Thermodynamics Fundamentals PDF

Thermodynamics: fundamentals Thermodynamics describes the macroscopic state of a (microscopically) complex system through a small number of macroscopic variables (e.g. pressure, temperature), the so-called state-variables, and through the so-called thermodynamic potentials. This subject can to...

Thermodynamics: fundamentals Thermodynamics describes the macroscopic state of a (microscopically) complex system through a small number of macroscopic variables (e.g. pressure, temperature), the so-called state-variables, and through the so-called thermodynamic potentials. This subject can touch an enormously wide range of phenomena (i.e. efficiency of a heat engine, heat pumps and refrigerators, taking chemistry on the way, and reaching as far as processes of life) Thermodynamics summarises the properties of energy and its transformation from one form to another Do not think that thermodynamics is “only” about steam engines: it is about almost everything Universe System What we are interested in (i.e. a block of iron, a beaker of water, an engine, a human body, etc.) Surroundings The remainder of the Universe outside the system (where we stand to make observations about the system and infer its properties) The system is defined by its boundaries OPEN: matter and energy can be exchanged between system and surroundings (i.e. an open flasc) Surroundings System Matter Energy  CLOSED(DIATHERMIC): ONLY energy can be exchanged (i.e. a sealed bottle) Surroundings System Matter Energy  ISOLATED (ADIABATIC): neither matter or energy can be exchanged (i.e. a stoppered vacuum flasc) Surroundings System Matter Energy The properties of a system depend on the prevailing conditions Two classes: Extensive Properties Depending on the quantity of matter in the system (i.e. mass, volume, etc. – 2 Kg of iron occupy twice the volume of 1 Kg of iron) Intensive Properties Independent of the amount of matter present (i.e. temperature, density, etc. - The density of iron is 8.9 Kg.cm- 3 regardless of whether we have a 1 Kg block or a 2 Kg block) *State and Path Functions In Thermodynamics it is important to distinguish between two distinct classes of function. The Internal Energy (U) (and as we shall see later the Enthalpy, H and Entropy, S) are state functions whereas the heat (q) and work (w) are path functions. A state function is one whose value depends only on the state of the substance under consideration; it has the same value for a given state no matter how that state came about. In contrast, a path function depends on the path which the system takes in going between two states. Thermodynamics: fundamentals The most important thermodynamics 4 laws: 0th Law of Thermodynamics: All parts of system within a thermodynamic equilibrium have the same temperature. 1st Law of Thermodynamics: Conservation of energy. The inner energy U of a system can only be changed by heat supply to/extraction from or work done on/performed by the system. (in other words: energy can not be created or destroyed, but only transformed) 2nd Law of Thermodynamics: Direction of state-changes. Reversible processes have zero-change in entropy S. Irreversible (spontaneous) processes have a positive change in entropy. In other words: state-changes will follow the direction of maximum entropy(-change). 3rd Law of Thermodynamics (Nernst Theorem): Approaching zero-temperature (T → 0) , the entropy becomes constant (set to 0). That is, the absolute zero- temperature (T = 0) cannot be reached. 0th Law of Thermodynamics: All parts of system within a thermodynamic equilibrium have the same temperature. (I) If two bodies/systems having different temperature are brought in “contact” (i.e. interaction of any kind is possible, physical contact is not necessary), then the warmer body will get colder and the colder body will get warmer unti an equilibrium temperature has been reached. This is the basis of all thermometers. Temperature The (thermodynamic) temperature T is measured in Kelvin (K), named after Lord Kelvin (W. Thomson, 1824-1907) who defined the temperature-scale by the triple-point of water. The triple-point is the point at which the solid, liquid and gaseous state of a system is in equilibrium. 1K is the 273,16th part of the thermodynamic temperature of the triple- point of water (0.01°C). The Kelvin-scale is equivalent to the scale already proposed in 1742 by A. Celsius (1701- 1744). The Celsius Scale uses the melting- and the boiling-point of water (0 ̊ and 100 ̊, respectively) under normal atmospheric pressure to define temperature.  T The two scales are connected through: = − 273,15 C K Comment: the Fahrenheit-scale is not very practical today and use a different temperature scale – but it is still use in some countries (Cayman Islands, Palau, Bahamas and Belize and United States)...lower defining point was the freezing point of brine Temperature measurement (thermometers): Temperature-measurements can be done with a wide range of parameters. The only condition is that the parameter measured has a known dependence on the temperature. Thus, there is a wide number of thermometers possible: (a) Thermometers based on thermal expansion: Liquid- glass (mercury, alcohol) thermometer Vapour-pressure thermometer Spring thermometer Bimetal thermometer (b) Thermometers based on electrical propertise: Resistance thermometer Thermoelement (thermo-voltage) (c) Thermometers without mechanical and electrical contact: Pyrometer (electromagnetic radiation) Acustic thermometer (velocity of sound) Magnetic thermometer (magnetic susceptibility) Glass-fibre thermometer (refractive index) (i) Thermodynamics: - Definitions - temperature Syllabus - gas laws - thermodynamic potentials - changes of thermodynamic states (work- processes) - heat-conduction Thermal expansion The volume of a gas increases when heated at constant pressure - Charles’ Law When a gas is heated, the gas molecules move faster and hit the wall of the container violently. Thevolumeofgasmustincreaseto keep the pressure constant. So that the gas molecules hit the wall less frequently. Thermal expansion The experiments of J.A.C. Charles (1746 - 1823) and J.L. Gay- Lussac (1778 - 1823) have shown that the volume of gases depends on both pressure and temperature. The findings of Gay-Lussac lead to the so-called Gay- Lussac’s Law: The findings of Charles lead to the so-called Charles’s Law: V =const. T  p = const. T Thermal expansion J.A.C. Charles (1746 - 1823) and J.L. Gay-Lussac (1778 - 1823) showed that the volume of gases depends on both pressure and temperature. If the pressure is held constant, then the volume of the gas at the temperature T (in Celsius !!!) is: V(T)=V0 1+ T Gay-Lussac found that in the limit of zero-pressure all gases have the same volume- expansion-coefficient: 1  = 0.003661 K −1 = 273.15K A gas which is in this limit state is called an ideal gas. State-equation of an ideal gas The volume and the pressure of a gas in a fully closed system follows This is known as the Boyle-Mariotte’s Law which was found independently in 1662 by R. Boyle (1627-1691) and in 1679 by E. Mariotte (1620-1684). pV = const. State-equation of an ideal gas Taking together Gay-Lussac’s, Charles’ and Boyle- Mariotte’s Law one can easily derive the expression: pV = const. T which is the so-called ideal gas law (that is, the state- equation of the ideal gas). Note: Real gases follow this law better at lower pressure and higher temperatures. The reason for that is that in real gases there are interactions between the gas molecules. In other words: the ideal gas is a model-system consisting of non-interacting particles with negligible extension. State-equation of an ideal gas ideal gas law (also known as “state-equation of ideal gasses”). pV = const. T The pressure-volume dependence of a fixed amount of ideal gas at different temperatures. Each curve is a hyperbola (pV = constant) and is called an isotherm. State-equation of an ideal gas The constant in the ideal gas law can be determined from pV p V nn = T Tn where the index n denotes normal-conditions (Tn=273,15 K; pn = 101325 Pa) and From this follows: pV p pV =n× n Þ =nR TTnrn T And therefore Note: This is true for ideal gases. For real gases, every gas has its own individual gas- constant (the individual gas constant depends on the density of the gas). pV=n×R ×T Thus for “n” of particles: Vn=n·Vmn. It therefore follows: State-equation of an ideal gas According to Avogadro’s law (A. Avogadro, 1776-1856) an ideal gas always needs the same volume at a given temperature and pressure. This is independent from the type of gas. 1 mol of gas under normal-conditions occupies a volume of Vmn=22,41383 dm3/mol. é Jù pV=n×R×T with R=8,31441 ê ú ë m o l K û Furthermore, the number of particles in 1 mol is given by the Avogadro constant NA = 6,022045·1023:  R é Jù pV=N×k ×T with k o =1,380662×10-23 N ê ë Kú û BB A where N is the number of particles and kB the Boltzmann- constant. State-equation of an ideal gas Standard Temperature and Pressure (STP) The standard condition of temperature and pressure is known as STP. The universal value of STP is 1 atm (pressure) and 0 o C. In STP, 1 mole of gas will occupy a volume of up to 22.4 L. Units of P, V and T The table below lists the different units for each property. Factor Pressure P Volume Moles Temperature Gas Constant Variable Units atm Torr Pa mmHg V L m3 n mol T K R* Values of R 0.082057 L atm mol-1 K-1 62.364 L Torr mol-1 K-1 8.3145 m3 Pa mol-1 K-1 8.3145 J mol-1 K-1* Deviations from Ideal Gas Behaviour: Nature of Gas Deviations from ideality can be described by the COMPRESSION FACTOR, Z (sometimes called the compressibility). Z = pV/(nRT) = pVm/(RT) For ideal gases Z = 1 Attractive forces vary with nature of gas At High Pressures repelling forces dominate Z= Real gas – Van der Waals equation ideal gas : PV = nRT (P + x)(V − y)= nRT   n   RT  a  2 2 P+a  (V−nb)=nRTorP= −  V  V −b V mm Johannes Diderik van der Waals got the Noble price in physics in 1910 Chemistry for the Life Sciences Lecture 2 Prof. Valeria Nicolosi School of Chemistry PI at Centre for Research on Adaptive Nanostructures and Nanodevices (CRANN) e-mail: [email protected] Characterisation & Processing of Advanced Materials Group http://chemistry.tcd.ie/cpam/ The laws of thermodynamics The 1st law of thermodynamics: The kinetic gas-theory has shown that heat is nothing else than a form of energy. This implies that there are many different forms of energy. The 1st law of thermodynamics states: The quantity of energy in a fully closed system remains constant. (The internal energy of an isolated system is constant) Work It is motion against an opposing force. Work is done when a force moves. WORK is a transfer of energy that utilizes or causes uniform motion of atoms in the surroundings. A force F moving a distance x does work equal to Fx. The most common form of work (and often the only one which we have to consider) is the work done against the surrounding pressure when volume, V increases. However many chemical changes also do work by releasing electrical energy or light and of course work can be done by physical processes e.g. motion against friction. (i.e. we do work when we raise a weight against gravity; we do work when we climb a ladder; we do work when cycling into the wind) With WORK (w) being a primary concept in thermodynamics, we need to denote the capacity of a system to do work. ENERGY: a measure of the capacity of a system to do work HEAT: Is the description of a process – not the name of a form of energy! It is the means by which energy is transferred from a hotter body to a cooler one in order to equalize their temperatures. Internal Energy (U) The Internal Energy is in very simple words the total of the kinetic energy due to the motion of molecules (translational, rotational, vibrational) and the potential energy (associated with the atoms within molecules or crystals). It is separated in scale from the macroscopic ordered energy associated with movement; it refers to the invisible microscopic energy on the atomic and molecular scale. For example, a room temperature glass of water sitting on a table has no apparent energy, either potential or kinetic. The INTERNAL ENERGY of a system (U) depends upon: *State and Path Functions In Thermodynamics it is important to distinguish between two distinct classes of function. The Internal Energy (U) (and as we shall see later the Enthalpy, H and Entropy, S) are state functions whereas the heat (q) and work (w) are path functions. A state function is one whose value depends only on the state of the substance under consideration; it has the same value for a given state no matter how that state came about. In contrast, a path function depends on the path which the system takes in going between two states. More formally, State functions are exact differentials: f U = dU = U − U fi i where f is and i represent final and initial states Path functions are inexact differentials: f q= dqq−q fi i,path Sign Conventions Both heat and work are signed quantities and it is necessary to define what the signs mean. +q: heat is absorbed by the system (an endothermic process) -q: heat is given out by the system (an exothermic process) For work, w is the work done on the system. It therefore follows that when a gas is compressed by an external force the work is positive (+w), whereas when a gas expands by pushing against an external force the work is negative (-w). It is sometimes helpful to think about the work done by the system. This is often given the symbol w’ and is simply the reverse of the work done on the system: w = −w A simple Gas Expansion We imagine an ideal piston (one with no mass that moves without friction) with an area A, which contains a gas at a pressure pint and where the external pressure is pext Under conditions where pint>pext the piston will move out (by a distance dx) and in doing so does work, dw’ (=-dw) against the external pressure. dx A pint pext Extra slide to answer your questions! Please remember that the sign of Pint and Pext are opposite! A pint pext The work done by the gas (the system) is therefore dw=forcedistance=pextAdx=pext dV The work done on the gas, dw is simply dw=−dw'=ddw−w=p=−−dwV w'='=−−pp ddVV ext exetxt int If pext= 0 (in vacuum) then clearly no work is done i.e. no work is done by a gas expanding into a vacuum. If pext= constant then we can calculate the work done by the piston when it expands from an initial volume, Vi to a final volume Vf by simply integrating:  Vf Vf dw= p dV=p dV=p (V−V)  ext ext ext f i VV ii ❖The above results are entirely general. Later we will look again at simple gas expansions under two specific conditions namely constant temperature (isothermal) and isolated (adiabatic). THE FIRST LAW OF THERMODYNAMICS “Energy cannot be created or destroyed but is just transformed from one form to another”. We have already seen that the “forms” of energy are heat (q), work (w) and internal energy (U). The first law can be expressed mathematically in the following way. If we take a system from state A to state B, then there is a definite change in the internal energy, ΔU: U=U −U BA Where, UB and UA are the internal energies of the system in the two states. This energy change can appear as heat and/or as work provided that their sum, q+w is equal to ΔU. Therefore the first law can also be expressed in the following way: U=q+w Q and W are path functions, therefore the partitioning of the internal energy into heat and work will depend on the path taken. If we look at infinitesimally small changes (reversible) in U and infinitesimally small amounts of q and w: dU = dq + dw If we now make the simplification that only PV work is done on the system: dU =dq−d(pV) We can now consider three possible expansion scenarios for the internal energy change of a system which does only PV work: - A change where the surroundings are a perfect vacuum (p=0) - A change at constant volume (“isochoric”), (dV=0) - A change where the surroundings are at constant pressure (dp=0) 1. A change in which the surroundings are a perfect vacuum, where no work is done (p=0): U = q + w dU=dq dw=−dw'=−pext dV 2. A change at constant volume (“isochoric”) where again no work is done (dV=0): Pf P0 Volume Therefore, in isochoric conditions: dU = dq - Exothermic reaction (-q): ΔU is negative (internal energy decreases) - Endothermic reaction (+q): ΔU is positive (internal energy increases) 3. A change where the surroundings are at constant pressure (“isobaric”): U =q+w dU=dq−pdV V0 Vf Volume This is particularly interesting as it enables us to predict the conditions under which maximum work is done The external pressure should be as high as possible in order to maximise the work However if the external pressure exceeds the internal pressure then the gas is compressed. Therefore to obtain the maximum expansion work the external pressure needs to be infinitesimally smaller than the internal pressure. The laws of thermodynamics Work and the First Law: Example C3H8(g) + 5 O2(g) → 3CO2(g) + 4H2O(l) at 298 K & 1 atm (1 atm = 101325 Pa) What is the work done by the system? For an ideal gas; pV = nRT (p = pex) n – no. of moles R – gas constant T = temperature V – volume p = pressure Pex h pex pressure (P) h is distance moved P A = area of piston The laws of thermodynamics Work and the First Law: Example C3H8(g) + 5 O2(g) → 3CO2(g) + 4H2O(l) at 298 K 1 atm  V= nRT/p or Vi = niRT/pex 6 moles of gas: Vi = (6 × 8.314 × 298)/ 101325 = 0.1467 m3 3 moles of gas: Vf = (3 × 8.314 × 298)/ 101325 = 0.0734 m3 work done = -pex × (Vf – Vi) = -101325 (0.0734 – 0.1467) = +7432 J Heat and work Work is a transfer of energy that can cause motion against an opposing force. We can easily visualise ways in which work could be obtained from a chemical reaction. One example is when a gas is produced e.g., Zn(s) + 2HCl(aq) ⇄ ZnCl2(aq) + H2(g) Hydrogen gas is produced when Zn reacts with hydrochloric acid. How much work is done when we throw 50 g of Zinc in HCl in a open beaker? (Molar mass of Zn = 65.409 g/mol) dw = -pexdV pV = n × Ri × T DV=Vf -Vi »Vf =nRT pex nRT 50g dw=-p =-nRT =- 8.3145JK-1 ́298K =-1.9kJ ex pex 65.4g / mol This is what you want to calculate dw = -pexdV You are treating ideal gasses, so you can apply the Ideal Gasses Law: The expansion volume: We can approximate that the final volume is much larger than the initial volume because at the start you do not have any gas, and after the reaction you have the production of a gas, which occupies a much larger volume. So you can basically consider the volume from the Ideal gasses law So you swap this in the original expression for work: dw = - pexdV pV = n × Ri × T DV=Vf -Vi »Vf =nRT pex nRT 50g dw=-p =-nRT =- 8.3145JK-1 ́298K =-1.9kJ ex pex 65.4g / mol Chemistry for the Life Sciences Lecture 3 Prof. Valeria Nicolosi School of Chemistry PI at Centre for Research on Adaptive Nanostructures and Nanodevices (CRANN) e-mail: [email protected] Characterisation & Processing of Advanced Materials Group http://chemistry.tcd.ie/cpam/ Heat Capacity How to measure dq – change in heat of a system? When we supply heat to an object its temperature rises and the relationship between the heat supplied, q and the temperature rise, dT is: dq=CdT C is the “heat capacity” The SI unit of heat capacity is joule per kelvin J/K. The “molar heat capacity” is the amount of heat required to raise one mole of substance through one degree (units JK-1Mol-1). Heat capacity: It is the amount of heat needed to raise the temperature of a certain mass 1 degree Celsius. Examples: Water 4.19 1 kJ/(kg K) What does that mean? That means it takes 4.2 joules of heat energy to raise one gram of water one degree Celsius. Consider the energy needed to heat 1.0 kg of water from 0 oC to 100 oC when the specific heat of water is 4.19 kJ/kg.K (kJ/kg.oC): Q = (4.19 kJ/kg.K) (1.0 kg) ((100 oC) - (0 oC)) = 419 (kJ) Oil, vegetable 1.67 1 J/(g K) alcohol 2.46 J/(gK) at 20°C-25°C The laws of thermodynamics The heat-capacity depend also on  - the pressure (not in the case of ideal gases), - the volume - the process-path. In other words, the heat-capacity always depends on the actual experimental set- up/environment which is measuring it. Two important experimental conditions are to set the volume or the pressure constant: a) constant volume - isochore heat-capacity (C ,c ,C ) v v mv b) constant pressure - isobar heat-capacity (Cp,cp,Cmp) For a process taking place at constant volume we know that the heat is equal to the internal energy change dU=dq =CdT VV which can be rewritten as a partial derivative* (T only) as:   U T V V C= which is the usual definition of the molar heat capacity at constant volume C is the “heat capacity” and is a measure of the increase in energy of a system when the Temperature is changed! Bomb calorimetry How to measure dq – change in heat of a system? The calorimeter directly gives dU (via qV ). Again: dU = dqV+dw = dqV - pexdV = dqV dU = qV Measure change in temperature, T after Performing an reaction. Measure rise in T ( = dT ). Here it is obvious that the heat absorbed during the reaction is a quantity that depends only on the initial and final states because U, P and V are all state functions. U = q+w =0 (V is constant) Constant Pressure Processes Processes at constant pressure are more common than those taking place at constant volume. Heat absorbed by a gas under constant pressure: -some of the heat will increase the internal energy - some of the heat will appear as work of expansion HEAT WORK ΔU It is therefore convenient to define the Enthalpy, H, which links the internal energy change under constant pressure and the work of expansion done as: H=U+pV If we express this as a differential form (differentiation of a product) we get: dH = dU + pdV + Vdp However we already know from the first law that for a process with only PV work : q = dU − w = dU + pdV Therefore: dH = dU + pdV + Vdp dH = q + Vdp becomes: We now impose a further condition of constant external pressure The Enthalpy change is equal to the heat change under constant (which is reasonable as the surroundings are much larger than the dp = 0 pressure conditions (heat absorbed or given by a chemical reaction!) system) so that: and hence: dH = qp The enthalpy function is therefore an essential tool for dealing with systems that are subject to constant pressure and are free to change their size The change in enthalpy can be identified with the energy supplied as heat! H, the enthalpy change is the heat absorbed or given out in a reaction occurring at constant pressure. heat absorbed (taken in) -ENDOTHERMIC: ΔH positive heat evolved (given out) - EXOTHERMIC: ΔH negative Heat Capacity at Constant Pressure We already know that the Heat Capacity by definition is the amount of heat required to raise a certain quantity of substance of a certain temperature: We just found that at constant pressure: or in partial differential form we can define the molar heat capacity at constant pressure as: dq=CdT dH = qp dH=CpdT 𝐶= 𝑃 𝑑𝐻 𝑑𝑇 𝐶= 𝑉 𝑑𝑈 𝑑𝑇 V P The Variation of Enthalpy with Temperature Suppose that we know the enthalpy of a substance at one temperature, H(T1) but we want to know its enthalpy at another temperature, H(T2). In practice this is quite common since data are often tabulated at one temperature and these may not correspond to the temperature of interest where values may be difficult to measure experimentally. We can use the definition of Cp to derive a relationship for variation of H with T since: 𝑑𝐻 𝑑𝑇 P Rearranging this gives: 𝐶= 𝑃 dH=CpdT We can now integrate both sides of this equation: T2 T2 dH=C dT (ifC p p isassumedtobeconstant) TT 11  H(T)−H(T)=CT−T 21p21 This is a very important practical relationship since it enables us to calculate H at any temperature provided we know its value at some other temperature and its heat capacity. In fact it turns out (as we shall see) that we can only measure changes in enthalpy rather than the enthalpies themselves. However a modified form of this equation can be used to convert ΔH values from one temperature to another. The assumption that Cp is a constant is often acceptable. However when greater accuracy is required a parameterised form of Cp is often used which is valid over a wide range of “normal” temperatures (Cp varies in fact with T). This parameterised form of Cp is given by:  c Cp =a+bT+ 2 T The equivalent integral to this new equation is now given by: T2 T2  dH= ( a + b T + c 1 / T 2 )d T TT 11 111  H(T)−H(T)=a(T−T)+ b(T2−T2)−c −  2121 21 2 1 2 TT The laws of thermodynamics STANDARD STATES Because we have no absolute values it is essential to find a baseline if we are to make any further progress. For this reason we define what we call standard states in which a substance is in its most stable form at 298 K and 1 atmosphere pressure. Elements in their standard states are assigned zero enthalpy and we can now define standard enthalpy changes STANDARD ENTHALPY CHANGE: ΔHo298: the enthalpy when reactants in their standard states are converted to products in their standard states. STANDARD ENTHALPY OF FORMATION: ΔHof: use elements in their standard states to form 1 mole of the substance in its standard state. We can also have standard enthalpies of combustion, solution, transition etc... Chemistry for the Life Sciences Lecture 4 Prof. Valeria Nicolosi School of Chemistry PI at Centre for Research on Adaptive Nanostructures and Nanodevices (CRANN) e-mail: [email protected] Characterisation & Processing of Advanced Materials Group http://chemistry.tcd.ie/cpam/ Physical and chemical change To begin to understand the complex structural changes that biological macromolecules undergo when heated or cooled, we need to understand how simpler physical changes occur. A phase is a specific state of matter that is uniform throughout in composition and physical state. i.e. The liquid and vapour states of water are two of its phases. The term ‘phase’ is more specific than ‘state of matter’ because a substance may exist in more than one solid form, each one of which is a solid phase Enthalpies of vaporization The vaporization of a liquid, such as the conversion of liquid water to water vapor when a kettle boils at 100°C, is an endothermic process (ΔH > 0) because heating is required to bring about the change. At a molecular level, molecules are being driven apart from each other and this process requires energy. One of the body’s strategies for maintaining its temperature at about 37°C is to use the endothermic character of the vaporization of water because the evaporation of perspiration requires energy and withdraws it from the skin. The energy that must be supplied as heat at constant pressure per mole of molecules that are vaporized under standard conditions (1 bar) is called the standard enthalpy of vaporization of the liquid and is denoted ΔVAPH°. For example, 44 kJ of heat is required to vaporize 1 mol H2O(l) at 1 bar and 25°C, so ΔVAPH° = +44 kJ mol−1. A thermochemical equation shows the standard enthalpy change (including the sign) that accompanies the conversion of an amount of reactant equal to its stoichiometric coefficient in the accompanying chemical equation (in this case, 1 mol H2O). If the stoichiometric coefficients in the chemical equation are multiplied through by 2, then the thermochemical equation would be written: 2 H2O(l) → 2 H2O(g) ΔH° = +88 kJ This equation means that 88 kJ of heat is required to vaporize 2 mol H2O(l) at 1 bar and at 298.15 K. There are some striking differences in standard enthalpies of vaporization: the value for water is 44 kJ mol−1 the value for methane, CH4 is only 8 kJ mol−1. Even allowing for the fact that vaporization is taking place at different temperatures, the difference between the enthalpies of vaporization means that water molecules are held together in the bulk liquid much more tightly than methane molecules. Enthalpies of fusion: Another common phase transition is fusion, or melting, as when ice melts to water. The change in molar enthalpy that accompanies fusion under standard conditions (pure solid at 1 bar) is called the standard enthalpy of fusion, ΔfusH°. Its value for water at 0°C is +6.01 kJ mol−1. Notice that the enthalpy of fusion of water is much less than its enthalpy of vaporization: in vaporization the molecules become completely separated from each other, whereas in melting the molecules are merely loosened without separating completely! Solid Liquid fusion Gas vaporisation The reverse of vaporization is condensation The reverse of fusion (melting) is freezing The molar enthalpy changes are, respectively, the negative of the enthalpies of vaporization and fusion. This is because the energy that is supplied (during heating) to vaporize or melt the substance is released when it condenses or freezes! Enthalpies of condensation and freezing: and in general: H2O(s) → H2O(l) ΔH° = +6.01 kJ H2O(l) → H2O(s) ΔH° = −6.01 kJ ΔforwardH° = ΔreverseH° Bond enthalpy: To understand bioenergetics at a molecular level we need to account for the flow of energy during chemical reactions as individual chemical bonds are broken and made. For example, for the dissociation of the hydroxyl radical OH(g), we have: HO(g) → H(g) + O(g) ΔH° = +428 kJ Th e corresponding standard molar enthalpy change is called the bond enthalpy A complication when dealing with bond enthalpies is that their values depend on the molecule in which the two linked atoms occur. For instance, the total standard enthalpy change for the atomization (the complete dissociation into atoms) of water: H2O(g) → 2 H(g) + O(g) ΔH° = +927 kJ It is not twice the O–H bond enthalpy in H2O, even though two O–H bonds are dissociated! There are in fact two different dissociation steps: 1) First an O–H bond is broken in an H2O molecule: H2O(g)→ HO(g)+H(g) ΔH°=+499kJ 2) Second, the O–H bond is broken: HO(g) → H(g) + O(g) ΔH°= +428 kJ The sum of the two steps is the atomization of the molecule. Thermochemical properties of fuels: We need to understand the molecular origins of the energy content of biological fuels, the carbohydrates, fats, and proteins. For example, the photosynthesis and the oxidation of organic molecules are the most important processes that supply energy to organisms. We will now go through a quantitative study of biological energy conversion by assessing the thermochemical properties of fuels. The consumption of a fuel in a furnace or an engine is the result of a combustion. An example is the combustion of methane in a natural gas flame: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔCH°= - 890 kJ According to the discussions in Lectures 2 and 3: Hm = Um + pVm For condensed phases (solid and liquids), pVm is so small that it may be ignored. For example, the molar volume of liquid water is 18 cm 3 mol- 1, and at 1.0 bar pVm = (1.0  105 Pa)  (18  10-6m3 mol-1) = 1.8 Pa m3 mol-1= 1.8 J mol-1 BUT, the molar volume of a gas, and therefore the value of pVm, is about 1000 times greater and cannot be ignored! For gases treated as perfect, pVm may be replaced by RT. Therefore, if in the chemical equation the difference (products – reactants) in the stoichiometric coefficients of gas phase species is Δνgas, we can write: ΔcH = ΔcU + ΔνgasRT Note that νgas (where ν is nu) is a dimensionless number. N.B. The properties of fuels and foods can also be commonly discussed in terms of their specific enthalpy (enthalpy of combustion divided by the mass of the sample - typically in kilojoules per gram) or the enthalpy density (enthalpy of combustion divided by the volume of the sample - typically in kilojoules per cubic decimeter). Example: The energy released at constant volume as heat by the combustion of the amino acid glycine is −969.6 kJ mol-1 at 298.15 K, so ΔcU = −969.6 kJ mol-1. From the chemical equation: NH2CH2COOH(s) + 9/4 O2(g) → 2 CO2(g) + 5/2 H2O(l) + 1/2 N2(g) Δνgas = (2 + 1/2) −9/4 = 1/4 Therefore, ΔcH = ΔcU + 1/4RT = −969.6 kJ mol−1 + 1⁄4 X (8.3145 × 10−3 kJ K−1 mol−1) X (298.15 K) = −969.6 kJ mol−1 + 0.62 kJ mol−1 = −969.0 kJ mol−1 The combination of reaction Enthalpies: Hess’s Law This fundamental law of Thermodynamics arises directly from the fact that the Enthalpy, H is a state function. This implies that the Enthalpy change for a process is the same no matter what pathway we take in going from the initial to the final state. (the heat of a whole reaction is equivalent to the sum of its steps) For example: C + O2 CO2 This can happen in 2 steps: C + 1⁄2O2 CO + 1⁄2 O2 C + CO + O2 i.e. C + O2 CO CO2 CO + CO2 CO2 ΔH°=-110.5kJ ΔH°=-283.0 kJ ΔH°=-393.5kJ ΔH°=-393.5kJ We can also see how these steps add together via an “enthalpy diagram” C + 1⁄2O2 CO ΔH°=-110.5kJ CO i.e. + 1⁄2 O2 CO2 ΔH°=-283.0 kJ C + O2 CO2 ΔH°=-393.5kJ C + O2 ΔH°=-110.5 kJ CO ΔH°=-393.5 kJ ΔH°=-283.0 kJ CO2 Reactants Intermediate Products Standard Enthalpies of formation We need to simplify even further the process of predicting reaction enthalpies of biochemical reactions. The standard reaction enthalpy, Δ rH, is the difference between the standard molar enthalpies of the reactants and the products, with each term weighted by the stoichiometric coefficient, n (nu), in the chemical equation kilojoules per mole. Kirchhoff’s Equation We need to know how to predict the reaction enthalpy of a biochemical reaction at one temperature from its value at other temperatures. Suppose we want to know the enthalpy of a particular reaction at body temperature, 37°C, but have data available for 25°C, or suppose we want to know whether the oxidation of glucose is more exothermic when it takes place inside an Arctic fi sh that inhabits water at 0°C than when it takes place at mammalian body temperatures. Initially we assume that the heat capacity does not vary with temperature in the range T1 to T2 in which case we have already shown that: H(T)−H(T)=CT −T 21p21 Entropy and the second law of thermodynamics Some processes occur naturally and some don’t! Gases expand to fill a vacuum, hot objects cool to the temperature of their surroundings and chemical reactions proceed in one direction rather than the other. Therefore it is reasonable to assume that some aspect of the world determines this direction of spontaneous change. Reversible and Irreversible Processes The above discussion leads to the important idea of thermodynamic reversibility. A reversible change in thermodynamics is one that can be reversed by an infinitesimal modification of a variable. A system is said to be in equilibrium with its surroundings if such an infinitesimal change in the conditions in opposite directions results in opposite changes in its state. When the internal and external pressures are equal the system is in equilibrium. If the external pressure is reduced infinitesimally then the gas inside the piston expands slightly which can be reversed by an infinitesimal increase in the external pressure. This is therefore a reversible thermodynamic change. Conversely, if the external pressure is significantly lower than the internal pressure then an infinitesimal increase in the external pressure does not reverse the expansion; the system is not in equilibrium and the process of expansion is irreversible. Reversible processes: - Are infinitely slow - Are at equilibrium - Do maximum work Irreversible processes: - Go at finite rate - Are not at equilibrium - Do less than the maximum work The first law tells us which changes are permissible (only those changes where the internal energy of an isolated system remains constant are allowed). The second law, through another state function (the entropy) identifies which of these are spontaneous. Careful, thermodynamics is completely silent on the rate at which a spontaneous change in fact occurs! Some spontaneous processes (such as the conversion of diamond to graphite) may be so slow that the tendency is never realised in practice! The thermodynamic definition of entropy is motivated by the idea that a change (energy is dispersed in a disordered fashion) depends on how much energy is transferred as heat which stimulates disordered motion in the surroundings. In contrast work stimulates a uniform motion of the atoms in the surroundings and hence does not change the entropy of the system) We therefore define the entropy as: (3.9) in which qrev is the heat change for a reversible path between the two states of the system. dS = dqrev T For our initial encounter with the concept, we can identify entropy with disorder: - if matter and energy are distributed in a disordered way (as in a gas for example) then entropy is high - if matter and energy are distributed in an ordered way (as in a crystal for example) then entropy is low Spontaneous Processes One of the most important questions that we can answer using Thermodynamics is what criterion decides the direction of a spontaneous process It is attractive to think that all spontaneous processes lead to either a lowering of the internal energy or to an increase in the disorder of the system. However practical experience shows that there are many processes which appear to contradict this. There are processes where the entropy decreases (e.g. burning a magnesium ribbon where the products have low entropy but the gas consumed has a high entropy). A more sophisticated approach considers the entropy change of the system and the surroundings together (the Universe) and the second law of Thermodynamics states that “In a Spontaneous Process the Entropy of the Universe Increases”. As the Universe is simply the System + its Surroundings the Entropy change of the Universe can be separated into the Entropy change of the system, Ssys and the Entropy change of the surroundings Ssurr Suniv = Ssys + Ssurr (4.0) We now need to consider how we determine each of these The Entropy change of the System is simply: qrev,sys T = (4.1) It is important to note that even if the process we are looking at is NOT reversible we need to calculate the heat change that would be observed if the process took place reversibly. S sys sys The laws of thermodynamics N.B. dS can be evaluated only by carrying out a reversible process Because S is a state function it value does not depend on whether the process is reversible or irreversible but it can only be evaluated by measuring dqrev during a reversible process – using the equality. _________________________________ The simplest place to apply the equation is in an isolated system because here dq = 0 so that the three possibilities are: ΔS > 0 ΔS = 0 ΔS < 0 spontaneous (irreversible) process reversible process process not feasible i.e., when a natural (spontaneous) process occurs in an isolated system there is an increase in entropy – the entropy increases to a maximum at equilibrium. The Variation of Entropy with Temperature As with Enthalpy values, it is convenient to have a method for converting Entropies from one temperature to another At constant pressure the heat change is equal to the Enthalpy change and hence (dqrev=dH): dq dH dS = = TT CpdT T dS = T2 T2 dH=C dT (ifC isassu p p T2 dS=C T dT p T2 1 TT 11 by definition (dH=CdT) T 2 S(T)−S(T)=C ln 21p 1 T  1 T  T 1 H (T ) − H (T ) = C T − T  21p21 m The Entropy change accompanying a phase Change At the temperature of the phase change the two phases are in equilibrium and therefore any transfer of heat between the system and its surroundings is reversible and hence at constant pressure: q = Htransition Therefore the entropy change for the phase transition is given as: S = trans Htrans T trans The Gibbs Function The discussions of spontaneous change in the previous sections require an analysis of changes in both the system and the surroundings. It is however, also possible to define functions for the system itself which decrease during a spontaneous process and reach a minimum at equilibrium. These functions are known as the Gibbs Function (for constant pressure processes), which is defined as G = H −TS Example The laws of thermodynamics Equilibrium at the boiling point between liquid and vapour ΔH >0 latent heat of vaporization absorbed by the liquid Δ S > 0 because the gaseous state is much more disordered ΔG = ΔH - T ΔS ΔG = 0 at equilibrium at sufficiently high T (i.e., when T reaches the boiling point) T ΔS becomes equal to ΔH ΔHvaporization = Tboiling point x ΔS vaporisation so that ΔG is zero and there is equilibrium – any further heating and the liquid boils. This is why there are definite boiling points for liquids e.g., water boils at 373 K. The Master Equations: series of useful relationships from 1st and 2nd laws We know from the first law that for a process in which the only work done is that due to gas expansion against constant external pressure: dU =dq−pdV which for a reversible process (see equation 3.9: ) dq =TdS rev can be written as: dU = TdS − pdV This is often known as the First Thermodynamic Master Equation as it combines the 1st and 2nd laws The second Master Equation is developed by starting from the definition of H and taking its complete differential: H = U + PV  dH=dU+pdV+VdP we now substitute for dU from the first Master Equation to give the second Master Equation as: dU dH = TdS − pdV + pdV +VdP dH =TdS+VdP The final (and most useful) master equation is found by taking the complete differential of G and substituting for dH from the 2nd mater equation G = H −TS dG = dH −TdS − SdT dH dG = TdS +Vdp −TdS − SdT  dG=Vdp−SdT The Variation of the Gibbs Function with Temperature and Pressure We can use the 3rd Master Equation to simply derive expressions for the variation of G with T and p. Recalling the 3° Master Eq.:: dG = Vdp − SdT impose the condition of constant temperature (dT = 0) Hence: Because all volumes are positive, the Gibbs energy increases when the pressure increases! dG=Vdp Alternatively starting from the same final Master Equation but this time under a condition of constant pressure (so that Vdp = 0)... Recalling the 3° Master Eq.:: dG = Vdp − SdT We find: dG = −SdT This tells us that, because entropy is positive, an increase in temperature results in a decrease in G. Moreover, for a given change of temperature, the change in Gibbs energy is proportional to the entropy. For a given substance, because the entropy of the gas phase is greater than that for a condensed phase, the Gibbs energy falls more steeply with temperature for a gas than for a condensed phase. The entropy of the liquid phase of a substance is greater than that of its solid phase, so the slope is least steep for a solid. Phase Diagrams To prepare for being able to describe phase transitions in biological macromolecules, first we need to explore the conditions for equilibrium between phases of simpler substances. The phase diagram of a substance is a map showing the conditions of temperature and pressure at which its various phases are thermodynamically most stable. The boundaries between regions in a phase diagram, which are called phase boundaries, show the values of p and T at which the two neighbouring phases are in dynamic equilibrium. If we consider a sample at a and cool this a constant pressure the sample remains as a gas until the temperature reaches b when liquid appears. Here two phases are in equilibrium. Lowering the T further takes the system to a one phase liquid region and within this the temperature can be varied. Further cooling takes the system to the point d at when ice appears. What is Equilibrium? This is not Equilibrium! Equilibrium two opposing processes occurring at the same rate a system at equilibrium is in balance During a game, players enter and leave. Always the same number of players on field. H2O(l) ⇄ H2O(g) Equilibrium Phase Boundaries: The pressure of a vapor that is in equilibrium with its condensed phase is called the vapor pressure of the substance. All liquids and solids have a tendency to evaporate into a gaseous form, and all gases have a tendency to condense back to their liquid or solid form – dynamically! It is an indication of a liquid's evaporation rate. It relates to the tendency of particles to escape from the liquid. A substance with a high vapour pressure at normal temperatures is often referred to as volatile. Vapor pressure increases with temperature because, as the temperature is raised, more molecules have sufficient energy to leave their neighbours in the liquid and escape... Location of Phase Boundaries: Thermodynamics provides us with a way of predicting the location of the phase boundaries and relating their location and shape to the thermodynamic properties of the system. For instance, the shape of the vapor pressure curve (the liquid–vapor boundary) is related to the enthalpy of vaporization of the liquid. At a given temperature, the pressure corresponding to equilibrium is the vapor pressure of the liquid. If we change the temperature, the vapor pressure changes to a different value. That is, there is a relation between the change in temperature, dT, and the accompanying change in vapor pressure, dp. If we were considering the equilibrium between a solid and a liquid, the focus would be different: in this case we would typically be interested in the change in melting point as the pressure is increased. The relation between dT and dp that ensures that in either case the two phases remain dp S −S in equilibrium is given by the Clapeyron equation for the slope of the phase boundary at   =mm dT V −V This is because, To proceed we need to make some additional assumptions and to consider specific cases. 1. Solid → Gas and Liquid → Gas In these cases we can safely assume that the molar volume of gas is much greater than that of either the liquid or solid. any temperature: dp(V −V)=dT(S −S mmm mm dp Sm dT Vm = qrev Hm TT Sm = = or... Next, we suppose that the vapor behaves as a perfect gas and write its molar volume as Vm(g) = RT/p. Then: and therefore: which defines the Clausius – Claperon equation (for boundaries with gas). 2. Solid → Liquid H dp= m dT TVm p2 dp= m p Vm T T 1 1 H T 1 2 H T p −p = m ln 2 21 V T m1 Why are we studying all this???? Proteins and biological membranes can exist in ordered structures stabilized by a variety of molecular interactions, such as hydrogen bonds and hydrophobic interactions. However, when certain conditions are changed, the helical and sheet structures of a polypeptide chain may collapse into a random coil and the hydrocarbon chains in the interior of bilayer membranes may become more or less flexible. These structural changes may be regarded as phase transitions in which molecular interactions in compact phases are disrupted at characteristic transition temperatures to yield phases in which the atoms can move more randomly. Example: Tm varies linearly with the fraction of C–G base pairs, at least in this range of composition. The equation of the line that fits the data is: Tm/K = 325 + 39.7f It follows that Tm = 341 K for 40.0 per cent C–G base pairs (at f = 0.400). Phase transitions of biological membranes: All lipid bilayers undergo a transition from a state of high to low chain mobility at a temperature that depends on the structure of the lipid. At physiological temperature, the bilayer exists as a liquid crystal, in which some order exists but the chains writhe. There is sufficient energy available at normal temperatures for limited bond rotation to occur and the flexible chains to writhe around. However, the membrane is still highly organized in the sense that the bilayer structure does not come apart At lower temperatures, the amplitudes of the writhing motion decrease until a specific temperature is reached at which motion is largely frozen. The bilayer is said to exist as a gel. Chemistry for the Life Sciences Lecture 6 Prof. Valeria Nicolosi School of Chemistry PI at Centre for Research on Adaptive Nanostructures and Nanodevices (CRANN) e-mail: [email protected] Characterisation & Processing of Advanced Materials Group http://chemistry.tcd.ie/cpam/ Let’s recall the 3rd Master Equation: dG = Vdp − SdT impose the condition of constant temperature (dT = 0) dG=Vdp For an ideal gas where: nRT  p2 p2 dG= dp p p1 p1 G(p )−G(p)=nRT  nRT p V= we can integrate: 2 p 1  dp=nRTln p 2 21 pp p1 1 p1= the standard pressure - p0=1bar G(p1)=G(p0)=G0= Gibbs Energy at Standard Pressure for 1 mole (n = 1) all quantities become molar quantities which are indicated with a subscript m. Key relationship: Molar Gibbs Energy G (p)=G +RTln mm 0 p p0 Dealing with Mixtures We now leave pure materials and the limited but important changes they can undergo and examine mixtures. A simple example: -two gases A and B separated by a partition with both gases at the same temperature and pressure, p - there are nA moles of A and nB moles of B - The partition is removed completely and the gases mix but the resultant pressure is still p and the temperature does not change. n A nB AB pp p The crucial thing that changes on mixing is the partial pressure of each component which for mixtures of ideal gases (and ideal solutions - see later) is given by the mole fraction of the component xi multiplied by the total pressure: pi =xiptot Where: ni x= ; n =n i tot i n i tot Therefore for gas A (and similarly for gas B): n n pA=Ap pB=Bp nA +nB  nA +nB  We have already seen that the Molar Gibbs Energy of a pure gas varies as: p p0 G (p)=G0 +nRTln mm If we are considering ideal gases then there is no difference between a pure gas at a pressure p and the same gas at a partial pressure pi = p in a mixture. for 1 mole: this shows that the Gibbs energy falls as the partial pressure falls so that the Gibbs energy of A in the mixture is less than its Gibbs Energy before mixing. Hence the lowering of partial pressures that takes place on mixing leads pi G (p)=G +nRTln 0 m,i i m,i p0 Before Mixing: p to a reduction in the Gibbs Energy for each gas and hence of the mixture After Mixing: pi‹p as a whole; this is why gases mix! Ideal gases mix spontaneously! Solutions The previous sections have dealt with equilibria between phases with only one component. To extend these arguments to two component systems (solutions). We shall consider only homogeneous mixtures, or solutions, in which the composition is uniform however small the sample. The component in smaller abundance is called the solute and that in larger abundance is the solvent. Let’s consider first nonelectrolyte solutions, where the solute is not present as ions! Examples are sucrose dissolved in water, sulfur dissolved in carbon disulfide, and a mixture of ethanol and water. To assess the spontaneity of a biological process involving a mixture, we need to know how to compute the contribution of each substance to the total Gibbs energy of the mixture. The most important partial molar property for our purposes is the partial molar Gibbs energy, GJ,m, of a substance J, which is the contribution of J (per mole of J) to the total Gibbs energy of a mixture. If we know the partial molar Gibbs energies of two substances A and B in a mixture of a given composition, then we can calculate the total Gibbs energy of the mixture by using: G = nAGA,m + nBGB,m To gain insight into the significance of the partial molar Gibbs energy, consider a mixture of ethanol and water:  - Ethanol has a particular partial molar Gibbs energy when it is pure (and every molecule is surrounded by other ethanol molecules) - Ethanol has a different partial molar Gibbs energy when it is in an aqueous solution of a certain composition (because then each ethanol molecule is surrounded by a mixture of ethanol and water molecules). The partial molar Gibbs energy is so important in chemistry that it is given a special name and symbol is given. From now on, we shall call it the chemical potential and denote it μ (mu): G = nAμA + nBμB We saw before that the molar Gibbs energy of a pure substance is the same in all the phases at equilibrium. We can now use the same argument: a system is at equilibrium when the chemical potential of each substance has the same value in every phase in which it occurs. Solutions The previous sections have dealt with gas mixtures, now let’s extent to liquids and solids. The component in smaller abundance is called the solute and that in larger abundance is the solvent. Ideal Solutions Non-ideal Solutions Ideal Solutions Consider a pure liquid in a closed vessel. The pressure of a vapour in equilibrium with the liquid is called “the vapour pressure” of the substance All liquids and solids have a tendency to evaporate into a gaseous form, and all gases have a tendency to condense back to their liquid or solid form. It is an indication of a liquid's evaporation rate. It relates to the tendency of particles to escape from the liquid. A substance with a high vapour pressure at normal temperatures is often referred to as volatile. Phase boundaries A pure liquid, A has a vapour pressure, p* (where the * denotes a pure substance) at a particular temperature. If we now add another pure liquid, B with a different vapour pressure then the vapour pressure of the mixture will be different from that of either pure liquid. Raoult’s law provides a means for defining ideal behaviour. The ratio of the partial vapour pressure of each component to its vapour pressure as a pure liquid (pA/pA*) is approximately equal to the mole fraction of the A in the liquid mixture): p = X p* AAA and p = X p* BBB Adding these two contributions we obtain the total vapor pressure: p=pA +pB =xApA* +xBpB* This can also be expressed graphically: Non-Ideal Solutions – Henry’s Law In practice most real solutions do not obey Raoult’s law across the entire composition range as shown below: Nonetheless, even in these cases the law is obeyed increasingly closely for the component in excess (the solvent) as it approaches purity. The law is therefore a good approximation for the solvent if the solution is very diluted The English chemist William Henry found experimentally that , for real solutions at low concentrations, although the vapour pressure of the solute (called also “activity”) is proportional to its molar fraction, the constant of proportionality is not the vapour pressure of the pure substance. Henry’s Law: pB = activity pB =a=XB We have to note that  (activity coefficient= correction for interactions between the different molecules) is not a constant and we need to know its value at each concentration. An ideal-dilute solution is one in which the solvent obeys Raoult’s Law but the solute obeys Henry’s Law!! γ Real solutions: activities No actual solutions are ideal, and many solutions deviate from ideal- dilute behaviour as soon as the concentration of solute rises above a small value. In thermodynamics we try to preserve the form of equations developed for ideal systems so that it becomes easy to step between the two types of system. This is the thought behind the introduction of the activity, “a”, of a substance, which is a kind of effective concentration. The Free Energy of Mixing We are now in a position to examine further the thermodynamics of solution. In the first instance we examine how the Gibbs energy changes when two components are mixed Initially we need to recall some key results obtained earlier. We have already shown that the free energy of mixing of an ideal gas, A and another ideal gas, B is given by: Gmix =Gmixed −Gseparated =naRTlnxA +nbRTlnxB XA,XB are ‹ 0 ln XA,XB are ‹ 0 ΔGMIX ‹ 0 In all proportions The thermodynamic criterion for spontaneous change at constant temperature and pressure is ΔG < 0. At constant temperature and pressure, a reaction mixture tends to adjust its composition until its Gibbs energy is a minimum. a) very little of the reactants convert into products before G has reached its minimum value, and the reaction ‘does not go’. c) a high proportion of products must form before G reaches its minimum and the reaction ‘goes’. b) at equilibrium the reaction mixture contains substantial amounts of both reactants and products Chemistry for the Life Sciences Lecture 7 Prof. Valeria Nicolosi School of Chemistry PI at Centre for Research on Adaptive Nanostructures and Nanodevices (CRANN) e-mail: [email protected] Characterisation & Processing of Advanced Materials Group http://chemistry.tcd.ie/cpam/ Reaction rates THE LAW OF MASS ACTION (Guldberg & Waage 1864) The rate of a chemical reaction is proportional to the active masses of the reacting substances. “Active masses” we may, for the moment, regard as concentrations of the participating species or partial pressures in the case of gases. Reaction A + B ⇄ products Here:  - “k” is a constant (called the rate constant) - Square brackets [ ] = “concentration” Rate of reaction = k[A][B] H2O+CO → H2+CO2 Chemical Equilibrium Rate of reaction = k[A][B] Reaction rates A + B ⇄ C +D Rate forward (as shown) k1 [A][B] = Rf Rate backward (as shown) k2 [C][D] = Rb At equilibrium Rf = Rb the equilibrium constant [C][D] k1 [A][B] k2 = =K Equilibrium From this we may write that in general for the reaction aA + bB ⇄ cC + dD where K is the equilibrium constant For each reaction, there is only one value for Keq at a specific temperature. Characteristics of equilibrium: Dynamic: balance of reversible reactions All reactants and products are present (both reactions can occur) Move to equilibrium is spontaneous – if disturbed then it returns to the same equilibrium point. Represents a compromise between H (change in enthalpy taking the enthalpy to a minimum S (change in entropy) taking the entropy to a maximum. constant only at a specific temperature Keq = [C]c[D]d [A]a[B]b Expressions for Equilibrium Constants Examples: [NH3 ]2 [N2 ][H2 ]3 [PCl3 ][Cl2 ] [PCl5 ] [CO][H 2 ]3 [CH4 ][H2O] N2(g) + 3H2(g) ⇄ 2NH3(g); K = PCl5(g) ⇄ PCl3(g) + Cl2(g); CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g); K = K= - Value of equilibrium constant (Keq) shows the extent to which reactants are converted into products. Keq < 1: Reactants are favored at equilibrium Keq > 1: Products are favored at equilibrium The expression of equilibrium constant depends on how the equilibrium equation is written. For example, for the following equilibrium: H2(g) + I2(g) ⇄ 2 HI(g); [HI]2 Kc = [H2 ][I2 ] For the reverse reaction: 2HI(g) ⇄ H2(g) + I2(g); [H2 ][I2 ] Kc ' = =1/Kc [HI]2 And for the reaction: HI(g) ⇄ 1⁄2H2(g) + 1⁄2I2(g); [H2][I2] Kc"= = Kc' = [HI]2 1 Kc Reaction rates Write an equilibrium constant expression for: For reaction 2A ⇄ C + D Write as A + A ⇄ C + D Rate forward (as shown) : Rate backward (as shown) : K = [C][D] [A]2 N2(g) + 3H2(g) ⇄ 2NH3(g) [NH3]2 Keq = k1 [A][A] k2 [C][D] , so that ___________ [N2] [H2]3 Write the equilibrium constant expression for N2(g) + O2(g) 2NO(g) [NO]2 [N2 ][O2 ] Keq = [0.0035]2 [0.2][0.15] Keq = Equilibrium Calculate the value of Keq if [N2] = 0.20 mol/L, [O2] = 0.15 mol/L, and [NO] = 0.0035 mol/L. Keq = 4.1 x 10-4 What does the value of Keq tell you about the equilibrium? Keq < 1: Reactants are favored at equilibrium Expression and Values of Equilibrium Constant Using Partial Pressures So far we have dealt with concentrations. For gases the concentration is proportional to the partial pressure at a fixed temperature! Imagine two gasses mixing: n A nB pp AB p PV = nRT Solving for n/V (which is the molar concentration of the gas) we have: n/V = P/RT i.e., the molar concentration of the gas is equal to its partial pressure divided by RT – and RT is constant at a given temperature. Consider the following reaction involving gases: 2SO 2(g) + O2(g) ⇄ 2SO3(g) Kp = (P )2 SO3 2 (P ) (P ) SO2 O2 The laws of thermodynamics RELATIONSHIP BETWEEN FREE ENERGY CHANGE AND THE EQUILIBRIUM CONSTANT There is a relationship between the free energy change for a reaction ΔG and the equilibrium constant for that reaction, K Where ΔGo is the standard free energy change and may be written ΔGo = ΔHo - TΔSo and ΔHo and ΔSo are standard enthalpy and entropy changes and may be determined experimentally. Note: The magnitude of the equilibrium constant depends on the way in which the chemical reaction is written so that the stoichiometric numbers are known. For this reason the value for any equilibrium constant should be accompanied by a balanced chemical equation. ΔGo =-RTlnK Equilibrium From this we may write that in general for the reaction aA + bB ⇄ cC + dD K = [C]c[D]d where K is the [A]a[B]b equilibrium constant A system has only one value for Keq at a specific temperature, however there are unlimited number of equilibrium positions. Keq < 1: Reactants are favored at equilibrium Keq > 1: Products are favored at equilibrium Le Châtelier’s Principle When factors that influence an equilibrium are altered, the equilibrium will shift to a new position that tends to minimize those changes. or If a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. Factors that influence equilibrium: Concentration, temperature, and partial pressure (for gaseous) Example: If a reactant or product is added to a system at equilibrium, the system will shift away from the added component. If a reactant or product is removed, the system will shift toward the removed component Le Châtelier’s Principle CONCENTRATION EFFECTS : Once the equilibrium constant for a given equilibrium mixture is determined, we can calculate what will happen to any mixture of the substances concerned at the same temperature. Example: AgCl -> Ag+(aq) + Cl-(aq) K = 2.8 x 10-10 pure solid AgCl does not appear so K = [Ag+] [Cl-] So, if we put some AgCl into water it will dissolve so that [Ag+] = [Cl-] 1 x 10-5 m/L (and this concentration, as we have already said, is extremely small). Suppose we now add AgNO3 which is soluble in water to give Ag+ and NO3- - this will have the affect of increasing [Ag+] so that [Ag+][Cl-] now > 2.8 x 10-10 The fact that some of the Ag+ has come from AgCl and some from AgNO3 does not matter – the ions do not have memories - what matters is the value of the total [Ag+] in the solution. To retain equilibrium, [Ag+] and [Cl-] must decrease and this will be accomplished by the precipitation of more solid AgCl. This means that the addition of Ag+ ions (using AgNO3) causes some AgCl to precipitate from solution. This is called the common ion effect. Le Châtelier and the equilibrium law The response to changes in an equilibrium can be explained via the equilibrium law Consider C2H4(g) + H2O(g) ⇄ C2H5OH(g) [C2H5OH] , Sample value K= What happens if C2H5OH is added? [0.150] [0.0222] [0.0225] 300 = [C2H4] [H2O] Now [C2H5OH] > 0.150 HOWEVER, remember... K does not change for a given temperature To reestablish equilibrium the reaction goes back s Le Châtelier’s Principle Improve yield: Water removal in Esterification Le Châtelier’s Principle Changes in Pressure (Volume) 1. For a gas, decreasing volume of container increases pressure; particles have less space, collide more frequently. 2. System will respond by trying to relieve stress (decrease pressure.) 3. Forward reaction speeds up. 4. Over time, forward reaction slows down and reverse reaction speeds up. 5. At equilibrium, forward and reverse reactions occur at same rate, new equilibrium position established. 6. Equilibrium has shifted right, value of Keq unchanged. Pressure and equilibrium Pressure will increase if volume decreases, his will cause a shift in equilibrium... C2H4(g) + H2O(g) C2H5OH(g) K= [C2H5OH] [C2H4] [H2O] [0.150] [0.0222] [0.0225] , 300 = If volume is reduced, for example, by half, we will have: [0.300] [0.0444] [0.0450] For K to stay = 300, we must have a shift to the right = 150 N2O4(g) 2NO2(g) light dark (a) The syringe with a total volume of 15 mL contains an equilibrium mixture of N2O4 and NO2; the red-brown color is proportional to the NO2 concentration. 2. (b) If the volume is rapidly decreased by a factor of 2 to 7.5 mL, the initial effect is to double the concentrations of all species present, including NO2. Hence the color becomes more intense. 3. (c) With time, the system adjusts its composition in response to the stress as predicted by Le Châtelier’s principle, forming colorless N2O4 at the expense of red- brown NO2, which decreases the intensity of the color of the mixture. Pressure and equilibrium Consider the reaction: 2SO2(g) + O2(g) ⇄ 2SO3(g), 1. The total moles of gas decreases as reaction proceeds in the forward direction. 2. If pressure is increased by decreasing the volume (compression), a forward reaction occurs to reduce the stress. 3. Reactions that result in fewer moles of gas favor high pressure conditions. Equilibrium Determine whether the following reactions favor high or low pressures? 1. 2SO2(g) + O2(g) ⇄ 2 SO3(g); 2. PCl5(g) ⇄ PCl3(g) + Cl2(g); 3. CO(g) + 2H2(g) ⇄ CH3OH(g); 4. N2O4(g) ⇄ 2 NO2(g); 5. H2(g) + F2(g) ⇄ 2 HF(g); The Effect Temperature on Equilibrium Consider the following endothermic reaction: CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g), Ho = 205 kJ 1. Endothermic reaction absorbs heat heat is a reactant; 2. If heat is added to increase the temperature, it will cause a net forward reaction. 3. If heat is removed to reduce the temperature, it will cause a net reverse reaction. 4. Endothermic reactions favor high temperature condition. THE EFFECT OF TEMPERATURE on Equilibrium The equilibrium constant K has a specific value at a given temperature. If the temperature is changed then K will also change according to the relationship d(ln K) = H dT RT2 This relationship (Van 't Hoff equation) is given by thermodynamics – we will not derive it here. So qualitatively we can draw up the table below The Effect Temperature on Equilibrium If H is positive: endothermic (heat written as reactant) If H is negative: exothermic (heat written as product) N2O4(g) 55.3 kJ + N2`O4(g) light ⇄ 2NO2(g) ⇄ 2NO2(g) dark H = 55.3 kJ An increase in temperature will cause the forward reaction to take place in order to absorb the added heat (Le Chatelier) A new equilibrium will be established at the higher temperature PNO2 will increase; PN2O4 will decrease The gas mixture will become more brown Calculating Equilibrium Constant Example of exothermic reaction H2(g) + I2(g) ⇄ 2HI(g) H = -51.0 kJ [HI]2 [H2 ][I2 ] Keq = An increase in temperature favors the endothermic reverse reaction. When the system returns to equilibrium, the hydrogen iodide concentration will be smaller and the concentrations of hydrogen and iodine larger. The equilibrium constant will be changed: at 445°C, Keq = 64 at 490°C, Keq = 45.9 Equilibrium Exercise  Determine whether the following reactions favors high or low temperature? 1.2SO2(g) + O2(g) ⇄ 2 SO3(g); 2.CO(g) + H2O(g) ⇄ CO2(g) + H2(g); 3.CO(g) + Cl2(g) ⇄ COCl2(g); 4.N2O4(g) ⇄ 2 NO2(g); 5.CO(g) + 2H2(g) ⇄ CH3OH(g); Ho = -180 kJ Ho = -46 kJ Ho = -108 kJ Ho = +57 kJ Ho = -270 kJ Equilibrium Remember the Gibbs Free Energy state function? T=absolute temperature and is the weighting factor between H and S. G is the criterion for the feasibility of a chemical reaction i.e., G for a reaction negative means reaction is thermodynamically favourable G is also related to K as: where K is the equilibrium constant for a chemical reaction. G = H – TS G = -RTlnK. CATALYSIS Thermodynamics cannot tell us anything concerning the rate at which equilibrium is attained. So it is possible to have a reaction in which the products are expected to be formed in virtually a 100% yield but in which the forward rate is so slow that when the reactants are brought together nothing noticeable happens e.g., 2H2(g)+ O2(g) 2H2O(l) at 1 atm. and 298 K A catalyst is a chemical that affects the rate of a chemical reaction. It does not affect the position of equilibrium because it affects the forward and backward reactions equally. Le Chatelier’s Principle ignores the presence of catalysts Chemistry for the Life Sciences Lecture 8 Prof. Valeria Nicolosi School of Chemistry PI at Centre for Research on Adaptive Nanostructures and Nanodevices (CRANN) e-mail: [email protected] Characterisation & Processing of Advanced Materials Group http://chemistry.tcd.ie/cpam/ Ionic Equilibria Proton transfer equilibria: An enormously important biological aspect of chemical equilibrium is that involving the transfer of protons (hydrogen ions, H+) between species in aqueous environments, such as living cells. Even small drifts in the equilibrium concentration of H+ can result in disease, cell damage, and death! Brønsted–Lowry theory of acid and bases: An acid is a proton donor and a base is a proton acceptor. The proton, H+, is highly mobile and acids and bases in water are always in equilibrium with their deprotonated and protonated counterparts and hydronium ions (H3O+). Thus, an acid HA, such as HCN, immediately establishes the equilibrium: 𝐻 𝑂+ [𝐴−] 3 𝐾= 2 A base B, such as NH3, immediately establishes the equilibrium: 𝐻𝐴 [𝐻 0] 𝐻𝐵+ [𝑂𝐻−] 𝐾= 2 𝐵 [𝐻 0] Even in the absence of added acids and bases, proton transfer occurs between water molecules, and the autoprotolysis equilibriumis always present 𝐻𝐵+ [𝑂𝐻−] 𝐾= 2 As you will be familiar later on, the hydronium ion (H 3O+) concentration is commonly expressed in terms of the pH, which is defined formally as: ,where the logarithm is to base 10. 𝐵 [𝐻 0] Notice: - the higher the pH, the lower the concentration of hydronium ions H3O+ in the solution! - a change in pH by 1 unit corresponds to a 10-fold change in their molar concentration. Protonation and deprotonation: The protonation and deprotonation of molecules are key steps in many biochemical reactions, and we need to be able to describe procedures for treating protonation and deprotonation processes quantitatively. Going forward, all the solutions we consider are so dilute that we can regard the water present as being a nearly pure liquid and therefore as having unit activity. (a) The strengths of acids and bases When we set aH2O = 1 for all the solutions we consider, the resulting equilibrium constant is called the acidity constant, Ka, of the acid HA: 𝐾= 𝐴 𝐻 𝑂+ [𝐴−] 3 [𝐻𝐴] The value of the acidity constant indicates the extent to which proton transfer occurs at equilibrium in aqueous solution. - The smaller the value of Ka (for instance 10-8 compared with 10-6) and therefore the larger the value of pKa (for instance, 8 compared with 6), the lower is the concentration of deprotonated molecules. Most acids have Ka < 1 (and usually much less than 1), with pKa > 0, indicating only a small extent of deprotonation in water. These acids are classified as weak acids. A few acids, most notably, in aqueous solution, HCl, HBr, HI, HNO3, H2SO4, and HClO4, are classified as strong acids and are commonly regarded as being completely deprotonated in aqueous solution, Ka > 1. The corresponding value of pKb = −log Kb. A strong base is fully protonated in solution in the sense that Kb > 1 A weak base (like Ammonia, NH3) is not fully protonated in water, in the sense that Kb < 1 (and usually much less than 1). The autoprotolysis constant for water, Kw, is obtained in a similar way by setting the activity of water it its ‘pure’ value: [𝐻𝐵+][𝑂𝐻−] 𝐾= 𝑏 𝐾𝑎 [𝐵] [𝐻3𝑂+][𝐵] [𝐻𝐵+] = [𝐻3𝑂+][𝐵] [𝐻𝐵+][𝑂𝐻−] 𝑎𝑏 [𝐻𝐵+] [𝐵] 𝑤 𝐾𝐾= X =𝐾 - Ka increases as Kb decreases to maintain a product equal to the constant Kw - As the strength of a base decreases, the strength of its conjugate acid increases and vice versa If we take the negative common logarithm of both side: Or, another way to look at it: where pOH = −log aOH− How do we calculate the pH of a solution of a weak acid? Organize the necessary work into a table with columns headed by the species present in the mixture (ignoring H2O) and, in successive rows write: The initial molar concentrations of the species, ignoring any contributions to the concentration of H3O+ or OH- from the autoprotolysis of water. The changes in these quantities that must take place for the system to reach equilibrium. The resulting equilibrium values. Example: Acetic acid lends a sour taste to vinegar and is produced by aerobic oxidation of ethanol by bacteria in fermented beverages, such as wine and cider: CH3CH2OH(aq) + O2(g) ⇄ CH3COOH(aq) + H2O(l). Estimate the pH of: (a) 0.15M CH3COOH(aq) and (b) 1.5  10-4 M CH3COOH(aq). Ka = 1.8  10-5 We draw up an equilibrium table based on the proton transfer equilibrium CH3COOH(aq) + H2O(l) ⇄ H3O+(aq) + CH3CO2-(aq). Species CH3COOH H3O+ CH3CO2- Initial concentration(mol dm−3) 0.15 0 0 Change to reach equilibrium (mol -x +x +x dm−3) Equilibrium concentration (mol dm−3) 0.15-x x x (a) Th e value of x is found by inserting the equilibrium concentrations into the expression for the acidity constant: if x