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Gases Chemistry 120 Properties of Gases Kinetic Molecular Theory of Gases Pressure Boyle’s and Charles’ Law The Ideal Gas Law Gas reactions Partial pressures Gases Chemistry 120 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 1 2 Li 3 3 Na 11 4 K 19 5 Rb 37 6 Cs 55 He 2 Room Temperature Gases 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 Gases 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 Chemistry 120 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 Gases Chemistry 120 Gases Chemistry 120 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 Chemistry 120 Gases as an ensemble of particles The attractive forces between liquids and solids are very strong LiF: M.p.: 848°C Solid Liquid B. p.: 1676°C 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 Gases Gases as an ensemble of particles Chemistry 120 The structures of liquids and solids are well ordered on a microscopic level CaCl2 Ethanol, C2H5OH Gases Gases as an ensemble of particles Chemistry 120 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 Chemistry 120 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 Gases Gases as an ensemble of particles Chemistry 120 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 Chemistry 120 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 Chemistry 120 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 Gases Gases as an ensemble of particles Chemistry 120 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 Chemistry 120 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 Chemistry 120 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 Gases Pressure Measurement Chemistry 120 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 Chemistry 120 Let the height difference between the two Hg levels be ∆h Then the gas pressure is given by Pgas = P0 + ∆h As P = Force = F = mg where g = 9.81 ms-2 Area A A Gases Pressure Measurement Chemistry 120 How is the height difference related to the pressure? As density, ρ = m V Then m=ρV The volume of the column of mercury is V = A.∆h And so m = ρ V = ρ A.∆h 6 Gases Pressure Measurement Chemistry 120 The pressure above the baseline pressure P0 is therefore Pgas = mg =ρgA.∆h = ρg∆h A A Gases Gases as an ensemble of particles Chemistry 120 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 Gas Laws Chemistry 120 The factors that control the behavior of a gas are • The nature of the gas 7 The Gas Laws Chemistry 120 The factors that control the behavior of a gas are • The nature of the gas • The quantity of the gas - n The Gas Laws Chemistry 120 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 Gas Laws Chemistry 120 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 The Gas Laws Chemistry 120 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 The Gas Laws Chemistry 120 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 The Gas Laws Chemistry 120 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 The Gas Laws Chemistry 120 Boyle’s Law Robert Boyle experimented with gases in Oxford in 1660. 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. The Gas Laws Chemistry 120 Boyle’s Law Mathematically, PV = a constant as long as n and T are constant The Gas Laws Chemistry 120 Boyle’s Law Mathematically, PV = a constant, k or P=k V as long as n and T are constant. 10 The Gas Laws Chemistry 120 Boyle’s Law A graph of Boyle’s data shows this relationship: PV = k The Gas Laws Chemistry 120 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 The Gas Laws Chemistry 120 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 The Gas Laws Chemistry 120 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. The Gas Laws Chemistry 120 Charles’ Law Mathematically, we express this as V = k’T And a graph of Charles’ Law is a straight line: The Gas Laws Chemistry 120 Charles’ Law As the temperature is lowered, the volume decreases……. 12 The Gas Laws Chemistry 120 Charles’ Law The graph intersect the temperature axis and the zero volume occurs at –273.16oC, the lowest possible temperature. Chemistry 120 The Gas Laws The Combined Gas Law for a Perfect Gas Combining Boyle’s Law, Charles’ Law and Avogadro’s Law, V=k P and V = k’T and V = k”n we can say that V ∝ nT P Chemistry 120 The Gas Laws The Combined Gas Law for a Perfect Gas V ∝ nT P Or V =K nT P Rearranging we find PV = a constant nT 13 Chemistry 120 The Gas Laws The Combined Gas Law for a Perfect Gas The constant is termed the Universal Gas Constant, R, and takes the value R = 8.314 Jmol-1K-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 The Gas Laws Using the Combined Gas Law Chemistry 120 If the quantity of gas is the same, then changes in pressure, temperature or volume can be calculated easily as P1V1 = n = P2V2 RT1 RT2 Or P1V1 = P2V2 T1 T2 The Gas Laws Using the Combined Gas Law Chemistry 120 The advantage of this expression is that the units do not matter; the units used for P1 ,V1, and T1 will be returned in the calculation for P2 ,V2, and T2. 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 m3; the value for R is then 8.314 Jmol-1K-1 14 The Gas Laws The absolute temperature scale Chemistry 120 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 oC through T/K = T/oC + 273.16 Chemistry 120 The Gas Laws 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 The Gas Laws Example: Molar volumes Chemistry 120 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 0oC and 1 atm pressure occupies 22.4 L 15 The Gas Laws Example: Volumes and moles Chemistry 120 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. The Gas Laws Partial pressures Chemistry 120 In a mixture of gases, we can measure the total pressure of the mixture – PTotal 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 The Gas Laws Partial pressures Chemistry 120 So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture: 16 The Gas Laws Partial pressures Chemistry 120 So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture: PTotal = P1 + P2 + P3 + P4 + ........... The Gas Laws Partial pressures Chemistry 120 So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture: PTotal = P1 + P2 + P3 + P4 + ........... As PV = nRT then The Gas Laws Partial pressures Chemistry 120 So the total pressure Ptotal is written as the sum of all the individual pressures of the components of the gas mixture: PTotal = P1 + P2 + P3 + P4 + ........... As PV = nRT then nTotalRT = n1RT + n2RT + n3RT + n4RT + ........... 17 The Gas Laws Partial pressures Chemistry 120 So PTotal = P1 + P2 + P3 + P4 + ........... nTotalRT = n1RT + n2RT + n3RT + n4RT + ........... The Gas Laws Partial pressures Chemistry 120 So PTotal = P1 + P2 + P3 + P4 + ........... nTotalRT = n1RT + n2RT + n3RT + n4RT + ........... nTotal = n1 + n2 + n3 + n4 + ........... The Gas Laws Partial pressures Chemistry 120 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 The Gas Laws Partial pressures Chemistry 120 We can also write the fraction of the total pressure that is due to one of the component: PTotal = P1 + P2 + P3 + P4 + ........... The Gas Laws Partial pressures Chemistry 120 We can also write the fraction of the total pressure that is due to one of the component: PTotal = P1 + P2 + P3 + P4 + ........... nTotal = n1 + n2 + n3 + n4 + .......... The Gas Laws Partial pressures Chemistry 120 We can also write the fraction of the total pressure that is due to one of the component: PTotal = P1 + P2 + P3 + P4 + ........... nTotal = n1 + n2 + n3 + n4 + .......... P1 = n1RT 19 The Gas Laws Partial pressures Chemistry 120 We can also write the fraction of the total pressure that is due to one of the component: PTotal = P1 + P2 + P3 + P4 + ........... nTotal = n1 + n2 + n3 + n4 + .......... P1 = n1RT = n1 So, P1 PTotal n1 + n2 + n3 + n4 + .......... The Gas Laws Partial pressures P1 = PTotal Chemistry 120 n1 n1 + n2 + n3 + n4 + .......... The fraction on the RHS is called the mole fraction and is written as x1 so we can write P1 = Or n1 PTotal n1 + n2 + n3 + n4 + .......... P1 = x1 PTotal The Gas Laws Partial pressures Chemistry 120 The sum of the partial pressures for all the components and for the mole fractions is 1 ΣiPi = Ptotal and Σixi = 1 20 Thermochemistry Energy Chemistry 120 Energy is defined as the ability to do work. There are several forms of energy Kinetic energy – energy due to motion EK = 1/2mv2 Potential energy – the energy due to the position of a particle in a field e.g. Gravitational, electrical, magnetic etc. Thermochemistry Energy Chemistry 120 The unit of energy is the Joule (J) and 1 J = 1 kgm2s-2 Thermochemistry 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. Chemistry 120 Thermochemistry 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 The system can exchange matter and energy with the surroundings Closed The system can exchange energy only with the surroundings Isolated There is no exchange of matter or of energy with the surroundings 21 Chemistry 120 Thermochemistry 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 Chemistry 120 Thermochemistry 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. Chemistry 120 Thermochemistry 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 Thermochemistry Energy is the capacity to do work Chemistry 120 but what is work? Thermochemistry Energy is the capacity to do work Chemistry 120 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. Thermochemistry Energy is the capacity to do work Chemistry 120 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=Fxd where w is the work, F is the force and d is the distance. Work is measured in Joules. 23 Thermochemistry PV work Chemistry 120 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=Fxd As Thus P=F A w = PAd and as Ad = Vfinal – Vinitial = ∆V Then w = P∆V then F = PA Thermochemistry PV work Chemistry 120 By convention, the work done when a gas expands is negative, Thus w = - P∆V for an expanding gas Thermochemistry State Functions Chemistry 120 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 1023 particles, defining the state of a system uniquely is experimentally impossible in an absolute sense. 24 Thermochemistry State Functions and U Chemistry 120 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. Thermochemistry State Functions and U Chemistry 120 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. Thermochemistry Internal Energy, U and State Functions Chemistry 120 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 Thermochemistry Internal Energy, U and State Functions Chemistry 120 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 Thermochemistry Internal Energy, U and State Functions Chemistry 120 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. Chemistry 120 Thermochemistry 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 Chemistry 120 Thermochemistry 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 Thermochemistry Internal energy rises: Chemistry 120 q>0 w>0 Internal energy drops: q<0 w<0 Thermochemistry The First Law of Thermodynamics Chemistry 120 Energy can be exchanged but cannot be created or destroyed. It is a statement of the Law of Conservation of Energy ∆U = Ufinal – Uinitial = q + w 27 Thermochemistry Chemical applications of the 1st Law Chemistry 120 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. Thermochemistry Chemical applications of the 1st Law Chemistry 120 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) Chemistry 120 Thermochemistry Reactions at constant pressure and constant volume At constant volume, ∆V = 0 and so ∆UV = qV - P∆V ∆UV = qV + 0 = qV When the system can do PV work, i.e. a system at constant pressure, ∆UP = qP - P∆V where w = - P∆V 28 Chemistry 120 Thermochemistry 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 ∆UP = qP - P∆V ∆UP + P∆V = qP We term qP the enthalpy of the reaction qP = ∆H = ∆UP + P∆V Chemistry 120 Thermochemistry 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. Chemistry 120 Thermochemistry 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 Chemistry 120 Thermochemistry We can characterize reactions as endothermic or exothermic using the enthalpy, ∆H. If the enthalpy change is negative, the reaction is exothermic and heat is given out by the system Reactants ∆H < 0, negative Products H Chemistry 120 Thermochemistry 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 Products ∆H >0, positive Reactants H Chemistry 120 Thermochemistry 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 Chemistry 120 Thermochemistry 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 6.1 – 6.5 (pp 235 – 245) BRING THEM TO THE NEXT DISCUSSION PERIOD Thermochemistry Heat Capacities Chemistry 120 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. Thermochemistry Heat Capacities Chemistry 120 The specific heat is the heat per gram of sample and the molar heat capacity is the heat capacity per mole. 31 Chemistry 120 Thermochemistry 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 1 gram though 1 K The molar heat 1 mole through 1 K The units of heat capacity are Jg-1K-1 (specific heat) or Jmol-1K-1 (molar heat) Chemistry 120 Thermochemistry 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 = mCs∆T where m = mass For the molar heat q = nCm∆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 Chemistry 120 Thermochemistry 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 Chemistry 120 Thermochemistry 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. Thermochemistry Heat capacity measurements Chemistry 120 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. Thermochemistry Heat capacity measurements Chemistry 120 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, Tsample = Tfluid and so we know ∆T for the sample and for the fluid. 33 Thermochemistry Heat capacity measurements Chemistry 120 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, Tsample = Tfluid and so we know ∆T for the sample and for the fluid. We also know Cfluid and therefore we know qfluid, the heat transferred into the fluid - q = Cfluid∆Tfluid Thermochemistry Heat capacity measurements Chemistry 120 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, Tsample = Tfluid and so we know ∆T for the sample and for the fluid. We also know Cfluid and therefore we know qfluid, the heat transferred into the fluid - q = Cfluid∆Tfluid As this is the only source of heat in the calorimeter, we know qfluid and ∆Tsample, so we can calculate Csample Thermochemistry Example Chemistry 120 15.5g of alloy A has a temperature of 98.9 oC. It is placed into a calorimeter containing 25 g of water at 22.5oC. Thermal equilibrium is achieved at 25.7 oC. What is the heat capacity of A? 34 Thermochemistry Example Chemistry 120 15.5g of alloy A has a temperature of 98.9 oC. It is placed into a calorimeter containing 25 g of water at 22.5 oC. Thermal equilibrium is achieved at 25.7 oC. What is the heat capacity of A? Cwater = 4.18 Jg-1K-1 1. Calculate qwater 2. qwater = - qA from conservation of energy 3. Calculate CA from qA Thermochemistry Example Chemistry 120 15.5g of alloy A has a temperature of 98.9 oC. It is placed into a calorimeter containing 25 g of water at 22.5 oC. Thermal equilibrium is achieved at 25.7 oC. What is the heat capacity of A? Cwater = 4.18 Jg-1K-1 1. Calculate qwater: ∆Twater = Tfinal – Tinitial = (25.7 – 22.5) oC = 3.2 oC qwater= 25 x 4.18 x 3.2 = 334 J Note: qwater is positive as heat is entering the water Thermochemistry Example Chemistry 120 15.5g of alloy A has a temperature of 98.9 oC. It is placed into a calorimeter containing 25 g of water at 22.5 oC. Thermal equilibrium is achieved at 25.7 oC. What is the heat capacity of A? Cwater = 4.18 Jg-1K-1 1. qwater = 334 J 2. qwater = - qA thus qA = - 334 J 35 Thermochemistry Example Chemistry 120 15.5g of alloy A has a temperature of 98.9 oC. It is placed into a calorimeter containing 25 g of water at 22.5 oC. Thermal equilibrium is achieved at 25.7 oC. What is the heat capacity of A? Cwater = 4.18 Jg-1K-1 1. qwater = 334 J 2. qwater = - qA thus qA = - 334 J 3. qA = mCA∆TA ∆TA = Tfinal – Tinitial = (25.7 – 98.9) oC = -73.2 oC Thermochemistry Example Chemistry 120 15.5g of alloy A has a temperature of 98.9 oC. It is placed into a calorimeter containing 25 g of water at 22.5 oC. Thermal equilibrium is achieved at 25.7 oC. What is the heat capacity of A? Cwater = 4.18 Jg-1K-1 1. qwater = 334 J 2. qwater = - qA thus qA = - 334 J 3. qA = mCA∆TA; ∆TA = -73.2 oC CA = qA/m∆TA = -334/(15.5 x –73.2) = 0.29 Jg-1K-1 Thermochemistry Chemistry 120 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 Thermochemistry Chemistry 120 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 Thermochemistry Chemistry 120 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. Thermochemistry Chemistry 120 Hess’ Law of Summation Example The combustion of C results in the formation of CO2 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 CO2. 37 Thermochemistry Chemistry 120 Hess’ Law of Summation ∆Hf(CO) CO + 1/2O2 Cgraphite + O2 ∆Hcombustion(CO) ∆Hf(CO2) CO2 Thermochemistry Chemistry 120 Hess’ Law of Summation ∆Hf(CO) CO + 1/2O2 Cgraphite + O2 ∆Hcombustion(CO) ∆Hf(CO2) CO2 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. Thermochemistry Chemistry 120 Hess’ Law of Summation ∆Hf(CO) Cgraphite + O2 ∆Hf(CO2) CO + 1/2O2 ∆Hcombustion(CO) CO2 ∆Hf(CO2) = ∆Hf(CO) + ∆Hcombustion(CO) 38 Thermochemistry Chemistry 120 Hess’ Law of Summation ∆Hf(CO) Cgraphite + O2 ∆Hf(CO2) CO + 1/2O2 ∆Hcombustion(CO) CO2 ∆Hf(CO2) = ∆Hf(CO) + ∆Hcombustion(CO) ∆Hf(CO) = ∆Hf(CO2) - ∆Hcombustion(CO) Thermochemistry Hess’ Law of Summation C 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 Chemistry 120 ∆Hf(CO) graphite CO + 1/2O2 + O2 ∆Hcombustion(CO) ∆Hf(CO2) CO2 ∆Hf(CO2) = ∆Hf(CO) + ∆Hcombustion(CO) ∆Hf(CO) = ∆Hf(CO2) - ∆Hcombustion(CO) Thermochemistry Hess’ Law of Summation C 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 Chemistry 120 ∆Hf(CO) graphite + O2 ∆Hf(CO2) CO + 1/2O2 ∆Hcombustion(CO) CO2 ∆Hf(CO2) = ∆Hf(CO) + ∆Hcombustion(CO) ∆Hf(CO) = ∆Hf(CO2) - ∆Hcombustion(CO) ∆Hf(CO2) = - 393.5 kJ ∆Hcombustion(CO) = - 283.0 kJ ∆Hf(CO2) = ∆Hf(CO2) - ∆Hcombustion(CO) 39 Thermochemistry Hess’ Law of Summation C 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 Chemistry 120 ∆Hf(CO) graphite + O2 CO + 1/2O2 ∆Hcombustion(CO) ∆Hf(CO2) CO2 ∆Hf(CO2) = ∆Hf(CO) + ∆Hcombustion(CO) ∆Hf(CO) = ∆Hf(CO2) - ∆Hcombustion(CO) ∆Hf(CO2) = - 393.5 kJ ∆Hcombustion(CO) = - 283.0 kJ ∆Hf(CO2) = (- 393.5) – (- 283.0) = -110.5 kJ Thermochemistry Chemistry 120 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 105 Pa) and the temperature of interest. Thermochemistry Chemistry 120 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 Thermochemistry Chemistry 120 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. Thermochemistry Chemistry 120 Standard enthalpies of formation and reaction For the reaction ∆Hdimerization(NO2) 2NO2 N2O4 We can construct a Hess’ cycle: Thermochemistry Chemistry 120 Standard enthalpies of formation and reaction For the reaction ∆Hdimerization(NO2) 2NO2 N2O4 ∆Hdimerization(NO2) We can construct a Hess’ cycle: 2NO2 ∆Hf(NO2) N2O4 /2∆Hf(NO2) 1 1 /2N2+ O2 41 Thermochemistry Chemistry 120 Standard enthalpies of formation and reaction For the reaction ∆Hdimerization(NO2) 2NO2 N2O4 ∆Hdimerization(NO2) We can construct a Hess’ cycle. Note that we must include the stoichiometry in the calculation. 2NO2 ∆Hf(NO2) N2O4 /2∆Hf(NO2) 1 1 /2N2+ O2 42