Chapter 3: Second Law of Thermodynamics - PDF

Limitations of 1st Law of Thermodynamics i. It does not make distinctions between the processes that occur spontaneously and non spontaneously. ii. It does not tell us the direction of flow of heat. iii. Important distinction between 1st and 2nd is that it is impossible to convert internal energy c...

Limitations of 1st Law of Thermodynamics i. It does not make distinctions between the processes that occur spontaneously and non spontaneously. ii. It does not tell us the direction of flow of heat. iii. Important distinction between 1st and 2nd is that it is impossible to convert internal energy completely into work done. The total energy of the universe is constant: U = q + w Esys + Esurr = Euniv = 0 Esys = -Esurr or 2  The 2nd Law of Thermodynamics, can be used to classify Thermodynamic Processes into 3 Types: ❖ Irreversible Processes, or Natural Processes or Spontaneous Processes) ❖ Reversible Processes ❖ Impossible Processes 3  Spontaneous Processes Spontaneous Processes : ❑ It proceeds automatically without the need to do work on the system. ❑ The opposite of every Spontaneous Process is a Non-Spontaneous Process ❑ Non-Spontaneous Process that can only proceed if external work is done on the system. 12  The 2 nd Law of Thermodynamics The 2 nd Law of Thermodynamics says that: ❖ The opposite of the irreversible processes cannot proceed automatically without an external work is done on it. 13  Impossible Processes ❑ “Impossible Processes”  Process which Cannot Occur Naturally, because they would violate the 2nd Law of Thermodynamics. 14  Impossible Processes impossible processes that do not violate first law ✓ A cup of coffee does not get warm in a cooler room by absorbing heat from environment. ✓ It is impossible to construct a machine which operates in a cycle converts heat into equivalent amount of work. 15  Impossible Processes Example 1: “Free Compression” of a Gas! Initially, the valve is open & gas molecules are uniformly distributed in the 2 containers. Valve Gas Open Gas After some time, all gas molecules are gathered in the right container & the left container is empty. Gas Vacuum Valve Open 16  Reversible process The system undergoing the reversible process can be returned to its initial conditions along the same path. i. Reversible processes are equilibrium processes. ii. The system always remains in thermodynamic equilibrium. iii. Heat would not net flow into or out of a system. iv. The system would not do work against its surroundings. 17 Reversible Processes Reversible Process, the system undergoes changes such that the system plus it’s surroundings can be put back in their original states by exactly reversing the process. Reversible Process, changes proceed in infinitesimally small steps, so that the system is infinitesimally close to equilibrium at every step. This is obviously an idealization & can never happen in a real system! 18  The 2nd Law of Thermodynamics, Clausius’ statement: for Refrigerators Energy will not flow spontaneously from a colder body to a warmer body without any work having been done to accomplish this flow (Non Spontaneous). 19  The 2nd Law of Thermodynamics Kelvin Statement: It is impossible to take heat from a single source and is totally converted into work done. 20  Heat Engine ❖ Any device that converts heat partially into mechanical work is called Heat Engine. Heat engine absorb heat from hot reservoir and uses partially to do mechanical work W and the remaining heat is rejected as Qc to the cold reservoir. QH = W out + QC W out = QH - QC 21  Heat Engine The existence of such machine will be the violation of the 2nd law of thermodynamics. And Clausius’ statement becomes true that Energy does not flow spontaneously from a cold object to a hot object. 22  Refrigerator Refrigerator works on the reverse principle of heat engines, i.e. they absorb energy Qc from colder region and expel an amount of energy Qh to hotter region. But this is only accomplished by the help of some work that is to be done on the engine to transfer this heat. 23  Heat Engine Heat engines convert heat into mechanical work. Efficiency of a heat engine, 𝜼: ❖ is the ratio of the energy you get out ( in the form of work) to the energy you put in. W out = QH- QC 𝜼= = = 1- 𝜼= 24  Heat Engine Example: Maximum Efficiency: Determine the maximum efficiency of a heat engine with a high-temp reservoir of 1200 K, and a low-temp reservoir of 400 K. Answer 𝜼= = = 0.67 = 67 % 25  Exercise: Heat Engine Solution 26 Exercise: Heat Engine 27  Heat Engine For 100% efficiency Qc = 0 It is impossible that Qh = W So Kelvin plank’s statement becomes true that It is impossible for a heat engine absorbs heat from a reservoir, to convert all heat into work done. 28  The 2nd Law of Thermodynamics Heat flows naturally from a warmer object to a colder object and cannot flow from a colder object to a warmer object without doing work on the system. Heat energy cannot be completely transferred into mechanical work. Nothing 100% efficient. 29  Entropy ΔS and the Second Law of Thermodynamics Entropy is a measure of the disorder of a system. This gives us yet another statement of the second law: Natural processes tend to move toward a state of greater disorder. 37  Entropy is a state function Entropy is a state function. the overall change in a state function (from the initial state to the final state and then back to the initial state again) is zero (e.g., ΔS = Sf – Si = 0). 38  Statement of the 2nd Law of Thermodynamics ❖ All natural system tend toward a higher level of disorder (entropy). ∆S ≥ 0. ❖ The only way to decrease the entropy of a system is to do work on it. ❖ “In any spontaneous process, there is always an increase in the entropy of the universe.” 39  Another Statement of the 2nd Law of Thermodynamics ❑ “The entropy of the universe does not change for Reversible Processes”. ❑ “The entropy of the universe increases for Irreversible ( Spontaneous) Processes” For Reversible (ideal) Processes: For Irreversible (real, spontaneous) Processes: 40 The 2nd Law of Thermodynamics, 41  Entropy and the Second Law of Thermodynamics The change in entropy S when an amount of heat Q is added is : For Reversible process: 42  Spontaneous Processes & Entropy Entropy can be viewed as a measure of the randomness or disorder of the atoms & molecules in a system. 2nd Law of Thermodynamics Total Entropy always increases in a spontaneous process! So, Microscopic Disorder also increases in a spontaneous process! 43  Factors Affecting Entropy 1. Entropy depends on temperature. For any substance, S° increases as temperature increases. 2. Entropy depends on the physical state of a substance. S° increases as the phase changes from solid to liquid to gas. 3. The formation of a solution increases entropy. 4. Entropy increases as atomic size increases. 5. Gas Expansion increases S. 44 Entropy 45  Entropy Exercise: ΔvapS°is 85 J K−1 mol−1 Solution: 46 Exercises: Entropy 47  Exercises: Entropy EXAMLE: 48  Entropy For a reversible change the energy lost as heat from the system is acquired by the surroundings, This change is the negative of the change in the system, so we can conclude that ΔStot = 0 49  Entropy For the Free isothermal expansion Means, (ΔU) = 0 (w = 0), then q = 0. Consequently, ΔSsur = 0, and the total entropy change is given by In this case, ΔStot > 0, as we expect for an irreversible process 50  Exercises: Entropy Solution 51 Entropy 52  The standard molar entropy So is of a substance, measured for a substance in its standard state in units of J/mol·K. The definition of a standard state includes: 1 atm for gases. 1 Molar for solutions. the pure substance for solids and liquids. 53  The standard reaction entropy ΔSo: Define as the difference between the molar entropies of the pure, separated products and the pure, separated reactants. 54  Exercises: Entropy Solution 55  Exercises: Entropy Solution 56  Entropy Heating: OR 𝐶 =𝐶 -R 57 Exercise: Entropy 58 Solution 59 THIRD LAW OF THERMODYNAMICS 60  The Third Law of Thermodynamics The Third Law of Thermodynamics T = 0 K is often referred to as “Absolute Zero” 61  Another Statement of The Third Law of Thermodynamics “The entropy of a true equilibrium state of a system at T = 0 K is zero.” (Strictly speaking, this is true only if the quantum mechanical ground state is non-degenerate. If it is degenerate, the entropy at T = 0 K is a small constant, not 0!) This is Equivalent to: The Third Law of Thermodynamics “It is impossible to reduce the temperature of a system to T = 0 K using a finite number of processes.” 62 The Third Law of Thermodynamics 0 0 63  The Third Law of Thermodynamics The number of degree of disorder entropy (S) of the system. S = k ln W where k is the proportionality (Boltzmann's) constant k= 1.38064852 × 10-23 m2 kg s-2 K-1, W is the number of accessible microstates. 64  The Third Law of Thermodynamics S = k ln W A perfect crystal has zero entropy at absolute zero. Ssys = 0 at 0 K A “perfect” crystal has all its particles in alignment at absolute zero, the particles have minimum energy, so there is only one microstate. S = k ln (W) = k ln (1) = 0 65  Gibbs Free Energy, G Gibbs Free, G: is a measure of the energy available to the system to do useful work. G°is expressed in kJ or kJ/mol G°is state function G°is extensive property 68  Gibbs Free Energy, G The Gibbs free energy (G) combines the enthalpy and entropy of a system; G = H – TS. The free energy change (G) is a measure of the spontaneity of a process and of the useful energy available from it. Gsys = Hsys - T Ssys G < 0 for a spontaneous process G > 0 for a nonspontaneous process G = 0 for a process at equilibrium 69  Gibbs Free Energy, G G or can be calculated using the Gibbs equation: G°sys = H°sys - T S°sys G or can also be calculated using values for the standard free energy of formation of the components. G°r = m G°products – n G°reactants 70 Gibbs Free Energy, G 71 Gibbs Free Energy, G 72 73  Thermochemistry Thermochemistry = The study of the energy transferred as heat during the course of chemical reactions is called thermochemistry Exothermic process = release of heat ΔH < 0. Endothermic process = Absorption of heat ΔH > 0 43  Standard enthalpy changes The standard enthalpy change, ΔH° is a change in enthalpy for a process in which the initial and final substances are in their standard states. The standard state of a substance at a specified temperature is its pure form at 1 bar. Examples (i) The standard state of liquid ethanol at 298 K is pure liquid ethanol at 298 K and 1 bar. (ii) The standard state of solid iron at 500 K is pure iron at 500 K and 1 bar. 44  Standard enthalpy changes ΔH° for a reaction (or a physical process) at specified T: ∆𝒓𝒆𝒂 𝑯° = 𝑯° 𝑷𝒓𝒅𝒖𝒄𝒕𝒔 − 𝑯° 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 Example: The standard enthalpy of vaporization, ΔvapH°, is the enthalpy change per mole when a pure liquid at 1 bar vaporizes to a gas at 1 bar, as in: H2O(l) → H2O(g), ΔvapH° (373 K) = +40.66 kJ mol−1 45  Standard enthalpy changes Enthalpies of physical change The standard enthalpy change that accompanies a change of physical state is called the standard enthalpy of transition and is denoted ΔtrsH°. Standard enthalpies of fusion and vaporization at the transition temperature, ΔtrsH° (kJ mol−1) H2O(s) → H2O(l) ΔfusH°(273 K) = +6.01 kJ mol−1 H2O(l) → H2O(g) ΔvapH°(373 K) = +40.66 kJ mol−1 46  Standard enthalpy changes Enthalpies of physical change Different types of enthalpies of transitions encountered in thermochemistry 47  Standard enthalpy changes Enthalpies of physical change H is a state function => ΔH is independent of the path between the two states. Example: conversion of a solid to a vapour (i) Direct path (one step) H2O(s) → H2O(g) ΔsubH° (ii) Indirect path (two steps) H2O(s) → H2O(l) ΔfusH° H2O(l) → H2O(g) ΔvapH° Overall: H2O(s) → H2O(g) ΔsubH° = ΔfusH° + ΔvapH° ΔH° (A→B) = - ΔH° (B→A) 48  Standard enthalpy changes Enthalpies of physical change The lattice enthalpy (ΔHL) is the change in standard molar enthalpy for this process. MX(s) → M+(g) + X−(g) Experimental values of ΔHL are obtained by using a Born– Haber cycle. It is a closed path of transformations starting and ending at the same point, one step of which is the formation of the solid compound from a gas of widely separated ions. At T = 0 => ΔHL = ΔUL At normal temperatures, ΔHL and ΔU differ by only a few kilojoules per mole, and the difference is normally neglected. 49  Standard enthalpy changes Enthalpies of physical change Example: A typical Born–Haber cycle, for potassium chloride Because the sum of these enthalpy changes is equal to zero, we can infer from 89 + 122 + 418 − 349 − ΔHL/(kJ mol−1) + 437 = 0 Thus, ΔHL = +717 kJ mol−1 50  Standard enthalpy changes Enthalpies of chemical change There are two ways of reporting the change in enthalpy that accompanies a chemical reaction: (i) CH4(g) + 2 O2(g)→CO2(g) + 2 H2O(l) ΔH° = −890 kJ (ii) CH4(g) + 2 O2(g)→CO2(g) + 2 H2O(l) ΔrH° = −890 kJ mol−1 In general, for a chemical reaction: Reactants → Products ∆𝒓 𝑯° = ෍ 𝝂𝑯°𝒎 − ෍ 𝝂𝑯°𝒎 𝑷𝒓𝒐𝒅𝒖𝒄𝒕𝒔 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 where 𝑯°𝒎 molar enthalpies of the species are multiplied by their (dimensionless and positive) stoichiometric coefficients, 𝝂 51  Standard enthalpy changes Hess’s law Hess’s law: The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided. Example: Using Hess’s law The standard reaction enthalpy for the hydrogenation of propene is −124 kJ mol−1. CH2=CHCH3(g) + H2(g) → CH3CH2CH3(g) ΔrH° = −124 kJ mol−1 The standard reaction enthalpy for the combustion of propane is −2220 kJ mol−1. CH3CH2CH3(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ΔrH° = −2220 kJ mol−1 Calculate the standard enthalpy of combustion of propene ? 52  Standard enthalpy changes Hess’s law Answer Add additional data. The combustion reaction we require is C3H6(g) + 9/2O2(g)→3 CO2(g) + 3 H2O(l) This reaction can be recreated from the following sum: ΔrH°/(kJ mol−1) C3H6(g) + H2(g) → C3H8(g) −124 C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) −2220 H2O(l) → H2(g) + 1/2O2(g) +286 C3H6(g) + 9/2 O2(g) → 3 CO2(g) + 3 H2O(l) −2058 53  Standard enthalpies of formation The standard enthalpy of formation, ΔfH°, of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference states, i.e., the most stable state at the specified temperature and 1 bar. Example ΔfH° of liquid benzene at 298 K, for example, refers to the reaction below is +49.0 kJ mol−1 6 C(s, graphite) + H2(g)→C6H6(l) Notes ΔfH° of elements in their reference states are zero at all temperatures. ΔfH°(H+, aq) = 0 54  Thermochemistry: Standard enthalpies of formation Reaction enthalpy in terms of enthalpies of formation 𝜟𝒓 𝑯° = ෍ 𝝂𝜟𝒇 𝑯° − ෍ 𝝂𝜟𝒇 𝑯° 𝑷𝒓𝒐𝒅𝒖𝒄𝒕𝒔 𝑹𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔 Example Consider the reaction : 2 HN3(l) + 2 NO(g) → H2O2(l) + 4 N2(g) The standard enthalpy of the reaction is calculated as follows: ΔrH° = {ΔfH°(H2O2,l) + 4ΔfH°(N2,g)} − {2ΔfH°(HN3,l) + 2ΔfH°(NO,g)} = {−187.78 + 4(0)} kJ mol−1 − {2(264.0) + 2(90.25)} kJ mol−1 = −896.3 kJ mol−1 55  The temperature dependence of reaction enthalpies The standard enthalpies may be calculated from heat capacities and the reaction enthalpy at some other temperature. Kirchhoff’s law 𝑻𝟐 𝜟𝒓 𝑯° 𝑻𝟐 = 𝜟𝒓 𝑯° 𝑻𝟏 + න 𝜟𝒓 𝑪°𝒑 𝒅𝑻 𝑻𝟏 𝚫𝐫 𝐇 ° 𝐓𝟐 = 𝚫𝐫 𝐇 ° 𝐓𝟏 + (𝐓𝟐 − 𝐓𝟏 )𝚫𝐫 𝐂𝐩° where Kirchhoff’s law Δ𝑟 𝐶𝑝° = ෍ ° 𝜈𝐶𝑝,𝑚 − ෍ ° 𝜈𝐶𝑝,𝑚 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑅𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 56  The temperature dependence of reaction enthalpies Example: Using Kirchhoff’s law The standard enthalpy of formation of H2O(g) at 298 K is −241.82 kJ mol−1. Estimate its value at 100°C given the following values of the molar heat capacities at constant pressure: H2O(g): 33.58 J K−1 mol−1; H2(g): 28.82 J K−1 mol−1; O2(g): 29.36 J K−1 mol−1. Assume that the heat capacities are independent of temperature. Answer Δ𝑟 𝐻 ° 𝑇2 = Δ𝑟 𝐻 ° 𝑇1 + (T2 −T1 )𝜟𝒓 𝑪°𝒑 57  The temperature dependence of reaction enthalpies Example: Using Kirchhoff’s law To proceed, write the chemical equation, identify the stoichiometric coefficients, and calculate Δ𝑟 𝐶𝑝° from the data. The reaction is 1 H2 (g) + 2 O2 (g) → H2 O(g) so ° ° ° 𝟏 ° 𝜟𝒓 𝑪𝒑 = 𝑪𝒑,𝒎 𝑯 𝐎, 𝒈 − 𝑪𝒑,𝒎 𝑯 , 𝒈 + 𝑪𝒑,𝒎 𝑶 , 𝒈 𝟐 𝟐 𝟐 𝟐 = −9.92 J K −1 mol−1 It then follows that 𝜟𝒓 𝑯° 𝟑𝟕𝟑 𝑲 = −241.82 kJ mol−1 + 75 K × −9.92 J K −1 mol−1 = −242.6 kJ mol−1 58