Thermodynamics vs. Kinetics

Thermodynamics vs. Kinetics
Domain of Kinetics Rate of a reaction depends on the pathway from reactants to products. Thermodynamics tells us whether a reaction is spontaneous based only on the properties of reactants and products. Copyright © Cengage Learning. All rights reserved

Spontaneous Processes and Entropy
Thermodynamics lets us predict the direction in which a process will occur but gives no information about the speed of the process. A spontaneous process is one that occurs without outside intervention. Copyright © Cengage Learning. All rights reserved

What should happen to the gas when you open the valve?
CONCEPT CHECK! Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant. What should happen to the gas when you open the valve? The gas should spread evenly throughout the two bulbs. Copyright © Cengage Learning. All rights reserved

Calculate ΔH, ΔE, q, and w for the process you described above.
CONCEPT CHECK! Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant. Calculate ΔH, ΔE, q, and w for the process you described above. All are equal to zero. All are equal to zero. Since it is a constant temperature process, H = 0 and E = 0. The gas is working against zero pressure (evacuated bulb) so w = 0. E = q + w, so q = 0. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Consider 2.4 moles of a gas contained in a 4.0 L bulb at a constant temperature of 32°C. This bulb is connected by a valve to an evacuated 20.0 L bulb. Assume the temperature is constant. c) Given your answer to part b, what is the driving force for the process? Entropy Some students are probably aware of the concept of entropy. This is a good introduction to the concept. Copyright © Cengage Learning. All rights reserved

The Expansion of An Ideal Gas Into an Evacuated Bulb
Copyright © Cengage Learning. All rights reserved

Entropy The driving force for a spontaneous process is an increase in the entropy of the universe. A measure of molecular randomness or disorder. A measure of how many micro states can exist. Copyright © Cengage Learning. All rights reserved

Entropy Thermodynamic function that describes the number of arrangements that are available to a system existing in a given state. Nature spontaneously proceeds toward the states that have the highest probabilities of existing. Copyright © Cengage Learning. All rights reserved

The Microstates That Give a Particular Arrangement (State)

Positional Entropy A gas expands into a vacuum to give a uniform distribution because the expanded state has the highest positional probability of states available to the system. Therefore: Ssolid < Sliquid << Sgas

Interactive Example 17.1 - Positional Entropy
For each of the following pairs, choose the substance with the higher positional entropy (per mole) at a given temperature Solid CO2 and gaseous CO2 N2 gas at 1 atm and N2 gas at 1.0×10–2 atm 12

Predict the sign of ΔS for each of the following, and explain:
CONCEPT CHECK! Predict the sign of ΔS for each of the following, and explain: The evaporation of alcohol The freezing of water Compressing an ideal gas at constant temperature Heating an ideal gas at constant pressure Dissolving NaCl in water + – a) + (a liquid is turning into a gas) b) - (more order in a solid than a liquid) c) - (the volume of the container is decreasing) d) + (the volume of the container is increasing) e) + (there is less order as the salt dissociates and spreads throughout the water) Copyright © Cengage Learning. All rights reserved

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. The total energy of the universe is constant, but the entropy is increasing. Suniverse = ΔSsystem + ΔSsurroundings Copyright © Cengage Learning. All rights reserved

ΔSsurr ΔSsurr = +; entropy of the universe increases
ΔSsurr = -; process is spontaneous in opposite direction ΔSsurr = 0; process has no tendency to occur Copyright © Cengage Learning. All rights reserved

Entropy Changes in the Surroundings (ΔSsurr)
ΔSsurr is determined by flow of energy as heat Exothermic process increases ΔSsurr Important driving force for spontaneity Endothermic process decreases ΔSsurr Impact of transfer of energy as heat to or from the surroundings is greater at lower temperatures Copyright © Cengage Learning. All rights reserved 16 16

CONCEPT CHECK! For the process A(l) A(s), which direction involves an increase in energy randomness? Positional randomness? Explain your answer. As temperature increases/decreases (answer for both), which takes precedence? Why? At what temperature is there a balance between energy randomness and positional randomness? Since energy is required to melt a solid, the reaction as written is exothermic. Thus, energy randomness favors the right (product; solid). Since a liquid has less order than a solid, positional randomness favors the left (reactant; liquid). As temperature increases, positional randomness is favored (at higher temperatures the fact that energy is released becomes less important). As temperature decreases, energy randomness is favored. There is a balance at the melting point. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! Describe the following as spontaneous/non-spontaneous/cannot tell, and explain. A reaction that is: Exothermic and becomes more positionally random Spontaneous Exothermic and becomes less positionally random Cannot tell Endothermic and becomes more positionally random Endothermic and becomes less positionally random Not spontaneous Explain how temperature affects your answers. a) Spontaneous (both driving forces are favorable). An example is the combustion of a hydrocarbon. b) Cannot tell (exothermic is favorable, positional randomness is not). An example is the freezing of water, which becomes spontaneous as the temperature of water is decreased. c) Cannot tell (positional randomness is favorable, endothermic is not). An example is the vaporization of water, which becomes spontaneous as the temperature of water is increased.. d) Not spontaneous (both driving forces are unfavorable). Questions "a" and "d" are not affected by temperature. Choices "b" and "c" are explained above.

ΔSsurr The sign of ΔSsurr depends on the direction of the heat flow.
The magnitude of ΔSsurr depends on the temperature. $50 to a millionaire or a high school student Copyright © Cengage Learning. All rights reserved

ΔSsurr Heat flow (constant P) = change in enthalpy = ΔH
If the reaction is exothermic: ΔH has a negative sign ΔSsurr is positive since heat flows into the surroundings Copyright © Cengage Learning. All rights reserved

Interactive Example 17.4 - Determining ΔSsurr
In the metallurgy of antimony, the pure metal is recovered via different reactions, depending on the composition of the ore For example, iron is used to reduce antimony in sulfide ores

Interactive Example 17.4 - Determining ΔSsurr (Continued)
Carbon is used as the reducing agent for oxide ores: Calculate ΔSsurr for each of these reactions at 25°C and 1 atm

Interactive Example 17.4 - Solution
For the sulfide ore reaction, ΔSsurr is positive since this reaction is exothermic, and heat flow occurs to the surroundings, increasing the randomness of the surroundings

Interactive Example 17.4 - Solution (Continued)
For the oxide ore reaction, In this case ΔSsurr is negative because heat flow occurs from the surroundings to the system

All quantities refer to the system
Free Energy (G) ΔG = ΔH – TΔS (at constant T and P) All quantities refer to the system Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! A liquid is vaporized at its boiling point. Predict the signs of: w q ΔH ΔS ΔSsurr ΔG Explain your answers. – + As a liquid goes to vapor, it does work on the surroundings (expansion occurs). Heat is required for this process. Thus, w = negative; q = H = positive. S = positive (a gas is more disordered than a liquid), and Ssurr = negative (heat comes from the surroundings to the system); G = 0 because the system is at its boiling point and therefore at equilibrium. Copyright © Cengage Learning. All rights reserved

EXERCISE! The value of ΔHvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C. Calculate ΔS, ΔSsurr, and ΔG for the vaporization of one mole of this substance at 72.5°C and 1 atm. ΔS = 132 J/K·mol ΔSsurr = -132 J/K·mol ΔG = 0 kJ/mol S = 132 J/K·mol Ssurr = -132 J/k·mol G = 0 kJ/mol Copyright © Cengage Learning. All rights reserved

Spontaneous Reactions
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Effect of ΔH and ΔS on Spontaneity

Interactive Example 17.5 - Free Energy and Spontaneity
At what temperatures is the following process spontaneous at 1 atm? What is the normal boiling point of liquid Br2?

The vaporization process will be spontaneous at all temperatures where ΔG°is negative Note that ΔS°favors the vaporization process because of the increase in positional entropy, and ΔH°favors the opposite process, which is exothermic These opposite tendencies will exactly balance at the boiling point of liquid Br2, since at this temperature liquid and gaseous Br2 are in equilibrium (ΔG°= 0)

Interactive Example 17.5 - Solution (Continued 1)
We can find this temperature by setting ΔG°= 0 in the following equation:

At temperatures above 333 K, TΔS°has a larger magnitude than ΔH°, and ΔG°is negative Above 333 K, the vaporization process is spontaneous The opposite process occurs spontaneously below this temperature At 333 K, liquid and gaseous Br2 coexist in equilibrium

Summary of observations (the pressure is 1 atm in each case) T > 333 K The term ΔS°controls, and the increase in entropy when liquid Br2 is vaporized is dominant T < 333 K The process is spontaneous in the direction in which it is exothermic, and the term ΔH°controls

T = 333 K The opposing driving forces are just balanced (ΔH°= 0), and the liquid and gaseous phases of bromine coexist This is the normal boiling point

Entropy Changes and Chemical Reactions
Positional probability determines the changes that occur in a chemical system Fewer the molecules, fewer the possible configurations Copyright © Cengage Learning. All rights reserved 40 40

Figure 17.6 - Entropy of Water
H2O molecule can vibrate and rotate in several ways Freedom of motion leads to a higher entropy for water 41

Entropy Changes in Reactions That Involve Gaseous Molecules
Change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products If the number of product molecules is greater than the number of reactant molecules: Positional entropy increases ΔS is positive Copyright © Cengage Learning. All rights reserved 42 42

Interactive Example 17.6 - Predicting the Sign of ΔS°
Predict the sign of ΔS°for each of the following reactions Thermal decomposition of solid calcium carbonate Oxidation of SO2 in air

CONCEPT CHECK! Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The bond energy of AB is much greater than that of either reactant. Predict the signs of: ΔH ΔSsurr ΔS ΔSuniv Explain. Since the average bond energy of the products is greater than the average bond energies of the reactants, the reaction is exothermic as written. Thus, the sign of H is negative; Ssurr is positive; S is close to zero (cannot tell for sure); and Suniv is positive. – Copyright © Cengage Learning. All rights reserved

Third Law of Thermodynamics
The entropy of a perfect crystal at 0 K is zero. The entropy of a substance increases with temperature. Copyright © Cengage Learning. All rights reserved

Standard Entropy Values (S°)
Represent the increase in entropy that occurs when a substance is heated from 0 K to 298 K at 1 atm pressure. ΔS°reaction = ΣnpS°products – ΣnrS°reactants Copyright © Cengage Learning. All rights reserved

2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information:
EXERCISE! Calculate ΔS° for the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) Given the following information: S° (J/K·mol) Na(s) H2O(l) NaOH(aq) H2(g) ΔS°= –11 J/K [2(50) + 131] – [2(51) + 2(70)] = –11 J/K ΔS°= –11 J/K Copyright © Cengage Learning. All rights reserved

AP Learning Objectives, Margin Notes and References
LO 5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored by determination of (either quantitatively or qualitatively) the signs of both Ho and So, and calculation or estimation of Go when needed. LO 5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs free energy. LO 5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable. LO 5.16 The student can use Le Châtelier’s principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product. LO 5.17 The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction. LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions.

Standard Free Energy Change (ΔG°)
The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states. ΔG° = ΔH° – TΔS° ΔG°reaction = ΣnpG°products – ΣnrG°reactants Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! A stable diatomic molecule spontaneously forms from its atoms. Predict the signs of: ΔH° ΔS° ΔG° Explain. The reaction is exothermic, more ordered, and spontaneous. Thus, the sign of H is negative; S is negative; and G is negative. – – – Copyright © Cengage Learning. All rights reserved

Consider the following system at equilibrium at 25°C.
CONCEPT CHECK! Consider the following system at equilibrium at 25°C. PCl3(g) + Cl2(g) PCl5(g) ΔG° = −92.50 kJ What will happen to the ratio of partial pressure of PCl5 to partial pressure of PCl3 if the temperature is raised? Explain. The ratio will decrease. S is negative (unfavorable) yet the reaction is spontaneous (G is negative). Thus, H must be negative (exothermic, favorable). Thus, as the temperature is increased, the reaction proceeds to the left, decreasing the ratio of partial pressure of PCl5 to the partial pressure of PCl3. Copyright © Cengage Learning. All rights reserved

Interactive Example 17.11 - Calculating ΔG°
Methanol is a high-octane fuel used in high-performance racing engines Calculate ΔG °for the following reaction: The following energies of formation are provided: 52

Use the following equation:

Interactive Example 17.11 - Solution (Continued)
The large magnitude and the negative sign of ΔG°indicate that this reaction is very favourable thermodynamically

LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions. Additional AP References LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”)

Free Energy and Pressure
System under constant P and T proceeds spontaneously in the direction that lowers its free energy Free energy of a reaction system changes as the reaction proceeds Dependent on the pressure of a gas or on the concentration of species in solution Equilibrium - Point where free energy value is at its lowest Copyright © Cengage Learning. All rights reserved 56 56

Free Energy and Pressure (Continued 1)
For ideal gases: Enthalpy is not pressure-dependent Entropy depends on pressure due to its dependence on volume At a given temperature for 1 mole of ideal gas: Slarge volume > Ssmall volume Or, Slow pressure > Shigh pressure Copyright © Cengage Learning. All rights reserved 57 57

Q - Reaction quotient T - Temperature in Kelvin R - Universal gas constant ( J/K·mol) ΔG°- Free energy change at 1 atm ΔG - Free energy change at specified pressures Copyright © Cengage Learning. All rights reserved 59 59

Interactive Example 17.13 - Calculating ΔG°
One method for synthesizing methanol (CH3OH) involves reacting carbon monoxide and hydrogen gases Calculate ΔG at 25°C for this reaction where carbon monoxide gas at 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol

To calculate ΔG for this process, use the following equation: First compute ΔGₒ from standard free energies of formation

One might call this the value of ΔG°for one round of the reaction or for 1 mole of the reaction Thus, the ΔG°value might better be written as –2.9×104 J/mol of reaction, or –2.9×104 J/mol rxn Use this value to calculate the value of ΔG

ΔG°= – 2.9×104 J/mol rxn R = J/K·mol T = = 298 K Note that the pure liquid methanol is not included in the calculation of Q

Note that ΔG is significantly more negative than ΔG°, implying that the reaction is more spontaneous at reactant pressures greater than 1 atm This result can be expected from Le Châtelier’s principle

Sketch graphs of: G vs. P H vs. P
CONCEPT CHECK! Sketch graphs of: G vs. P H vs. P ln(K) vs. 1/T (for both endothermic and exothermic cases) G vs. P: A natural log graph (levels off as P increases). H vs. P: no relationship (slope of zero). lnK vs 1/T: endothermic - straight line, negative slope exothermic - straight line, positive slope Copyright © Cengage Learning. All rights reserved

The Meaning of ΔG for a Chemical Reaction
A system can achieve the lowest possible free energy by going to equilibrium, not by going to completion. Copyright © Cengage Learning. All rights reserved

LO 5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical reaction can produce large amounts of product for certain sets of initial conditions. LO 6.25 The student is able to express the equilibrium constant in terms of Go and RT and use this relationship to estimate the magnitude of K and, consequently, the thermodynamic favorability of the process. Additional AP References LO 5.18 (see Appendix 7.11, “Non-Spontaneous Reactions”) LO 6.25 (see Appendix 7.11, “Non-Spontaneous Reactions”)

Change in Free Energy to Reach Equilibrium

Interactive Example 17.15 - Free Energy and Equilibrium II
72

Calculate ΔG° from ΔG° = ΔH°– TΔS°

Therefore, ΔG° = ΔH°– TΔS° = (– 1.652×106 J) – (298 K) (– 543 J/K) = – ×106 J

Therefore, K = e601 This is a very large equilibrium constant The rusting of iron is clearly very favourable from a thermodynamic point of view

Reversible and Irreversible Processes
Reversible process: Universe is exactly the same as it was before a cyclic process Irreversible process: Universe is different after a cyclic process All real processes are irreversible Characteristics of a real cyclic process Work is changed to heat Entropy of the universe increases Copyright © Cengage Learning. All rights reserved 77 77

All real processes are irreversible.
Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy. All real processes are irreversible. First law: You can’t win, you can only break even. Second law: You can’t break even. As we use energy, we degrade its usefulness. Copyright © Cengage Learning. All rights reserved

Thermodynamics vs. Kinetics

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