Gases. Properties of Gases Kinetic Molecular Theory of Gases Pressure Boyle s and Charles Law The Ideal Gas Law Gas reactions Partial pressures.

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1 Gases Properties of Gases Kinetic Molecular Theory of Gases Pressure Boyle s and Charles Law The Ideal Gas Law Gas reactions Partial pressures Gases Properties of Gases All elements will form a gas at some temperature Most small molecular compounds and elements are either gases or have a significant vapor pressure. 1 H He 1 Room Temperature Gases 2 2 Li 3 3 Na 11 4 K 19 5 Rb 37 6 Cs 55 Be 4 Mg 12 Ca 20 Sr 38 Ba 56 Sc 21 Y 39 Lu 71 Ti 22 Zr 40 Hf 72 V 23 Nb 41 Ta 73 Cr 24 Mo 42 W 74 Mn 25 Tc 43 Re 75 Fe 26 Ru 44 Os 76 Co 27 Rh 45 Ir 77 Ni 28 Pd 46 Pt 78 Cu 29 Ag 47 Au 79 Zn 30 Cd 48 Hg 80 B 5 Al 13 Ga 31 In 49 Tl 81 C 6 Si 14 Ge 32 Sn 50 Pb 82 N 7 P 15 As 33 Sb 51 Bi 83 O 8 S 16 Se 34 Te 52 Po 84 F 9 Cl 17 Br 35 I 53 At 85 Ne 10 Ar 18 Kr 36 Xe 54 Rn 86 Gases Properties of Gases As the temperature rises, all elements form a gas at some point. In the following diagram, Blue represents solids Green represents liquids Red represents gases At O K, all elements are solids At 6000 K, all are gases 1

2 Gases Gases Properties of Gases Gases have no shape and no volume. They take the volume and shape of the container Their densities are low usually measured in gl -1 The atoms or molecules of the gas are far further apart than in a solid or a liquid. Gases Gases as an ensemble of particles The attractive forces between liquids and solids are very strong LiF: M.p.: 848 C B. p.: 1676 C Solid Liquid Liquid Gas In a gas, the forces between particles are negligible and as there are no attractive forces, a gas will occupy the volume of the container. 2

3 Gases Gases as an ensemble of particles The structures of liquids and solids are well ordered on a microscopic level CaCl 2 Ethanol, C 2 H 5 OH Gases Gases as an ensemble of particles In a gas, there is no order and all the properties of the gas are isotropic all the properties of the gas are the same in all directions. Gas particles are distributed uniformly throughout the container. They can move throughout the container in straight line trajectories. Gases Gases as an ensemble of particles The directions of the motions of the gas particles are random and The velocities form a distribution there is a range of possible velocities around an average value. The trajectories of the gas particles are straight lines and there are two possible fates for a gas molecule... 3

4 Gases Gases as an ensemble of particles A gas particle can collide with the walls of the container Or another gas molecule When this happens, the gas particle changes direction. Gases Gases as an ensemble of particles Kinetic energy can be transferred between the two colliding particles one can slow down and the other speed up but the net change in kinetic energy is zero. These collisions are termed elastic, meaning that there is no overall change in kinetic energy. Gases The average kinetic energy for a given gas is determined by the temperature alone and the width and peak maximum is also determined by the temperature. The Maxwell-Boltzmann distribution for He 4

5 Gases Gases as an ensemble of particles The force exerted by the gas particles on the walls of the container gives rise to the pressure of the gas. We define pressure as the force exerted per unit area: P = Force = F Area A The unit of pressure is the Pascal (Pa) 1 Pa = 1 Nm -2 In practice, the Pascal is too small - kpa or GPa Gases Pressure measurement Pressure is also measured in several other non SI units: In industry: Pounds per square in (p.s.i.) In research: Pascal, atmosphere, bar, Torr Gases Pressure conversion factors Atmospheric pressure = 101,325 Pa 1 Atmosphere = 101,325 Pa = 1 bar 1 Atmosphere = 101,325 Pa = 1 bar = 760 Torr = 760 mmhg = 14.7 p.s.i. 5

6 Gases Pressure Measurement Pressure is measured using a manometer or barometer either one containing Hg or an electronic gauge A mercury manometer is a U tube connected to the gas vessel, with the other end either evacuated or open to the atmosphere. The measurement of the height difference between the mercury levels on both sides of the U gives the pressure... Gases Pressure Measurement Let the height difference between the two Hg levels be h Then the gas pressure is given by P gas = P 0 + h As P = Force = F = mg where g = 9.81 ms -2 Area A A Gases Pressure Measurement How is the height difference related to the pressure? As density, ρ = m V Then m = ρ V The volume of the column of mercury is V = A. h And so m = ρ V = ρ A. h 6

7 Gases Pressure Measurement The pressure above the baseline pressure P 0 is therefore P gas = mg =ρga. h = ρg h A A Gases Gases as an ensemble of particles Kinetic energy can be transferred between the two colliding particles one can slow down and the other speed up but the net change in kinetic energy is zero. These collisions are termed elastic, meaning that there is no overall change in kinetic energy. The factors that control the behavior of a gas are The nature of the gas 7

8 The factors that control the behavior of a gas are The nature of the gas The quantity of the gas - n The factors that control the behavior of a gas are The nature of the gas The quantity of the gas - n The pressure - P The factors that control the behavior of a gas are The nature of the gas The quantity of the gas - n The pressure - P The temperature - T 8

9 The factors that control the behavior of a gas are The nature of the gas The quantity of the gas - n The pressure - P The temperature - T The volume of a gas -V These laws apply to a perfect gas or and ideal gas. All gases behave as ideal gases at ordinary temperatures and pressures. The qualities of an ideal gas are: Zero size to the gas particles We assume that the volume of the container is very much larger than the total volume of the gas molecules No attractive forces between atoms These laws apply to a perfect gas or and ideal gas. All gases behave as ideal gases at ordinary temperatures and pressures. At low temperatures and high pressures gases deviate from ideality. The ideal gas laws are based on three interdependent laws Boyle s Law, Charles Law and Avogadro s Law. 9

10 Boyle s Law Robert Boyle experimented with gases in Oxford in He discovered that the product of the volume and the pressure of a gas is a constant, so long as the quantity of gas and the temperature are constant. Boyle s Law Mathematically, PV = a constant as long as n and T are constant Boyle s Law Mathematically, PV = a constant, k or P = k V as long as n and T are constant. 10

11 Boyle s Law A graph of Boyle s data shows this relationship: PV = k Boyle s Law A graph of 1 / P as the abscissa and V as the ordinate. V = k P The graph shows a straight line of slope k Boyle s Law As the pressure rises, 1 / P becomes smaller and the graph passes through the origin. This implies that at infinitely large pressure, the volume of a gas is zero. We know that molecules and and atoms have a definite volume, so Boyle s law must fail at very high pressures. 11

12 Charles Law Jacques Charles was a Feench scientist and aeronaut who discovered (1787) that all gases expand by the same amount when the temperature of the gas rises by the same amount. Charles Law Mathematically, we express this as V = k T And a graph of Charles Law is a straight line: Charles Law As the temperature is lowered, the volume decreases. 12

13 Charles Law The graph intersect the temperature axis and the zero volume occurs at o C, the lowest possible temperature. The Combined Gas Law for a Perfect Gas Combining Boyle s Law, Charles Law and Avogadro s Law, V = k and V = k T and V = k n P we can say that V nt P The Combined Gas Law for a Perfect Gas V nt P Or Rearranging we find V =K nt P PV = a constant nt 13

14 The Combined Gas Law for a Perfect Gas The constant is termed the Universal Gas Constant, R, and takes the value R = Jmol -1 K -1 So the Universal Gas Law is written as PV = nrt This relationship applies to all gases as long as they fulfill the conditions for near ideal behavior not at high pressure and not at low temperature Using the Combined Gas Law If the quantity of gas is the same, then changes in pressure, temperature or volume can be calculated easily as P 1 V 1 = n = P 2 V 2 RT 1 RT 2 Or P 1 V 1 = P 2 V 2 T 1 T 2 Using the Combined Gas Law The advantage of this expression is that the units do not matter; the units used for P 1,V 1, and T 1 will be returned in the calculation for P 2,V 2, and T 2. However, if you have to use PV = nrt, you must use the correct units which are consistent with R. The easiest way is to convert all temperatures to K, all pressures to Pa and all volumes to m 3 ; the value for R is then Jmol -1 K -1 14

15 The absolute temperature scale From Charles Law, the decrease in volume per unit temperature is always the same and therefore there must be a minimum temperature that can be reached. This is absolute zero O K, and is the zero point for the absolute temperature scale. The temperature in K is related to the temperature in o C through T/K = T/ o C Example: Molecular Mass determinations If we know the mass of gas in a sample of known volume, pressure and temperature, then we can calculate the relative molecular mass as we can calculate n. As n = m then, PV = mrt, so RMM = mrt RMM RMM PV RMM = mrt PV Example: Molar volumes From Avogadro s Law, equal quantities of gas occupy equal volumes. The volume of one mole of gas is therefore independent of the nature of the gas, as long as the gas behaves as ideal. One mole of a perfect gas at 0 o C and 1 atm pressure occupies 22.4 L 15

16 Example: Volumes and moles When we react solids or liquids, the easiest way is to measure the mass of the sample and then convert to moles by dividing by the relative molecular mass. For gases, the easiest way is to measure the pressure or the volume, as the densities of gases are so low. For these calculations, you must use the same temperatures and pressures for each gas. Partial pressures In a mixture of gases, we can measure the total pressure of the mixture P Total and therefore we can use PV = nrt to determine the total number of moles of gas present. As the mixture contains more than one gas, we can write the contribution of the pressure of each gas to the total pressure Partial pressures So the total pressure P total is written as the sum of all the individual pressures of the components of the gas mixture: 16

17 Partial pressures So the total pressure P total is written as the sum of all the individual pressures of the components of the gas mixture: P Total = P 1 + P 2 + P 3 + P Partial pressures So the total pressure P total is written as the sum of all the individual pressures of the components of the gas mixture: P Total = P 1 + P 2 + P 3 + P As PV = nrt then Partial pressures So the total pressure P total is written as the sum of all the individual pressures of the components of the gas mixture: P Total = P 1 + P 2 + P 3 + P As PV = nrt then n Total RT = n 1 RT + n 2 RT + n 3 RT + n 4 RT

18 Partial pressures So P Total = P 1 + P 2 + P 3 + P n Total RT = n 1 RT + n 2 RT + n 3 RT + n 4 RT +... Partial pressures So P Total = P 1 + P 2 + P 3 + P n Total RT = n 1 RT + n 2 RT + n 3 RT + n 4 RT +... n Total = n 1 + n 2 + n 3 + n Partial pressures So the pressures of each component of the gas mixture correlate with the number of moles of the gas component of the mixture a simple extension of Avogadro s Law. 18

19 Partial pressures We can also write the fraction of the total pressure that is due to one of the component: P Total = P 1 + P 2 + P 3 + P Partial pressures We can also write the fraction of the total pressure that is due to one of the component: P Total = P 1 + P 2 + P 3 + P n Total = n 1 + n 2 + n 3 + n Partial pressures We can also write the fraction of the total pressure that is due to one of the component: P Total = P 1 + P 2 + P 3 + P n Total = n 1 + n 2 + n 3 + n P 1 = n 1 RT 19

20 Partial pressures We can also write the fraction of the total pressure that is due to one of the component: P Total = P 1 + P 2 + P 3 + P n Total = n 1 + n 2 + n 3 + n P 1 = n 1 RT So, P 1 = n 1 P Total n 1 + n 2 + n 3 + n Partial pressures P 1 = n 1 P Total n 1 + n 2 + n 3 + n The fraction on the RHS is called the mole fraction and is written as x 1 so we can write Or P 1 = n 1 P Total n 1 + n 2 + n 3 + n P 1 = x 1 P Total Partial pressures The sum of the partial pressures for all the components and for the mole fractions is 1 and Σ i P i = P total Σ i x i = 1 20

21 Energy Energy is defined as the ability to do work. There are several forms of energy Kinetic energy energy due to motion E K = 1 / 2 mv 2 Potential energy the energy due to the position of a particle in a field e.g. Gravitational, electrical, magnetic etc. Energy The unit of energy is the Joule (J) and 1 J = 1 kgm 2 s -2 is the study of chemical energy and of the conversion of chemical energy into other forms of energy. It is part of thermodynamics the study of the flow of heat. Thermochemically, we define the system as the part of the universe under study and the surroundings as everything else. Systems come in three forms: Open Closed Isolated The system can exchange matter and energy with the surroundings The system can exchange energy only with the surroundings There is no exchange of matter or of energy with the surroundings 21

22 Matter is continually in motion and has an internal energy that is composed of several different types There is Translation Rotation Vibration Potential between molecules and inside molecules. The internal energy is written as U Matter is continually in motion and has an internal energy that is composed of several different types There is Translation Rotation Vibration Potential between molecules and inside molecules. The internal energy is written as U The internal energy is directly connected to heat and the transfer of heat. Heat is the transfer of internal energy between the surroundings and the system or between systems. The direction of the heat flow is indicated by the temperature heat flows along a Temperature gradient from high temperature to low temperature. When the temperature of the system and that of the surroundings are equal, the system is said to be in thermal equilibrium 22

23 Energy is the capacity to do work but what is work? Energy is the capacity to do work but what is work? Work is the action of a force over a distance. To be able to do work, we must be able to exert a force over a distance. During this process, energy is expended. Energy is the capacity to do work but what is work? Work is the action of a force over a distance. To be able to do work, we must be able to exert a force over a distance. During this process, energy is expended. w = F x d where w is the work, F is the force and d is the distance. Work is measured in Joules. 23

24 PV work When a gas expands against an external pressure, for example in a cylinder, against a constant weight (weight being a force...) the work done can be written as w = F x d As P = F A then F = PA Thus w = PAd and as Ad = V final V initial = V Then w = P V PV work By convention, the work done when a gas expands is negative, Thus w = - P V for an expanding gas State Functions The state of a system is defined by the precise conditions of the system: The quantity and type of matter present The temperature and pressure The molecular structure of the system As 1 mole = 6.02 x particles, defining the state of a system uniquely is experimentally impossible in an absolute sense. 24

25 State Functions and U The internal energy, U, of a system is a function of the state of the system. Although we cannot measure the absolute state of a system, we can measure changes in the state of the system in a relative way, by measuring the work and the heat that takes place during a chemical change. As U is a function of the state of the system, it does not depend on the way the state of the system is prepared it is independent of the path. State Functions and U U is therefore a state function of the system. It depends only on the present state of the system and not on the previous history or the path by which the system was prepared. Because we have no measure of the state of a system, or of the internal energy, we can only measure the change in the state, through the observation of work and transfers of heat into and out of the system. Internal Energy, U and State Functions Energy, and therefore the capacity to do work is present in all matter. This internal energy is stored in translational, rotational, vibrational and potential forms or modes in the material. The exact distribution of energy defines the state of the system, together with external variables such as pressure, temperature. 25

26 Internal Energy, U and State Functions U is a function of the state of the material only, not of the history of the sample or the path taken to prepare the state of the sample. Heat is the transfer of energy between the surroundings and the sample - the symbol for heat is q Work is the result of a force acting over a distance - the symbol for work is w Internal Energy, U and State Functions Heat and work are the only two ways of changing the internal energy of a system. Temperature is defined by the direction of the flow of heat, which is always from high temperature to low temperature. When the the temperature of the system and the surroundings are the same, the system is at thermal equilibrium with it s surroundings. The sign conventions of thermochemistry When the internal energy of the system rises, this energy change has a positive sign. - The energy of the system rises when heat is absorbed - The energy of the system rises when work is done on the system e.g. a gas is compressed - in these cases, q is positive w is positive 26

27 The sign conventions of thermochemistry When the internal energy of the system lowers, this energy change has a negative sign. - The energy of the system lowers when heat is leaves the system - The energy of the system rises when the system does work e.g. a gas expands against an external pressure - in these cases, q is negative w is negative Internal energy rises: q > 0 w > 0 Internal energy drops: q < 0 w < 0 The First Law of Thermodynamics Energy can be exchanged but cannot be created or destroyed. It is a statement of the Law of Conservation of Energy U = U final U initial = q + w 27

28 Chemical applications of the 1 st Law Any chemical change can be characterized as an Endothermic change or an Exothermic change. In an exothermic reaction, internal chemical energy is converted into heat, which leaves the system if the system is not isolated or causes the temperature to rise if the system in isolated. Chemical applications of the 1 st Law In an endothermic reaction, heat is required to drive the chemical reaction and in an isolated system, the temperature will fall. In an non-isolated system, heat is absorbed from the surroundings. Exothermic T rises (isolated) q negative (non-isolated) Endothermic T falls (isolated) q positive (non-isolated) Reactions at constant pressure and constant volume At constant volume, V = 0 and so U V = q V - P V U V = q V + 0 = q V When the system can do PV work, i.e. a system at constant pressure, U P = q P - P V where w = - P V 28

29 Most reactions take place at constant pressure and therefore we define a new function, which is a state function in the same way that U is a state function Rearranging U P = q P - P V U P + P V = q P We term q P the enthalpy of the reaction q P = H = U P + P V Enthalpy is an extensive property one that depends on the quantity of the material present in the reaction. This follows directly from the fact that the enthalpy is the heat generated by a reaction there is more energy released from 1000 kg of methane when it burns than from 1 g. Enthalpies and internal energies are measured in kj mol -1 and the stoichiometry of a reaction is directly applicable to the enthalpy half the quantity of the reaction results in half the enthalpy change taking place. 29

30 We can characterize reactions as endothermic or exothermic using the enthalpy, H. If the enthalpy change is negative, the reaction is Reactants exothermic and heat is given out by the system H < 0, negative Products H We can characterize reactions as endothermic or exothermic using the enthalpy, H. If the enthalpy change is negative, the reaction is endothermic and heat is absorbed by the system H >0, positive Products Reactants H Using the enthalpy, we can account for the heat entering a reaction at constant pressure in the same way that we account for the products and reactants in a reaction. In an endothermic reaction, the energy absorbed by the system can be considered as a reactant. Conversely, an exothermic reaction, one which evolves heat, has the energy as a product. 30

31 Enthalpies and internal energies are measured in kj mol -1 and the stoichiometry of a reaction is directly applicable to the enthalpy half the quantity of the reaction results in half the enthalpy change taking place. DO EXCERCISES AND EXAMPLES (pp ) BRING THEM TO THE NEXT DISCUSSION PERIOD Heat Capacities When a definite quantity of energy is absorbed by materials, the temperature rises.with different materials, the temperature rise, T, is different. The quantity of energy required to raise a quantity of material by 1 K is termed the heat capacity. Mathematically, C = q T where C is the heat capacity, q is the heat. Heat Capacities The specific heat is the heat per gram of sample and the molar heat capacity is the heat capacity per mole. 31

32 Specific Heats, Molar Heats and Calorimetry The heat capacity is the quantity of heat required to raise a given quantity of a substance by 1 K The specific heat The molar heat 1 gram though 1 K 1 mole through 1 K The units of heat capacity are Jg -1 K -1 (specific heat) or Jmol -1 K -1 (molar heat) Specific Heats, Molar Heats and Calorimetry To calculate the heat transferred to a sample we use q = quantity x heat capacity x T For the specific heat q = mc s T where m = mass For the molar heat q = nc m T where n = no. of moles Make sure that the units of the heat capacity matches the units of quantity that is in the heat equation Specific Heats, Molar Heats and Calorimetry To measure the heat capacity, a calorimeter is used. A calorimeter measures heat transfers, heats of reaction or heats of dissolution. 32

33 Specific Heats, Molar Heats and Calorimetry In principle, they consist of an insulated chamber and an accurate way of measuring temperature (a thermocouple or thermometer). Insulation ensures that the only heat involved in the temperature rise is that inside the calorimeter. Heat capacity measurements A sample with a known temperature is placed into a fluid of known heat capacity and known temperature and allowed to come to thermal equilibrium. Heat capacity measurements A sample with a known temperature is placed into a fluid of known heat capacity and known temperature and allowed to come to thermal equilibrium. At thermal equilibrium, T sample = T fluid and so we know T for the sample and for the fluid. 33

34 Heat capacity measurements A sample with a known temperature is placed into a fluid of known heat capacity and known temperature and allowed to come to thermal equilibrium. At thermal equilibrium, T sample = T fluid and so we know T for the sample and for the fluid. We also know C fluid and therefore we know q fluid, the heat transferred into the fluid - q = C fluid T fluid Heat capacity measurements A sample with a known temperature is placed into a fluid of known heat capacity and known temperature and allowed to come to thermal equilibrium. At thermal equilibrium, T sample = T fluid and so we know T for the sample and for the fluid. We also know C fluid and therefore we know q fluid, the heat transferred into the fluid - q = C fluid T fluid As this is the only source of heat in the calorimeter, we know q fluid and T sample, so we can calculate C sample Example 15.5g of alloy A has a temperature of 98.9 o C. It is placed into a calorimeter containing 25 g of water at 22.5 o C. Thermal equilibrium is achieved at 25.7 o C. What is the heat capacity of A? 34

35 Example 15.5g of alloy A has a temperature of 98.9 o C. It is placed into a calorimeter containing 25 g of water at 22.5 o C. Thermal equilibrium is achieved at 25.7 o C. What is the heat capacity of A? C water = 4.18 Jg -1 K Calculate q water 2. q water = - q A from conservation of energy 3. Calculate C A from q A Example 15.5g of alloy A has a temperature of 98.9 o C. It is placed into a calorimeter containing 25 g of water at 22.5 o C. Thermal equilibrium is achieved at 25.7 o C. What is the heat capacity of A? C water = 4.18 Jg -1 K Calculate q water : T water = T final T initial = ( ) o C = 3.2 o C q water = 25 x 4.18 x 3.2 = 334 J Note: q water is positive as heat is entering the water Example 15.5g of alloy A has a temperature of 98.9 o C. It is placed into a calorimeter containing 25 g of water at 22.5 o C. Thermal equilibrium is achieved at 25.7 o C. What is the heat capacity of A? C water = 4.18 Jg -1 K q water = 334 J 2. q water = - q A thus q A = J 35

36 Example 15.5g of alloy A has a temperature of 98.9 o C. It is placed into a calorimeter containing 25 g of water at 22.5 o C. Thermal equilibrium is achieved at 25.7 o C. What is the heat capacity of A? C water = 4.18 Jg -1 K q water = 334 J 2. q water = - q A thus q A = J 3. q A = mc A T A T A = T final T initial = ( ) o C = o C Example 15.5g of alloy A has a temperature of 98.9 o C. It is placed into a calorimeter containing 25 g of water at 22.5 o C. Thermal equilibrium is achieved at 25.7 o C. What is the heat capacity of A? C water = 4.18 Jg -1 K q water = 334 J 2. q water = - q A thus q A = J 3. q A = mc A T A ; T A = o C C A = q A /m T A = -334/(15.5 x 73.2) = 0.29 Jg -1 K -1 Bomb Calorimetry For reactions which generate gas, the P V work makes a significant contribution and the quanitiy we will measure in an open calorimeter is the enthalpy. We cannot easily measure the P V work in this case. We can measure U in a bomb calorimeter one where the volume change is zero and therefore V = 0. The calorimeter is calibrated using a known sample. 36

37 Hess Law of Summation If we wish to determine the heat of reaction or formation of a compound which is not stable, cannot be isolated or cannot be measured for some reason, we use Hess Law to determine this quantity. Hess law states that the the heat of reaction is constant and is not determined by the path of the reaction. We know this as U (and H) is a state function Hess Law of Summation Practically, if we can find a cycle of reactions that is measureable, then we can derive the unmeasurable quantity as we know the total sum of all the energy changes in the cycle. Hess Law of Summation Example The combustion of C results in the formation of CO 2 in a bomb calorimeter. The heat of formation of CO is therefore hard to measure. We can measure the heat of combustion of CO and that of C both to give CO 2. 37

38 Hess Law of Summation H f (CO) C graphite + O 2 CO + 1 / 2 O 2 H f (CO 2 ) H combustion (CO) CO 2 Hess Law of Summation H f (CO) C graphite + O 2 CO + 1 / 2 O 2 H f (CO 2 ) H combustion (CO) CO 2 Of the reactions in this cycle, the heats of combustion of CO and C are known, but the heat of formation of CO from C is not. Hess Law of Summation H f (CO) C graphite + O 2 CO + 1 / 2 O 2 H f (CO 2 ) H combustion (CO) CO 2 H f (CO 2 ) = H f (CO) + H combustion (CO) 38

39 Hess Law of Summation H f (CO) C graphite + O 2 CO + 1 / 2 O 2 H f (CO 2 ) H combustion (CO) CO 2 H f (CO 2 ) = H f (CO) + H combustion (CO) H f (CO) = H f (CO 2 ) - H combustion (CO) Hess Law of Summation Using the lower equation and the values for the heats of combustion of CO and C, we can calculate the unknown heat in the cycle H f (CO) C graphite + O 2 CO + 1 / 2 O 2 H f (CO 2 ) CO 2 H combustion (CO) H f (CO 2 ) = H f (CO) + H combustion (CO) H f (CO) = H f (CO 2 ) - H combustion (CO) Hess Law of Summation Using the lower equation and the values for the heats of combustion of CO and C, we can calculate the unknown heat in the cycle H f (CO) C graphite + O 2 CO + 1 / 2 O 2 H f (CO 2 ) CO 2 H combustion (CO) H f (CO 2 ) = H f (CO) + H combustion (CO) H f (CO) = H f (CO 2 ) - H combustion (CO) H f (CO 2 ) = kj H combustion (CO) = kj H f (CO 2 ) = H f (CO 2 ) - H combustion (CO) 39

40 Hess Law of Summation Using the lower equation and the values for the heats of combustion of CO and C, we can calculate the unknown heat in the cycle H f (CO) C graphite + O 2 CO + 1 / 2 O 2 H f (CO 2 ) CO 2 H combustion (CO) H f (CO 2 ) = H f (CO) + H combustion (CO) H f (CO) = H f (CO 2 ) - H combustion (CO) H f (CO 2 ) = kj H combustion (CO) = kj H f (CO 2 ) = ( ) ( ) = kj Standard enthalpies of formation and reaction Just as we cannot determine the absolute value for the internal energy of a system and so concentrate on the change in internal energy, so we cannot fix an absolute zero-point for reaction and formation enthalpies. We chose the Standard state of a material as that at 1 bar pressure (1 bar = 1 x 10 5 Pa) and the temperature of interest. Standard enthalpies of formation and reaction The standard enthalpy of formation of an element in the standard state is defined as zero. Using these two facts, we can calculate the heats of formation and, through Hess cycles, the heats of reaction for all substances. 40

41 Standard enthalpies of formation and reaction When we combine different reactions, we must take account of the stoichiometry of the reaction. Remember that H can be thought of as a product of reaction and must be combine with the correct stoichiometry. Standard enthalpies of formation and reaction For the reaction H dimerization (NO 2 ) 2NO 2 N 2 O 4 We can construct a Hess cycle: Standard enthalpies of formation and reaction For the reaction H dimerization (NO 2 ) 2NO 2 N 2 O 4 We can construct a Hess cycle: H dimerization (NO 2 ) 2NO 2 N 2 O 4 H f (NO 2 ) 1 / 2 H f (NO 2 ) 1 / 2 N 2 + O 2 41

42 Standard enthalpies of formation and reaction For the reaction H dimerization (NO 2 ) 2NO 2 N 2 O 4 We can construct a Hess cycle. Note that we must include the stoichiometry in the calculation. H f (NO 2 ) H dimerization (NO 2 ) 2NO 2 N 2 O 4 1 / 2 N 2 + O 2 1 / 2 H f (NO 2 ) 42