CH1101 Physical Chemistry Tutorial 1. Prof. Mike Lyons.

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1 CH111 Physical Chemistry Tutorial 1. Prof. Mike Lyons. CH111 Section A Annual 1

2 Internal Energy Units: Joules J Internal Energy (U) : total kinetic & potential energy of system. e.g. Gas in container with (n,p,v,t). Gas molecules translate, rotate & vibrate so U = E trans + E rot + E vib First law Thermodynamics relates change in internal energy of system U to energy changes arising from transfers of heat q and work w. U = q + W

3 First Law of Thermodynamics: Internal energy U or E. Change in internal energy ( U or E) : energy change as system goes from an initial state with energy U i to a final state with energy U f. Hence U = U f U i. We suppose that a closed system undergoes a process by which it passes from a state A to a state B. Then the change in internal energy U = U B U A is given by U = q +W. Kotz, section 5.4. pp.-4 A Initial state B Final State Remember: analogy to fund transfer & how Bank A/c balance is effected. + q + U + W System Surroundings U = q + W Heat transfer to system We area led therefore to the important conclusion that the energy of an isolated system is constant. In an isolated system there is no transfer of heat or work and so q = and W =. Hence U = hence U = constant. Increase in internal energy Work done on system This is a mathematical statement of the First Law of Thermodynamics.

4 Spontaneous processes and entropy Kotz, Ch.19, pp Discussion on energy dispersal Very good. A process is said to be spontaneous if it occurs without outside intervention. Spontaneous processes may be fast or slow. Thermodynamics can tell us the direction in which a process will occur but can say nothing about the speed or the rate of the process. The latter is the domain of chemical kinetics. There appears to be a natural direction for all physical and chemical processes. A ball rolls down a hill but never spontaneously rolls back up a hill. Steel rusts spontaneously if exposed to air and moisture. The iron oxide in rust never spontaneously changes back to iron metal and oxygen gas. A gas fills its container uniformly. It never spontaneously collects at one end of the container. Heat flow always occurs from a hot object to a cooler one. The reverse process never occurs spontaneously. Wood burns spontaneously in an exothermic reaction to form CO and H O, but wood is never formed when CO and H O are heated together. At temperatures below C water spontaneously freezes and at temperatures above C ice spontaneously melts.

6 The First Law of thermodynamics led to the introduction of the internal energy, U. The internal energy is a state function that lets us assess whether a change is permissible: only those changes may occur for which the internal energy of an isolated system remains constant. The law that is used to identify the signpost of spontaneous change, the Second Law of thermodynamics, may also be expressed in terms of another state function, the entropy, S. The entropy (which is a measure of the energy dispersed in a process) lets us assess whether one state is accessible from another by a spontaneous change. The First Law uses the internal energy to identify permissible changes; the Second Law uses the entropy to identify the spontaneous changes among those permissible changes.

7 The characteristic common to all spontaneously occurring processes is an increase in a property called entropy (S). Entropy is a state function. This idea form the basis of the Second Law of Thermodynamics. The change in the entropy of the universe for a given process is a measure of the driving force behind that process. In simple terms the second law of thermodynamics says that energy of all kinds in the material world disperses or spreads out if it is not hindered from doing so. In a spontaneous process energy goes from being more concentrated to being more dispersed. Entropy change measures the dispersal of energy: how much energy is spread out in a particular process or how widely spread out it becomes at a specific temperature. Second law of Thermodynamics The second law of thermodynamics states that a spontaneous process is one that results in an increase in the entropy of the universe, S universe >, which corresponds to energy being dispersed in the process. S total = S system + S surroundings

Entropy measures the spontaneous dispersal of energy : How much energy is spread out in a process, or how widely spread out it becomes at a specific temperature.

8 Entropy measures the spontaneous dispersal of energy : How much energy is spread out in a process, or how widely spread out it becomes at a specific temperature. Mathematically we can define entropy as follows : entropy change = energy dispersed/temperature. In chemistry the energy that entropy measures as dispersing is motional energy, the translational, vibrational and rotational energy of molecules, and the enthalpy change associated with phase changes. Energy transferred as heat Under reversible conditions. S system SS system = system = qrev T H phasechange T Entropy units : J mol -1 K -1 Note that adding heat energy reversibly means that it is added very slowly so that at any stage the temperature difference between the system and the surroundings is infinitesimally small and so is always close to thermal equilibrium.

9 What entropy is not and what it is. Entropy is not disorder. Entropy is not a measure of disorder or chaos. Entropy is not a driving force. The diffusion, dissipation or dispersion of energy in a final state as compared with an initial state is the driving force in chemistry. Entropy is the index of that dispersal within a system and between the system and its surroundings. In short entropy change measures energy s dispersion at a stated temperature. Energy dispersal is not limited to thermal energy transfer between system and surroundings ( how much situation). It also includes redistribution of the same amount of energy in a system ( how far situation) such as when a gas is allowed to expand adiabatically (q = ) into a vacuum container resulting in the total energy being redistributed over a larger final total volume. Entropy measures the dispersal of energy among molecules in microstates. An entropy increase in a system involves energy dispersal among more microstates in the system s final state than in its initial state.

10 Possible ways of distributing two packets of energy between four atoms. Initially one atom has quanta and three with zero quanta. There are 1 different ways (1 microstates) to distribute this quantity of energy Microstate 1 ways (1 microstates) to distribute quanta of energy among 6 atoms.

11 Entropy in the context of Molecular Thermodynamics. Entropy measures the dispersal of energy among molecules in microstates. An entropy increase in a system involves energy dispersal among more microstates in the system s final state than in its initial state. S = k W B ln S = S = k = k final B B S initial { ln W ln W } W ln W final final initial initial W = number of accessible Microstates k B = Boltzmann constant Microstates with greatest energy dispersion = 1.38 x1-3 J K -1 are most probable.

12 A 1 W electric heater is used to heat gas in a cylinder for 1 min. The gas expands from 5 cm 3 to 1.4 dm 3 against a pressure of 1.1 atm. What is the change in internal energy of the gas? Strategy Calculate the energy transferred as heat from the power of the heater and the time for which it operates. Then determine the energy transferred as work as the gas expands using w = -P ext V. Finally, use U = q + W, to determine the change in the internal energy as the sum of the energy transferred as heat and as work. It will be necessary to pay close attention to the units and signs of the various quantities. Solution Power P = de/dt = rate of supply of energy. P = 1 W = 1 Js -1 Time t = 1 min = 6 s. Energy transferred as heat: q = (1 Js -1 ) x (6 s) = 6 x 1 3 J = 6 kj. When gas expands work done by system is negative, W = - P ext V. P ext = 1.1 atm = (1.1 atm)x(1135 Pa atm -1 ) = 111,458 Pa (or Nm ). Change in volume V = V final V initial = 1.4x1-3 m 3-5x1-6 m 3.9 m 3. q = + 6 kj Hence, work done by gas: W = -(111,458 Nm - ) x(.1 m 3 ) = - 4 Nm = - 4 J = -.4 kj Change in internal energy U = q + W = + 6 kj.4 kj = kj U = kj W = -.4 kj

13 Calculate the entropy change when 1. mol of water at 373 K vaporizes to steam. You may assume that the enthalpy change vap H for water is 4.7 kj mol -1. Use entropy change = vap S = q rev /T = vap H/T b Now T = T b = 373 K and vap H = kj mol -1 = + 4,7 J mol -1. Hence, vap S = vap H /T b = + 4,7 J mol -1 /373 K = + 19 JK -1 mol -1.

14 CH111 Section A Annual 1

15 Chemical Equilibrium Many reactions do not proceed to completion. Instead a definite quantity of reactants and products are present in reaction vessel after a long reaction time has elapsed (the state of equilibrium). Countless experiments with chemical systems have shown that in a state of equilibrium, the concentrations of reactants and products no longer change with time. This apparent cessation of activity occurs because under such conditions, all reactions are microscopically reversible. We look at the dinitrogen tetraoxide/ nitrogen oxide equilibrium which occurs in the gas phase. N O 4 (g) colourless Kinetic analysis. Rate equations R = k R = k [ NO4 ] [ NO ] NO (g) brown R = R k[ N O ] k 4 = [ NO ] eq [ NO ] eq k = = K [ N O ] k 4 Equilibrium constant enables one to Calculate the composition of a reaction Mixture (amount of reactants remaining & products formed) at equilibrium. Concentrations vary with time concentration eq [ NO ] t [ N O ] Equilibrium constant Concentrations time invariant [ NO ] eq 4 t [ NO4 ] eq Kinetic regime time Equilibrium: NO t N O 4 t

THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B p P + q Q the following is a CONSTANT (at a given T) Equilibrium constant K = p [ P] [ Q] eq a [ A] [ B] eq Product

16 THE EQUILIBRIUM CONSTANT For any type of chemical equilibrium of the type a A + b B p P + q Q the following is a CONSTANT (at a given T) Equilibrium constant K = p [ P] [ Q] eq a [ A] [ B] eq Product concentrations q eq b eq The size of the equilibrium constant indicates whether the reactants or the products are favoured. Reactant concentrations If K is known, then we can predict concentrations of products or reactants in the reaction mixture at equilibrium and hence the yield of the reaction. Reactants favoured when K c is small Reactants and products are in almost equal abundance when K c near unity Products favoured when K c is large

17 Relationship between Gibbs Energy and Equilibrium Constant. We now derive an expression which relates the change in Gibbs energy for a reaction as a function of the composition of the reaction mixture at any stage in the reaction. aa + bb pp + qq Activity = generalised concentration We can define Gibbs energy in terms of the activity a k of the species k. G k = G k + RT ln a k Hence after some algebra and simplification the change in Gibbs energy for reaction can be computed. r r r G ( pg + qg ) ( ag bg ) G = + G = = p r P Q ( GP + RT ln ap ) + q( GQ + RT ln aq ) a( G + RT ln a ) b( G + RT ln a ) G A a + RT ln a A p P a A. a. a q Q b B A B = r G B B + RT ln Q If the reaction is allowed to proceed To equilibrium then we replace Q by The equilibrium constant K and set r G = by definition. r r G G + RT ln K = = RT ln K ( pg + qg ) ( ag bg ) rg = P Q A + B Reaction quotient Q

Gibbs energy change for reaction mixture a A + b B c C + d D Q < K Q > K Under non-equilibrium conditions Gibbs energy change is : G = G = G + RT + RT ln ln c [ C] [ D] a [ A] [ B] Q d b Expression

18 Gibbs energy change for reaction mixture a A + b B c C + d D Q < K Q > K Under non-equilibrium conditions Gibbs energy change is : G = G = G + RT + RT ln ln c [ C] [ D] a [ A] [ B] Q d b Expression shows how G varies with composition of reaction mixture. Q = reaction quotient [ C ] c [ D ] d Q = [ A] a [ B] b Q defines reactant and product concentration ratio (i.e. reaction composition) at any stage in chemical transformation. When G = at constant T and P we have equilibrium. Hence Q = K c. c [ C] [ D] [ A] eq a[ B] d G RT ln eq G = RT ln K = + b eq eq K G = exp RT = G + RT ln K

product favoured. The reaction is said to be spontaneous when r G is negative. If Q/K > 1 then Q > K and r G is positive.

19 = + rg rg RT ln Q rg = RT ln K G = RT ln K + RT ln Q r = RT (ln Q ln K) This is the most useful form of the equation for interpretation. Q rg = RT ln K If Q/K <1 then Q < K and r G is negative, the reaction tends to proceed in forward direction and the reaction is said to be product favoured. The reaction is said to be spontaneous when r G is negative. If Q/K > 1 then Q > K and r G is positive. Here the reaction does not proceed spontaneously in the forward direction and is said to be reactant favoured. If Q = K then since ln 1 = r G = and the reaction is at equilibrium.

20 Reaction Gibbs energy r G = G ξ p, T extent of reaction = ξ K = 1 K << 1 K >> 1 G = r G = r G = r ξ = ξ = 1 r G is the slope of the G versus ξ graph at any degree of advancement ξ of the chemical reaction.

21 The thermodynamic equilibrium constant K is 6.46 at 373 K for the reaction N O 4 (g) NO (g). 1 mol of N O 4 is introduced into a sealed container at 373K. The final pressure in the container is 1.5 bar when the system comes to equilibrium. Find the composition of the mixture at equilibrium. Note that 1 bar = 1 x 1 5 Pa. Strategy. We use the stoichiometry of the reaction to derive expressions for the mole fraction, And hence partial pressure of each component in terms of the initial amount of N O 4 And the proportion that has reacted. We use these expressions for partial pressures to generate an expression that relates the equilibrium constant to the proportion of N O 4 which has reacted. The latter expression is then rearranged and solved for proportion of N O 4 reacted. If 1 mol N O 4 is present initially and a fraction α reacts then amount of N O 4 present at equilibrium can be determined. nn O = = = 4 n 1 + = = n 1 + ( NO ) ( ) n = n + n = 1 α + α = 1+ α X X total N O NO N O NO n n N O total NO total 1 α α α α 1 α pno = X 4 NO p = 4 p 1+ α α pno = X NO p = p 1+ α K NO p p p 1 pno p p p 4 4 = = N O 1 α mol N O 4 (g) NO (g) For every molecule of N O 4 that reacts molecules of NO are formed. N O NO ( α ) N O 4 NO { ( ) } ( 1 α ) p ( 1 α ) p = 1.5 bar K = 6.46 p = 1bar n = α mol Standard pressure p = 1 bar P = final pressure α p 1 + α 1 4α p 4α p K = = = { + } p ( 1 α )( 1 + α ) p 1 α p K 1 = 4α p p K α K = 4α p p α α n n K 1 = = 4 p 4 p K p K p =.518 α =.518 = = 1 α = 1.7 =.8 mol = α =.7 = 1.44 mol Equilibrium Composition

Chapter 16. Thermodynamics. Thermochemistry Review. Calculating H o rxn. Predicting sign for H o rxn. Creative Commons License

Chapter 16. Thermodynamics. Thermochemistry Review. Calculating H o rxn. Predicting sign for H o rxn. Creative Commons License Chapter 16 Thermodynamics GCC CHM152 Creative Commons License Images and tables in this file have been used from the following sources: OpenStax: Creative Commons Attribution License 4.0. ChemWiki (CC