses on heat and work as forms of energy transfer.
• Energy transformations obey fundamental laws, such as conservation of energy.
• Chemical reactions involve energy changes that can be quantified.
• Thermodynamics helps predict reaction spontaneity and equilibrium.
• It is essential for understanding engines, fuels, and biological processes.
• 📌 Thermodynamics: Study of energy changes and transformations.
• 📌 Heat (q): Energy transferred due to temperature difference.
• 📌 Work (w): Energy transferred when a force moves an object.

System and Surroundings

In thermodynamics, the universe is divided into two parts: the system and the surroundings. The system is the specific portion of the universe chosen for study, such as a chemical reaction mixture or a gas in a cylinder. Everything outside the system is called the surroundings. The system and surroundings together make up the universe. Systems can be classified into three types based on energy and matter exchange: open, closed, and isolated systems. An open system can exchange both matter and energy with its surroundings, for example, a boiling pot of water open to the air. A closed system can exchange energy but not matter, such as a sealed container that can be heated or cooled. An isolated system cannot exchange either energy or matter, like a perfectly insulated thermos flask. Understanding these distinctions is important because the laws of thermodynamics apply differently depending on the system type. The boundary between the system and surroundings can be real or imaginary and may be fixed or movable. For example, a piston-cylinder assembly has a movable boundary that allows volume change. The concept of the universe as system plus surroundings is fundamental and is expressed mathematically as: Universe = System + Surroundings.

• System is the part of the universe under study; surroundings are everything else.
• Universe = System + Surroundings (formula).
• Open system exchanges both matter and energy with surroundings.
• Closed system exchanges only energy, not matter.
• Isolated system exchanges neither energy nor matter.
• Boundaries can be fixed or movable, real or imaginary.
• 📌 System: Part of the universe chosen for study.
• 📌 Surroundings: Everything outside the system.
• 📌 Open system: Exchanges matter and energy.