Energetics of Chemical Processes (10 L) - University of Colombo - First Semester, May 2024 PDF

UNIVERSITY OF COLOMBO DEPARTMENT OF CHEMISTRY General and Physical Chemistry First Semester, May 2024 Energetics of Chemical Processes (10 L) Prepared and conducted By Professor K. M. Nalin de Silva Learning Outcomes: What determines th...

UNIVERSITY OF COLOMBO DEPARTMENT OF CHEMISTRY General and Physical Chemistry First Semester, May 2024 Energetics of Chemical Processes (10 L) Prepared and conducted By Professor K. M. Nalin de Silva Learning Outcomes: What determines the position of equilibrium? Why do some reactions go to completion? Clearly understand the heats of reaction Understand the standard state of reactions and standard enthalpies Understand how enthalpy varies with temperature Clear understanding of heat capacities (constant pressure and Volume) Understand the First Law of Thermodynamics (Heat and Work) Understand the various interpretations of the First Law Understand the Reversibility : Key concept in energetics Explain the failure of the First Law in relation to spontaneity Understand how distribution of energy affect the spontaneity of reactions Understand the necessity of the Second Law of Thermodynamics Clear understanding of Entropy (disorder, Chaos, Randomness, etc..) Understand the importance of Universal Entropy and Spontaneity Derivation of the Second Law of Thermodynamics Disguised form of Universal Entropy : Gibbs Function (also Helmholtz function) Understand the Master Equations and Pressure dependence of Gibbs Function Understand the Gibbs Function and Equilibrium Reference: P.W. Atkins – Physical Chemistry (Oxford University Press)2 INTRODUCTION AND HEATS OF REACTION The aims of chemical thermodynamics Thermodynamics is the study of transformations of energy. It is the “book keeping” side of chemistry, addressing such questions as why some reactions produce heat while others need heat to go. The fundamental theme is “what determines the position of equilibrium” i.e. Why do some reactions go to completion (i.e. have large equilibrium constants) while others do not (i.e. have small equilibrium constants. e.g.   HF 2  H2 + F2 2HF K =  (  10 48 )  H 2 F2 eq  N 2O5 2  −21 2N2 +5O2 2N2O5 K =  (  10 )  N 2  O2  eq 2 5  products  Eq. Constant K =    reactantseq (concentrations raised to appropriate power) We shall see that the position of the equilibrium is always such as to maximize the global entropy: Second Law – that the entropy of the universe increases in any spontaneous process – is the heart of thermodynamics. Suppose we have a chemical reaction, such as the industrial preparation of ammonia by the Haber process N2(g) + 3H2(g) 2NH3(g) We might ask a no. of questions about this reaction 1. What is the heat evolved 2. Does the reaction go – in other words what is the equilibrium constant K and 3. If the equilibrium is unfavourable, how can we change the conditions (e.g. temperature) in order to favour the products?  These questions are all aspects of the more fundamental question “what is it that drives a chemical reaction towards a particular position of equilibrium” We shall start with the first law (the conservation of energy) which deals with the enthalpy, H and internal energy, U and The second law, which relate to entropy S and then see how the two laws are combined to give a new quantity, the Gibbs free energy, G. The central equation that we are leading up to is And we shall see that this equation is central not only to chemistry and the chemical industry but also to biology  Note however that thermodynamics is only concerned with energetics and the position of equilibrium; it has nothing whatever to say about how fast that the equilibrium is reached: this question is the domain of kinetics. Thermodynamics may predict that a reaction will definitely go because the energy of the products is less than that of the reactants (G pext so that the natural dz tendency is for the piston to move out. The system (gas + cylinder + piston) will then radius = r do work on the surroundings as the piston pushes against the force due to the external pressure. force pressure= or F = − pext. A area  So the gas is expanding against a force –pext.A (negative sign because it acts downwards). If the piston is pushed up through a distance dz, then the work done by the system will be dw = -F(z)dz = -(-pext.A)dz but Adz is simply dV, the volume swept out by the piston, so we have dw = pext.dV or dw = -pext.dV We see that in expansion (dV  0) the system does work (dw’  0), but in compression (dV  0), the system has work done on it (dw’  0). Expansion against a constant external pressure If the external pressure is constant, then for a gas expanding from volume Vi to Vf Vf Vf w' = p Vi ext dV = pext  dV = pext (V f − Vi ) Vi w' = pext V Note that the work done depends on the external pressure, not the pressure of the gas in the cylinder (the work is determined by the force the gas is pushing against). Note also that we do not write w’ as work is not a state function – whereas the volume is a state function, and the change in volume V depends only on the initial and final volumes.  Example: Calculate the work done when 1 mole of a gas expands from 5 dm3 to 10 dm3 against a constant pressure of 1 atmosphere. w’ = pextV = (101325 Nm-2)[(10 – 5) dm3  10-3 m3 dm-3] w’ = 506.6 J Reversible isothermal expansion The concept of a reversible process is central to the development of thermodynamics: A reversible process is one that can be reversed by an infinitesimal modification of a variable. For example, in the case of the gas confined by a piston, suppose that pext is only infinitesimally less than pint: the gas expands, but if pext were increased only infinitesimally it would rise above pint and the gas would be compressed. The expansion is therefore reversible, the piston moves out infinitely slowly, and at each stage the system is in equilibrium (pext = pint). But if pext differs measurably from pint then changing pext infinitesimally does not reverse the direction of the process, i.e. expansion is irreversible: the piston moves out at finite speed, and to reverse the process we would first have to supply energy to stop the piston moving and then supply extra force to push the piston back in – and these are not infinitesimal changes.  In fact all natural processes are irreversible: only processes occurring infinitely slowly or processes which have come to equilibrium can be regarded as reversible. For a reversible process the system must be in equilibrium with its surroundings at each stage, and in the case of the gas expansion this means we must adjust pext continuously so that at each stage it exactly balances pint. pint Pint dropping  Vf Vf w' = p Vi ext dV = p Vi int dV For an ideal gas pint can be computed from the gas law pV = nRT, where R is the gas constant, n is the number of moles of gas and T is the temperature. Imposing the additional condition that the process is isothermal (T constant), Vf nRT 1 pint = w' = nRT  dV V Vi V Vf  w' = nRT ln  reversible  Vi   Example: Calculate the work done when 1 mole of a gas reversibly expands from 5 dm3 to 10 dm3 at 25 0C. w’ = (1 mol) (8.314 J K-1 mol-1)(298.15 K) ln (10 dm3/5 dm3) = 2478  ln (2) = 1718 J The work done in a reversible change is a maximum More about heat: U, H, CV and Cp Internal energy and constant volume processes If we consider a process taking place at  dq  C =  constant volume, no work of expansion is done,  dT  dw = 0, and thus by First Law (dU = dq + dw) we have dU = (dq)V Thus we can identify U with the heat absorbed in a constant volume process. const. V  U  heat CV =    T V Enthalpy and constant pressure processes The constant volume heat capacity CV is not the same as the constant pressure heat capacity Cp. CV applies to processes in a closed container, by contrast Cp applies to processes in open vessels. (or in vessels confined by a movable piston).  Here the pressure remains constant at atmospheric, and any gases produced will expand to equalize the pressure, thus pushing back the piston (or in an open vessel, pushing back the atmosphere). Extra heat must be supplied so that the system can do work of expansion, so in general Cp  CV At constant pressure we require (dq)p to substitute into above equation to get Cp. Can we identify this with some thermodynamic function, as in the case of constant volume, where (dq)V = dU ? We can if we introduce a new state function, the enthalpy H: H = U + pV This function has exactly the required properties: dH = dU + d(pV) = dU + pdV + Vdp But at constant p, Vdp = 0 So dH = dU + pdV = (dq)p Extra heating needed for work Heat required at constant of expansion Volume (dq)V Therefore just as we can identify U with the heat supplied at constant volume, qV, so we can identify H with the heat supplied at constant pressure, qp. We therefore have   H  Cp =  const. p  T  p heat Since most chemical reactions ‘on the bench’ take place under constant pressure conditions, i.e. in open vessels, it is natural and convenient to use H rather than U. By using H instead U we automatically take account of the extra energy needed for the work of expansion. H = U + pV = U + nRT (for a perfect gas) So H = U + nRT Where n is the change in the no. of moles of gas during the reaction  The Second Law of Thermodynamics What determines the direction of spontaneous change Some things happen spontaneously, others do not. Heat will flow spontaneously from a hot body to a cooler one but the opposite is never observed even though it does not violate the conservation of energy. Chemical reactions run one way rather than the reverse, e.g. burning diomand gives hot CO2, but hot CO2 does not spontaneously form diamond. A gas expands to fill the available volume: it doesn’t spontaneously contract to a smaller volume. Actually we can make heat flow from cold to hot (as in refrigeration), we can make diamonds, and we can compress gas, but these things do not happen spontaneously: we have to do work to bring them about. But what determines the direction of spontaneous change? It is not energy which determines the direction of change. Things do NOT “tend to a lower energy” We know from the First Law that the total energy of the Universe is constant! But the energy of the system tend to a lower value? Again NO.  When a change occurs, the total energy remains constant but is parceled out differently. The direction of change is determined by the distribution of energy: in spontaneous change, things tend to a state in which the energy is more chaotically dispersed. A change is spontaneous if it leads to greater randomness and chaos in the Universe as a whole. To see how energy tends to get more chaotically dispersed, consider a ball bouncing on a floor. The ordered directional motion of the ball gets converted to the random thermal motion of the floor molecules.  A ball resting on a warm floor will not spontaneously start bouncing: for this to happen, (1) the heat of the floor would have to accumulate in the ball (2) all the ball molecules would have to move in the same direction – upwards – simultaneously. This is possible (and does not contravene the First Law) but very, very improbable. Things move from an ordered state to a disordered state. The Universe moves from a state of lower probability to a state of higher probability.  Spontaneous Processes ◼Example of common Spontaneous processes: – Aging If a process is – Objects fallingspontaneous in one – Time direction, then under – sun rise the same conditions –Ink mixing the reverse process is non-spontaneous. –Chem. Exams –i.e., We do not grow young, gas does not contract, H2O does not freeze at room temperature, time does not go backwards and exams are not canceled. –‡ Note under different conditions however, the reverse process can occur. i.e., Liquid freezes to solid at 0°C. The State of Things to Come  Entropy and the Second Law The degree of chaos, randomness, or dispersal of energy is measured by a state function called the entropy, S. The second Law of thermodynamics can then be stated as: The entropy of the Universe increases in a spontaneous process The First Law uses internal energy to say which changes are permissible: we must have Uuniv = 0 The Second Law uses entropy to say which changes are spontaneous: we must have Suniv  0.  Entropy and irreversibility The classic example of an irreversible process is a falling weight generating heat on the floor. it loses its potential energy (ability to do work) degraded to thermal motion of the floor molecules process is irreversible because the weight will not suddenly rise from the hot floor. As the weight drops the quality of energy is degraded : the work required to lift the weight to a height in the first place is degraded to heat. Irreversible processes generate entropy and reversible processes do not generate entropy Reversible processes occur without degrading the quality of energy, and so can be reversed. The irreversibility of processes involving the conversion of work into heat means that we can give another statement of the second law: A ball on the hot floor will not start bouncing because this involves a decrease in entropy: the quality of energy cannot spontaneously improve from random thermal motion (heat) to directional motion (work). One can get some work from heat (as in the power stations) but one cannot convert the heat totally into work  The thermodynamic (macroscopic) definition of entropy We have established qualitatively what entropy means – but how can we get a quantitative measure of it? Entropy is definitely related to the amount of heat generated i.e extent to which energy has been degraded. Therefore S  q But in fact the amount of extra disorder caused by a given amount of heat q depends on the temperature: if it is already very hot, a bit of extra heat doesn't create much more disorder, but if the temperature is very low, the same amount of heat will cause a dramatic increase in disorder. It is therefore more appropriate to write q dq S = or dS = T T  However, this equation is still not quite right. S is a state function S for a given process = Sf - Si q clearly depends on the path In order to obtain a consistent value of S path must be specified. Specify the reversible path qrev S = T It is important to note that the entropy change of the system is defined in terms of the heat absorbed by the system if that change is brought about reversibly, dqrev syst , even if the change itself is not reversible. We will now justify this choice of definition by showing that it lead to the Second Law syst syst Heat absorbed in a reversible process is a maximum i.e. dqrev is always larger than dqirrev U = q + w = q - w work done in a reversible process is maximum, so if w’ is largest for a reversible path, then q must also be largest for a reversible path, to keep U fixed for all paths. syst  syst dq rev dq irrev  It therefore follows that for an irreversible change syst dqsyst dq rev  T T where dqsyst is the heat change involved in the actual (irreversible) process. And hence dqsyst ds syst  T This is called the Clausius Inequality. The inequality applies to irreversible processes and the equality applies to reversible processes.  We will now use the Clausius Inequality to show that dSuniv does indeed increase, as required by the Second Law, when we define entropy using a reversible as in (1) The entropy change of the Universe is that of the system plus that of the surroundings: dsUni = dssys + dssurr syst surr dq dq ds univ = rev + rev T T Now the key point is that the surroundings, being “the rest of the Universe”, are so large that they constitute an effectively infinite heat sink, i.e. if some heat is transferred to the surroundings from the system the temperature of the surroundings will not rise appreciably.  Since the surroundings are so large as to be unaffected by the transfer of heat, such a transfer can always be considered to be reversible from the point of view of the surroundings. We can surr surr therefore write dq = rev dq Now since dqsurr = -dqsyst, we therefore have syst dqrev dq surr ds univ = + T T And since dqrev syst  dqsyst , using Clausius Inequality we have syst dqrev dq syst dq syst ds univ = − = ds − syst T T T dSuniv  0 in any spontaneous process  Engines and Fridges Thermodynamics grew out of the study of ways of improving the efficiency of heat engines, which are devices for converting heat into work (e.g. steam engines, jet engines). We shall see that the efficiency of engines is limited by the Second Law: not all of the heat can be converted into work, only some of it. Hot Reservoir (Th) −q S h = q Th w q’ + q' S c = Cold Sink (Tc) Tc q' q S univ = − ; q − q' = w Tc Th We may still have Suniv > 0, even if q’ < q, provided Tc < Th. Clearly more work can be extracted if the temperature difference is larger.  Whereas the engine enables one to obtain work by transferring heat from a hot body to a colder one, the refrigerator uses work to transfer heat from cold body (the fridge) to a hotter one (the kitchen). Kitchen (Th) q+w w q Refrigerator(TC) Heat transfer from cold to hot is not spontaneous because it involves a decrease in entropy, so we must do work to achieve it. Removal of heat q from the cold body produces a decrease in entropy, Sc = -q / Tc, that is greater than the corresponding increase in entropy of the hot body, Sh = +q / Th. So the process is impossible unless we do some work w to produce a greater increase in entropy of the hot body, Sh = (q + w) / Th. Clearly the greater the temperature difference the more work we have to supply to effect this unnatural process.  More about Entropy Evaluation of the thermodynamic entropy As reversible processes take place infinitely slowly it is not at all convenient to try to measure qrev: instead we use a method based on the measurement of heat capacities. We have already shown that the heat (dq)p supplied at constant pressure can be identified with dH. This is equally true for reversible and irreversible processes: Since H is a state function the value of dH is the same for all paths provided no extra non-expansion work is being done (NB dH already includes any work of expansion). Therefore we can write dqrev dH ds = = at constant p T T From the definition of Cp, we have dH = CpdT, Substituting for dH and integrating from T = 0 to the required temperature T T' T' C p dT  dS =  0 0 T dT We have written Cp(T) to emphasize that the heat capacity depends on temperature. The integral can be evaluated by measuring Cp as a function of temperature. In addition, we must include any entropy change due to phase transitions such as melting or boiling. At a phase change the two phases are in equilibrium, so the process is reversible and we may identify qrev with Hphase change. Thus H phase change S phase change = T phase change  The full expression for the entropy of a gas at a temperature T’ is therefore H melt H boil Tmelt Tboil T' C p ( solid ) C p (liq.) C p ( gas) S (T ' ) = S (0) + 0 T dT + Tmelt +  Tmelt T dT + Tboil +  Tboil T dT Assume that S(0) = 0 (Third Law), and hence find absolute entropies, S(T), which are tabulated. From this we can find the entropy change of any reaction: rS = LSL + MSM + …..- ASA - BSB -….. Generally absolute entropies are tabulated at one temperature (using 298 K) and we need to know how to convert these to different temperatures. We can use T2 Cp S (T2 ) = S (T1 ) +  dT T1 T To obtain entropies at T2 from entropies at T1. Note the similarity to the Kirchoff equation  The importance of UNIVERSAL entropy Increasing entropy is thus the signpost of the direction of natural change. Endothermic reactions are driven by a large increase in the entropy of the system. Exothermic reactions are driven by an increase in entropy in the surroundings (to which heat is given out). Thus the Second Law does permit local abatements of chaos provided greater disorder is generated elsewhere. Suniv = Ssyst + Ssurr  0  The Gibbs function: the universal entropy in disguise! We thus have a criterion for deciding if a process is spontaneous: we simply look at the sign of Suniv. But this is not very convenient because we have to consider both system and surroundings. surr dqrev dSuniv = dSsyst + dSsurr  0, we can write ds surr = T From the definition of the entropy. Surroundings large, therefore dq surr Now from the conservation of energy we must have dqsurr = -dqsyst ds surr = T dq surr dq syst ds surr = dS surr =− T T At constant pressure dq = dH, so  dH syst dS surr =− and the second law becomes T dH syst dS univ = dS syst − 0 T dH syst − TdS syst  0 This version of the Second Law can be made less cumbersome if we define a new state function, the Gibbs function G = H - TS Sometimes called the “Gibbs free energy”.  This function has the required properties: dG = dH – TdS - SdT dG = dH – TdS at constant pressure or G = H - TS We now drop the subscript “syst” because all functions now refer to the system, and the Second Law becomes, Second Law: G  0 at constant T and P for spontaneous processes i.e. examination of the sign of G will reveal if a process is spontaneous or not. H S G Spontaneity -ve +ve -ve Spontaneous at all temperatures +ve -ve +ve Non-spontaneous at all temperatures -ve -ve +ve or Non-spontaneous at high –ve temperatures. Spontaneous at low temperatures +ve +ve +ve or Non-spontaneous at low –ve temperatures Spontaneous at high temperatures  The master equations dU = TdS – pdV (du = dq + dw) dH = TdS + Vdp (H = U + pV) dG = Vdp – SdT (G = H – TS) dA = -pdV - SdT (A = U – TS) The pressure dependence of the Gibbs function  G  dG = Vdp − SdT   = V  p T G2 p2 p2 nRT G dG = p Vdp = p p dp 1 1 1  p2  G2 − G1 = nRT ln   p1  We can use this to relate the Gibbs function G of an ideal gas at a general pressure p to the Gibbs function G0 at a standard pressure of p0 = 1 bar  G = G0 + nRT ln p 0 G G = + RT ln p n n  = μ + RT ln p 0  - Chemical potential, is the molar Gibbs function  The Gibbs Function and Equilibrium The condition for equilibrium Reactions go in the direction of increasing universal entropy, Suniv  0, which corresponds to decreasing Gibbs function of the system, G  0. If we plot the Gibbs function as the composition changes from reactants to products we find the equilibrium composition (minimum G) somewhere in between. Forward reaction occur G < 0 G = 0 G > 0 Reverse reaction occur Spontaneous nonspontaneous in in forward forward direction direction G < 0 G > 0 Consider a reaction AA +   +  LL +  +  GA = G +  A RT ln p A 0 A GB = G +  B RT ln pB 0 B GL = G +  L RT ln pL 0 L GM = G +  M RT ln pM 0 M For the given reaction, change in Gibbs function is given by substituting (GL0 + GM0 ) −  ( M ln p M +  L ln p L ) −   rG =  0  + RT   (G A + GB )  0 ( A ln p A +  B ln p B )    ( p L ) ( pM )  L M  r G =  r G + RT ln  0 B   ( p A ) ( pB )  A To say that rG varies during the reaction is the same as saying that the slope of the “Gibbs hill” varies during the reaction. The reaction will “run down the Gibbs hill” until it reaches equilibrium at the point of zero slope, rG =0  ( p L ) ( pM ) L M  r G = − RT ln  0 B   ( p A ) ( pB )  A   r G = − RT ln K p 0  ( p L ) L ( pM ) M  where K p =  B   ( p A ) ( pB )  A is the equilibrium constant of the reaction. For real gases we replace pressure by fugacity and for reactions in solutions we replace pressure by the corresponding activity.