CHEM 217 Introduction to Thermodynamics PDF

CHEM 217 INTRODUCTION TO THERMODYNAMICS 1  SYSTEM, BOUNDARY, SURROUNDINGS A system is that part of the universe which is under thermodynamic study and the rest of the universe is surroundings. The real or imaginary surface separating the system from the surroun...

CHEM 217 INTRODUCTION TO THERMODYNAMICS 1  SYSTEM, BOUNDARY, SURROUNDINGS A system is that part of the universe which is under thermodynamic study and the rest of the universe is surroundings. The real or imaginary surface separating the system from the surroundings is called the boundary 2  TYPES OF THERMODYNAMIC SYSTEMS ✓ An isolated system is one that cannot transfer neither matter nor energy to and from its surroundings ✓ A closed system is one which cannot transfer matter but can transfer energy in the form of heat, work and radiation to and from its surroundings. ✓ An open system is one which can transfer both energy and matter to and from its surroundings 3  Homogeneous and Heterogeneous Systems ✓Homogeneous System. When a system is uniform throughout Examples are : a pure single solid, liquid or gas, -made of one phase only. ✓Heterogeneous system is one which consists of two or more phases. It is not uniform throughout. Example: ice in contact with water. ✓A phase is defined as a homogeneous, physically distinct and mechanically separable portion of a system 4  Intensive and Extensive Properties ❖Intensive Properties A property which does not depend on the quantity of matter present in the system Examples: pressure, temperature, density, and concentration ❖Extensive Properties A property that does depend on the quantity of matter present in the system Examples: volume, number of moles, enthalpy, entropy, and 5 Gibbs’ free energy  State of a System A thermodynamic system is said to be in a certain state when all its properties are fixed. The fundamental macroscopic properties which determine the state of a system are pressure (P), temperature (T), volume (V), mass and composition. 6  Thermodynamics The study of the flow of heat or any other form of energy into or out of a system as it undergoes a physical or chemical transformation First law of thermodynamics: Energy of the universe is constant – Known as the law of conservation of energy – Energy cannot be created or destroyed; it can only be converted from on form to another. Thermochemistry is the study of heat changes associated with chemical reactions. 7  Work and Energy Energy is the capacity of a system to do work (and ultimately, to lift a weight). If a system can do a lot of work, we say that it possesses a lot of energy the total store of energy in a system is called its internal energy, U or E. We cannot measure the absolute value of the internal energy of a system, because that value includes the energies of all the atoms, their electrons, and the components of their nuclei. Work is the transfer of energy to a system by a process that is equivalent to raising or lowering a weight. For work done on a system, w is positive, for work done by a system w is negative. The internal energy of a system maybe changed by doing work: When energy is transferred to a system by doing work and in the absence of other changes, ∆U = w. 8  The First Law of Thermodynamics The total internal energy of an isolated system is constant. if an isolated system has a certain internal energy at one instant and we inspect it again later, then we shall find that it still has exactly the same internal energy, no matter how much time has passed. In practice, is not to truly isolate a chemical reaction from its surroundings. Thus is important to measure the energy that leaves or enters a system. Therefore we measure change in the internal energy(E or U) of system. ΔE = E final - E Initial or ΔE = EProducts - EReactants ΔESystem + ΔESurrounding = 0 ΔESystem = - ΔESurrounding 9  The First Law of Thermodynamics heat transfer in heat transfer out (endothermic), +q (exothermic), -q SYSTEM ∆E = q + w w transfer in w transfer out (+w) (-w) 10  Work and Energy The work required to move an object a certain distance against an opposing force is calculated by multiplying the opposing force by the distance moved against it Work = opposing force X distance moved Unit is the joule, J, with 1 J = 1 kgm2s-2 11  EXPANSION WORK Types of work associated with a chemical process – Work done by a gas through expansion – Work done to a gas through compression Example - Motion of a car – In an automobile engine, heat from the combustion of gasoline expands the gases in the cylinder to push back the piston, and this results in the motion of the car 12  Deriving the Equation for Work Consider a gas confined to a cylindrical container with a movable piston – F is the force acting on the piston of area A – Pressure is defined as force per unit area F P= A 13  Deriving the Equation for Work – Consider that the piston moves a distance of Δh Work = force  distance = F  h Work = F  h = P  A  h – Volume of the cylinder equals the area of the piston times the height of the cylinder Change in volume ΔV resulting from the piston moving a distance Δh is: V = final volume − initial volume = A  h 14  Deriving the Equation for Work (Continued 2) – Substitute the expression derived for ΔV into the expression for work Work = P  A  h = PV Since the system is doing work on the surroundings, the sign of work should be negative – ΔV is a positive quantity since volume is increasing – Therefore, w = − PV 15  Deriving the Equation for Work For a gas expanding against an external pressure P, w is a negative quantity as required – Work flows out of the system When a gas is compressed, ΔV is a negative quantity (the volume decreases) – This makes w a positive quantity (work flows into the system) 16  Calculating the work of reversible isothermal expansion of gas dw = - PΔV At each stage of the expansion pressure is related to volume by the ideal gas law. PV = nRT = P = nRT/V 𝑛𝑅𝑇 dw = - ⅆ𝑣 𝑣 w = -nRT ln (Vfinal / Vinitial ) 17  Examples Calculate the work associated with the expansion of a gas from 46 L to 64 L at a constant external pressure of 15 atm Ans: w = − PV ΔV = 64 L – 46 L = 18 L w = -15 atm x 18 L = -270 L.atm 1L.atm = 101.325J -270 L.atm = 270 x 101.325 = -27,357.75J = -27.4kJ 18 Calculating the work done when a gas expands Suppose a gas expands by 500. mL (0.500 L) against a pressure of 1.20 atm and no heat is exchanged with the surroundings during the expansion. (a) How much work is done in the expansion? (b) What is the change in internal energy of the system? Ans: w = -PΔV = - 1.20 atm x 0.500 L = -0.600 L.atm = 1 L.atm = 101.325J 0.600 =.600 x 101.325J = -60.8J (B) No heat ⸫ w = ΔU 19 ΔU = -60.8J  Calculating the work of isothermal expansion of gas A piston confines 0.100 mol Ar(g) in a volume of 1.00 L at 25°C. Two experiments are performed. (a) The gas is allowed to expand through an additional 1.00 L against a constant pressure of 1.00 atm. (b) The gas is allowed to expand reversibly and isothermally to the same final volume. Which process does more work? Ans: (a) Irreversible path w = - PΔV = - 1.00 atm x 1.00 L = 1.00 L.atm = -101.3 J (b) Reversible w = -nRT ln (Vfinal / V initial) -1 -1 w = - 0.100 mol x 8.3145J.K mol x 298K x ln (2/1) = -172J Thus, gas does more work in the reversible 20  CHEM 217 ENTHALPY 21  HEAT It is the energy transferred as a result of a temperature difference. Energy flows as heat from a high- temperature region to a low-temperature region. if the system is cooler than its surroundings, energy flows into the system from the surroundings. Heat is the transfer of energy as a result of a temperature difference. When energy is transferred as heat and no other processes occur, ∆E = q. When energy enters a system as heat, q is positive; when energy leaves a system as heat, q is negative 22  The Measurement of Heat Calorimetry Science of measuring heat – Based on observations of temperature change when a body absorbs or discharges energy in the form of heat Calorimeter: Device used to determine the heat associated with a chemical reaction 23 Copyright © Cengage Learning. All rights reserved  Heat Capacity (C) heat absorbed C= increase in temperature Specific heat capacity: Energy required to raise the temperature of one gram of a substance by one degree Celsius – Units - J/ °C · g or J/K · g – Heat capacity(C) is related to its mass(m) and specific heat capacity(Cs) as C= m x Cs – q = m x Cs x ΔT 24  Heat Capacity (C) Molar heat capacity: Energy required to raise the temperature of one mole of a substance by one degree Celsius – Units - J/ °C · mol or J/K · mol – Molar Heat capacity(C) is related to its mole(n) and specific molar heat capacity(Cm) as C= n x Cm – q= n x Cm x ΔT Heat capacities of metals are different from that of water – Takes less energy to change the temperature of a gram of a metal by 1°C than for a gram of water 25  The Specific Heat Capacities of Some Common Substances 26  Constant-Pressure Calorimetry Example - Coffee-cup calorimeter Atmospheric pressure remains constant during the process Used to determine enthalpy changes in reactions that occur in a solution – ΔE = q + w = – = qp -PΔV – qp= ΔE + PΔV – At constant pressure qp is called the heat of reaction or enthalpy change(ΔH) qp = ΔH – ΔH = ΔE + PΔV – H = E + PV – Δ H = Hfinal - Hinitial – ΔH = H products - H reatants 27  Constant-Pressure Calorimetry When two reactants at the same temperature are mixed: – An exothermic reaction warms the solution – An endothermic reaction cools the solution Calculation of heat for a neutralization reaction – Energy (as heat) released by the reaction = Energy (as heat) absorbed by the solution = Specific heat capacity × mass of solution × change in temperature = m x Cs x ΔT Heat of a reaction is an extensive property – Depends entirely on the amount of substance 28  Constant-Pressure Calorimetry When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0˚C is mixed with 1.00 L of 1.00M Na2SO4 solution at 25.0˚C in a calorimeter, the white solid BaSO4 forms, and the temperature of the mixture increases to 28.1˚C. Cₛ = 4.18J/˚C.g ,Density = 1.0g/mL What is the enthalpy change(ΔH) per mol. Ans: The ions present are Ba²⁺, NO₃⁻, Na⁺ and SO₄²⁻ Na⁺ and NO₃⁻ ions spectator ions since NaNO₃ is very soluble in water. Net reaction is Ba²⁺ (aq) + SO₄²⁻ (aq) = BaSO4 (s) 1 mol each q H₂O + q rxn = 0 q H₂O = -q rxn q = m x Cₛ x ΔT = ΔT = 28.1 – 25 = 3.1˚C H₂O mass = if 1g = 1mL 2L = (1g x 2L)/1mL = 2000g q cal = 2000g x 4.18J/˚C.g x 3.1˚C = 25916J = -26.0kJ = -26.0kJ/mol Temperature increases therefore is exothermic reaction and enthalpy is negative. 29  Constant-Volume Calorimetry Used in conditions when experiments are to be performed under constant volume ΔE = q + w q = ΔE + PΔV ΔV = 0 No work is done since V must change for PV work to be performed Thus qv = ΔE Bomb calorimeter – Weighed reactants are placed within a rigid steel container and ignited – Change in energy is determined by the increase in temperature of the water and other parts – q system = q cal + q rxn = 0 – qcal = -qrxn – qcal = Ccal ΔT 30  Heat Transfers at Constant Pressure Energy changes at constant pressure is called the enthalpy, H: H = E + PV where U, P, and V are the internal energy, pressure, and volume of the system. Enthalpy is a state function by noting that U (from the first law), P, and V are all state functions, and so H = E +PV must be a state function, too a change in the enthalpy of a system is equal to the heat released or absorbed at constant pressure. Every chemical reaction is accompanied by a change in energy, and commonly that energy is released as heat This entire expression is called a thermochemical equation, and consists of a chemical equation together with a statement of the reaction enthalpy, the corresponding enthalpy change. CH4(g) + 2 O2(g) →CO2(g) 2 H2O(l) ∆H= -890. kJ 31  Determining a reaction enthalpy from experimental data When 0.113 g of benzene, C H , burns in excess oxygen in a calibrated constant-pressure 6 6 calorimeter with a heat capacity of 551 J・(°C), the temperature of the calorimeter rises by 8.60°C. Write the thermochemical equation for the reaction 2 C H (l) + 15 O (g) →12 CO (g) + 6 H O(l) 6 6 2 2 2 qcal = C calΔT = 551J.˚C x 8.60˚C = 4738.6J = 4.7kJ Mole = mass/Molar mass = 0.113g/78.2g/mol = 1.445x10⁻³ mol ΔH = q p qcal = -qrxn = - 4738.6J ΔH =- 4738.6J/ 1.445x10⁻³ mol = -3279307.96J/mol Because C H is 2 mol , ΔH = 2 x 3279307.96J/mol = - 6558615.9J/mol = -6558.6kJ/mol 6 6 2 C H (l) + 15 O (g) →12 CO (g) + 6 H O(l) 6 6 2 2 2 ΔH = -6558.6kJ/mol 32  Heat capacities at constant volume and pressure At constant volume, heat transfer is interpreted as ∆U; at constant pressure, it is interpreted as ∆H. C= q Cv = ΔE C p = ΔH ΔT ΔT ΔT Molar heat capacities are these divided by the moles: Cv, m C p, m The constant-volume and constant-pressure heat capacities of solids are similar, same for liquids but not gas. Because gas expands more when heated, more energy is lost as work when gas is heated at constant pressure. H = ΔE + PV replace PV by nRT Thus, H = ΔE + nRT Cp = ΔH = ΔE + nRT = ΔE + nR ΔT ΔT ΔT 33 Cp = Cv + nR C p,m = Cv,m +R  Relating the enthalpy change and internal energy change for a chemical reaction A constant-volume calorimeter showed that the heat generated by the combustion of 1.000 mol glucose molecules in the reaction C H O (s) + 6 O (g)→ 6 CO (g) + 6 H O(g) is 2559 kJ at 298 K, and so ∆E = 2559 6 12 6 2 2 2 kJ. What is the change in enthalpy for the same reaction? ΔH = ΔE + Δn(gas) RT Δn(gas) = 12 – 6 = 6 mol ΔH = - 2559kJ + [ 6 x 8.3145J.K⁻¹ mol⁻¹ x 298K] - 2559kJ + 14914.5J - 2559kJ + 14.9kJ = -2544.1kJ 34  Calculate the work, heat, and ΔE for the expansion of an ideal gas Suppose that 1.00 mol of ideal gas molecules maintained at 292 K and 3.00 atm expands from 8.00 L to 20.00 L and a final pressure of 1.20 atm by two different paths. (a) Path A is an isothermal, reversible expansion. (b) Path B has two parts. In step 1, the gas is cooled at constant volume until its pressure has fallen to 1.20 atm. In step 2, it is heated and allowed to expand against a constant pressure of 1.20 atm until its volume is 20.00 L and T 292 K. Determine for each path the work done, the heat transferred, and the change in internal energy (w, q, and ΔE). (a) w =-nRT ln(Vfinal / Vinitial ) = - 1.00 mol x 8.3145J.K⁻¹.mol⁻¹ x 292K ln(20.00/8/00) = -2224.6J = - 2.22kJ ΔE = q + w = 0. Because process is isothermal, ΔE (of this ideal gas) is zero. q = -w = + 2.22kJ (b) S1 cools at constant volume, so no work is done so w = 0. S2 w = -PΔV = -1.2 x 12 = -14.4 L.atm = -1459.08J =-1.46kJ Total work = w = 0 + (-1.46kJ) = -1.46kJ ΔE = q + w = 0 q = -w = +1.46kJ In summary q w ΔE For reversible path + 2.22kJ -2.22kJ 0 For irreversible path + 1.46kJ - 1.46kJ 0 35  Standard Reaction Enthalpies The standard reaction enthalpy, , is the reaction enthalpy when reactants in their standard states change into products in their standard states. For example, for reaction CH (g) + 2 O2(g)→CO2(g) + 2 H2O(l) ∆H = -890. kJ, the 4 value signifies that the heat output is 890. kJ when1 mol CH (g) as pure methane gas at 4 1 bar is allowed to react with pure oxygen gas at1 bar, giving pure carbon dioxide gas and pure liquid water, both at 1 bar. Thus: Standard reaction enthalpies refer to reactions in which the reactants and products are in their standard state, the pure form at 1 bar; they are usually reported for a temperature of 298.15 K. 36  Combining Reaction Enthalpies: Hess’s Law Enthalpy is a state function; therefore, its value is independent of the path between given initial and final states. Hesses law: the overall reaction enthalpy is the sum of the reaction enthalpies of the steps into which the reaction can be divided Importance of Hess’s Law: The enthalpy change of some chemical reactions cannot be determined directly because: the reactions cannot be performed/controlled in the laboratory the reaction rates are too slow and the reactions may involve the formation of side products But the enthalpy change of such reactions can be determined indirectly by applying Hess’s Law. 37  Combining Reaction Enthalpies: Hess’s Law To use Hess’s law to find the enthalpy of a given reaction find a sequence of reactions with known reaction enthalpies that adds up to the reaction of interest. Step 1 Select one of the reactants in the overall reaction and write down a chemical equation in which it also appears as a reactant. Step 2 Select one of the products in the overall reaction and write down a chemical equation in which it also appears as a product. Add this equation to the equation written in step 1 and cancel species that appear on both sides of the equation. Step 3 Cancel unwanted species in the sum obtained in step 2 by adding an equation that has the same substance or substances on the opposite side of the arrow. Step 4 Once the sequence is complete, combine the standard reaction enthalpies 38  Characteristics of Enthalpy Changes and Hess Law If a reaction is reversed, the sign of ΔH is also reversed Magnitude of ΔH is directly proportional to the quantities of reactants and products in a reaction – If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH is multiplied by the same integer Work backward from the required reaction – Use the reactants and products to decide how to manipulate the other given reactions at your disposal Reverse any reactions as needed to give the required reactants and products Multiply reactions to give the correct numbers of reactants and 39 products  Using Hess’s law Consider the synthesis of propene C₃H₈ , a gas used as camping fuel: 3C(gr) + 4H₂(g) C₃H₈(g) It is difficult to measure enthalpy change directly. However, standard enthalpies of combustion reactions are easy to measure. Calculate the standard enthalpy of this reaction from this data. (a)C₃H₈(g) + 5O₂(g) 3CO₂(g) + 4H₂O(l) ΔH˚ = -2220kJ (b)C(gr) + O₂(g) CO₂(g) ΔH˚ = -394kJ (c)H₂(g) + 1/2O₂(g) H₂O(l) ΔH˚ = -286kJ Soln: 3CO₂(g) + 4H₂O(l) C₃H₈(g) + 5O₂(g) ΔH˚ = +2220kJ 3C(gr) + 3O₂(g) ) 3 CO₂(g) ΔH˚ = 3 x -394kJ = -1182kJ 4H₂(g) + 2O₂(g) 4 H₂O(l) ΔH˚ = 4 x -286kJ = -1144kJ 3C(gr) + 4H₂(g) C₃H₈(g) ΔH˚ = -106kJ 40  Standard Enthalpies of Formation The standard enthalpy of formation, of a substance is the standard reaction enthalpy per mole of formula units for the formation of a substance from its elements in their most stable form, 2 C(gr) + 3 H2(g) + 12O2(g) → C2H5OH(l) ∆H° –277.69 kJ the standard enthalpy of formation of an element in its most stable form is zero. 41  Using standard enthalpies of formation to calculate a standard enthalpy of reaction Amino acids are the building blocks of proteins, which have long chainlike molecules. They are oxidized in the body to urea, carbon dioxide, and liquid water. Is this reaction a source of heat for the body? Predict the standard enthalpy of reaction for the oxidation of the simplest amino acid, glycine (NH2CH2COOH), a solid, to solid urea (H2NCONH2), carbon dioxide gas, and liquid water: 2NH2CH2COOH(s) + 3 O2(g) →H2NCONH2(s) + 3CO2(g) + 3 H2O(l) NH2CH2COOH(s) : ΔH˚ = -532.9kJ.mol⁻¹ , H2NCONH2(s) : ΔH˚= f -333.51kJ.mol⁻¹ , f CO₂ : ΔH˚f = -393.51kJ.mol⁻¹ , H₂O: ΔH˚f = -285.83kJ.mol⁻¹ Soln: ΔH˚ = ⅀nΔH˚(products) f - ⅀nΔH˚(reactants) f ⅀[ΔH˚(H2NCONH2(s))+ f 3 ΔH˚(CO2(g) f + 3ΔH˚(H2O(l))] f - ⅀ [2ΔH˚(NH2CH2COOH(s)) f + 3ΔH˚(O2(g) f [-333.51kJ.mol⁻¹ + 3x -393.51kJ.mol⁻¹ + 3x -285.83kJ.mol⁻¹] - [2x -532.9kJ.mol⁻¹ + 3 x 0] [-333.51kJ.mol⁻¹ -1180.53kJ.mol⁻¹ - 857.49kJ.mol⁻¹] – [ -1065kJ.mol⁻¹ +0] -2371.53kJ.mol⁻¹ + 1065.8kJ.mol⁻¹ = -1305.73kJ.mol⁻¹ 42 Enthalpy change of formation of CO(g) (Born-Haber cycle) cannot be Hc [graphite] = -393 kJ mol-1 determined Hc [CO(g)] = -283.0 kJ mol-1 directly due Hf [CO(g)] to further C(graphite) + ½O2(g) CO(g) oxidation of + ½O2(g) + ½O2(g) CO to CO2. H1 H2 The reaction CO2(g) cannot be Hf [CO(g)] + H2 = H1 controlled. Hf [CO(g)] = H1 - H2 = -393 - (-283 ) = -110 kJ mol-1 43  Hf [C4H10(g)] 4C(graphite) + 5H2(g) C4H10(g) + 4O2(g) 5Hc [H2(g)] Hc [C4H10(g)] 4Hc [graphite] + 2.5 O2(g) + 6.5 O2(g) 4CO2(g) + 5H2O(l) Hf [C4H10(g)] = 4Hc[C(graphite)] + 5Hc[H2(g)] - Hc[C4H10(g)] = [4(-393)+ 5(-286) – (−2877)] kJ mol−1 = −125 kJ mol−1 44  Enthalpy level diagram Relate substances together in terms of enthalpy changes of reactions Enthalpy level diagram for the oxidation of C(graphite) to CO2(g) C(graphite) + O2(g) 1. Draw the enthalpy level of elements. Enthalpies of Hf[CO(g)] = −110 kJ Enthalpy / kJ mol−1 elements are arbitrarily taken as zero. Hc[graphite] = CO(g) + 1/2 O2(g) 2. Higher enthalpy levels are drawn −393 kJ above that of elements Hc[CO(g)] = −283 kJ 3. Lower enthalpy levels are drawn CO2(g) below that of elements Route 1 Route 2 45  Standard enthalpy change of combustion Hc The enthalpy change when one mole of the substance undergoes complete combustion (reacts with oxygen) under standard conditions. C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) Hc [C2H5OH(l)] = -1368 kJ mol−1 46  Substance Hc (kJ mol-1) C (diamond) -395.4 C (graphite) -393.5 C(diamond) + O2(g) 1.9 C(graphite) + O2(g) Enthalpy −395.4 Hf [diamond] −393.5 = +1.9 kJ mol−1 CO2(g) Reaction coordinate 47  C(graphite) + 2 O2(g) → CO(g) 1 (a) Incomplete combustion H = Hf [CO(g)]  Hc [graphite] (b) 2H2(g) + O2(g) → 2H2O(l) H = 2  Hc [H2(g)] = 2  Hf [H2O(l)] (c) C(graphite) + O2(g) → CO2(g)   H =Hf [CO2(g)] = H c [graphite] (d) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) H= Hc [CH4(g)]  Hf [CO2(g)] Not formed from  2  Hf [H2O(l)] elements 48  Standard enthalpy changes of neutralization Standard enthalpy change of neutralization (Hneut) is the enthalpy change when one mole of water is formed from the neutralization of an acid by an alkali under Enthalpy level diagram for the standard conditions. neutralization of a strong acid and a strong alkali e.g. H+(aq) + OH-(aq) → H2O(l) Hneut = -57.3 kJ mol-1 49  Acid Alkali Hneu HCl NaOH -57.1 HCl KOH -57.2 HCl NH3 -52.2 HF NaOH -68.6 NH3(aq) + H2O(l) NH4+(aq) + OH−(aq) H1 > 0 H+(aq) + OH-(aq) + Cl−(aq) → H2O(l) + Cl−(aq) H2 = -57.3 NH3(aq) + HCl(aq) → NH4Cl (aq) Hneu = H1 + H2 = −52.2 kJ mol−1 HF(aq) H+(aq) + F−(aq) H1 < 0 H+(aq) + OH-(aq) + Na+(aq) → H2O(l) + Na+(aq) H2 = -57.3 HF(aq) + NaOH(aq) → NaF(aq) + H2O(l) Hneu = H1 + H2 = −68.6 kJ mol−1 ø 50  Standard enthalpy change of solution Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is dissolved in a specified number of moles of solvent (e.g. water) under standard conditions. NaCl(s) + 10H2O(l) → Na+(aq) + Cl-(aq) Hsoln[NaCl(s)]= +2.008 kJ mol-1 dilution NaCl(aq) NaCl(aq) H > 0 51  Standard enthalpy change of solution Standard enthalpy change of solution (Hsoln) is the enthalpy change when one mole of a solute is dissolved to form an infinitely dilute solution under standard conditions. concentration → 0 NaCl(s) + water → Na+(aq) + Cl-(aq) Hsoln= +4.98 kJ mol-1 52  Standard enthalpy change of solution NaCl(s) + water → Na+(aq) + Cl-(aq) LiCl(s) + water → Li+(aq) + Cl-(aq) Hsoln= +4.98 kJ mol-1 Hsoln= -37.2 kJ mol-1 ø + 4.98 kJ mol−1 Enthalpy level diagram for Enthalpy level diagram for the the dissolution of NaCl dissolution of LiCl in water 53  Standard enthalpy change of solution Salt Hsoln(kJ mol-1) NH3 −35.5 NaOH −43.1 HCl −73.0 H2SO4 −74.0 LiCl −37.2 NaCl +4.98 NaNO3 +21.0 NH4Cl +22.6 54  Bond Dissociation Enthalpy/Energy Bond dissociation enthalpy/energy is the enthalpy change for breaking one mole of a bond in a particular environment with the reactants and products in the gaseous state under standard conditions. B.D.E. / kJ mol−1 Cl – Cl(g) → Cl(g) + Cl(g) 242 55  Bond Dissoication Enthalpy/Energy 1.B.D.E. values are always > 0 Bond breaking processes are endothermic B.D.E. values depend on the environment of the bond Cl – Cl(g) → Cl(g) + Cl(g) B.D.E. = 242 kJ mol−1 Br – Br(g) → Br(g) + Br(g) B.D.E. = 193 kJ mol−1 I – I(g) → I(g) + I(g) B.D.E. = 151 kJ mol−1 O = O(g) → O(g) + O(g) B.D.E. = 498 kJ mol−1 N  N(g) → N(g) + N(g) B.D.E= 945 kJ mol−1 CH3−H(g) → CH3(g) + H(g) B.D.E. = 423 kJ mol−1 CH2−H(g) → CH2(g) + H(g) B.D.E. = 480 kJ mol−1 CH−H(g) → CH(g) + H(g) B.D.E. = 425 kJ mol−1 C−H(g) → C(g) + H(g) B.D.E. = 334 kJ mol−1 56  Bond Enthalpy/Energy Bond enthalpy/energy is the average of the bond dissociation enthalpies taken from a large number of species containing a particular chemical bond. CH3−H(g) → CH3(g) + H(g) B.D.E. = 423 kJ mol−1 CH2−H(g) → CH2(g) + H(g) B.D.E. = 480 kJ mol−1 CH−H(g) → CH(g) + H(g) B.D.E. = 425 kJ mol−1 C−H(g) → C(g) + H(g) B.D.E. = 334 kJ mol−1 The average B.D.E. values of C – H bond in CH4 ¼( 423 + 480 + 425 + 334) KJmol⁻¹ Bond enthalpy of C – H bond = E(C – H) = 413 kJ mol−1 An average B.D.E. values from CH4, C2H6, C3H8…etc. 57  𝑜Ӎ Estimation of B.E. from ΔHfoӍ and ΔHatm 4E(C – H) CH4(g) C(g) + 4H(g) ΔHof Ӎ [CH4 (g)] 𝑜Ӎ ෍ ΔHatm C(graphite) + 2H2(g) ΔHof Ӎ [CH4 (g)] + 4E(C –H) = 𝑜Ӎ ෍ ΔHatm 4E(C – H) = [715 + 4  218 − (−75)] kJ mol−1 = 1662 kJ mol−1 E(C – H) = 415.5 kJ mol−1 58  C(g) + 4H(g) 4218 kJ mol−1 Enthalpy / kJ mol−1 C(g) + 2H2(g) 4E(C – H) 715 kJ mol−1 C(graphite) + 2H2(g) −75 kJ mol−1 CH4(g) 59  Example 1 oӍ ΔHatm [C2 H6 (g)] = 6E(C – H) + E(C – C) C2H6(g) 2C(g) + 6H(g) oӍ ΔHatm [compounⅆ] = the enthalpy change when one mole of the compound in gaseous state is broken down into its constituent atoms in gaseous state under standard conditions. oӍ ΔHatm [element] = the enthalpy change when one mole of atoms of the element in gaseous state are formed from the element in its normal and most stable state under standard conditions. C(graphite) → C(g) 60  Example 1(b) oӍ ΔHatm [C2 H6 (g)] = 6E(C – H) + E(C – C) C2H6(g) 2C(g) + 6H(g) ΔHfoӍ [C2 H6 (g)] 𝑜Ӎ ෍ ΔHatm 2C(graphite) + 3H2(g) ΔHof Ӎ [C2 H6 (g)] + 6E(C – H) + E(C – C) 𝑜Ӎ = ෍ ΔHatm E(C – C) = [2715 + 6218 − (−85) −6415.5] kJ mol−1 = 330 kJ mol−1 61  E(C – C) in C2H6 E(C – C) in C2H6 The discrepancy is due to the fact that the average B.D.E. values of Estimated value Experimental value C – H bonds in CH4 and C2H6 are 330 kJ mol−1 386 kJ mol−1 not the same. H H H H H C C C C H 4C(g) + 10H(g) H H H H Ӎ ΔHoatm [C4 H10 (g)] = 5165 kJ mol−1 = 3E(C – C) +10E(C – H) H H H H H H C C C C C H 5C(g) + 12H(g) H H H H H Ӎ ΔHoatm [C5 H12 (g)] = 6337 kJ mol−1 = 4E(C – C) +12E(C – H) Solving (1) and (2) 62 E(C – C) = 348 kJ mol−1 E(C – H) = 412 kJ mol−1  Use of bond enthalpies to estimate enthalpy changes of reactions Assumption : The bond enthalpies are transferable from one molecule to another. oӍ ΔHreaction = ෍ E(bonⅆs broken) − ෍ E(bonⅆs formeⅆ) Bond forming is always exothermic 63  Example 1 Given : E(Cl−Cl) = 242 kJ mol−1 E(C−H) = 413 kJ mol−1 E(C−Cl) = 339 kJ mol−1 E(H−Cl) = 431 kJ mol−1 Estimate the H of the following reaction. CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) Bond broken / B.E.(kJ mol−1) Bond formed / B.E.(kJ mol−1) C – H 413 C – Cl 339 Cl – Cl 242 H – Cl 431 oӍ ΔHreaction = (242 + 413) – (339 +431) kJ = −115 kJ 64  Example 2) N2H4(g) + 2F2(g) → N2(g) + 4HF(g) Bond broken / B.E. (kJ mol−1) Bond formed / B.E. (kJ mol−1) N–N 163 NN 945 4N – H 4390 4H – F 4565 2F – F 2158 oӍ ΔHreaction = (163+ 4390+2158)– (945+4565) kJ = −1166 kJ Highly exothermic, used as rocket fuel 65  Example 3 CH4(g) + H2O(g) → CO(g) + 3H2(g) Bond broken / B.E. (kJ mol−1) Bond formed / B.E. (kJ mol−1) 4C – H 4413 CO 1072 2O – H 2463 3H – H 3436 oӍ ΔHreaction = (4413 + 2463) – (1072 +3436) kJ = 198 kJ 66  Example 4 2O3(g) → 3O2(g) Bond broken / B.E. (kJ mol−1) Bond formed / B.E. (kJ mol−1) 2O – O 2146 3O = O 3498 2O = O 2498 oӍ ΔHreaction = (2146 + 2498) – (3498) kJ = −206 kJ 67  Using mean bond enthalpies to estimate the enthalpy of a reaction Estimate the enthalpy of the reaction between gaseous iodoethane and water vapor: CH CH I(g) + H O(g) → CH CH OH(g) + HI(g) 3 2 2 3 2 Bond broken / B.E. (kJ mol−1) Bond formed / B.E. (kJ mol−1) C–I 238 C -O 360 O–H 463 H–I 299 oӍ ΔHreaction = (238 + 463) – (360 + 299) kJ 701 - 659 = 42kJ 68  Chem 217 Entropy 1  Spontaneous Change A process is said to be spontaneous if no external “forces” are required to keep the process going, although external “forces” may be required to get the process started (Ea). The process may be a physical change or a chemical change Spontaneity : The state of being spontaneous Spontaneous : self-generated natural happening without external influence external boundary external internal external 2 external  Spontaneous physical change: E.g. condensation of steam at 25°C Spontaneous chemical change: E.g. burning of wood once the fire has started Exothermicity is NOT the only reason for Exothermic → the spontaneity of a process spontaneous ? Some spontaneous changes are endothermic Endothermic → not spontaneous ? E.g. Melting of ice Dissolution of NH4Cl in water 3  Exothermic ? Spontaneous ? Change (Yes / No) (Yes / No) Burning of CO at Yes Yes 25°C Melting of ice Condensation of Yes Yes steam at 25°C Melting of ice at No Yes 25°C Dissolution of NH4Cl in water No Yes Dissolution of ammonium at 25°C nitrate in water 4  Entropy and Disorder Entropy is a measure of the randomness or the degree of disorder (freedom) of a system Solid Liquid Gas Entropy increases 5  ENTROPY AND DISORDER Entropy is a measure of disorder; According to the second law of thermodynamics, the entropy of an isolated system increases in any spontaneous process. Entropy is a state function. Entropy changes(S) Entropy change means the change in the degree of disorder of a system S = Sfinal - Sinitial Second Law of Thermodynamic: The Entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process 6  Positive entropy change (S > 0) Increase in entropy Sfinal > Sinitial Example: Ice (low entropy) ⎯→ Water (high entropy) S = Swater – Sice = +ve Negative entropy change (S < 0) Decrease in entropy Sinitial > Sfianl Example: Water (high entropy) ⎯→ Ice (low entropy) 7 S = Sice – Swater = -ve  Changes S 2CO(g) + O2(g) → 2CO2(g) -ve H2O(g) → H2O(l) -ve H2O(s) → H2O(l) +ve NH4Cl(s) → NH4Cl(aq) +ve 8  Changes S 2 C(s) + O2(g) → 2 CO2(g) +ve 2SO2(g) + O2(g) → 2SO3(g) -ve Diffusion of a drop of ink in +ve water diamond → graphite +ve 9  ENTROPY CHANGES dS =dq rev dqrev = CdT, dS =CdT T T q is the energy transferred as heat and rev. on q signifies that the energy must be transferred reversibly. T is the absolute Temp. which indicates the temp of the system and the surroundings must be similar At constant temp, from Boyles law, Pressure is At constant volume ΔE = q + w inversely related to volume ΔE = 0 for isothermal expansion of an ideal gas V₂ = P₁ q = -w qrev = - wrev w = - nRT ln(V₂/V₁) V₁ P₂ ΔS = Δq = -w = nRT ln(V₂/V₁) = nR ln(V₂/V₁) ΔS = nR ln (P₁/P₂) T T 10 Calculating the entropy change due to an increase in temperature A sample of nitrogen gas of volume 20.0 L at 5.00 kPa is heated from 20.°C to 400.°C at constant volume. What is the change in the entropy of the nitrogen? The molar heat capacity of nitrogen at constant volume, C , is 20.81 J・K⁻¹・mol⁻¹ V,m Ans: C = n xC V V,m ΔS = C ln (T₂ /T₁) = 0.0410 mol x 20.81 J・K⁻¹・mol⁻¹ T₂ = 400 + 273.15K = 673K =0.85321 J・K⁻¹ T₁ = 20 + 273.15 = 293K ΔS = C ln (T₂ /T₁) ln(673/293) = 0.831572 = 0.85321 J・K⁻¹ x 0.831572 C = n xC V V,m = + 0.7095 PV = nRT + because disorder 5.00kPa x 20.0 L = 0.0410 mol increases with 8.3145L.kPa.K⁻¹.mol⁻¹ x 293K increase temp. 11  Calculating the change in entropy when an ideal gas expands isothermally What is the change in entropy of the gas when 1.00 mol N2(g) expands isothermally from 22.0 L to 44.0 L? ΔS = nR ln (V₂/V₁) =1.00 mol x 8.3145 J.K⁻¹.mol⁻¹ x ln(44.0/22.0) = + 5.76 J.K⁻¹ Calculate the change in entropy when the pressure of 0.321 mol O₂(g) is increased from 0.300 atm to 12.00 atm at constant temperature ΔS = nR ln (P₁/P₂) =0.321 mol x 8.3145 J.K⁻¹.mol⁻¹ x ln(0.300/12.00) = - 9.85 J.K⁻¹ 12  Calculating the change in entropy when both temperature and volume change In an experiment, 1.00 mol Ar(g) was compressed suddenly (and irreversibly) from 5.00 L to 1.00 L by driving in a piston (like a big bicycle pump), and in the process its temperature was increased from 20.0°C to 25.2°C. What is the change in entropy of the gas? Cm= 12.47 J.K⁻¹.mol⁻¹ ΔS = nR ln (V₂/V₁) =1.00 mol x 8.3145 J.K⁻¹.mol⁻¹ x ln(1.00/5.00) = - 13.4 J.K⁻¹ ΔS = C ln (T₂ /T₁) C = n x Cm ΔS = n x Cm ln (T₂ /T₁) = 1.00 x 12.47 J.K⁻¹.mol⁻¹ ln(298.4 K/293.2 K) = + 0.22 J.K⁻¹ Add the entropy changes for the two steps ΔS = ΔS1 + ΔS2 = - 13.4 J.K⁻¹ + 0.22 J.K⁻¹ 13 =-13.2 J.K⁻¹  Standard Molar Entropies In theory it is possible for a material to have a perfect order, therefore we should be able to set a scale of entropy S 0 whilst T 0. The perfect crystal. We can determine the change in entropy ΔS from a perfect entropy to temperature of interest. T ΔS = ʃ CpDt The third law of thermodynamics: The entropy of 0 T a perfect ordered crystalline at 0 K is zero. Thus, the entropy of any substance at a temperature above 0 K is greater than zero Standard Reaction Entropy To calculate the change in entropy that accompanies a reaction, we need to know the molar entropies of all the substances taking part; then we calculate the difference between the entropies of the products and those of the reactants. More specifically, the standard reaction entropy, S°, is the difference between the standard molar entropies of the products and those of the reactants, taking into account their stoichiometric coefficients 14 ∆S° = ∑nSm°(products) - ∑nSm°(reactants)  Calculating the standard reaction entropy Calculate the standard reaction entropy of N₂(g) + 3 H₂ (g)→2 NH₃ (g) at 25°C. S˚(NH₃) = 192.4 J.K⁻¹, S˚(N₂) = 191.6 J.K⁻¹ ,S˚(H₂) = 192.4 J.K⁻¹ ∆S° = ∑nSm°(products) - ∑nSm°(reactants) = 2 x 192.4 J.K⁻¹ - [191.6 J.K⁻¹ + 3 x 192.4 J.K⁻¹] = -198.9 J.K⁻¹ 15  GLOBAL CHAGES IN ENTROPY Some processes seem to defy the second law. For example, water spontaneously freezes to ice at low temperatures, and cold packs for athletic injuries spontaneously become ice cold, even on warm days, when the ammonium nitrate inside the pack dissolves in the water that it contains. The dilemma is resolved when we realize that the second law refers to an isolated system. THE SURROUNDINGS ΔS Total = ΔSSystem + ΔSsurroundings 16  Consider an isolated system which has no exchange of energy and matter with its surroundings S1 S2 Spontaneous, S = S2 – S1 > 0 S2 S1 Not spontaneous, S = S1 – S2 < 0 17  A spontaneous process taking place in an isolated system is always associated with an increase in entropy (i.e. S > 0) In closed system, The spontaneity of a process depends on both H and S. The driving force of a process is a balance of H and S. Ice (less entropy) ⎯→ Water (more entropy) H is +ve  not favourable S is +ve  favourable Considering both S & H 18 the process is spontaneous  Spontaneity depends on temperature Spontaneous Changes H S (Yes / No) H2O(g) → H2O(l) at 25°C -ve -ve Yes H2O(g) → H2O(l) at 110°C -ve -ve No H2O(s) → H2O(l) at 25°C +ve +ve Yes H2O(s) → H2O(l) at -10°C +ve +ve No 19  The Overall Change in Entropy As we have already emphasized, to use the entropy to judge the direction of spontaneous change, we must consider the change in the entropy of the system plus the entropy change in the surroundings: If Stot is positive (an increase), the process is spontaneous. ΔS universe = ΔSsystem + ΔSsurrounding > 0 If Stot is negative (a decrease), the reverse process is spontaneous. ΔS universe = ΔSsystem + ΔSsurrounding < 0 If Stot 0, the process has no tendency to proceed in either direction ΔSuniverse = ΔSsystem + ΔSsurrounding = 0 Second Law of Thermodynamic: The Entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process 20  Calculating the total entropy change for the expansion of an ideal gas Calculate S, Ssurr, and Stot for (a) the isothermal, reversible expansion and (b) The isothermal, free expansion of 1.00 mol of ideal gas molecules from 8.00 L to 20.00 L at 292 K.Explain any differences between the two paths. ΔS = nR ln (V₂/V₁) = 1.00 mol x 8.3145 J.K⁻¹ x (20.0/8.00) = + 7.6 J.K⁻¹ ΔE = 0 q = -w because the heat flows in the surrounding is equal to the heat flows out of the system ΔSsurrounding = q surrounding = -nRT ln V₂ = - 7.6 J.K⁻¹ T T V₁ ΔSTot/universe = ΔSsys + ΔSsurr = 7.6 J.K⁻¹ - 7.6 J.K⁻¹ = 0 B) No work on free expansion , w = 0 , ΔE = 0, Then q = 0 qsurr = 0 ΔS surr = qsurr / T = 0 ΔS Tot/universe = +7.6 J.K⁻¹ + 0 = + 7.6 J.K⁻¹ 21  Asses whether the combustion of magnesium is spontaneous at under standard conditions given 2 Mg(s) + O₂ (g) 2 MgO (s) ΔS˚ = -217 J.K⁻¹ ΔH˚ = -1202 kJ ΔS˚surr = - ΔH˚ = 1202000 J = + 4.03 x 10³ J.K⁻¹ T 298 K ΔSTot = ΔSsys + ΔS surr = -217 J.K⁻¹ + 4.03 x 10³ J.K⁻¹ = + 3.81 x 10³ J.K⁻¹ 22  Equilibrium When a chemical reaction mixture reaches a certain composition, the reaction seems to come to a halt. A mixture of substances at chemical equilibrium has no tendency either to produce more products or to revert to reactants. At equilibrium, reactants are still forming products, but products are decaying at a matching rate into reactants and there is no net change of composition. The common characteristic of any kind of dynamic equilibrium is the continuation of processes at the microscopic level but no net tendency for the system to change in either the forward or the reverse direction. That is, neither the forward nor the reverse process is spontaneous The general criterion for equilibrium in thermodynamics is ∆Stot = 0 23  Focusing on the System ΔS Tot/universe = ΔS sys + ΔS surr ΔSsurr = -ΔH/T ΔS Tot/universe = ΔSsys - ΔH At constant temperature and pressure T Gibbs free Energy which is define as G = H - TS ΔG = ΔH - TΔS ΔG = ΔH - ΔS - ΔG = ΔS - ΔH - ΔG = ΔSTot/universe T T T T T ΔG = - TΔS Tot/universe At constant temperature and pressure Thus, at constant temperature and pressure, the direction of spontaneous change is the direction of decreasing Gibbs free energy Importance Provided the temperature and pressure are constant, we can predict the spontaneity of a process solely in terms of the thermodynamic properties of the system. 24  Summary of factors that affect spontaneity Enthalpy Change Entropy Change Spontaneous Exothermic (ΔH < 0) Increase (ΔS > 0) Yes , ΔG < 0 Exothermic (ΔH < 0) Decrease (ΔS < 0) Yes, if TΔS < ΔH , ΔG < 0 Endothermic (ΔH > 0) Increase (ΔS > 0) Yes, if TΔS > ΔH , ΔG < 0 Endothermic (ΔH > 0) Decrease (ΔS < 0) No, ΔG > 0 Effects of Temperature on ΔG Case Results ΔS positive, ΔH negative Spontaneous at all temperatures ΔS positive, ΔH positive Spontaneous at high temperatures (where exothermicity is relatively unimportant) ΔS negative , ΔH negative Spontaneous at low temperature (where exothermicity is relatively unimportant) ΔS negative, ΔH positive Process not spontaneous at any temperature (reverse process is spontaneous at all temperatures) 25  Deciding whether a process is spontaneous Calculate the change in molar Gibbs free energy, ∆Gm, for the process H2O(s) → H2O(l) at 1 atm and (a) 10.°C; (b) 0.°C. Decide for each temperature whether melting is spontaneous or not. Treat ∆Hfus and ∆Sfus as independent of temperature. ∆H fus = 6.01 kJ.mol⁻¹ ΔSfus = 22.0 J.K⁻¹.mol⁻¹ ΔG = ΔH - TΔS = 6.01 kJ.mol⁻¹ - 283 K x 22.0J.K⁻¹.mol⁻¹ 6.01 kJ.mol⁻¹ - 6.23 kJ.mol⁻¹ = - 0.22 kJ.mol⁻¹ ΔG = ΔH - TΔS = 6.01 kJ.mol⁻¹ - 273 K x 22.0 J.K⁻¹.mol⁻¹ = 6.01 kJ.mol⁻¹ - 6.01 kJ.mol⁻¹ = 0 26  Gibbs Free Energy of Reaction The decrease in Gibbs free energy as a signpost of spontaneous change and ∆G = 0 as a criterion of equilibrium are applicable to any kind of process, provided that it is occurring at constant temperature and pressure. Because chemical reactions are our principal interest in chemistry, we now concentrate on them and look for a way to calculate. ∆G for a reaction ΔG = ⅀ n Gₘ (products) - ⅀ n Gₘ (reactants) The standard Gibbs free energy of reaction, G°, is defined like the Gibbs free energy of reaction but in terms of the standard molar Gibbs energies of the reactants and products ΔG˚ = ⅀ n Gₘ˚(products) - ⅀ n Gₘ˚reactants Gf° (the “standard free energy of formation”), of a substance is the standard Gibbs free energy of reaction per mole for the formation of a compound from its elements in their most stable form. The most stable form of an element is the state with the lowest Gibbs free energy 27 Calculating a standard Gibbs free energy of formation from enthalpy and entropy data Calculate the standard Gibbs free energy of formation of HI(g) at 25°C from its standard molar entropy and standard enthalpy of formation ΔH˚f (HI (g) = 26.48 kJ ΔS˚(HI (g) = 206.6 J.K⁻¹.mol⁻¹ ΔS˚(I (g) = 116.1 J.K⁻¹.mol⁻¹ ΔS˚(H (g) = 130.7 J.K⁻¹.mol⁻¹ ΔG˚ = ΔH - TΔS˚ ½ H₂(g) + ½ I₂(g) HI (g) ΔS˚ = ⅀ n Sₘ˚(products) - ⅀ n Sₘ˚(reactants) 1 x 206.6 J.K⁻¹.mol⁻¹ - ½ x 130.7 J.K⁻¹.mol⁻¹ + ½ x 116.1 J.K⁻¹.mol⁻¹ = +83.2 J.K⁻¹.mol⁻¹ G˚ = ΔH - TΔS˚ = 1 x 26.48 kJ - 298 K x 0.0832 kJ.mol⁻¹ = + 1.69kJ 28  The Dependence of Free Energy on Pressure ΔG = ΔG° + RT ln Q Where Q is the reaction quotient aA+bB↔ cC+dD If Q > K the rxn shifts towards the reactant side – The amount of products are too high relative to the amounts of reactants present, and the reaction shifts in reverse (to the left) to achieve equilibrium If Q = K equilibrium If Q < K the rxn shifts toward the product side – The amounts of reactants are too high relative to the amounts of products present, and the reaction proceeds in the forward direction (to the right) toward equilibrium  PCc PDd   PCc PDd  compare Q =  a b  K =  a b  29  PA PB any conditions  PA PB equilibrium  ΔG = ΔG° + RT ln Q – Where Q is the reaction quotient aA+ b B↔c C + d D If Q < K the rxn shifts towards the product side If Q = K equilibrium If Q > K the rxn shifts toward the reactant side At Equilibrium conditions, ΔG = 0 ΔG° = -RT ln K NOTE: we can now calculate equilibrium constants (K) for reactions from standard ΔGf functions of formation c [C] [D] d ΔG = cΔG (C) + dΔG (D) o o o K= a b r f f 30 [A] [B] − aΔG (A) − bΔG (B) o f o f  Calculate the equilibrium constant for this reaction at 25C. 3NO(g) N2O(g) + NO2(g) ΔG˚ = -105 kJ mol⁻¹ Use - ΔG ° = RT ln K -(-105,000 J mol-1 ) −ΔG ° ln K = -1 -1 = 42 ln 𝐾 = (8.3145 J K mol )(298.15 K) 𝑅𝑇 K = e 42 = 2 x 1018 [ PN 2O ]1[ PNO2 ]1 K= = 1.8x1018 [ PNO ]3 31  ΔG = ΔG° + RT ln Q Where Q is the reaction quotient aA+bB↔ cC+dD Criteria for Spontaneity in a Chemical Reaction Spontaneous Equilibrium Non- Conditions Processes Processes spontaneous Processes ΔSuniv > 0 ΔSuniv = 0 ΔSuniv < 0 All conditions ΔGf < 0 ΔGf = 0 ΔGf > 0 Constant P and T QK Constant P and T 32