Chemical Reactions as Sources of Energy Part 1: Thermodynamics Contents Standard enthalpies and standard Gibbs energies of formation 4 Calculating enthalpy changes and Gibbs energy changes for reactions 5 Standard Gibbs energy and Gibbs energy 12 Effect of Δ r G for a reaction on the position of equilibrium 14 Relationship between G and H 16 Getting energy out of combustion 16 Part 2: Electrochemistry Part 3: Energy and the hydrogen economy Part 1: 1
Chemical reactions can be used as a source of energy either to provide heat or to do work. Your body burns compounds such as sugars (glucose C 6 H 12 O 6, for example) to obtain heat to maintain your body temperature and to do work; building the chemicals your body requires, pumping the blood round your body, lifting your feet Chemical reactions are also used to provide the heating in your home and to make electricity to do work; pumping the water round central heating pipes and radiators or vibrating the membrane in the loudspeakers of a TV. The amount of energy available from a reaction can be calculated using the science of thermodynamics. Three important thermodynamic properties used in chemistry are listed below with brief definitions. If you are not familiar with enthalpy, Gibbs energy and entropy, you should look them up in a text book before you go on. Check also that you understand the meaning of the words in italics. Enthalpy (H): this is a measure of the energy stored within a system. Only changes in enthalpy can be measured. Enthalpy is defined so that a change in enthalpy during any process equals the heat input, ΔH positive (or output, ΔH negative) when that process is carried out under a total external pressure of one bar. Gibbs energy (G): this is an alternative measure of the energy stored within a system. Again only changes can be measured. It is defined so that for any spontaneous change, the Gibbs energy change is negative (ΔG < 0). The Gibbs energy change is also a measure of the total amount of work you can get out of a process if it is carried out under a total external pressure of one bar. Entropy (S): this is a measure of the number of ways the atoms and energy within a system can be re-arranged while, as far as you can tell, the system remains unchanged. The entropy of a system, not just entropy changes, can be determined. For a system, the thermodynamic quantities are related by G = H - T S and for any change ΔG = ΔH - T ΔS where T is the absolute temperature (in kelvin). Part 1: 2
To summarize for a chemical process: Enthalpy changes provide information about the heat output (or input) resulting from that chemical process; Entropy changes define what processes are spontaneous (for a spontaneous process the total entropy change for the system and its surroundings must be positive); The Gibbs energy change, combining enthalpy and entropy changes, defines what processes are spontaneous in terms of the system alone. The chemical unit of energy is a joule, symbol J (named after the Lancashire brewer and scientist James Prescott Joule who did a lot of work investigating the relationship between heat and work). What is a joule? Joules are pretty small so they re often counted in thousands (kilojoules: 1000 J = 1 kj). A 60 watt light bulb uses about 200 kj of energy in one hour. Charging a mobile phone for one hour uses about 8 kj. A simple chemical reaction commonly used to provide energy is the combustion of the gas methane, CH 4. The reaction is CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) The amount of energy (as heat or work) that we can get from this reaction can be worked out from information given in tables of thermodynamic data. Since the enthalpies and Gibbs energies of compounds cannot be found, the values given in tables are the changes in enthalpy (and Gibbs energy) during the formation of a compound from its elements. For example, for carbon dioxide gas, this is the enthalpy (or Gibbs energy) change for the process elements compound graphite solid + oxygen gas carbon dioxide gas C(s) + O 2 (g) CO 2 (g) Part 1: 3
Standard enthalpies and standard Gibbs energies of formation Some thermodynamic properties for the compounds involved in methane combustion are given in the table. Δ f H (at 298 K)/kJ mol -1 Δ f G (at 298 K)/kJ mol -1 CH 4 (g) -75-51 O 2 (g) 0 0 CO 2 (g) -393-394 H 2 O(g) -242-228 What do the symbols mean? Δ means the change in The subscript f means during formation of the compound from its elements H is the symbol for enthalpy G is the symbol for Gibbs energy The superscript means that every compound in the formation reaction is in its standard state. What is a standard state? For a gas, its standard state is the gas at one bar pressure. For a solid (or liquid), its standard state is simply the pure solid (or liquid). For a substance dissolved in water, its standard state is the substance at a concentration of one mole per cubic decimetre of solution. Note that (despite what some school text books say) there is no standard temperature. That means that whenever you see a table of thermodynamic data, it should always give a temperature. Often that temperature is 298 K (25 ºC). Remember too that enthalpy and Gibbs energy changes apply to changes where the external pressure (of the atmosphere) remains constant at one bar. What is a bar? The standard pressure used to be one atmosphere. Unfortunately when one atmosphere is translated into SI units, it becomes 101 325 N m -2 (newtons per square metre). It was therefore decided that the standard pressure would be better defined as a simpler multiple of the SI unit: hence 1 bar = 100 000 N m -2. One bar is a pressure about one per cent lower than atmospheric pressure at sea level. Part 1: 4
Calculating enthalpy changes and Gibbs energy changes for reactions From the table of thermodynamic properties for formation, the change in that property for the reaction (such as the burning of methane) can be worked out. For example, the amount of heat you can get out of a reaction is given by sum of formation enthalpies of products minus sum of formation enthalpies of reactants. To work out how much work you could get out of a reaction (in theory), the equation is exactly the same except that Gibbs energy replaces enthalpy: sum of formation Gibbs energies of products minus sum of formation Gibbs energies of reactants. Why do I say in theory? Because in practice it s very difficult to get all the energy out of a reaction as work. The equations can be written in more mathematical forms and Δ r H = Δ f H (products) - Δ f H (reactants) Δ r G = Δ f G (products) - Δ f G (reactants) What do the symbols r and Σ mean? The subscript r means during reaction (in the same way as f meant during formation ). means the sum of. For example, Δ f G (products) means the sum of the standard Gibbs energies of formation of all the products of the reaction. Before you go on to the next page, calculate (1) the amount of heat you can get from burning one mole of methane (the enthalpy change for the reaction) and (2) the maximum amount of work you could get from burning one mole of methane (the Gibbs energy change for the reaction). The balanced chemical equation is given next to the picture of the gas flame on page 3. Part 1: 5
CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) (1) Δ r H = Δ f H (all products) - Δ f H (all reactants) = [-393 + 2x(-242)] [-75 + 2x(0)] Why have the H 2 O and O 2 values been multiplied by two? Because in the equation at the top of the page, there are two moles of H 2 O and two moles of O 2. Δ r H = -877 + 75 kj mol -1 Δ r H = - 802 kj mol -1 (2) Δ r G = Δ f G (all products) - Δ f G (all reactants) = [-394 + 2x(-228)] [-51 + 2x(0)] Δ r G = -850 + 51 kj mol -1 Δ r G = - 799 kj mol -1 The minus sign means that, if you let the products return to the same temperature as the reactants were at the start, energy comes out of the reaction; about the same amount of energy as radiated from the 60 W light bulb during four hours. Here the two quantities (Δ r H and Δ r G ) are almost the same. Carry out the same calculation but assume that the water produced is liquid rather than vapour. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(liq) Δ f H (liquid H 2 O at 298 K)/kJ mol -1 = - 286 Δ f G (liquid H 2 O at 298 K)/kJ mol -1 = - 237 Part 1: 6
CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(liq) Δ r H = Δ f H (all products) - Δ f H (all reactants) = [-393 + 2x(-286)] [-75 + 2x(0)] Δ r H = - 890 kj mol -1 Δ r G = Δ f G (all products) - Δ f G (all reactants) = [-394 + 2x(-237)] [-51 + 2x(0)] Δ r G = - 817 kj mol -1 The amounts of energy you can get out of the reaction are greater than before because you are getting a bit out of the H 2 O(g) H 2 O(liq) process. The enthalpy change and the Gibbs energy change are no longer the same. It appears that you can get more energy out as heat than you can get out as work. Why are the two amounts of energy slightly different? The enthalpy change shows the amount of heat we could get out of the reaction; the Gibbs energy change, the maximum amount of work. Imagine the mixture before the reaction. You started with three moles of gas (about 22 litres of methane and 44 litres of oxygen). After the reaction, you have only one mole of gas (about 22 litres of carbon dioxide) and two moles of liquid water (less than 0.04 litres). If the pressure of the outside world stays constant (which in the world we live in, it does, at about one atmosphere) then the reaction mixture will shrink it will be squashed by the outside world. To do this squashing, the outside world must do a bit of work; and this work appears as additional heat output. The extra 70 kj of heat doesn t actually come from the reaction at all but from the world outside (in the jargon of thermodynamics from the surroundings ). Part 1: 7
Example 1 Now let s suppose we burn a mole of bottled gas (propane) vapour at 500 K. Calculate the standard enthalpy change and the standard Gibbs energy change for this reaction. Note that the first thing you will have to do is write a balanced equation for the combustion of propane in oxygen (at this temperature the H 2 O will have to be gas). C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O(g) Δ f H (at 500 K)/kJ mol -1 Δ f G (at 500 K)/kJ mol -1 C 3 H 8 (g) - 116 + 35 O 2 (g) 0 0 CO 2 (g) - 394-395 H 2 O(g) - 244-219 Example 2 * * * Calculate the standard enthalpy change and standard Gibbs energy change when solid caesium sulphate (Cs 2 SO 4 ) is dissolved in one litre of water. in water Cs 2 SO 4 (s) 2 Cs + (aq) + SO 2-4 (aq) Is the dissolution process exothermic or endothermic? Is the dissolution process spontaneous? Δ f H (at 298 K)/kJ mol -1 Δ f G (at 298 K)/kJ mol -1 Cs 2 SO 4 (s) -1442-1323 Cs + (aq) -259-292 SO 2-4 (aq) -909-744 Part 1: 8
Answer 1 C 3 H 8 (g) + 7 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g) Δ r H (at 500 K)/kJ mol -1 = [(-394 x 3) + (-244 x 4)] - (-116) Δ r H (at 500 K)/kJ mol -1 = - 2042 Δ r G (at 500 K)/kJ mol -1 = [(-395 x 3) + (-219 x 4)] - (+35) Δ r G (at 500 K)/kJ mol -1 = - 2096 It appears that you can get more energy out as work than you can get out as heat despite the reduction in volume. Here the bit of heat you get out of the atmosphere squashing the reaction mixture is more than compensated for by the big difference between Δ f H (at 500 K) and Δ f G (at 500 K). This difference becomes even bigger as molecule size gets larger and temperatures get higher. For example, benzene at this temperature Δ f H [benzene(g) at 500 K] = 74 kj mol -1 ; Δ f G [benzene(g) at 500 K] = 164 kj mol -1 Δ f H [benzene(g) at 900 K] = 64 kj mol -1 ; Δ f G [benzene(g) at 900 K] = 240 kj mol -1 Discussion question: from what you know about entropy (and the relationship between ΔH and ΔG, can you explain why Δ f G values for formation of large molecules should differ so much from Δ f H values with rising temperature? Additional questions (with example answers) involving the calculation of enthalpy changes and Gibbs energy changes can be found at http://saps.glam.ac.uk/chemistryforourfuture/home.htm Part 1: 9
Answer 2 Δ r H (at 298 K)/kJ mol -1 = [(-259 x 2) + (-909)] - (-1442) Δ r H (at 298 K)/kJ mol -1 = + 15 Δ r G (at 298 K)/kJ mol -1 = [(-292 x 2) + (-744)] - (-1323) Δ r G (at 298 K)/kJ mol -1 = - 5 The process is endothermic; the reaction (the dissolution process) takes in heat from the surroundings. If you were holding the beaker, it would feel cold. The Gibbs energy change, however, is negative indicating that caesium sulphate does dissolve in water. The enthalpy change indicates that the ion-water interactions are less strong than the interactions in the solid. On its own, this would suggest that caesium sulphate will not dissolve in water. The ions in solution, however, have a lot more freedom of movement and are distributed in a much larger volume than the ions in the solid. The entropy of the products (the dissolved ions) is therefore much bigger than the entropy of the reactants (the ions in the solid). This large entropy change more than compensates for the unfavourable enthalpy change. Here, the TΔS term is more important than the ΔH term in determining the sign of r G r G = r H - T r S Part 1: 10
Standard Gibbs energy and Gibbs energy If you pour a standard solution of potassium nitrate into a lake, the potassium and nitrate ions will spread out until the whole lake becomes a very dilute solution of potassium nitrate. In a similar way, if you open a flask containing methane at one bar pressure, the methane will diffuse out of the flask and spread out until it is found throughout the atmosphere. These are known as spontaneous processes; they happen every time. At the start, the enthalpy and Gibbs energy of the potassium nitrate and the methane are standard; at the finish, they are not. For example, there will have been a change in the enthalpy and Gibbs energy of the methane. Since there is no measurable heat input or output resulting from this expansion so the enthalpy has not, in fact, changed even though you can t call it standard any more. The Gibbs energy has, however, changed. By the time the methane has occupied the room you re sitting in, its Gibbs energy has gone down by about 10 kj mol -1. You would write Δ expansion G = - 10 kj mol -1 Again, this is not a standard Gibbs energy change because the product (the spread out gas) is not in its standard state of one bar pressure. The important thing to notice is not the size of this Gibbs energy change but the fact that it is negative. This is true of all spontaneous processes. Or If its Gibbs energy change is negative, a process is spontaneous. A process will be spontaneous if its Gibbs energy change is negative. How is this applied to a chemical reaction? Take as a very simple example, the isomerization of trans-but-2-ene to cisbut-2-ene trans-but-2-ene(g) cis-but-2-ene(g) for which I have found thermodynamic properties at 400 K. compound S / J K -1 mol -1 f H / kj mol -1 f G / kj mol -1 trans-but-2-ene 325.1-17.2 89.1 cis-but-2-ene 327.1-13.8 91.7 Using the figures in the final column, calculate r G process. Do this before looking at the next page. / kj mol -1 for this Part 1: 11
You should get r G = + 2.6 kj mol -1 (or + 2600 J mol -1 ) You might be tempted to say that this reaction is not spontaneous, it won t go. Unfortunately things are not quite so simple. It is true to say that trans-but-2- ene at one bar pressure will not go completely to cis-but-2-ene. Gibbs energy and equilibrium Imagine you have trans-but-2-ene at one bar pressure in a flask (standard trans-but-2-ene). A bit of it reacts. The total pressure remains at one bar. The Gibbs energy per mole of the trans-but-2-ene will go down a bit because its pressure has gone down. You ve made a bit of cis-but-2-ene. Its partial pressure is small so its Gibbs energy per mole is a lot less than the standard Gibbs energy per mole. So maybe you can make a little bit of cis-but-2-ene. This is best shown on a diagram. Start at the left, pure reactant trans-but-2-ene. As you convert some trans-but- 2-ene into product cis-but-1-ene, the Gibbs energy of the mixture goes down. Reaction is spontaneous. As you follow the curve of Gibbs energy, reaction continues until the minimum of the curve is reached (the diagram shows that this point is reached when about a quarter of the reactant has been converted into product). Once the mixture reaches this point, it can go neither forward nor back because either way requires the Gibbs energy to go up again (ΔG +ve). This is the position of equilibrium for this reaction. At this composition, the Gibbs energy per mole of trans-but-2-ene equals the Gibbs energy per mole of cis-but-2-ene. Part 1: 12
Equilibrium can then be defined as Or Equilibrium is the reaction mixture composition where the Gibbs energy is a minimum. Equilibrium is the reaction mixture composition where ΔG = 0. Effect of Δ r G for a reaction on the position of equilibrium Take the reaction trans-but-2-ene(g) cyclobutane(g) compound at 400 K S / J K -1 mol -1 f H / kj mol -1 f G / kj mol -1 cyclobutane 290.3 19.3 139.6 The figures for trans-but-2-ene(g) were given in the earlier table. Before you go on to the next page calculate r G. Part 1: 13
You should get a value of r G = + 50.4 kj mol -1 (or + 50 400 J mol -1 ) The diagram below shows how the Gibbs energy of the trans-but-2-ene, cyclobutane reaction mixture varies with composition (100 % trans-but-2-ene on the left, 100 % cyclobutane on the right). The equilibrium composition is the lowest point on the curve. In this case, the lowest point is the extreme left of the graph (100 % trans-but-2-ene). A large positive value of r G cyclobutene will take place. indicates that no reaction of trans-but-2-ene to In summary large negative r G small negative r G small positive r G large positive r G reaction goes to completion equilibrium mixture mostly products but some reactant remains equilibrium mixture mostly reactants but some product produced no reaction takes place. But how large is large? At temperatures around 300-400 K, small can be taken as less than 15 kj mol -1 and large, greater than 15 kj mol -1. The same borderline is not suitable at all temperatures. Since combustion usually takes place at high temperatures, the reactions taking place, and the compounds appearing among the products, may be more varied. For example, exhaust from petrol engines contains partially oxidized hydrocarbons, carbon monoxide and oxides of nitrogen. Part 1: 14
Relationship between G and H The two quantities are related by G = H - T S and in the case of standard values G = H - T S T is the temperature in kelvin and ΔS is the entropy change for the process taking place. Getting energy out of combustion The chemical reactions we use to produce energy are chosen because they produce large amounts of energy. Under those circumstances, the differences between Δ r H and Δ r G for a particular reaction are relatively small. So to compare different fuels, it doesn t matter much whether we look at Δ r H or Δ r G particularly since we waste a fair amount of the energy in turning it into a useful form, such as electricity. Working out the energy change per mole is one thing, but sometimes we need to make other comparisons: for example, which fuel gives me the best output per gram? Compare the heat output per gram when the compounds given in the table are burnt to produce carbon dioxide and liquid water. The data for carbon dioxide and water are given on pages 3 and 5. compound hydrogen methane ethanol octane glucose H 2 (g) CH 4 (g) C 2 H 5 OH(liq) C 8 H 18 (liq) C 6 H 12 O 6 f H / kj mol -1 0-85 -235-208 -1273 The answers are given in the table on the following page. Part 1: 15
Standard enthalpies of combustion per mole (note the subscript c to indicate combustion) and enthalpies of combustion per gram for the compounds in the question on the previous page. compound hydrogen methane ethanol octane glucose H 2 (g) CH 4 (g) C 2 H 5 OH(liq) C 8 H 18 (liq) C 6 H 12 O 6 f H / kj mol -1 0-85 -235-208 -1273 c H / kj mol -1-242 -802-1277 -5113-2536 molar mass 2 16 46 114 180 c H / kj g -1-121 -50-28 -45-14 The enthalpies of combustion are all negative so, since the question asks for heat outputs, the final answer would be heat output / kj g -1 +121 +50 +28 +45 +14 Some more thermodynamics problems, with example answers, can be found at http://saps.glam.ac.uk/chemistryforourfuture/home.htm Part 1: 16