Entropy, the Second and Third Law of Thermodynamics By Doba Jackson, Ph.D. Associate Professor of Chemistry and Biochemistry Huntingdon College

Outline of Chpt 3 Sec 3.1-2: The Universe has a natural direction of change Sec 3.3: Heat Engines & Second Law Sec 3.4: Introducing Entropy Sec 3.5: Entropy to calculate natural direction of change

Outline of Chpt 3 Sec 3.6: Absolute Entropies and the Third Law of Thermodynamics Sec 3.7: Standard Molar Entropies Sec 3.7 con’t: Entropy changes in chemical reactions

The Universe has a Natural Direction of change Why ?

Reminder of microscopic definition of Heat and Work Energy released as HeatEnergy produces useful work

Direction of Spontaneous change Bouncing ball demonstrates the direction of spontaneous change. System: The Ball Surroundings: The Floor - The internal energy of the ball depletes after each collision with the floor. - Some energy is converted to work on the ball which causes the ball to rise. - Other energy is dispersed into the floor (as heat).

Direction of Spontaneous change The bouncing ball spontaneously bounces toward a state of lower energy until all of its energy has been dissipated into the floor. System: The Ball Surroundings: The Floor - By the first law, total energy for the ball colliding with the floor is conserved. -However, ordered motion (work) is gradually converted to disordered motion (heat)

Sec 5.2: Heat Engines and the Second law of Thermodynamics

Argument for new Thermodynamic law Saudi Carnot (in 1824) argued for a new law based upon many studies on how to improve a heat engine.

Argument for new Thermodynamic law Saudi Carnot (in 1824) argued for a new law based upon many studies on how to improve a heat engine. Carnot Cycle Cyclic Process: A-B: Isothermal Expansion B-C: Adiabatic Expansion C-D: Isothermal Compression D-A: Adiabatic Compression

Carnot Cycle A-B: Isothermal Expansion B-C: Adiabatic Expansion C-D: Isothermal Compression D-A: Adiabatic Compression

Net balance of energy Net ∆U: State Function Net Work (W):

Net Heat (q): Heat Work T AB = T h T CD = T c

Carnot and others were trying to make the engine more efficient Heat engine Efficiency Only way to reach 100% efficiency is to make Tc = 0.

Carnot tried to improve the heat engine by coupling a second engine capable of working in reverse This was a failure!! Carnot’s Theorems 1- All reversible heat engines operating between two reservoirs are equally efficient as a Carnot engine operating between the same reservoirs. 2- All irreversible heat engines operating between two reservoirs are equally efficient as a Carnot engine between the same reservoir.

Problem 1: A turbine in a steam power plant takes in steam from a boiler at 520*C and exhaust it into a condenser at 100*C. What is the maximum possible efficiency?

Sec 5.3: Introducing Entropy

Entropy Later Lord Kelvin postulated that the ratio of heat to temperature was a state function and called it entropy.

Second Law of Thermodynamics Typical Engine Hypothetical Engine

Spontaneous nature of heat Spontaneous Not Spontaneous Refrigeration

Introduction to Entropy Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence. Spontaneous processes are favored by a decrease in H (negative ∆H). favored by an increase in S (positive ∆S). Nonspontaneous processes are favored by an increase in H (positive ∆H). favored by a decrease in S (negative ∆S). Entropy (S): The amount of molecular randomness in a system.

Enthalpy term is larger than Entropy Entropy term is larger than Enthalpy

Sec 5.4, 5.5: Calculating Entropy changes

Calculating Entropy changes Adiabatic, Reversible process Mathematical Statement of the Second Law The sum of the ratio of heat to temperature for any reversible cyclic process will be zero.

Calculating Entropy changes of an Ideal Gas Isothermal Expansion, reversible

Calculating Entropy changes of an Ideal Gas Constant Volume, reversible

Calculating Entropy changes of an Ideal Gas Constant Pressure, reversible

Calculating Entropy changes when a phase change occurs Phase changes occur under constant temperature and pressure conditions

Calculate the change in entropies of the system, surroundings and the universe when a sample of nitrogen gas of mass 14 g at 298 K and 1.00 bar doubles its volume in (a) an isothermal reversible expansion (b) an isothermal irreversible expansion against a constant external pressure of P=0 (c) an adiabatic reversible expansion.

Sec 5.6: The Clausius Inequality

Types of Expansion Work Expansion against a constant pressure, irreversible If external pressure does, not vary, this expression can be integrated

Types of Expansion Work Expansion is reversible In this case, the external pressure P ex exactly equals the internal pressure P at each stage of the expansion. The integral cannot be evaluated because temperature is not a constant throughout the integration.

Clausius Inequality Reversible Irreversible An irreversible process produces less work and heat than a reversible process

Reversible, Nonspontaneous Reversible Process: The thermodynamic process in which a system can be changed from its initial state to its final state then back to its initial state leaving all thermodynamic variables for the universe (system + surroundings) unchanged. A truly reversible change will: - Occur in an infinite amount of time - All variables must be in equilibrium with each other at every stage of the change Irreversible Reversible

Irreversible, Spontaneous Irreversible Process: The thermodynamic process in which a system that is changed from its initial state to its final state then back to its initial state will change some thermodynamic variables of the universe. A truly irreversible change will: - Occur in an finite amount of time - All variables will not be in equilibrium with each other at every stage of the change Irreversible Reversible

Clausius Inequality leads to equations for spontaneity Reversible Irreversible Constant Volume Helmholtz Energy

Clausius Inequality leads to equations for spontaneity Reversible Irreversible Constant Pressure Gibbs Energy

Sec 5.7: The Change in Entropy of the surroundings and Universe

Problem 5.7: One mole of an Ideal gas at 300 K is reversibly and isothermally compressed from a volume of 25.0 L to a volume of 10.0 L. Because the water bath’s thermal reservoir in the surroundings is very large, the temperature remains relatively constant at 300 K. Calculate ∆S, ∆S surr ∆S tot.

Problem 5.7: One mole of an Ideal gas at 300 K is isothermally compressed at a constant external pressure from a volume of 25.0 L to a volume of 10.0 L. At the end of the compression, the internal and external pressures are equal. Calculate ∆S, ∆S surr ∆S tot. Since Entropy is a state function, the initial and final states of the system are the same, so the entropy change of the system is the same. However, the surroundings and total entropy will change because the process is irreversible.

Problem 5.7: One mole of an Ideal gas at 300 K is isothermally compressed at a constant external pressure from a volume of 25.0 L to a volume of 10.0 L. At the end of the compression, the internal and external pressures are equal. Calculate ∆S, ∆S surr ∆S tot. Work and heat are not state functions so they will be different for the irreversible process. We can calculate the entropy of the surroundings by first calculating the Work done by the surroundings. Since ∆U = 0, isothermal process

Problem 5.7: One mole of an Ideal gas at 300 K is isothermally compressed at a constant external pressure from a volume of 25.0 L to a volume of 10.0 L. At the end of the compression, the internal and external pressures are equal. Calculate ∆S, ∆S surr ∆S tot. The total entropy change of the universe is positive even when the entropy of the system is negative.

Problem 5.9: Calculate ΔS, ΔS tot, Δs sur when a volume of 150 g of CO initially at 273 K and 1.00 bar increases by a factor of two in: (a) Adiabatic reversible expansion (b) expansion against a P ext =0 (c) isothermal reversible expansion Take C p,m as a constant at J/K*mol and state whether each process is spontaneous. The temperature of the surroundings is 273 K for each process.

Entropy change during Temperature change Heat required to change the temperature At Constant Pressure At Constant Volume Assume a Constant Heat Capacity Over the temperature range

Sec 5.7: Absolute Entropies and the Third Law of Thermodynamics

Entropy in Chemical Thermodynamics We rely on calorimetry to measure the internal energy (∆U) enthalpy (∆H). We can also measure Entropy (∆S) for all Chemical and Physical Changes.

Third Law of Thermodynamics Third Law: The entropy of a perfect crystalline substance is zero at T=0 At T=0, all thermal motion has been quenched and in a perfect crystal, all atoms are in a uniform array.

Standard Entropies (Third Law Entropies) Perfect crystalline S m (0) = 0

Standard Molar Entropies are used to determine reaction enthalpies (Δ r S º ) H 2 O (l) H 2(g) + ½ O 2(g) ReactantsProducts ∆ r S° = { (1)S m (H 2 O) } – { (1)S m (H 2 ) + (1/2)S m (O 2 ) } = kJ/mol Reaction entropies (Δ r Sº) can be determined by the difference of the product entropies and the reactant entropies. Example: