Thermodynamic

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1 Thermodynamics

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2 Thermodynamics is the branch of science which studies the transformation of energy from one form to another. Chemical thermodynamics looks at the energy transformations which occur as a result of chemical reactions. How much heat is evolved during a chemical reaction ("thermo chemistry")? What determines the direction of spontaneous chemical change? if you put a flame to a 2:1 mixture of H2 and O2 there is an almighty bang and the reaction proceeds spontaneously to products, i.e. without any further supply of external energy. FOR EXAMPLE What is Thermodynamics ? The questions that chemical thermodynamics asks are:

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3 Why does it not happen that water spontaneously goes to H2 and O2 when we put a lit taper to it under the same conditions? What determines the extent of chemical change? Not all reactions result in all of the reactants being converted to products. Some reactions will proceed only so far before the state of chemical equilibrium is achieved. Chemical equilibrium is said to exist if no further tendency for overall chemical change is observed - the ratios of the concentrations of products to reactants remain fixed. So chemical thermodynamics attempts to answer the important question, "What determines the position of chemical equilibrium?".

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4 Because we are interested in the heat given off by reactions it is very important to define the system that we are studying. In this way we can be sure exactly what we mean when we say, for example, that so many joules of heat have been given off - this amount of heat energy has left the system. Note that the system may be defined by physical boundaries, like a test-tube or a bomb calorimeter, or the system may simply be defined by a set of cartesian coordinates specifying a particular volume in space. Once we've defined the system as the particular region of space that we are making measurements on, the definition of the surroundings follows naturally - the surroundings are everything that is not the system. One way of summarizing what we have just said: 4.The System and The Surroundings.

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5 Internal Energy, U We are going to be talking about our chemical system losing and gaining energy, so it is logical to introduce a quantity which represents the total energy of the system. This is known as the internal energy, U. As with all types of energy, the unit of U is the joule. Now, the internal energy is defined as the total energy of our system - chemical, nuclear, heat, gravitational, any other type of energy you can think of - the sum of all the system's energy. A little thought will convince ourselves that it would be impossible to actually measure the total internal energy of our system. So why define a quantity which we cannot measure? Well, although we cannot measure the absolute value of the internal energy of a system, we can measure changes in the internal energy. And since thermodynamics is all about changes in energy this makes the change in internal energy of a system a very useful experimental quantity. Internal Energy, U

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6 The internal energy U of a system may change in 3 different ways. The internal energy U of a system will change if : heat passes into or out of the system; work is done on or by the system; mass enters or leaves the system. It is possible to define various types of system depending on which of the three mechanisms for changing U are possible. With a closed system, no transfer of mass is possible: internal energy may only change due to heat and work. With an isolated system, no change in the internal energy is possible: heat, work and mass transfer are all impossible. With an open system, the internal energy may change due to transfer of heat, mass and work between system and surroundings. ('Work transfer' above should be taken to mean that work is done on or by the system.) 5. Changes in Internal Energy

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7 Let's now try to express the change in a system's internal energy in a scientific manner. First of all we have to make a couple of definitions. The quantity q is defined as the heat absorbed by the system from the surroundings. The quantity w is defined as the amount of work done on the system by the surroundings. Both q and w are measured in joules. Note well that in both the definitions above the word "or" does not appear once. The quantity q is not defined as "the heat absorbed or given out by the system" and w is not defined as "the work done on or by the system by or on the surroundings". Note, however, that the functions q and w can deal algebraically with heat being given out from the system or work being done by the system on the surroundings. If q is a negative number, we interpret this as heat being transferred from the system to the surroundings. A negative heat absorbed is equivalent to a positive amount given out. Similarly, a negative value for the quantity w corresponds to the system doing work on the surroundings. The First Law of Thermodynamics

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8 Now let us make a slight simplification by considering a closed system i.e. one that allows no matter to pass in or out. We can easily go back at any point and amend our equations to take mass transfer into account. Now, for a closed system, the only way in which the internal energy of the system can change is through heat and work. Most of you will be familiar with the statement of the First Law of Thermodynamics in the form: energy can neither be created nor destroyed, only changed from one form into another. You should be able to convince yourself that, for a closed system, the First Law of Thermodynamics may be expressed as: the total change in the internal energy must equal the sum of the heat absorbed by the system and the work done on the system. This is expressed in equation form in the slide.

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9 All functions which refer to a system may be categorized as either "state functions" or "path functions". Consider a system changing from an initial state to a final state. A state function is one whose change on going from initial to final is independent of the route taken. Conversely, if the change in a function is dependent on the route taken, then the function is known as a path function. Some familiar state functions are volume, V, pressure, P, and temperature, T. For example, if we take a state from 0oC to 100oC, the change in the temperature is +100oC whether we go straight up the temperature scale or we first cool the system for a few degrees then take the system to the final temperature. The change in temperature is independent of the route taken so temperature is a state function. State and Path Functions

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10 Position, for example, is another state function. If, at the end of the day I make my way home from my office directly to my flat, my position will have changed in exactly the same overall amount had I first stopped in at Ye Olde Transporter, visited Fitzgerald's, looped back to the Museum Vaults, skipped over to the Royalty, hiked to the Ivy House, slogged up to the Rosedene, nipped into the office, and then headed for my flat. I'll leave you to ponder whether sobriety is a state or path function. Here's some work for you to do, are the following state or path functions? Click on the appropriate button. Internal energy, U state path heat absorbed, q state path work done on system, w state path

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11 The animation shows a metal block falling down an incline. (Take the block plus slide as the system.) The potential energy which the block possesses initially due to its raised position is dissipated mainly as heat as the block finally ends up at the foot of the incline. No work has been done. Reload the page to see the animation again. Now consider the case where we attach a rope and pulley system to the block. As the block falls it is now capable of lifting something - a rabbit, for example. Different Paths

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12 Because of the attached mass, the block in this case moves more slowly and less heat is dissipated as friction. Instead this energy has been used to do work, to raise Eek's potential energy. Eek himself is now capable of doing work by virtue of his new-found potential energy. Note that in the two experiments that have just been looked at, the initial and final states have been the same: the block stationary at the top of the incline to the block stationary at the bottom of the incline. The change in the internal energy in both cases has been the same. The system loses the potential energy of the block (strictly speaking only after system has cooled down to original T). However, q and w were different in each case. In the first experiment q was large and w was small. In the second experiment q was small and w was large. In both cases the sum of q and w were the same and equal to the change in the internal energy (the First Law). The two experiments illustrate the fact that q and w are both path functions. 8a. The Block Does Work

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13 If the process is "isothermal", then the temperatures of the inital and final states are the same. That is, the total kinetic energy of all the gas molecules does not change. Effectively, all that has been done is to arrange that the gas particles are contained in a larger volume. Hence, the internal energy change of the isothermal expansion must be zero. From the First Law 0 = q + w therefore q = - wWhat this last equation says is that in order to do an amount of work w on the surroundings an equal amount of energy in the form of heat must be absorbed by the system. The net change in energy is zero. Isothermal Expansion of an Ideal Gas against a Fixed External

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14 10.Reversible, Isothermal Expansion of an Ideal Gas 11. Measurement of Internal Energy Change 12. The Constant Volume Bomb Calorimeter If you've ever wondered how people calculate the calorific values of cream buns and bowls of Frosties, then wonder no longer - the bomb calorimeter's your very man.

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15 So, we know that the heat absorbed at constant volume is equal to the change in internal energy. Most chemical reactions, however, are carried out not at constant volume, but at constant pressure. That is, chemical reactions are normally carried out in flasks or test tubes under the (relatively) constant pressure of the atmosphere. So, it would be useful if there existed some state function whose change equalled the heat absorbed by a system at constant pressure. Enthalpy, H , is that state function. Enthalpy is defined as follows, H = U + PV where U is the internal energy, P is the pressure and V is the volume of the system. The change in enthalpy is equal to the heat absorbed by the system at constant pressure (see first equation on slide). 13. A Useful New State Function - Enthalpy

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16 Now let's look at the change in this expression as enthalpy changes by an infinitesimally small amount dH The changes in the 4 quantities involved are H -> H + dH U -> U + dU P -> P + dP V -> V + dV If the 4 substitutions - H + dH for H etc - are made in H = U + PV , and the products of an infinitesimally small quantity with another are ignored, you should be able to show that dH = dU + PdV + VdP If not, press here - At constant pressure, dP, the change in P is zero, so the last equation becomes dH = dU + PdV After integrating both sides we get the second equation on the slide. In words this is: the change in enthalpy at constant pressure is equal to the change in the internal energy plus the product of the pressure times the change in volume of the system.

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17 14. Conversion between Internal Energy and Enthalpy

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18 It is convenient to define a set of standard conditions to which thermodynamic data can be referred. A reaction is said to take place under standard conditions when both reactants and products are in what is called their standard states. A substance is in its standard state if it is pure, unmixed and in its most stable form at 1 bar pressure. Any thermodynamic quantity referring to standard conditions has the superscript symbol shown in the slide. In conversation we would say "delta H nought" is the standard enthalpy of the reaction. Note that temperature is not part of the definition of standard conditions, but it is obviously essential to quote the temperature to which any thermodynamic quantity applies. Normally values are quoted for 298 K. Unless stated otherwise, this will be the temperature to which data in this programme refer. The reason why temperature is not included in the definition of standard conditions is simply that many reactions will not take place at a specific temperature. Standard Conditions

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19 8. Enthalpies of Physical Change (i) Enthalpy of Vaporization

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20 17. Enthalpies of Physical Change (ii) Enthalpy of Fusion and Freezing 18. Enthalpies of Physical Change (iii) Enthalpy of Ionization

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21 19. Enthalpies of Physical Change (iv) Enthalpy of Atomization 20. Enthalpies of Chemical Change (i) Enthalpy of Combustion 21. Enthalpies of Chemical Change (i) Enthalpy of Formation

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22 22. Enthalpies of Chemical Change (iii) Enthalpy of Hydrogenation

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23 Here are two ways of expressing Hess's Law.The enthalpy of a reaction is equal to the sum of the enthalpies of the reactions into which it may be divided. (Hess's Law is a consequence of enthalpy being a state function, and all other state functions behave in an analogous way.)Hess's law is useful because it allows us to calculate enthalpy changes of reactions that might be difficult to perform from enthalpies of reactions that aren't. We'll also see that if we know the enthalpies of formation of all reactants and products involved in any reaction, Hess's Law allows us to calculate the reaction enthalpy. 24. Hess's Law

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24 Now let's say that we want to determine the enthalpy change associated with the transformation of diamond into carbon. Now that's not so easy a reaction. I recall that when Superman was strapped for cash he used his heat vision and super-strength to create high pressure and temperature conditions and transform lead from his pencil into a diamond. However, unless your name's Lois Lane you won't have access to Superman in the lab. Luckily, Hess's Law provides an alternative

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25 Two easy experiments to carry out are the combustion of the two substances. (It is physically easy to burn a diamond if not psychologically.) So we can measure the enthalpy of these two reactions (see opposite). What is that rabbit up to? Now reverse the second reaction and note that if we reverse the direction of the reaction we must reverse the sign of the enthalpy change (heat out one way, heat in the other way).Now let's consider a two-step reaction. Start off with 1 mol of diamond and combust to form one mole of carbon dioxide; then take this mole of carbon dioxide and form one mole of graphite and one mole of oxygen.

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26 What is the overall reaction? Well, we started with 1 mol of oxygen and we finish with 1 mol of oxygen, so the oxygen has not participated in the overall reaction. This means that the overall reaction is simply the conversion of 1 mol of diamond to 1 mol of graphite, the desired reaction. If we get out 395.40 kJ in the first reaction and we take in 393.51 kJ in the second there is a net output of 1.89 kJ. That is, the enthalpy of the conversion of diamond to graphite is -1.89 kJ mol-1.

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27 Let's recall that the standard enthalpy of formation of a substance is the enthalpy change observed when one mole of a substance is formed from its constituent elements, with reactants and products in their standard states. Also let's recall that Hess's Law says that the enthalpy of a reaction is independent of the route taken. Now, we may describe any reaction going directly from reactants to products as a two-stage process. The first is disassembling all the reactant molecules into their constituent elements. The second is assembling the product molecules from their constituent elements (produced by disassembling the reactants). 24. Hess's Law: Reaction Enthalpies from Enthalpies of Formation.

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28 The enthalpy change taking place when one mole of a substance is disassembled into constituent elements is equal to minus the enthalpy of formation, by definition. The enthalpy change taking place when one mole of a substance is assembled from its elements is simply the enthalpy of formation, by definition. With a bit of reflection it will be seen that the overall reaction enthalpy will be equal to the sum of all the enthalpies of formation of the products minus the sum of all the enthalpies of formation of the reactants. (We must, of course, take into account the stoichiometric coefficients of the chemical reaction: if one reactant appears as two moles in a balanced chemical equation then its contribution to the overall enthalpy of the reaction must be minus two times the enthalpy of formation of the molecule.) Any pure elements involved in the reaction as reactants or products will make no contribution to the calculation of the reaction enthalpy because the enthalpy of formation of any element is zero.

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29 THANK YOU