UNIT 5 Thermodynamics Free Energy Electrochemistry Internal Energy FIRST LAW OF THERMODYNAMICS: ΔE = q + w The signs of thermodynamic variables are important. • ΔE > 0 means the internal energy of the system increased. • q > 0 means heat was transferred into the system. • w > 0 means work was done on the system. w>0 q>0 ΔE State Functions Functions with a fixed value for a given P, T, V, and component composition are called state functions. E is a STATE FUNCTION. Enthalpy Enthalpy is a state function that accounts for heat flow in chemical changes occurring at constant pressure. The enthalpy H is defined in terms of the internal energy, pressure, and volume of a system: H = E + PV When pressure is constant, ΔH = ΔE + PΔV H is a STATE FUNCTION. Enthalpy When a chemical reaction is performed under conditions of constant pressure, the enthalpy can be used to determine whether the reaction is exothermic or endothermic: ΔH = qp Summary 1. Work and heat are ways to transfer energy. 2. First Law of Thermodynamics: “The energy of the universe is conserved.” The first law shown in terms of the system: ΔE = q + w Preview For chemical reactions carried out at constant temperature and pressure, the spontaneity of the reaction is given by the Gibbs Free Energy G ΔG = ΔH - TΔS where ΔS is the change system. in entropy of the Entropy Entropy (S) is a state function that is related to the degree of randomness of a system. S is a STATE FUNCTION. Entropy - FYI On a molecular level, the entropy is defined as follows: S = k ln W where k is Boltzmann constant (the gas constant R divided by Avogadro’s number) and W is the number of available microstates. We will concentrate on changes in S, not on its absolute value. Standard Molar Entropies S° THE STANDARD MOLAR ENTROPY S° is for the substance in its standard state. The standard state of any substance is its pure form at 1 bar pressure and the temperature of interest (often 25°C). 1. Unlike Δf H°, S° of an element in its standard state is NOT zero at 298 K. 2. S° for a gas is greater than S° for a liquid or solid. 3. S° generally increases with increasing molar mass. 4. S° generally increases with increasing numbers of atoms in the formula of a substance. Comparing Entropies Which of each pair has greater entropy? a) 1 mol of NaCl(s) at 25°C or 1 mol of HCl(g) at 25°C? 1 mol of HCl(g) at 25°C b) 2 mol of HCl(g) at 25°C or 1 mol of HCl(g) at 25°C? 2 mol of HCl(g) at 25°C c) 1 mol of HCl(g) at 25°C or 1 mol of Ar(g) at 25°C? 1 mol of HCl(g) at 25°C d) 1 mol of N2(g) at 25°C or 1 mol of N2(g) at 84 K? 1 mol of N2(g) at 25°C Predicting ΔS for a Reaction Entropy generally increases for the following processes: 1. Liquids or solutions formed from solids. 2. Gases formed from liquids or solids. 3. An increase in the number of moles of gas as the result of a reaction. Predict the entropy change in each of the following isothermal reactions. a) CaCO3(s) CaO(s) + CO2(g) ΔS>0 b) N2(g) + 3H2(g) 2NH3(g) ΔS<0 c) N2(g) + O2 (g) 2NO(g) ΔS≈0 Using S° Data to Find ΔS° for a Reaction This is done in exactly the same way we calculated heats of reaction ΔrxnH°. ΔrxnS° = ΣmS° (products) – ΣnS°( reactants) (m and n represent the stoichiometric coefficients) We get the values for S°(25°C) from Appendix C. ΔS of a Reaction 2H2(g) + O2(g) 2H2O(l) ΔS° = S°final – S°initial ΔS°(25°C) = ? This is the same as ΔS° = S°products – S°reactants ΔS°(25°C) = 2S°(H2O(l)) - 2S°(H2(g)) - S°(O2(g)) ΔS°(25°C) = 2(69.91) - 2(130.58) - 205.0 ΔS°(25°C) = -326.3 Units are not . Note: The only time entropy is zero is for a perfect crystal at 0 K. Entropy is an Extensive Property 2H2(g) + O2(g) 2H2O(l) ΔS°(25°C) = -326.3 Burning two moles of H2 in O2 (under conditions of constant P) would decrease the entropy of the system by 326.3 J/K for every mole of O2 or every two moles of water or H2. How much does the entropy change when 0.275 g of H2 is ignited with O2 in a constant-pressure system? 0.275 g H2(g) 1 mol H2 -326.3 J/K = -22.3 2.016 g H2 2 mol H2 ΔS = -22.3 The randomness of the system decreases. Second Law of Thermodynamics For any process, the entropy of the universe either increases or does not change. ΔSuniv ≥ 0 ΔSuniv = ΔSsys + ΔSsurr The entropy of the system is the one in which we are most interested. First Law of Thermodynamics For any process, the energy of the universe does not change. It is conserved. ΔEuniv = ΔEsys + ΔEsurr = 0 Second Law of Thermodynamics For any process, the entropy of the universe either increases or does not change. It is NOT conserved! ΔSuniv = ΔSsys + ΔSsurr ≥ 0 Second Law of Thermodynamics The second law of thermodynamics says the universe is headed toward randomness. This expresses the notion that there is an inherent direction in which processes occur. For the universe OR for an isolated system (a system which does not exchange energy or matter with its surroundings), ΔS = 0 for a reversible process, and ΔS > 0 for a spontaneous, irreversible process. Reversible Processes A reversible process is a special way in which the state of a system can change. In a reversible process, the system and surroundings can be restored to their original state by exactly reversing the change. Reversible processes tend to be small, slow changes made to a system at equilibrium. Reversible Processes Reversible Process - Example #1 Ice melting at 0°C and 1 bar. Ice and water are in equilibrium under these conditions. Adding heat slowly to ice at 0°C will cause it to melt. The process can be undone and the ice reformed by slowly removing the same amount of heat. + heat + heat ice, 0°C water, 0°C - heat - heat Entropy - FYI An often-used definition of entropy is in terms of heat: ≡ where dqrev is the amount of heat going into the system for a reversible process. (To get ΔS, the equation must be integrated.) ΔS for a Phase Change This definition means that the change in entropy ΔS for a phase change (which is an isothermal process) is ∆ ∆ = = For melting, ∆ = For boiling, ∆ = ∆ ∆ Irreversible Processes An irreversible process is one in which the system and surroundings cannot be restored to their original state by exactly reversing the change. dropping a vase and breaking it reacting hydrogen and oxygen to form water burning a match Irreversible = Spontaneous Spontaneous chemical reactions are irreversible. Systems not at equilibrium tend to undergo spontaneous, irreversible changes. A reaction being spontaneous is NOT the same as a reaction being fast! Spontaneous reactions can be slow. Spontaneous Processes A chemical reaction carried out at constant temperature and pressure is spontaneous if the change in its Gibbs Free Energy is less than 0. ΔG < 0: process is spontaneous ΔG = 0: process is at equilibrium ΔG > 0: process is nonspontaneous (but the reverse reaction is spontaneous) Gibbs Free Energy G ≡ H - TS The change in Gibbs free energy is what we want to know: ΔG = ΔH – Δ(TS) For a process that takes place at constant T and P (like a lot of open beaker lab reactions) ΔG = ΔH – TΔS Gibbs Free Energy If only P-V work is performed, the change in Gibbs Free Energy ΔG is the maximum work* that can be obtained from a chemical reaction at constant temperature T. ΔG = ΔH – TΔS *It is the maximum P-V work for an ideal gas, actually. More generally true is that ΔG = wnon-PV for constant T and P. Gibbs Free Energy There two ways to obtain ΔG: 1. ΔrxnG = ΣmG (products) – ΣnG( reactants) (m and n represent the stoichiometric coefficients) 2. ΔG = ΔH – TΔS Standard Gibbs Free Energy G° The standard Gibbs Free Energy G° applies to substances in their standard states at a specified T. There are two ways to calculate ΔG°. 1. ΔrxnG° = ΣmG° (products) – ΣnG°( reactants) (m and n represent the stoichiometric coefficients) 2. ΔG° = ΔH° – TΔS° Standard Gibbs Free Energy of Formation ΔfG° These are the values that are tabulated in Appendix C. The same rules that apply to ΔfH° apply to ΔfG°… Free Energies of Formation • The free energy of formation applies to the formation of a compound from its elements. **The state of each element MUST be specified.** • THE STANDARD FREE ENERGY OF FORMATION is designated ΔfG° and is for the formation of 1 mole of the compound in its standard state from its elements in their standard states. • The standard state of any substance is its pure form at 1 bar pressure and the temperature of interest (often 25°C). • ΔfG° for an element in its standard state is 0 at all temperatures. Calculating ΔG° for a Reaction Calculate ΔG° for the combustion of methane using both methods. 1. ΔrxnG° = ΣmΔfG° (products) – ΣnΔfG° ( reactants) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) Look up ΔfG° in Appendix C. ΔrxnG° = ? Calculating ΔG° for a Reaction Calculate ΔG° for the combustion of methane using both methods. 2. ΔG° = ΔH° – TΔS° CH4(g) + 2O2(g) CO2(g) + 2H2O(l) Look up ΔfH° and S° in Appendix C. ΔrxnG° = ? Calculating ΔG° for a Reaction Calculate ΔG° for the combustion of methane using both methods. CH4(g) + 2O2(g) CO2(g) + 2H2O(l) Is the reaction spontaneous under standard conditions at 25°C? What drives the reaction at 25°C? Is there a T at which the reaction is no longer spontaneous under standard conditions?