Chemical Thermodynamics Chapter 1 (PDF)

**CHAPTER (1)** **Basic Concepts and First Law of Thermodynamics** T**he study of the flow of heat or any other form of energy into or out of a system, as it undergoes a physical or chemical transformation, is called** Thermodynamics. In studying and evaluating the flow of energy into or out of a...

**CHAPTER (1)** **Basic Concepts and First Law of Thermodynamics** T**he study of the flow of heat or any other form of energy into or out of a system, as it undergoes a physical or chemical transformation, is called** Thermodynamics. In studying and evaluating the flow of energy into or out of a system it will be useful to consider changes in certain properties of the system. These properties include temperatures, pressure, volume, and concentration of the system. Measuring the changes in these properties from the initial state to the final state, can provide information concerning changes in energy and related quantities such as heat and work. **The Three Empirical Laws.** The study of thermodynamics is based on three broad generalizations derived from well-established experimental results. These generalizations are known as **the first, second and Third law of thermodynamics.** These laws have stood the test of time and are independent of any theory of the atomic or molecular structure. The discussion of these laws will be the subject of our study. Application of Thermodynamics 1. Most of the important laws of Physical Chemistry including the Van\'t, Hoff law of lowering of vapour pressure, phase Rule and the Distribution Law, can be derived from the laws of thermodynamics. 2. It tells whether a particular physical or chemical change can occur under a given set of conditions of temperature, pressure and concentration. 3. It also helps in predicting how far a physical or chemical change can proceed until the equilibrium conditions are established. **Limitations of Thermodynamics** 1. Thermodynamics is applicable to **macroscopic systems** consisting of matter in bulk and not to **microscopic systems** of individual atoms or molecules. *It ignores the internal structure of atoms ad molecules.* 2. Thermodynamics does not bother about the **time factor.** That is, it does not tell anything regarding the rate of a physical change or a chemical reaction. It is concerned only with the initial and the final states of the system. **THERMODYNAMIC TERMS AND BASIC CONCEPTS** An important part of the study of thermodynamic is a few terms and definitions, which must be understood clearly. **SYSTEM, BOUNDARY, SURROUNDINGS** **A system is that part of the universe which is under thermodynamic study and the rest of the universe is surroundings.** The real or imaginary surface separating the system from the surroundings is called the **Boundary.** In experimental work, a specific amount of one or more substance constitutes the system. Thus 200 g of water contained in a beaker constitutes a thermodynamic system. The beaker and the air in contact, are the surroundings. Similarly, I mole of oxygen confined in a cylinder fitted with a piston, is a thermodynamic system. The cylinder, the piston and all other objects outside the cylinder, form the surroundings. Here the boundary between the system (oxygen) and the surroundings (cylinder and piston) is clearly defined. **HOMOGENEOUS AND HETEROGENEOUS SYSTEMS** **When a system is uniform throughout, it is called a** *Homogenous system.* Examples are: a pure single solid, liquid or gas, mixtures of gases, and true solution of a solid in a liquid. A homogeneous system is made of one phase only. A phase is defined as *a homogenous, physically distinct and mechanically separable position of a system.* **A heterogeneous system is one which consists of two or more phases.** In other words it is not uniform throughout. Examples of heterogeneous systems are: Ice in contact with water, ice in contact with vapour, etc. Here ice, water, and vapour constitute separate phases. **TYPES OF THERMODYNAMIC SYSTEMS** There are three types of thermodynamic systems, depending on the nature of the boundary. If the boundary is closed or sealed, no matter can pass through it. If the boundary is insulated, no energy (say heat) can pass through it. **(1) Isolated system.** When the boundary is both sealed and insulated, no interaction is possible with the surroundings. Therefore, **An isolated system is one that can transfer neither matter nor energy to and from its surroundings.** Let us consider a system 100 mL of water in contact with its vapour in a closed vessel which is insulated. Since the vessel is sealed, no water vapour (matter) can escape from it. Also, because the vessel is insulated, no heat (energy) can be exchanged with the surroundings. A substance, say boiling water, contained in a *thermos flask,* is another example of an isolated system. **(2) Closed system.** Here the boundary is sealed but not insulated. Therefore, **A closed system is one which cannot transfer matter but can transfer energy in the form of heat, work and radiation to and form its surroundings.** A specific quantity of hot water contained in a sealed tube, is an example of a closed system. While no water vapour can escape from this system, it can transfer heat through the walls of the tube to the surroundings. A gas contained in a cylinder fitted with a piston constitutes a closed system. As the piston is raised, the gas expands and transfers heat (energy) in the form of work to the surroundings. **(3) Open System is one which can transfer both energy and matter to and from its surroundings.** Hot water contained in a beaker placed on laboratory table is an open system. The water vapour (matter) and also heat (energy) is transferred to the surroundings through the imaginary boundary. Zinc granules reacting with dilute hydrochloric acid to produce hydrogen gas, in a beaker, is another example of open system. Hydrogen gas escapes and the heat of the reaction is transferred to the surroundings. **What are Adiabatic systems?** *Those systems, in which no thermal energy passes into or out of the system, are said to be* **Adiabatic systems.** INTENSIVE AND EXTENSIVE PROPERTIES The macroscopic or bulk properties of a system (volume, pressure, mass, etc) can be divided into two classes. a. b. **A property which does not depend on the quantity of matter present in the system is known as** *Intensive Property.* Some examples of intensive properties are *temperature, density,* and concentration. If the overall temperature of a glass of water (our system) is 20°C, then any drop of water in that glass has a temperature of 20°C. Similarly, if the concentration of salt, NaCl, in the glass of water is 0.1 mole/liter, then any drop of water from the glass also has a salt concentration of 0.1 mole/liter. **A property that does depend on the quantity of matter present in the system is called an *Extensive property.*** Some examples of extensive properties is volume, number of moles, enthalpy, entropy, pressure and Gibb\'s free energy. Some of these properties are unfamiliar to you but these will be defined and illustrated later. By definition, the **Extensive properties are Additive,** while intensive properties are not. Let us consider the system \"a glass of water\". If we double the mass of water, the volume is doubled and so are the number of moles and the internal energy of the system. **\ Table 1.1 Common properties of System** **Intensive properties** **Extensive Properties** -------------------------- -------------------------- ----------------- Viscosity Pressure, Mass Freezing point Volume Internal energy Enthalpy Refractive index Entropy **STATE OF A SYSTEM** **A thermodynamic system is said to be in a certain state when all its properties are fixed.** The fundamental properties which determine the state of a system are pressure (*P),* temperature (*T),* volume (*V),* mass and composition. Since a change in the magnitude of such properties alters the state of the system, these are referred to as **State variables** or **state functions** or **Thermodynamic parameters.** It also stands to reason that a change of system from the *initial state* to the *final state* (2^nd^ state) will be accompanied by change in the state variables. It is not necessary to state all the properties (state variables) to define a system completely. For a pure gas, the composition is fixed automatically, as it is cent per cent. The remaining state variables *P, V, T* are interrelated in the form of an algebraic relationship called the **Equation of State**. Thus for one mole of a pure gas, the equation of state is *PV = RT*. Where *R* is gas constant. If of the three state variables (*P, V, T*), *P* and *T* are specified, the value of third (*V*) is fixed automatically and can be calculated, from the equation of state. The variables (*P* and *T)* which must be necessarily specified to define the state of a system, are designated as **Independent state variables.** The remaining state variable (*V)* which depends on the value of *p* and *T*, is called **Dependent state variable.** An important characteristic of a state variable (or state function) is that when the state of a system is altered, the change in the variable depends on the initial and final states of the system. For example, if we heat a sample of water from 0°C to 25°C, the change in temperature is equal to difference between the initial and final temperatures. *The way in which the temperature change is brought about has no effect on the result.* **EQUILIBRIUM AND NON-EQUILIBRIUM STATES** **A system in which the state variables have constant values throughout the system, is said to be in a state of thermodynamic equilibrium.** *Example.* Suppose we have a gas confined in a cylinder that has a frictionless piston. If the piston is stationary, the state of the gas can be specified by giving the values of pressure and volume. The system is then in a *state of equilibrium.* **A system in which the state variables have different values in different parts of the system, is said to be in a non-equilibrium state.** If the gas contained in a cylinder as stated above is compressed very rapidly by moving down the piston, it passes through states in, which pressure and temperature cannot be specified, since these properties vary throughout the gas. The gas near the piston is compressed and heated and that at the far end of the cylinder is not. The gas then would be said to be in non−equilibrium state. Thermodynamics is concerned only with equilibrium states. **The criteria for Equilibrium:** 1. *The temperature of the system must be uniform and must be the same as the temperature of the surroundings.* (**Thermal equilibrium).** 2. *The mechanical properties must be uniform through the system **(*****Mechanical equilibrium).** That is, no mechanical work is done by one part of the system on any other part of the system. 3. *The chemical composition of the system must be uniform, with no net chemical change (***chemical equilibrium).** If the system is heterogeneous, the state variables of each phase remain constant in each phase. **THERMODYNAMIC PROCESSES** **When a thermodynamic system change from one state to another, the operation is called a *process.*** These processes involve the change of conditions (temperature, pressure and volume). The various types of thermodynamic processes are: 1. **Isothermal process.** *Those processes, in which the temperature remains fixed, are termed* **Isothermal processes.** This is often achieved by placing the system in a *thermostat* (a constant temperature bath). For an isothermal process *dT* = 0 2. **Adiabatic Process.** *Those processes in which no heat can flow into or out of the system, are called* **Adiabatic Processes.** Adiabatic conditions can be approached by carrying the process in an insulated container such as thermos bottle. High vacuum and highly polished surfaces help to achieve thermal insulation. For an adiabatic process *dq* = 0 3. **Isobaric process.** *Those processes which take place at constant pressure are called* **Isobaric processes.** For example, heating of water to its boiling point, and its vaporisation take place at the same atmospheric pressure. These changes are, therefore, designated as isobaric processes and are said to take place isobarically. For all isobaric processes *dp* = 0. **(4) Isochoric Processes.** Those processes in which the volume remains constant are known as **Isochoric Processes.\ **The heating of a substance in a non-expanding chamber is an example of isochoric process. For isochoric process *dV* = 0 **(5) Cyclic processes** When a system in a given state goes through a number of different processes and finally returns to its initial state, the overall process is called a **Cyclic** or **Cycle process.** For a cyclic process *dE* = 0, *dH* = 0. **REVERSIBLE AND IRREVERSIBLE PROCESSES** **A thermodynamic reverse process is one that takes place infinitesimally slowly and its direction at any point can be reversed by an infinitesimal change in the state of the system.** In fact, a reversible process is considered to proceed from the initial state to the final state through an infinite series of infinitesimally small stages. At the initial, final and all intermediate stages, the system is in equilibrium state. This is so because an infinitesimal change in the state of the system at each intermediate step is negligible. **When a process goes from the initial to the final state in a single step and cannot be carried in the reverse order, it is said to be an irreversible process.** Here the system is in equilibrium state in the beginning and at the end, but not at points in between. **Example.** Consider a certain quantity of a gas contained in a cylinder having a weightless and frictionless piston. The expansion of the gas can be carried by two methods illustrated in Fig.1.5. Let the pressure applied to the piston be *P* and this is equal to the internal pressure of the gas. Since the external and internal pressures are exactly counterbalanced, the piston remains stationary and there is no change in volume of the gas. Now suppose the pressure on the piston is decreased by an infinitesimal amount *dp*. Thus the external pressure on the piston being *P − dP,* the piton moves up and the gas will expand by an infinitesimally small amount. The gas will therefore, be expanded infinitely slowly *i.e*. by a thermodynamically *reversible process*. At all stages in the expansion of the gas, *dp* being negligibly small, the gas is maintained in a state of equilibrium throughout, if at any point of the process, the pressure is increased by *dp*, the gas would contract reversibly. On the other hand, the expansion is *irreversible* (see Fig1.5b). If the pressure on the piston is decreased suddenly. It moves upward rapidly in a single operation. The gas is in equilibrium state in the initial and final stages only. The expansion of the gas, in this case, takes place in an *irreversible* manner. **Differences between Reversible and Irreversible Processes.** Theses are listed below, particularly with reference to the expansion of a gas as cited above. 1. *Since a reversible process takes place by infinitesimal small steps, the process would take infinite time to occur.* Such a process is idealized and is true in principle only. On the other hand, an irreversible process takes finite time. **Thus all processes which actually occur are irreversible.** 2. *A reversible process is unreal* as it assumes the presence of a frictionless and weightless piston*.* An irreversible process is real and can be performed actually. 3. *A reversible process is in equilibrium state at all stages of the operation, while an irreversible process is in equilibrium state only at the initial and final stages of the operation.* **NATURE OF HEAT AND WORK** When a change in the state of a system occurs, energy is transferred to or from the surroundings. This energy may be transferred as heat or mechanical work. We shall refer the term \"work\" for mechanical work which is defined as *force X distance.* **Units of work. In** CGS system, the unit of work is erg which is defined as *the work done when a resistance of I dyne is moved through a distance of I centimeter.* Since the erg is so small, a bigger unit, the **joule** (J) is now used. 1 joule = 10^7^ ergs or 1 erg = 10^-7^ J We often use **kilojoules** (KJ) for large quantities of work. 1 KJ = 1000J **Units of Heat.** The unit of heat, which was used for many years, is **calorie** (cal). A calorie is defined as *that quantity of heat required to raise the temperature of one gram of water by 1°C in the vicinity of 15°C*. Since heat and work are interrelated, SI unit of heat is the joule (J). 1 Joule = 0.2390 calories or 1 Calorie = 4.184 J 1 Kcal = 4.184 KJ **Sign Convention of Heat.** The symbol of heat is q. if the heat flows from the surroundings into the system to raise the energy of the system; it is taken to be positive, +*q*. if heat flows from the system into the surroundings, lowering the energy of the system, it is taken to be negative, −*q*. **Sign Convention of work.** The symbol of work is *w*. If work is done on a system by the surroundings and the energy of the system is thus increased, it is taken to be positive, +*w*. If work is done by the system on the surroundings and energy of the system is decreased, it is taken to be negative, − *w.* Summary of Sign Conventions Heat *flows into the system,* q is + *Work is done on the system, w is +* *Heat flows out of the system, q is* -- *Work is done by the system, w is* -- **PRESSURE-VOLUME WORK** In physics, mechanical work is defined as force multiplied by the distance through which the force acts. In elementary thermodynamics, the only type of work generally considered is the work done in expansion (or compression) of a gas. This is known as Pressure. Volume work or *PV* work of Expansion work. Consider a gas contained in a cylinder fitted with a frictionless piston. The pressure (force per unit area) of the gas, *P*, exerts a force on the piston. This can be balanced by applying an equal but opposite pressure from outside on the piston. Let it be designated as *P~ext~*. It *is important to remember that it is the external pressure, P~ext~, and not the internal pressure of the gas itself, which is used in evaluating work.* This is true whether it be expansion or contraction. If the gas expands at constant pressure, the piston would move say through a distance *l.* we know that *Work = force* × *distance* (by definition) or *W = f ×...........*(1) since pressure is force per unit area, *f* = *P~ext~* × *A*....................................(2) Where A is the cross-section area of the piston, From (1) and (2), we have *W* = *P~ext~* × *A* × *l* = *P~ext~* × ∆*V* where ∆*V* is the increase in volume of the gas. Since the system (gas) is doing work on the surroundings (piston), it bears negative sign. Thus *W* = − *P~ex~*~t~ × ∆*V*. Proceeding as above, the work done in compression of a gas can also be calculated. In that case the piston will move down and sign of the work will be positive. *W* = *P~ext~* × ∆*V*. As already stated, work may be expressed in dynes-centimeters, ergs, or joules. *PV* work can as well be expressed as the product of pressure and volume units. *e.g.*, in litre-atmospheres. It may be noted that **the work done by a system is not a state function.** This is true for the mechanical work of expansion. We shall show presently that the *work is related to the process carried out rather than to the internal and final states.* This will be evident from a consideration of the *reversible expansion* and an *irreversible process.* **Example.** Calculate the pressure-volume work done when a system containing a gas expands from 1.0 liter to 2.0 liters against a constant external pressure of 10 atmospheres. Express the answer in calories and joules. **Solution:** *W* = --*P~ex~*~t~ (*V*~2~ -- *V*~1~) = -- (10 atm) (2 − 1 *)* = -- 10 *l* atm = -- (10 *l* atm) = -- 242 cal but 1 calorie = 4.184 J *W* = 1012.528 J **ISOTHERMAL REVERSIBLE EXPANSION WORK OF AN IDEAL GAS** Consider an ideal gas confined in a cylinder with a frictionless piston. Suppose it expands reversibly from volume *V*~1~ to *V*~2~ at a constant temperature. The pressure of the gas is successively reduced from *P~1~* to *P*~2~. The reversible expansion of the gas takes place in a finite number of infinitesimally small intermediate steps. To start with, the external pressure, *P~ext~* is arranged equal to the internal pressure of the gas, *P~gas~*, and the piston remains stationary. If *P~ext~* is decreased by an infinitesimal amount *dp* the gas expands reversibly and the piston moves through a distance *dl*. since *dp* is so small, for all practical purposes *P~ext~* = *P~gas~* = *P* The work done by the gas in one infinitesimal step *(dw*), can be expressed as *dw* = *P × A × dl* (*A* = cross-sectional area of piston)\ = *P* × *dV* Where *dV* is the increase in volume. The total amount of work done by the isothermal reversible expansion of the ideal gas from *V*~1~ to *V*~2~ is, therefore, by the ideal gas equation which integrates to give since **Note** *Isothermal compression work of an ideal gas may be derived similarly and it has exactly the same value with the sign changed.* Here the pressure on the piston, *P~ext~*, is increased by *dp* which reduces the volume of the gas. **ISOTHERMAL IRREVERSIBLE EXPANSION WORK OF AN IDEAL GAS** Suppose we have an ideal gas contained in a cylinder with a piston. This time the process of expansion of the gas is performed irreversibly i.e., by instantaneously dropping the external pressure, *P~ext~*, to the final pressure *P*~2~. The work done by the system is now against the pressure *P*~2~ throughout the whole expansion and is given by the expression. = *P~2~* (*V~2~ -- V~1~*) = *P~2~* dV **MAXIMUM WORK DONE IN REVERSIBLE EXPANSION** The isothermal expansion of an ideal gas may be carried either by the *reversible process or irreversible* as stated above. The reversible expansion is shown in Fig 1.8 (a) in which the pressure is falling as the volume increases. The reversible work done by the gas is given by the expression which is represented by the shaded area. If the expansion is performed irreversibly by suddenly reducing the external pressure to the final pressure *P~2~*, the irreversible work is given by Which is shown by the shaded area in Fig 1.8 (b). In both the processes, the state of the system has changed from *A* to *B* but the work done is much less in the irreversible expansion than in the reversible expansion. **Thus mechanical work is not a state function** *as it depends on the path by which the process is performed rather than on the initial and final states.* It is also important to note that **the work done in the reversible expansion of a gas is the maximum work that can be done by a system (gas) in expansion between the same initial (A) and final state (B).** This is proved as follows: We know that the work always depends on the external pressure, *P~ext~*; the larger the *P~ext~*, the more work done by the gas. But the *P~ext~* on the gas cannot be more than the pressure of the gas, *P~ext~*, or compassion will take place. Thus the largest value *P~ext~* can have without a compression taking place is equal to *P~gas~*. But an expansion that occurs under these conditions is the reversible expansion. *Thus, maximum work is done in the reversible expansion of a gas.* **Example (1).** One mole of an ideal gas at 25°C is allowed to expand reversibly at constant temperature from a volume of 10 litres to 20 litres. Calculate the work done by the gas in joules and calories. **Solution (1)** **Notes.** **(1)** In *x* = 2.303 log *x*. **(2)** The unite of *R* will determine the units of *w* in this expression. **(3)** The temperature must be expressed in degrees Kelvin. **Example** **(2)**. Find the work done when one mole of the gas is expanded reversibly and isothermally from 5 atm to 1 atm at 25C. **Solution (2):** **Example (3)**. Calculate the work done when 1 mole of an ideal gas at 25°C and 5 atm is allowed to expand irreversibly against the constant outside pressure of 1 atm until the internal pressure has been lowered to 1 atm. **Solution (3):** The final volume is five times larger i.e., *V*~2~ = 24.5 litres Thus *w* = -- *P* ∆*V* = -- *P* (*V*~2~ -- *V*~1~) = 1 atm (24.5 -- 4.9) = -- 19.6 *l*-atm Converted to joules, it is - 81.9 J. **Note.** It is evident from examples (2) and (3) that *W~irr~* is less than *W~rev~,* even though the initial and the final states of the ideal gas were the same. **INTERNAL ENERGY** A thermodynamic system, containing some quantity of matter, has within itself a definite quantity of energy. This energy includes not only the translation kinetic energy of the molecules but also other molecular energies such as rotational, vibrational energies. The kinetic and potential energy of the nuclei and electrons within the individual molecules also contribute to the energy of the system. **The total of all the possible kinds of energy of a system is called its** *Internal Energy.* *The internal energy of a* system, like temperature, pressure, volume, etc., is determined by the state of a system and is independent of the path by which it is obtained. Hence internal energy of a system is a **state Function.** For example, we consider the heating of one mole of liquid water from 0° to 100°C. the change in energy is 1.8 kcal, and is the same regardless of the form in which this energy is transferred to the water by heating, by performing work, by electrical means or in any other way. Since the value of internal energy of a system depends on the mass of the matter contained in a system, it is classed as an **Extensive Property.** **Symbol Representation of internal Energy and Sign Conventions.** The internal energy of a system is represented by the symbol *E* (Some books uses the symbol *U*). It is neither possible nor necessary to calculate the absolute value of internal energy of a system. In thermodynamics we are concerned only with the energy changes when a system changes from one state to another. If ∆*E* be the difference of energy of the initial state (*E~in~*) and the final state *(E~f~)*,we can write Δ*E* is + ve if *E~f~* is greater than *E~in~* and -- ve if *E~f~* is less than *E~in~*. A system may transfer energy to or from the surroundings as heat or as work, or both. **Units of Internal Energy.** The SI unit for internal energy of a system is the joule (J). Another unit of energy, which is not an SI unit is the calorie, 1 cal = 4.184 J. **FIRST LAW OF THERMODYNAMICS** The First law of thermodynamics is, in fact, an application of the broad principle known as *the law of conservation of Energy* to the thermodynamic systems. It states that: **The total energy of an isolated system remains constant, though it may change from one form to another.** When a system is changed from state *A* to state *B*, it undergoes a change in the internal energy from *E~A~* to *E~B~*, thus, we can write Δ *E* = *E~B~* -- *E~A~* This energy change is brought about by the evolution or absorption of heat and/or by work being done by the system. Because the total energy of the system must remain constant, we can write the **mathematical statement of the first law** as: Δ*E = q -- w*...........................................................(1) where *q* = the amount of heat supplied to the system and *w* = work done by the system. Thus first law may also by stated as: **The net energy change of a closed system is equal to the heat transferred to the system minus the work done by the system.** To illustrate the mathematical statement of the first law, let us consider the system \"expanding hot gas\" (see Fig 1.9). The gas expands against an applied constant pressure by volume Δ*V*. The total mechanical work done is given by the relation From (1) and(2), we can restate Δ *E = q − P* × Δ*V* **Example (1).** Find Δ*E*, *q* and w; if 2 moles of hydrogen at 3 atmos pressure expand isothermally at 50°C and reversibly to a pressure of 1 atmos. **Solution (1):** Since the operation is isothermal and the gas is ideal Δ*E* = 0 From the first law Δ*E* = *q* -- *w* **∴** *q* -- *w* = 0 when Δ*E* = 0 or *q* = *w* for a reversible process *w = nRT In (P*~1~*/ P*~2~*)* = − 2 × 1.987 × 323 × 2.303 × log 3 = − 1410 calories since *q* = *w*, *q* = 1410 calories **Example (2).** 1 g of water at 273°K is converted into steam at the same temperature. The volume of water becomes 1671 ml on boiling. Calculate the change in the internal energy of the system if the heat of vaporisation is 540 cal/g. **Solution (2):** As the vaporisation takes place against a constant pressure of 1 atmosphere, work done for an irreversible process, *w*, is *W = P (V~2~ -- V~1~)* = *nRT* = 41 cal/g now = 41 cal/g. since Δ *E* = *q -- w* (First law) = 540 -- 41 ∴ Δ*E* = 499 cal/g **Example (3).** A gas contained in a cylinder fitted with a frictionless piston expands against a constant external pressure of 1 atm. From a value of 5 litres to a volume of 10 litres. In doing so, it absorbs 400 J thermal energy from its surroundings. Determine Δ*E* for the process. **Solution (3):** Δ*E = q -- w*............................................... \....(1) here *q* = 100 J *W* = − P (*V*~2~ -- *V*~1~) = − (1) (10 − 5) = − 5.atm. = − 506 J substituting values in (1) Δ*E* = 400 J -- 506 J = −106 J **ENTHALPY OF A SYSTEM** In a process carried at constant volume (say in a sealed tube), the heat content of a system is the same as internal energy *(E)*, as no *PV* work is done. But in a constant-pressure process, the system (a gas) also expends energy in doing *PV* work. Therefore, *the total heat content of a system at constant pressure is equivalent to the internal energy E plus the PV energy.* This is called the **Enthalpy** (Greek *en =* in ; *thalpos = heat)* of the system and is represented by the symbol ***H***. Thus enthalpy is defined by the equation. **Enthalpy a function of state.** In the equation (1) above, *E, P, V* are all state functions. Thus *H*, the value of which, depends on the values of *E, P, V* must also be a function of state. Hence its value is independent of the path by which the state of the system is changed. **Change in Enthalpy.** If Δ*H* be the difference of enthalpy of a system in the final state (H~2~) and that in the initial state (*H*~1~), Substituting the values of *H*~2~ and *H*~1~, as from (1) and (2), we have Δ*H* = (*E*~2~ *+ P*~2~*V*~2~) *--* (*E*~1~ *+ P*~1~*V*~1~) = (*E*~2~ *-- E*~1~) *+* (*P~2~V~2~ -- P~1~V~1~*) *=* Δ*E +* Δ *PV* if *P* is constant while the gas is expanding, we can write Δ*H =* Δ*E + P*Δ*V* or Δ*H =* Δ *E + w* *(w* = work)......................(3) According to the first law. Δ*E* = *q -- w*.......................(4) where *q* = heat transferred. from equations (3) and (4) or Δ*H* = *q* when change in state occurs at constant pressure. This relationship is usually written as Δ*H* = *q~p~* Where subscript *p* mean constant pressure. Thus Δ*H* can be measured by measuring the heat of a process occurring at constant pressure. **Units and Sign Conventions.** Since Δ*H = H*~2~ *-- H*~1~ Δ*H* is positive if *H*~2~ \> *H*~1~ and the process or reaction will be endothermic. Δ*H* is negative if *H*~1~ \> *H*~2~ and the reaction will be exothermic. In case of a chemical reaction carried in the laboratory in an open vessel, Δ*H* = *H* products − *H* reaction = *q~p~* The heat of reaction at one atmosphere pressure is usually shown along with the equation. Thus, The quantity of heat 68.32 kcal on the right hand represents −Δ*H* of the reaction The units of Δ*H* are *kilocalories* (kcal) or kilojoules (kJ). **RELATION BETWEEN ΔH AND ΔE** Caloriefic values of many gaseous fuels are determined in constant volume calorimeters. These values are, therefore, given by the expression. *q~v~* = Δ*E* When any fuel is burnt in the open atmosphere, additional energy of expansion, positive or negative, against the atmosphere is also involved. The value of *q* thus actually realised, *i.e. q~p~* = Δ*H*, may be different from the equation If gases are involved in a reaction, they account for most of the volume change as the volumes of solids and liquids are negligibly small in comparison. Suppose we have *n*~1~ moles of gases before reaction, and *n*~2~ moles of gases after it. Assuming ideal gas behaviour, we have *PV*~2~ = *n*~2~*RT PV*~1~ = *n*~1~*RT* *P*(*V*~2~ −*V*~1~) = (*n*~2~*- n*~1~) *RT P*Δ*V* = Δ*nRT* or *P*Δ*V* = Δ*nRT* substituting in equation (1) we have, Δ*H* = Δ*E +* Δ*nRT* **Example.** *For the reaction* Δ*E =* −14.2 Kcal/mol at 25°C. Calculate Δ*H* for the reaction. **Solution:** Δ*H* = Δ*E +* Δ*nRT* Δ*n* = *n*~2~ *-- n*~1~ now *n*~2~ = 1 + 1 = 2 *n*~1~ = 1 *n*~2~ *-- n*~1~ = 2 − 1 = 1 **∴** Δ*H* = Δ*E +* 1 × 1.987 × 298/1000 = - 14.2+ 0.6 = 13.6 kcal /mole **MOLAR HEAT CAPACITIES** By heat capacity of a system we mean the capacity to absorb heat and store energy. As the system absorbs heat, it goes into the kinetic motion of the atoms and molecules contained in the system. This increased kinetic energy raises the temperature of the system. If *q* calories is the heat absorbed by mass *m* and the temperature rises from *T*~1~ to *T*~2~, the heat capacity (*C*) is given by the expression Thus heat capacity of a system is the heat absorbed by unit mass in raising the temperature by one degree (*k* or °C) When mass considered is 1 mole, the expression (1) can be written as where *c* is denoted as *Molar heat capacity.* **The molar heat capacity of a system is defined as the amount of heat required to raise the temperature of one mole of the substance (system) by 1k.** Since the heat capacity (*C*) varies with temperature, its true value will be given as where *dq* is a small quantity of heat absorbed by the system, producing a small temperature rise *dT*. Thus the *molar heat capacity* may be defined as **the ratio of the amount of heat absorbed to the rise in temperature.** **Units of Heat Capacity.** The usual units of the molar heat capacity are calories per degree per mole (cal K^-1^ mol^-1^), or joules per degree per mole (JK^-1^ mol^-1^), the latter being the SI unit. *Since heat is not a state function, neither is heat capacity.* It is, therefore, necessary to specify the process by which the temperature in raised by one degree. The two important types of molar heat capacities are those (1) at constant volume; and (2) at constant pressure. **Molar heat capacity at constant volume.** This molar heat capacity is related to a process occurring at constant volume. This is denoted by *C~v~* and may be defined as where *q~v~* is the amount of heat supplied per mole at constant volume. From (1) the heat required to raise the temperature of one mole of material from *T*~1~ to *T*~2~ at constant volume is this integrates to **Molar heat capacity at constant pressure.** The molar heat capacity at constant pressure, *C~p~*, is defined as The heat required to raise the temperature of one mole material from *T*~1~ to *T*~2~ for a process carried at constant pressure is this integrates to *q~p~ = C~p~ (T*~2~ *-- T*~1~*)* **Relation between C~p~­ and C~v~.** When the volume is held constant, no work is done by the system and thus the amount of heat absorbed, *q~v~*, is equal to the internal energy change. Thus from (1) In the case of a gas expanding at constant pressure, the heat supplied to the system is used to raise the internal energy of the system and also in doing work, Δ(*PV*). Here, *dH* = *q~p~* Where *dH* is a small change in enthalpy Thus from (2) for an ideal gas system *H* = *E + PV* substituting value from (3) and (4) in (5) and *RT* for *PV* since (*PV* = *RT*)*,* or C~p­~ = C~v~ + R or C~p~ - C~v~ = R Hence, **the difference between molar heat capacity of a gas at constant pressure (*C~p~*) and constant volume (*C~v~*) is equal to the gas constant R.** The value of *R* in cal K^-1^ mol^-1^ is 1.987 and the value in SI units is 8.314 JK^-1^ mol^-1^. **Note**. When one or more variables are held constant during the change of another, the derivatives are called *partial Derivatives* with respect to the changing variable. The d is replaced by *curly delta ()* and the constant quantities are written as suffix. Thus definitions of *C~v~* and *C~p~* become. **Example (1)** Calculate the amount of heat necessary to raise 213.5 g of water from 25°C to 100°C. Molar heat capacity of water is 18 cal/mol/.K. **Solution (1):** by definition or *q = C (T*~2~ *-- T*~1~*)* - for 1 mole *q* = *nC* (T~2~ -- T~1~) -for *n* moles...............(1) in the present case *n* = 213.5/18 *C* = 18 cal mole^-1^ K^-1^ *T*~2~ *-- T*~1~ = (373 -- 298)K substituting the value in (1) = 16.000 cals. **Example (2)** Three moles of an ideal gas (*C~v~* = 5 cal deg^-1^ mole^-1^) at 10.0 atm and 0 °C are converted to 2.0 atm at 50 °C. Find Δ*E* and Δ*H* for the change. **Solution (2)** *R* = 2 cal mole^-1^/deg *(a)* Δ*E* = *nCv dT* = 3 × 5 × (50 − 0) = 750 cals *(b)* Δ*H = nC~p~ dT= n (Cv + R) dT* = 3 × (5+ 2) × 50 = 1050 cals **JOULE-THOMSON EFFECT** Joule and Thomsan (later Lord Kelvin) showed that when a compressed gas is forced through a porous plug into a region of low pressure, there is appreciable cooling. **The phenomenon of producing lowering of temperature when a gas is made to expand adiabatically from a region of high pressure into a region of low pressure, is known as** *Joule-Thomson Effect* **or** *Joule-Kelvin Effect.* **Joule-Thomson Experiment.** The apparatus used by Joule and Thomson to measure the temperature change on expansion of a given volume of gas is illustrated in Fig1.10. An insulated tube is fitted with a porous plug in the middle and two frictionless pistons *A* and *B* on the sides. Let a volume *V*~1~ of a gas at pressure *P~1~ be forced through the porous plug by a show movement of piston A. the gas in the right*−*hand chamber is allowed to expand to volume V~2~ and pressure P~2~ by moving the piston B outward. The change in temperature is found by taking readings on the two thermometers.* *Most gases were found to undergo cooling on expansion through the porous plug. Hydrogen and helium were exceptions as these gases showed a warming up instead of cooling.* ***Explanation.** The work done on the gas at the piston **A** is **P~1~V~1~.** and the work done by the gas at the piston **B** is **P~2~V~2~**Hence the net work (w) done by the gas is* *W = P~2~ V~2~ -- P~1~V~1~* *ΔE = q -- w (First Law)* *But the process is adiabatic and, therefore, q = 0* ***∴** ΔE = E~2~ -- E~1~ = -- w = -- (P~2~V~2~ -- P~1~V~1~)* *or E~2~ -- E~1~ = -- (P~2~V~2~ -- P~1~V~1~)* *Rearranging.* *E~2~ + P~2~V~2~ = E~1~ + P~1~V~1~* *H~2~ = H~1~ or ΔH = 0* ***Thus the process in Joule-Thomson experiment takes place at constant enthalpy.*** ***Joule-Thomson Coefficient.** The number of degrees temperature change produced per atmosphere drop in pressure under constant enthalpy conditions on passing a gas through the porous plug, is called **Joule-Thomson Coefficient.** It is represented by the symbolμ. Thus* *If μ is positive, the gas cools on expansion; if μ is negative, the gas warms on expansion. The temperature at which the sign changes is called the **Inversion Temperature.** Most gases have positive Joule--Thomson coefficients, and hence they cool on expansion at room temperature. Most gases have positive Joule-Thomson coefficients, and hence they cool at room temperature. Thus liquefaction of gases is accomplished by a succession of Joule--Thomson expansions.* *The inversion temperature for H~2~ is -80°C. Above the inversion temperature, μ is negative. Thus at room temperature hydrogen warms on expansion. Hydrogen must first be cooled below -- 80°C (with liquid nitrogen) so that it can be liquefied by further Joule--Thomson expansion. So is the case with helium.* ***Explanation of Joule-Thomson Effect.** We have shown above that Joule--Thomson expansion of a gas is carried at constant enthalpy. But H = E + PV* *Since H remains constant, any increase in PV during the process must be compensated by decrease of E, the internal energy. This leads to a fall in temperature i.e., T~2~ \< T~1~. For hydrogen and helium PV decreases with lowering of pressure, resulting in increase of E and T~2~ \> T~1~. Below the inversion temperature, PV increases with lowering of pressure and cooling is produced.* ***ADIABATIC EXPANSION OF AN IDEAL GAS*** *A process carried in a vessel whose walls are perfectly insulated so that no heat can pass through them, is said to be **Adiabatic.** In such a process there is no heat exchange between a system and surroundings, and q = 0.* *According to the first law* *ΔE = q -- w = 0 -- w* *or Δ E = -- w...............\...\....(1)* *Since the work is done at the expense of internal energy, the internal energy decrease and the temperature falls.* *Consider 1 mole of an ideal gas at pressure P and a volume V. for an infinitesimal increase in volume dV at pressure P, the work done by the gas is -- pdV. The internal energy decreases by dE.* *According to equation (1)* *by definition of molar heat capacity at constant volume* *from (2) and (3) C~v~ dT=* − *pdV* *for an ideal gas P = RT/V* *and hence* *or* *Integrating between T~1~, T~2~ and V~1~, V~2~ and considering C~v~ to be constant,* *thus* *since R = C~p~ -- C~v~, this equation may be written as* *the ratio of C~p~ to C~v~ is often written as γ,* *and equation (4) thus becomes* *Replacing --ve sign by inverting V~2~/V~1~ to V~1~/V~2~* *We can eliminate the temperature by making use of the ideal. Gas relationship* *Equating the right-hand sides of equations (5) and (6)* ***Comparison between Isothermal and Adiabatic Expansions.** Boyle\'s law describes pressure-volume relations of an ideal gas under isothermal conditions (T, constant). This is similar to the relation derived for adiabatic expansion.* *PV = constant (Boyl\'s law)* *PV ^γ^ = constant (Adiabatic expansion)* *Y for an ideal monatomic gas = 1.67. The difference between the two processes is:* ***In an isothermal process, temperature of a system remains constant, while in an adiabatic process, temperature must change.*** ***Explanation.** In an isothermal process heat is absorbed to make up for the work done by the gas in expansion, and the temperature remains unchanged. On the other hand, adiabatic expansion takes place at the expense of internal energy which decreases and the temperature falls. For the same reason, the curve for the adiabatic process (Fig 1.11) is steeper than for the isothermal process.* ***WORK DONE IN ADIABATIC REVERSIBLE EXPANSION*** ***Step (1) Value of VdP from adiabatic equations:*** *For an adiabatic process* *differentiating it, we have* *dividing by, we get* ***Step (2) Value of VdP from ideal gas equation:*** *For 1 mole of an ideal gas PV = RT* *Complete differentiation gives PdV + VdP = RdT* ***Step (3) Substitution*** *Substituting the value of VdP from (1) in (2), we get* *RdT -- PdV =* − *γPdV* *or RdT = P (1* − *γ)dV* *or PdV =* *If there are n moles of a gas* ***Step 4. Integration:*** *Integrating from T~1~, V~1~ to T~2~, V~2~, with γconstant* *When T~2~ \> T~1~, w~max~ is negative because 1− γ is negative. This means that work is done by the gas. On the other hand, when T~2~ \< T~1~, w~max~ is positive which means that work is done on the gas.* ***Example.** Calculate w for the adiabatic reversible expansion of 2 moles of an ideal gas at 273.2°K and 20 atm. To a final pressure of 2 atm.* *Given C~p~ = 5 R/2, mole^-1^deg^-1^* *C~v~ =3 R/2 mole^-1^ deg^-1^* *R = 8.314 J mole^-1^ deg^-1^* ***Solution:*** ***Step 1.** To calculate the value of T~2~ the final temperature, using the equation...................................................(1)* *substituting the value of γ in (1) (T~2~/273.2)^5/3^ = (2/20)* *solving it, we get T~2~ = 108.8°K* ***Step 2.** To calculate maximum work under adiabatic conditions:* ***Alternative Solution*** *The work done under adiabatic conditions may be obtained by calculating decreases in internal energy.* *= − 2 × 3/2 × 8.314 (108.8 -- 273.2)* *= 4100 J = 4.1 kJ* *CHAPTER (2)* ***Introduction*** *It has already been pointed out in the previous chapter that every substance possesses a definite amount of energy known as its **intrinsic energy** or **internal energy** (E) which is a function only of its state and the value of which depends upon its chemical nature, quantity, temperature, pressure and volume. It is the resultant of various types of energies due to the electrostatic interactions within its molecules, position of the molecules. Motion of molecules, atomic vibrations, atomic rotations and chemical linkages. Its exact value cannot be determined but the **change in internal energy (ΔE)** can be accurately measured and it is the value of this change which is important from over point of view.* *It is noticed that energy in the form of heat is generally either evolved or absorbed as a result of a chemical change. This is due mostly to the breaking of bonds in the reactants and formation of new bonds in the products. **The study of thermal changes accompanying chemical transformations** is known as **Thermo-chemistry.** It covers the experimental determination of heat changes in the above processes as well as the calculation of these without recourse to experiment.* *The energy units in which heat changes are usually expressed are the **calorie (**cal), **kilocalorie (**Kcal = 1000 cal), **Joule (**J) and **kilojoules** (kJ). A kilocalorie is also written with a capital **\"C\"** whereas calorie is written with a small \"**c\" i.e.** the C = 1000c.* ***HEAT OF REACTION*** *Thermochemical measurements are made either at (a) constant volume or (b) constant pressure. The magnitude of changes observed under the two conditions are different for a change occurring at constant volume, heat evolved or absorbed at constant temperature is equal to the change in internal energy (ΔE) of the system because no external work is performed. However, at constant pressure, not only does the change in internal energy take place but work is also involved because of expansion or contraction. If ΔV be the change in volume in case of a reaction occurring at a constant temperature and a constant pressure P, the thermal effect observed will be the sum of change in internal energy (ΔE) and the work done in expansion or contraction PΔV i.e. (ΔE + PΔV). But (ΔE + PΔV) is equal to ΔH i.e. the change in enthalpy. It means that whereas the heat evolved or absorbed is a measure of the change in internal energy at constant volume. It gives at constant pressure the change in the enthalpy of the system.* *The **heat of reaction at constant volume** at a certain temperature is defined as the **change in internal energy (ΔE)** of the system when number of gram molecules of substances as specified by the chemical equation have completely reacted at constant volume.* *The **heat of reaction at constant pressure** at a certain temperature defined as the **difference in enthalpies (ΔH) of the products and the reactants** when number of gram molecules of the substances as indicated by the chemical equation have completely reacted at constant pressure.* *For reactions involving solids and liquids only, the change in volume (ΔV) is very small and the term PΔV is negligible. **For such reactions ΔH is equal to ΔE.** In case of gases, however, we must specify whether the reaction has taken place at constant volume or at constant pressure because in their case, the value of PΔV is appreciable. Most of such reactions are, however, studied at constant pressure and it is the change in enthalpy (ΔH) that is involved.* ***Exothermic and Endothermic Reactions.** Let us consider a general reaction at constant pressure:* If *H~A~*, *H~B~*, *H~C~* and *H~D~ be the enthalpies of A, B, C, and D respectively, the heat of reaction at constant pressure viz, ΔH is equal to the difference in enthalpies of the products and the reactants i.e. ΔH = (H~C~ + H~D~) -- (H~A~+ H~B~)* *or ΔH = (enthalpies of products) -- (enthalpies of reactants)* *ΔH may be either zero, negative or positive. When ΔH is zero, the enthalpies of the products and the reactants being the same, no heat is either evolved or absorbed. In case ΔH is negative, the enthalpy of products is less than that of the reactants and the difference in enthalpy is given out in the form of heat. Such reactions which are accompanied by the evolution of heat are known as **exothermic reactions.** When however, ΔH is positive, the enthalpy or heat content of products is more than the heat content of the reactants and an equivalent amount of heat is absorbed by the system from the surroundings. Such reactions as are accompanied by absorption of heat are called **endothermic reactions.*** ***Sign of ΔH and ΔE.** A negative sign of ΔH or ΔE shows that heat is evolved and the reaction is exothermic while a positive sign of ΔH or ΔE indicates that heat is absorbed and the reaction is endothermic.* ***Calculation of ΔH from ΔE and vice-versa.** The heat of reaction at constant pressure (ΔH) and heat of reaction at constant volume (ΔE) are related to each other by the reaction* *Where ΔV is the change in volume due to the expansion or contraction when measurement is done at constant pressure P. Though heats of reaction are usually measured at constant pressure, it is also sometime necessary to carry out the reaction at constant volume as for instance in the measurement of the heat of combustion in a bomb calorimeter. The above relationship can be used, if necessary, for the conversion of ΔH into ΔE and vice versa.* *consider a general reaction* *change in number of moles* *= (No. of moles of products) -- (No. of moles of reactants)* *= (C + d) -- (a + b) = Δn.............................(say)* *Let the volume occupied by one mole of the gas be V, then:* change in volume Δ*V* = change in No of moles × volume occupied by one mole of the gas = Δ*n* × *V* P. V = P (Δ*n* × V) = *PV*Δ*n*...............................(ii) but *PV* = *RT* (for one mole of gas) Putting RT in place of PV in the above equation *(ii)*, *P*. Δ*V* = *RT* Δ*n* Substituting the value of P. Δ*V* in equation (i)*,* we get Δ*H* = Δ*E* + Δ*nRT* It may be pointed out that while determining the value of Δ*n*, only the number of moles of gaseous reactants and products are taken into consideration. The value of gas constant *R* is taken either in calories or joules degree per mole and is 1.987 calories\ (approximately 2 calories) or 8.314 joules. **Example (1)** The heat of combustion of ethylene at 17°C and at constant volume is −332.19 Kcals. Calculate the heat of combustion at constant pressure considering water to be in liquid state (*R* = 2 cals). **Solution (1):** The chemical equation for the combustion of ethylene is C~2~H~4(g)~ + 3O~2(g)~ → 2CO~2(g)~ + 2H~2~O(*)* 1mole 3 moles 2moles negligible volume no. of moles of the product = 2 no of moles of the reactants = 4 **∴**Δ*n* = (2 -- 4) = -- 2 Δ*H =* Δ*E + R* Δ*nT* given that Δ*E* = -- 332.19 Kcals *T* = (273+ 17) = 290°K and R = 2 cals or 2 × 10^-3^ kcal **∴** Δ*H* = -- 332.19 + 2 × 10^-3^ × (--2) × 290 = -- 333.35 Kcals. **Example (2)**. The heat of combustion of carbon monoxide at constant volume and at 17°C is -- 283.3 KJ calculate its heat of combustion at constant pressure (*R* = 8.314 Joule degree^-1^ mole^-1^). CO (g) + ½ O~2~ (g) → CO~2~ (g) 1mole ½mole 1mole **Solution (2):** no. of moles of products = 1 no. of moles of reactants = 1½ *Δn* = no. of moles of products -- no. of moles of reactants = 1 -- 1.5 = -- 0.5 given that: Δ*E* = -- 283.3 kJ; *T* = (273 + 17) = 290°K and *R* = 8.314 joules or 8.314 × 10^-3^ kJ Substituting these values in the equation Δ*H =* Δ*E* + Δ*nRT* we get = -- 283.3 + \[-- ½ × (8.314 × 10^-3^) × 290\] = -- 283.3 -- 1.20 = -- 284.5 KJ ∴Heat of combustion of CO at constant pressure is -- 284.5 kJ **Example (3).** The heat of formation of methane at 25°C at constant pressure is --17890 cals. Calculate its heat of formation at constant volume. (*R* = 1.98 cals degree^-1^ mole^-1^) **Solution (3):** The thermochemical equation for the heat of formation of methane at constant pressure is: C(s) + 2H~2~ (g) → CH~4~ (g) Δ*H* = -- 17890cal Negligible 2moles 1mole Volume no. of moles of gaseous products = 1 no. of moles of gaseous reactants = 2 ∴ Δ*n* (*i.e.* change in the number of moles) = (1 -- 2) = --1 It is given that: Δ*H* = -- 17890 cals; *T* = (25 + 273) = 298°K and *R* = 1.987 Substituting the values in the equation Δ*H* = Δ*E +* Δ*nRT*, we have: --17890 = Δ*E* + (--1 × 1.987 × 298) = Δ*E* -- 592.1 Δ*E* = (--17890 + 592.1) cals = 17.297.9 Kcal ∴ Heat of formation of methane at constant volume is --17,297.9 kcals. **THERMOCHEMICAL EQUATIONS** There are a number of factors which affect the quantity of heat evolved or absorbed during a chemical or physical transformation. One of these factors has already been discussed *viz* whether the change occurs at constant pressure or constant volume. The other factors are: i. ii. iii. iv. An equation which indicates the amount of heat evolved or absorbed in the reaction or process is called a **Thermochemical equation.** It must essentially, (a) be balanced (b) give the values of Δ*E* or Δ*H* corresponding to the quantities of substances given by the equation (c) mention the physical state of the reactants and products. *For example,* H~2~ + ½ O~2~ → H~2~O Δ*H* = -68.32 Kcal Indicates that when 1 mole (= 2.016 g) of hydrogen reacts with 0.5 mole (= 16.0g) of oxygen 1 mole (= 18.016g) of water is formed and 68.32Kcal of heat are evolved at constant pressure. If 2 moles of hydrogen are burnt, the heat evolved would be (2 × 68.32) Kcals. This equation, however, *is not a complete thermochemical equation* because it does not tell us whether water is in the form of steam or liquid water. There is a difference in the value of ΔH if water is in the liquid or gaseous state. The two values of Δ*H* (at 25°C and 1 atm pressure i.e. under **standard conditions)** differ by the **heat of evaporation** of water. The complete thermochemical equation, therefore, would be in the form Similarly, the difference between the two heats of reaction is due to heat difference when one mole (12g) of carbon changes from diamond to amorphous variety and is termed as the **heat of transition**. Thus *heat of transition* may be defined as *the change in enthalpy when one gram mole of element changes from one allotropic form to another. The above transition may be represented as* *where -- 69 cal and* -- *4,300 cals are heats of transition of S(monoclinic), to S(rhombic) and white phosphorous to red phosphorus respectively.* ***VARIATION OF HEAT OF REACTION WITH TEMPERATURE*** *The heat of reaction changes with change in temperature of a gas due to variation in its specific heat. The equations*