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Chemical Thermodynamics Chapter 17
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Fig. 17-CO, p. 736 Gasoline Refinery
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Ch 17 Introduction Chemical Thermodynamics = the study of the energetics of chemical reactions Spontaneous reactions Processes that can occur without outside intervention Spontaneous processes may be fast or slow Many forms of combustion are fast Conversion of diamond to graphite is slow
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Ch 17 Introduction If a chemical reaction is spontaneous in one direction, then under the same conditions the reverse reaction is not spontaneous. That does not mean that the reverse reaction cannot proceed, only that it may require different conditions Ex: temperature, pressure, concentration
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17-1 Work and Heat Objectives To relate heat, work, and energy To calculate work from pressure-volume relationships
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17-1 Work and Heat The amount of heat absorbed or released in a chemical reaction is very important! Thermodynamics can: Evaluate the heat exchanged by a reaction Predict the maximum energy that can be produced Determine whether a reaction is feasible
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17-1 Work and Heat A review of some important definitions: System = matter of interest Surroundings = all other matter that is not part of the system In closed systems, energy can be exchanged with the surroundings but not matter The universe = the system + the surroundings State of the system = temperature, pressure, amounts, phases of substances
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p. 739
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17-1 Work and Heat State functions = properties that depend only on the state of the system and not in the manner in which the system arrives at the state Changes in state functions depend on the initial and final states of the system, not on the path by which it occurs Analogy: altitude of a climber = state function distance traveled is a path function
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17-1 Work and Heat Remember that the First Law of Thermodynamics states that ∆E = q + w where: ∆E = the change in energy q = heat w = work
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17-1 Work and Heat As noted in the previous slide, energy can be transferred via: Heat = thermal energy / random motion Can be measured by calorimetry Or, energy can be transferred via: Work = the application of force through a distance / directed motion w = force x distance In a system that contains a gas, a gas may expand against an opposing pressure exerted by the surroundings and do “PV work” w = -PΔV
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p. 739
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17-1 Work and Heat Note that the units of work are joules (J) Work = force x distance = N x m = joule (N = Newton = kgm/s 2 )
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17-1 Work and Heat If the system: Performs work on the surroundings, the sign of w is negative Receives work from the surroundings, the sign of w is positive Transfers heat to the surroundings, the sign of q is negative Absorbs heat from the surroundings, the sign of q is positive
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17-1 Work and Heat Q: Calculate the amount of work (in joules) done on the system when 6.30 x 10 3 L of an ideal gas (the system) initially at 1.00 atm are compressed at constant temperature and a pressure of 140 atm to a final volume of 45.0 L. The opposing pressure is the constant 140 atm applied to the system. Note: 1 Latm = 101.3 J w = -P ext ∆V A: w = + 8.87 x 10 7 J or + 8.87 x 10 4 kJ
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17-2 The First Law of Thermodynamics Objectives To relate heat, work, energy, and the first law of thermodynamics To distinguish between internal energy and enthalpy
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17-2 The First Law of Thermodynamics The First Law of Thermodynamics = energy cannot be created or destroyed, but can be converted from one form to another In a closed chemical system, energy is often transferred as heat and work
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17-2 The First Law of Thermodynamics Internal energy, E, is the total energy of the system and is expressed in joules A change in internal energy is represented by ΔE In a closed system, Δ E = q + w where q = heat absorbed by the system and w = work performed on the system
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17-2 Reminder! If the system: Performs work on the surroundings, the sign of w is negative Receives work from the surroundings, the sign of w is positive Transfers heat to the surroundings, the sign of q is negative Absorbs heat from the surroundings, the sign of q is positive
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17-2 The First Law of Thermodynamics Note that ΔE is a state function whereas w and q are path functions. Work and heat can be manipulated to provide either more heat or more work as needed even if ΔE is held constant. Ex: trying to maximize work done by a car engine while minimizing heat
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Fig. 17-1, p. 743 Work is a path function
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17-2 The First Law of Thermodynamics Let’s prove work is a path function…. If we change the volume of a system from 1.0 L to 10.0L at a final pressure of 1 atm the work done = -912 J If we expand in two steps: first to 5.0L at a pressure of 2 atm, then to 10.0L at a pressure of 1 atm then w 1 =-810 J and w 2 = - 507 J w T = -1317 J
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17-2 The First Law of Thermodynamics The first process provides less work but more heat The second process produces more work and generates less heat Therefore, processes can be performed in different ways to maximize the amounts of heat or work depending on your goals….
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17-2 The First Law of Thermodynamics Energy & Enthalpy Data about heats of reaction are tabulated as standard enthalpies of formation ( H f ), as described in Chapter 5. The equation that relates H to E is H = E + P V This equation is used to calculate the enthalpy of reaction from heats measured using constant-volume calorimetry.
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17-2 The First Law of Thermodynamics The difference between changes in enthalpy and internal energy is PV work and is significant only for reactions that involve gases. From the ideal gas law, PV = nRT, so H = E + nRT
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17-2 The First Law of Thermodynamics Calculate H at 25 C and 1.00 atm pressure for the following reaction. 2NO 2 (g) → N 2 O 4 (g) E = -54.7 kJ
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17-2 The First Law of Thermodynamics So, do all spontaneous processes have a negative ∆H? Let’s find out….
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17-3 Entropy and Spontaneity Objectives To define entropy and examine its statistical nature To apply the second law of thermodynamics to chemical systems To compare and contrast entropy changes in the system, surroundings, and universe To recognize that absolute entropies can be measured because the third law of thermodynamics defines a zero point
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p. 747 Based on these pictures, how would you define entropy?
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17-3 Entropy and Spontaneity Although many spontaneous reactions are accompanied by a release of energy, some are not….. Let’s look at the reaction of solid barium hydroxide with solid ammonium chloride…
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Ba(OH) 2 H 2 O (s) + NH 4 Cl (s) → BaCl 2 (aq) + 2NH 3 (g) +10H 2 O ( l )
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17-3 Entropy and Spontaneity Note that this endothermic reaction occurred spontaneously! How can that be? Because disorder also plays a measurably important role in predicting whether a reaction will be spontaneous…. Entropy (S) is the thermodynamic state function that describes the amount of disorder
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Entropy & Spontaneity The change in entropy can be the sole driving force for some processes. Let’s look at the diffusion of a gas….. Fig. 17-3, p. 747
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17-3 Entropy and Spontaneity The entropy of a substance increases when the solid forms a liquid and when the liquid forms a gas. Entropy generally increases when a condensed phase dissolves in a solvent. Entropy decreases when a gas dissolves in a solvent. Entropy increases as the temperature increases
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Positional Entropy Positional Entropy The probability of occurrence of a particular state depends on the a particular state depends on the number of ways (microstates) in which number of ways (microstates) in which that arrangement can be achieved that arrangement can be achieved S solid < S liquid << S gas S solid < S liquid << S gas
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17-3 Entropy and Spontaneity The Second Law of Thermodynamics = in any spontaneous process, the entropy of the universe increases
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Calculating Entropy Change in a Reaction Entropy is an extensive property (a function of the number of moles) Generally, the more complex the molecule, the higher the standard entropy value
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17-4 Free Energy Objectives To define free energy and relate the sign of a free energy change to the direction of spontaneous reaction To predict the influence of temperature on free energy
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17-4 Free Energy The change in free energy (∆G) is a quantitative indicator of spontaneity for all systems The only restrictions are that the process must occur under conditions of constant temperature and pressure
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For reactions at constant temperature: G 0 = H 0 - T S 0 Calculating Free Energy Method #1
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Calculating Free Energy: Method #2 An adaptation of Hess's Law: C diamond (s) + O 2 (g) CO 2 (g) G 0 = -397 kJ C graphite (s) + O 2 (g) CO 2 (g) G 0 = -394 kJ CO 2 (g) C graphite (s) + O 2 (g) G 0 = +394 kJ C diamond (s) C graphite (s) G 0 = C diamond (s) + O 2 (g) CO 2 (g) G 0 = -397 kJ -3 kJ
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Calculating Free Energy Method #3 Using standard free energy of formation ( G f 0 ): G f 0 of an element in its standard state is zero
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Table 17-1, p. 752
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H, S, G and Spontaneity Value of H Value of T S Value of G Spontaneity Negative PositiveNegativeSpontaneous Positive NegativePositiveNonspontaneous Negative ???Spontaneous if the absolute value of H is greater than the absolute value of T S (low temperature) Positive ???Spontaneous if the absolute value of T S is greater than the absolute value of H (high temperature) G = H - T S H is enthalpy, T is Kelvin temperature
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Table 17-2, p. 753
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See page 754 for an example problem
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17-5 Free Energy & the Equilibrium Constant Objectives To evaluate the effect of concentration on free energy To develop and use the relationship between standard free energy change and the equilibrium constant To determine the effect of temperature on the equilibrium constant To relate the change in free energy to the maximum useful work available from a reaction
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17-5 Free Energy & the Equilibrium Constant Concentration & Free Energy Take the reaction N 2 (g) + 3H 2 (g) 2NH 3 (g) ΔG° = -33 kJ at 25°C Spontaneous as written Can the reaction be reversed? How? Think Le Chatelier’s Principle ….
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17-5 Free Energy & the Equilibrium Constant Concentration & Free Energy Le Chatelier’s Principle qualitatively describes the effect of concentration on the direction a reaction will proceed to reestablish an equilibrium Calculations of free energy can describe the process quantitatively
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17-5 Free Energy & the Equilibrium Constant Concentration & Free Energy To date we have only calculated free energy under standard conditions 298K pure liquids or solids solutions with solute concentrations of 1M gases at a partial pressure of 1 atm What happens if we use non-standard conditions?
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17-5 Free Energy & the Equilibrium Constant Concentration & Free Energy Under non-standard conditions: ΔG = ΔG° + RTlnQ Reminder: For the reaction: aA + bB cC + dD Q = [C] c [D] d [A] a [B] b
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17-5 Free Energy & the Equilibrium Constant Calculate the Gibbs free energy change for the reaction of carbon dioxide and ammonia to form urea at 298K, when all gases are present at 0.1 atm pressure. CO 2 (g) + 2NH 3 (g) ↔ CO(NH 2 ) 2 (s) + H 2 O (l) First step = calculate ∆G° (see Table G p. A21) ∆G° = -7.30 kJ Second step = calculate Q Q = 1.0 x 10 3 Third step = solve for ∆G ∆G = + 9.8 kJ
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17-5 Free Energy & the Equilibrium Constant The Equilibrium Constant & Free Energy The equilibrium point occurs at the lowest value of free energy available to the reaction system At equilibrium Q = K eq and ΔG = 0. Therefore, at equilibrium: ΔG° = -RTlnK eq or K eq = e - ∆G°/RT Note that ΔG must be evaluated at the same temperature as the equilibrium constant
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17-5 Free Energy & the Equilibrium Constant The K eq for the formation of formaldehyde from hydrogen gas and carbon monoxide is 8.6 x 10 -7. Calculate the ∆G° at 298K. H 2 (g) + CO (g) ↔ CH 2 O (g) A: + 34.6 kJ
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17-5 Free Energy & the Equilibrium Constant The Gibbs free energy change for the following reaction is +163.2 kJ. Calculate the equilibrium constant for this reaction at 298K. 3/2 O 2 (g) ↔ O 3 (g) A: 2.47 x 10 -29
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17-5 Free Energy & the Equilibrium Constant Note that: If ∆G is < 0 or if Q < K eq, a reaction is spontaneous in the forward direction If ∆G is > 0 or if Q > K eq, a reaction is spontaneous in the reverse direction
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17-5 Free Energy & the Equilibrium Constant Temperature & the Equilibrium Constant Chemical reactions differ widely in their response to temperature Think about how we used Le Chatelier’s Principle to determine how a reaction would shift when temperatures were changed… Did we have the same predictions for endothermic and exothermic reactions? Can we now make a more quantitative analysis rather than a qualitative one?
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Temperature Dependence of K
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Table 17-3, p. 760
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17-5 Free Energy & the Equilibrium Constant Typically, ∆S° and ∆H° have smaller temperature dependencies than K eq and ∆G° If the value of the equilibrium constant is known at two different temperatures, an estimate of ∆H° can be made using the Clausius-Clapeyron Equation
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17-5 Free Energy & the Equilibrium Constant The Clausius-Clapeyron Equation ln = ∆H° 1 - 1 RT2T2 T1T1 K1K1 K2K2
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17-5 Free Energy & the Equilibrium Constant The vapor pressure of methyl alcohol is 100 torr at 21.2°C and 400 torr at 49.9°C. What is the standard enthalpy of vaporization of methyl alcohol? A: 38.2 kJ
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17-5 Free Energy & the Equilibrium Constant The change in free energy is the maximum useful work obtainable from a process at constant temperature and pressure w max = ΔG The amount of work obtained is always less than the maximum Ex: efficiency of a coal-fueled electrical generating plant is 34%
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Some extra notes not yet incorporated into my lecture follow…
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Entropy (S) A measure of the randomness or disorder The driving force for a spontaneous process is an increase in the entropy of the universe is an increase in the entropy of the universe Entropy is a thermodynamic function describing the number of arrangements that describing the number of arrangements that are available to a system are available to a system Nature proceeds toward the states that have the highest probabilities of existing have the highest probabilities of existing
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Second Law of Thermodynamics "In any spontaneous process there is always an increase in the entropy of the universe" "The entropy of the universe is increasing" For a given change to be spontaneous, S universe must be positive S univ = S sys + S surr
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Standard Free Energy Change G 0 is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states G 0 cannot be measured directly The more negative the value for G 0, the farther to the right the reaction will proceed in order to achieve equilibrium Equilibrium is the lowest possible free energy position for a reaction
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The Dependence of Free Energy on Pressure Enthalpy, H, is not pressure dependent Entropy, S entropy depends on volume, so it also depends on pressure S large volume > S small volume S low pressure > S high pressure
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Constant Volume Calorimetry (as opposed to constant pressure calorimetry described in chapter 5) Too difficult to measure heat produced in an open system during combustion reactions that produce gaseous products Use a closed system bomb calorimeter with thick walls (p. 750) & reactants under high pressure Measure the amount of heat released by knowing the heat capacity of the calorimeter ΔE = q + w but w = 0 since PΔV = 0 and qsurr = CΔT where C = heat capacity
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Using a combustion calorimeter with a heat capacity of 5.278 kJ/°C, combustion of a 0.629 g sample of ethane (C 2 H 6 ) causes the temperature to rise from 24.72°C to 30.87°C. What is ΔE for the combustion of one mole of ethane? A: -1.55 x 10 3 kJ/mol ethane
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