VIEWS: 33 PAGES: 25 CATEGORY: Current Events POSTED ON: 5/25/2010 Public Domain
Chemical Thermodynamics • Spontaneous Processes • Entropy and the Second Law of Thermodynamics • The Molecular Interpretation of Entropy • Entropy Changes in Chemical Reactions • Gibbs Free Energy • Free Energy and Temperature • Free Energy and the Equilibrium Constant 1 Spontaneous Processes First Law of Thermodynamics stated: Energy can not be created nor destroyed, it can be converted from one form to another or transferred from a system to surroundings (or vice versa) A reaction is spontaneous if; once it starts, it proceeds on its own, regardless of how fast the reaction occurs. Reactions that are spontaneous in one direction are non-spontaneous in the reverse direction 2 Natural change occurs in one direction only, for example the rusting of iron, melting of ice above 0 oC, a ball rolling downhill etc. A non spontaneous process can only be brought about by doing work, it won’t occur unless some external action is continually applied, e.g. pushing the ball back up the hill 3 To know if a reaction is going to be spontaneous or not we need to distinguish between reversible and irreversible reactions A (chemical) system at equilibrium is reversible, an infinitesimal change in the system (reactants or products) can reverse the direction Reversible processes are those that can reverse direction whenever there is an infinitesimal change in the property of the system A reversible process is one in which: 4 A chemical system that is not at equilibrium is irreversible, the reaction will proceed spontaneously until equilibrium is reached. In this case the system and surroundings are not BOTH returned to their original condition – IRREVERSIBLE process. I had to do work to return the system (gas) to its original volume. 5 Entropy Spontaneous reactions often give out heat ( exothermic ) but this isn’t always the case , so enthalpy alone is unable to account for the direction of spontaneous change. Entropy is the other thermodynamic quantity that accounts for the direction of spontaneous change. Entropy is nature’s driving force to move to a condition of maximum randomness or disorder. Entropy is given the symbol S Entropy is a state function, the change in entropy only depends on the initial and final states of the system 6 For an isothermal process (constant T), e.g. a phase change e.g. Phase Changes ΔS = qrev / T (absolute Temp.) Recall: qrev is the heat exchanged at constant pressure, the rev implies that the heat must be transferred reversibly (in infinitesimally small amounts) Ethanol boils at 78 oC. The molar enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy in the system when 68.3 g of ethanol at 1 atm is vaporized at the normal boiling point? Because S is a state function we can use the final and initial states to calculate ΔS for any isothermal process, not just those that are reversible, we imagine the process as reversible 7 Second Law of Thermodynamics Unlike energy, entropy is NOT conserved: The Universe wants to become more and more disordered, it wants its total entropy to increase continually! Reversible process ΔSuniv = ΔSsys + ΔSsurr = 0 Irreversible process ΔSuniv = ΔSsys + ΔSsurr > 0 The rusting of a car is a spontaneous process. During this process the entropy of the system (the car) decreases. What conclusion can be made about the Entropy of the surroundings 8 A Molecular interpretation of Entropy Why is greater randomness the driving consideration for spontaneity The relationship of Entropy to this disorder (randomness) or probability was studied by Boltzmann. Boltzmann was thinking about a mole of gas particles in random motion and then considered what state they would be in if you froze them all and took a picture. Their positions at that time he called a microstate To figure out the seemingly impossible problem of relating Entropy to how many different microstates there could possibly be in a system Boltzmann came up with a very simple equation k is Boltzmann’s constant 1.38 x 10-23J/K (the gas constant R, divided by Avogadro’s number) and W is the number of microstates associated with a particular state. What would be some of the motions of a water molecule? 9 ∆S = k ln Wfinal - k ln Wfinal Boltzmann’s equation demonstrates: Any change in a system that increases the number of microstates leads to an increase in the entropy Therefore any factors that can cause an increase in the number of microstates (motions and energies of the molecules) will increase the entropy These Factors are: Solid is sublimed to a vapour, molecules have more degrees of motion, and a greater range of Kinetic Energies Increase temperature further, Energies increase further 10 The more ‘stuff’ you have in your house, the messier it is! For example dissociating (NH4)2Cr2O7(s) to Cr2O3(s), N2(g) and 4H2O(g) Vaporizing water, the molecules spread out into a larger volume and have a greater range of motional energies (microstates) Predict whether the Entropy change will be positive or negative for the following processes: CO2(s) → CO2(g) 2SO2(g) + O2(g) → 2SO3(g) 11 So, if increasing the temperature increases the number of microstates could we decrease the temperature to a point where all motion stops and we have a single microstate? The Third Law Of Thermodynamics At 0K, all the units in the lattice have no thermal energy, no motion, 1 Microstate S = k lnW W = 1 lnW = 0 Therefore: S = 0 12 What happens to Entropy as we continue to heat Why is the Entropy of phase changes following the order Ssolid < Sliquid < Sgas 13 Entropy Changes in Chemical Reactions We can measure ΔH of a reaction by using calorimetry, however there is no such easy method for determining ΔS for a reaction Absolute values for Entropies, S, can be obtained Standard Molar Entropy So is defined for pure substances at 1 atm pressure and 298K Why? 14 The Entropy change for a reaction (system) can be calculated from the Standard Molar Entropies Using Standard Molar Entropies, calculate the Entropy Change ΔSo for the reaction: N2 (g) + 3H2(g) → 2NH3 (g) does the answer make sense? Why? Entropy Changes in the Surroundings The surroundings act as a big heat sink, the entropy change of the surroundings depends only on the heat gained or lost: For a constant pressure reaction (usual) then qp (heat exchanged at constant pressure) = ΔH, so the entropy change of the surroundings can be written as 15 Because Souniv is positive (increases) for any spontaneous reaction we can put together the equations for calculating ΔSsys with ΔSsurr to predict whether a reaction will be spontaneous What would be the entropy change (ΔSouniv) for the following reaction: CO(g) + 2H2(g) → CH3OH (l) at 298 K ΔHosys = -128.1 kJ Use you answer to predict whether the reaction would be spontaneous Prediction of Spontaneity depends on The sign of ΔH and ΔS The magnitude of ΔH, ΔS and the temperature (in KELVIN) A spontaneous reaction can be exothermic or endothermic 16 Gibbs Free Energy JW Gibbs came up with a new state function G, that connected entropy and enthalpy to predict whether a reaction occurring at a constant temperature would be spontaneous From the definition of ΔSuniv we can relate the state function G to spontaneity: ΔSuniv = ΔSsys + ΔSsurr = ΔSsys + (-ΔHsys/T) Multiply both sides by: –T to give: 17 Standard Free Energy The standard free energy change ΔGo is defined as the change in the free energy when reactants in their standard states are converted to products in their standard states: Conventions used in establishing Standard free energies Solid Pure solid Liquid Pure liquid Gas 1 bar pressure solution 1M elements ∆Gof in standard state = 0 18 Standard Free energy change can be calculated from the standard enthalpy change and the standard entropy change. ΔHo and ΔSo don’t change (significantly) with temperature, but ΔGo does depend on temperature and might even change sign as the temperature is changed. Calculate the standard free energy change ΔGo for this reaction at 25 oC and 500 oC. What do the calculated values tell you? Given that ΔHo = -92.38 kJ and ΔSo = -198.4 J/K N2(g) + 3H2(g) 2NH3(g) 19 Standard Free Energy of Formation The standard Gibbs free energy of formation is the change in free energy when of 1 mole of substance is formed from its component elements in their most stable (standard) form at 1 atm pressure and 298K. From these values we can calculate the standard free energy change of reactions at 298K G is a State Function, so: Calculate the standard Free energy change ΔGo for the reaction of glass being etched by hydrogen fluoride at 298 K. Use the answer to predict if the reaction happens spontaneously SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(g) ΔGfo SiO2 = -856 kJ/mol ΔGfo HF = -273 kJ/mol ΔGfo SiF4 = -1573 kJ/mol ΔGfo H2O = -224 kJ/mol 20 Free Energy and Temperature The entropy term (TΔS) is dependant on the absolute temperature and as such the change in free energy will also vary with temperature. As we saw, it is possible that the temperature could change the sign of ΔG and a reaction that is non spontaneous at one temperature could become spontaneous at another temperature, for example: The decomposition of sodium bicarbonate is non spontaneous at 298K . The enthalpy of the reaction is 135.4 kJ and the Entropy of reaction is 333 J/K Calculate the temperature above which the reaction would become spontaneous assuming ΔH and ΔS do not change with temperature 21 Recall: that when ΔG = 0 the reaction is at equilibrium and therefore: When ΔG = 0 slight adjustments in temperature can change the direction of spontaneity (see previous NaHCO3 example) Recall: a substance at its melting point and boiling point exists in equilibrium: ΔH = ΔS and so: |ΔH| = |TΔS| (isothermal) T The normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapour at 1 atm (bar) pressure. Given that ∆Ho and ∆So for CCl4 at 298K are 32.6 kJ and 95.0 J/K, calculate the normal boiling point of CCl4 in units of oC 22 Free Energy and the Equilibrium Constant We can use the value of ΔGo to calculate the value of ∆G under non standard conditions. For any chemical process the relationship between the standard free energy change and the free energy change under any other set of conditions at a given temperature is: The expression for Q is the same as the Equilibrium Constant [products]n [reactants]m Where m and n are the stoichiometric coefficients. But for Q, the reactants and products need not be in equilibrium. Therefore under non standard conditions we can calculate ΔG by calculating Q and using our value of ∆Go at the given temperature. 23 Calculate ΔG for this reaction at 298 K when Q = 9.3 x 10-3. Use the previously calculated value of ΔGo of -33.3 kJ N2(g) + 3H2(g) 2NH3(g) At equilibrium ΔG = 0 and Q = K (the equilibrium constant) therefore: If ΔGo is negative then ln K must be positive, i.e. K > 1. The larger the value of K the more negative the value of ΔGo which means products will be favoured in the equilibrium mixture. (For a given temperature). Use the value of Go at 25 oC to calculate the equilibrium constant K at 25 oC N2(g) + 3H2(g) 2NH3(g) And predict whether products or reactants will be favoured in the equilibrium mixture at this temperature 24 Driving non-spontaneous reactions This is done by coupling the non-spontaneous reaction with a spontaneous reaction for example: Cu2S(s) → 2Cu(s) + S(s) ΔGo = +86.2kJ S(s) + O2(g) → SO2(g) ΔGo = -300.4kJ What would be the standard free energy change ΔGo for the overall coupled reaction? Note: The 10th edition of Brown, US version of the course textbook uses 1 atm as the standard state and the 11th edition uses 1 bar. 1 bar is universally accepted. As these values are (to a close approximation) the same, these notes use bar and atm interchangeably. Atm is more commonly used as we have become accustomed to using this value in Gas Laws and thermochemistry. 25