honors chemistry unit 12: thermodynamics - notes ch 6, 9 ... · unit 12: thermodynamics - notes...

Unit 12: Thermodynamics - NotesHonors Chemistry Ch 6, 9.10, & 18

Section 6.1: The Nature of Energy and Types of Energy Objectives:

Identify, define, and explain: energy, work, radiant energy, thermal energy, chemical energy, gravitational potential energy, and the Law of Con-servation of Energy.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp.178.

THERMODYNAMICS - energy changes and transfers associated with chemical reactions and the tech-niques used to experimentally measure the energies

Energy: a system’s capacity to do work - or - ability to move an object against an opposing force

Three main classes of energy: kinetic (dynamic) energy - energy an object possesses by virtue of its motion potential (static) energy - energy an object possesses by virtue of its position radiant (electromagnetic) energy - energy a wave possesses by virtue of its frequency

Sub-classes: chemical energy - kinetic and potential energy associated with the breaking and formation of bonds thermal energy - kinetic and potential energy associated with random particle motion heat - thermal energy transferred between objects at different temperatures

Law of Conservation of Energy • Energy is neither created nor destroyed; its may be transferred from one part of the universe to another or it may be converted from one form to another. • The total quantity of energy in the universe is assumed constant.

Section 6.2: Energy Changes in Chemical Reactions Objectives:

Identify, define, and explain: heat, temperature, system, surroundings, closed system, open system, isolated system, exothermic process, and endothermic process. Classify processes as endothermic or exothermic.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp.178.

Heat: transfer of thermal energy between two different bodies that are at different temperatures Heat always flows from the body of higher thermal energy concentration to the body of lower thermal energy concentration. Heat is measured in joules (or calories or other unit of energy).

Temperature is a measure of the thermal energy of a body. Absolute temperature is directly proportional to the average kinetic energy of the molecules of a body. Temperature is measured in degrees Celsius or Kelvin (or other unit of temperature).

Temperature is NOT equal to thermal energy.

To study energy changes in chemical reactions, the universe is divided into the system (where the reaction/process occurs or the bodies involved in the reaction/process) and the surroundings (everywhere except the system.

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Three types of systems: • Open system - mass and energy in the system are ex-

changed with the surroundings • Closed system - only energy from the system is exchanged

with the surroundings • Isolated system - neither mass nor energy from the system

is exchanged with the surroundings

Exothermic versus Endothermic

Exothermic process - a system produces heat that is re-leased into the surroundings; heat is a product. This allows a system to achieve a lower energy state; this is energetically favorable (favorable from an enthalpy case).

Endothermic process - the process occurring in the sys-tem consumes heat from the surroundings; heat is a reactant. In this case, the system achieves a higher energy state; this is not favorable from an energetic (enthalpy) case.

6.2 Review of Concept Classify each of the following as an open system, closed system, or an isolated system. Justify your choice.

(a) Soup kept in a sealed container.

(b) A student reading in her/his dorm room.

(c) Air inside a tennis ball.

Section 6.3: Introduction to Thermodynamics Objectives:

Identify, Identify, define, and explain: thermodynamics, state of a system, state functions, First Law of Thermodynamics, internal energy, heat, work, and nonstate (path) functions. Restate the First Law of Thermodynamics. Classify properties as state functions or nonstate (path) functions. Name sign conventions for heat and work. Apply heat and work relationships for gas-phase problems. Define H in terms of E, P, and V. Calculate the internal energy, given thermochemical equations.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp. 181-187.

Thermodynamics involves changes in the state of a system which is defined by the values of al relevant macroscopic properties, for ex-ample composition, energy, temperature, pressure, and volume.

State function - properties determined by the state of the system, regardless of how that condition was achieved

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Ope

n

Closed

Isolated


Examples of state functions: energy, pressure, volume, temperature, en-thalpy, entropy, free energy, internal energy Usually designated by a capital letter: E, P, V, T, H, S, G, U

Path function (nonstate) functions - properties that depend on how the con-dition or state was achieved, in other words, they depend on the path taken. Usually designated by a lower case letter. Examples: heat (q) and work (w)

Internal Energy - the energy in a system arising from the relative positions and interactions of its parts. Internal energy has two components: kinetic energy and potential energy ⚛︎Kinetic energy component vs Potential energy component

Impossible to measure all the contributions to internal energy accurately. Cannot calculate the total energy of a system.

What gets measured/calculated? The CHANGE in the energy of a system!!!

First Law of Thermodynamics - energy can be converted from one form to another, but cannot be created nor destroyed. ΔUsystem + ΔUsurroundings = 0 or ΔUsystem = −ΔUsurroundiungs

Consider: S(s) + O2(g) → SO2(g)

Another form of the 1st Law ∆U = q + w q = heat exchange between system and surroundings q = m ΔT s

w = work done on (or by) the system w = F x d

Sign Convention for Work and Heat

Consider the change in a gas’ volume at constant pressure:

Remember, work (w) is not a state function, so ∆w ≠ wfinal − winitial. Example6.1

workdonebysystemonsurroundings (−) heatabsorbedbysystemfromsurroundings (+)

workdoneonsystembysurroundings (+) heatabsorbedbysurroundingsfromsystem (−)

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w = F × d but P = F

A= F

d2and V = d3

P× ΔV = F

d2× d3 = F × d = w so w = −PΔV

If ΔV > 0, then PΔV < 0 and w < 0


Practice Exercise 1: A gas expands from 264mL to 971mL at constant temperature. Calculate the work done (in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 4.00 atm.

Example6.2Practice Exercise 2: A gas expands and does P-V work on the surroundings equal to 279J. At the same time it absorbs 216J of heat from the surroundings. What is the change in energy of the system?

Section 6.3 - Review of Concept: Two ideal gases at the same temperature and pressure are placed in two equal-vol-ume containers. One container has a fixed volume while the other is a cylinder fitted with a weightless moveable piston. Initially, the gas pressures are equal to the external atmospheric pressure. The gases are then heated with a Bunsen burner. What are the signs of q and w for the gases under these conditions?

Section 6.4: Enthalpy of Chemical Reactions Objectives:

Identify, define, & explain: enthalpy, enthalpy of reaction, thermochemical equations, and change in enthalpy versus change in internal energy. Use thermochemical equations and stoichiometry to determine the heat change of a chemical reaction.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp.187-193.

Applying the 1st Law to processes at different conditions

Constant Volume - Most common laboratory case If ΔV = 0, then no P-V work will result, so . . . ΔU = q – P ΔV becomes ΔU = qv where the “v” sub-script indicating constant volume. So, at constant volume, change in internal energy = change in heat, ΔU = qv.

Constant Pressure Most reactions occur under constant pressure conditions – usually atmospheric pressure.

• System produces an increase in moles of gas –

• System consumes more moles of gas than are produced –

• No change in moles of gas –

For constant pressure situations:∆U = q + w = qp – P ∆V so qp = ∆U + P ∆Vwhere subscript “p” indicates constant pressure. And at constant pressure, ΔH = ΔU + P ΔV. So, at constant pressure, qp = ΔH.

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure. ΔH = H(products) − H(reactants)

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∆H = heat given released or absorbed during a reaction or process at constant pressure.

Thermochemical Equation • The stoichiometric coefficients always refer to the number of moles of a substance:

H2O(s) → H2O(l) ΔH = 6.01 kJ/mol • If you reverse a reaction, the sign of ΔH changes:

H2O(l) → H2O(s) ΔH = −6.01 kJ/mol • If you multiple both sides of the equation by a factor, n, then ΔH must change by the same factor, n:

2 H2O(s) → 2 H2O(l) ΔH = 2 × 6.01 kJ/mol = 12.0 kJ/mol • The physical states of all reactants and products must be specified in thermochemical equations:

Example6.3 Practice Exercise 3: Calculate the heat evolved when 266g of white phosphorus (P4) burns in air according to the following:

P4(s) + 5 O2(g) → P4O10(s) ΔH = −3013kJ/mol

Internal Energy versus Enthalpy Both ∆U and ∆H are associated with a reaction Both ∆U and ∆H measure energy changes But . . . The energy changes are measured under different conditions

Constant-volume: qv = ∆U Constant pressure: qp = ∆H

Example6.4Practice Exercise 4: What is ΔU for the formation of 1 mole of CO at 1.00 atm and 25oC?

C(graphite) + ½ O2(g) → CO(g) ΔH = −110.5 kJ/mol

Section 6.4 - Review of Concept: Which of the constant-pressure processes shown here has the smallest difference between ΔU and ΔH? (a) Water → water vapor (b) Water → ice (c) Ice → water vapor

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Is ∆H negative or positive?

Does it absorb or release heat?

Is it endothermic or exothermic?

Is ΔH > 0 or is ΔH < 0?


Section 6.5: Calorimetry - A Little Bit of a Review Objectives:

Identify, define, and explain: calorimetry, specific heat, and heat capacity. Perform calculations involving specific heat, mass, and temperature change. Sketch the main components of a bomb calorimeter. Determine the heats of reactions, given experimental data collected by calorimetry.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp.193-198.

⚛︎The specific heat (s) of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius.

⚛︎The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius.

Example6.5Practice Exercise 5: An iron bar has a mass of 869J cools from 94.0oC to 5.0oC. Calculate the heat (in kilojoules) by the metal. s (iron) = 0.444 J/goC

Constant-Volume Calorimetry Constant-Pressure Calorimetry

x

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Heat(q)absorbedorreleased:q=mxsxΔTq=CxΔT

ΔT=Tfinal−TiniBal

qsys=qwater+qbomb+qrxnqsys=0

qrxn=−(qwater+qbomb)qwater=mxsxΔTqbomb=CbombxΔT

ReacBonatConstantC∆H≠qrxnbutΔH≈qrxn

qsys=qwater+qcal+qrxnqsys=0

qrxn=−(qwater+qcal)qwater=mxsxΔTqcal=CcalxΔT

ReacBonatConstantP∆H=qrxn

heatgained+heatlost=0


Example6.6 Practice Exercise 6: A quantity of 1.922g methanol (CH3OH) was combusted in a constant-volume bomb calorimeter. Conse-quently, the temperature of the water rose by 4.20oC. If the heat capacity of the bomb plus water was 10.4 kJ/oC, calculate the molar heat of the combustion of methanol?

Example6.7 Practice Exercise 7: A 30.14g stainless steel ball bearing at 117.82oC is placed in a constant-pressure calorimeter containing 120.0mL of water as 18.44oC. If the specific heat of the ball bearing is 0.474 J/goC. Calculate the final temperature of the water. Assume the calorimeter to have negligible heat capacity.

Example6.8 Practice Exercise 8: A quantity of 4.00x102 mL of 0.600M HNO3 is mixed with 4.00x102 mL of 0.300M Ba(OH)2 in a con-stant-pressure calorimeter with negligible heat capacity. The initial temperature of both solutions is the same at 18.46oC. What is the final temperature of the solution? NOTE: For H+(aq) + OH−(aq) → H2O(l), the heat of neutralization is −56.2 kJ/mol.

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Section 6.5 Review of Concept: A 1-gram sample of aluminum and a 1-g sample of iron are heated from 50oC to 100oC. Which metal has absorbed the greater amount of heat? Aluminum Cp = 0.900 J/goC Iron Cp = 0.444 J/goC

Section 9.10: Bond Enthalpy Objectives:

Identify, define, and explain: Bond (dissociation) enthalpy. Use Lewis structures and bond energies to predict heats of reaction. Rationalize why ΔH for breaking chemical bonds is positive and the formation of chemical bonds is negative.


Bond Enthalpy or Bond Dissociation Enthalpy (Energy) • The enthalpy change required to break a particular bond in 1 mole of gaseous molecules • It is a measure of the stability of a molecule • The value for bond enthalpy is always POSITIVE since breaking bonds always requires an input of energy (endothermic). • Conversely, the formation of bonds releases energy (exothermic) since it is a stabilizing factor and has a NEGATIVE value.

x

x

x

Using Bond Enthalpies in Thermochemistry • Using average bond energies, it is possible to predict the approximate enthalpy of a reaction (∆Hrxn). • Breaking bonds always requires an INPUT of energy • Forming bonds always RELEASES energy • To find an approximate ∆Hrxn, count the total number of bonds broken and the total number of bonds formed in the reaction

and add all corresponding energy changes, taking into account if the energy is an input or a release.

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ΔHo = total energy input − total energy release ∆Ho = Σ BE(reactants) − Σ BE(products) just using the absolute values OR ∆Ho = Σ BE(reactants) + Σ BE(products) knowing that bond breaking (reactants) is (+) and bond formation (products) is (−).

Example9.13

Practice Exercise 9: For the reaction H2(g) + C2H4(g) → C2H6(g) . . . (a) estimate the enthalpy of reaction and (b) calculate the enthalpy of reaction using ∆Hof for each component in the reaction. Note: ΔHof C2H4 = 52.3kJ/mol & ΔHof C2H6 = -84.7kJ/mol

Section 9.10 Review of Concept: Why does ΔHorxn calculated using bond energies not always agree with that calculated using the ΔHof values?

Section 6.6: Standard Enthalpy of Formation and Reaction Objectives:

Identify, define, and explain: standard enthalpy of formation, standard state, standard enthalpy of reaction, Hess’ Law. Calculate standard enthalpy of reaction, given the standard enthalpy of formations for products and reactants. Apply Hess’s Law to determine ΔHorxn.


Standard Enthalpy of Formation = ΔHof established reference point for all enthalpy expressions ΔHof - heat change that results when one mole of a compound is formed from its elements in their standard state

Practice Exercise 10: Consider the following: (a) Consider the following equation and enthalpy change: C2H4(g) + H2(g) → C2H6(g) ΔH = −137 kJ Does −137 kJ

represent a molar heat of formation? Why or why not? Determine the ΔHform of this equation.

(b) Consider the following equation and enthalpy change: 2 Na(s) + Cl2(g) → 2 NaCl(s) ΔH = −822.2 kJ Does −822.2 kJ represent a molar heat of formation? Why or why not? Determine the ΔHform of this equation.

(c) Consider the following equation and enthalpy change: NO(g) → ½ N2(g) + ½ O2(g) ΔH = −90.4 kJ Does −90.4 kJ represent a molar heat of formation? Why or why not? Determine the ΔHform of this equation.

(d) Consider the following equation and enthalpy change: H2(g) + ½ O2(g) → H2O(g) ΔH = −244 kJ Does −244 kJ repre-sent a molar heat of formation? Why or why not? Determine the ΔHf of this equation?

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Compounds with high negative values of enthalpies of formations are very . . . ??? ΔHf of CO2 = -95.05 kcal/mol

Compounds with high positive values of enthalpies of formation are very . . . ??? ΔHf of HgC2N2O2 (mercury fulminate) = 64 kcal/mol

enthalpy of combustion (ΔHc or ΔHcomb): enthalpy change for the complete combustion of 1 mole of a substance; it is always negative (representing an exothermic reaction).

Standard conditions for thermodynamic processes are, by convention, 1 atm of pressure and 25oC.

Standard conditions are indicated by placing a o next to the quantity. ∆Hf = heat of formation but . . . ∆Hfo = standard heat of formation

KEY: The standard enthalpy of formation of any element in its standard/most stable form is ZERO.

ΔHof of O2(g) = 0 but ΔHof of O3(g) = 142 kJ/mol

ΔHof of C, graphite = 0 but ΔHof of C, diamond = 1.90 kJ/mol

Why does ΔHof = 0 for an element in its standard state?

Enthalpy changes for reactions can be obtained by simply subtracting the heats of formation of the reactants from the heats of for-mation of the products. Be sure to multiply the heats of formation by the coefficient of the compound involved.

For the following reaction: aA + bB → cC + dD ΔHorxn = [cΔHof (C) + dΔHof (D)] – [aΔHof (A) + bΔHof (B)]

In other words: ΔHorxn = Σ nΔHof (products) –Σ mΔHof (reactants) Example6.8

Practice Exercise 11: Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. Calculate the heat released (in kilojoules) per gram of the compound reacted with oxygen. The standard enthalpy of formation of benzene is 49.04 kJ/mol.

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Section 6.6 Review of Concept: Explain why reactions involving reactant compounds with positive ΔHof values are generally more exothermic than those with negative ΔHof values.

HESS’ LAW WORK - HERE

Section 18.1: The Three Laws of Thermodynamics Objectives:

State the three laws of thermodynamics.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp. 632.

• First Law: Energy can be converted from one form to another but it cannot be created nor destroyed. • Second Law: explains why chemical process tend to favor one direction over the other. • Third Law: extension of the second law – The entropy of a crystalline substance is zero and 0 K.

More to come on 2nd and 3rd laws.

Section 18.2: Spontaneous Processes Objectives:

Provide examples of spontaneous processes. Provide examples of endothermic spontaneous processes.

Text Reference: Chang and Goldsby (Chemistry: The Essential Concepts, McGraw-Hill, 2014) pp. 632-633. Spontaneous reaction (or process) - reaction that takes place without a continual supply of energy from outside the system • Goal in thermodynamics - predict the spontaneity of a reaction

Spontaneous Physical and Chemical Processes • A waterfall runs downhill • A lump of sugar dissolves in a cup of coffee • At 1 atm, water freezes below 0 oC and ice melts above 0 oC • Heat flows from a hotter object to a colder object • A gas expands in an evacuated bulb • Iron exposed to oxygen and water forms rust

______________________________ ______________________________

Enthalpy related to spontaneity . . .

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spontaneous

nonspontaneous


NOTE: Spontaneity is NOT related to the speed or rate of a chemical reaction or process. Thermodynamics indicates IF a reac-tion will occur, but indicate NOTHING about how fast the process occurs (that is kinetics).

Does a decrease in enthalpy mean a reaction proceeds spontaneously?

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

H2O(s) → H2O(l)

NH4NO3(s) → NH4+(aq) + NO3−(aq) a chemical cold pack

Section 18.3: Entropy Objectives:

Identify, define, and explain: entropy (S), microstates, & standard entropy. Define entropy in terms of microstates. Just why Sosteam > Sowater. Predict the sign of ∆S for common processes. Describe what is meant by ∆Suniverse, ∆Ssystem, and ∆Ssurroundings.


A second factor used to determine spontaneity is ENTROPY, symbolized S (unit J/mol-K).

PREVIOUSLY . . . Entropy was considered a measure of disorder or randomness of a system. The entropy (S) is a state function that increases in value as the disorder, or randomness of the system increases.

“Order” and “disorder” in chemical systems are discernible at the molecular level. Crystalline solids are highly ordered with molecules or ions occupying fixed lattice sites, and with a unit cell repeated identically over and over again. Liquids are less ordered than solids because the lattice has broken down, and molecules or ions have kinetic energy, and are in random motion. Gases are even more random than liquids. Upon vaporization, the volume increases almost 1000-fold and is more difficult to locate the position of any one molecule.

CURRENTLY . . . entropy is a measure of of how spread out or how dispersed the energy of a system is among the different possi-ble ways that system can contain energy. The greater the dispersal, the greater the entropy. Most processes are accompanied by a change in entropy. Dispersal of energy among the various states include energy states associated with translational, rotational, and vibrational motion of the molecules/atoms/ions. En-tropy is a state function.

The entropy of a substance increases as its molecules are distributed over an increasingly greater volume. For any given substance, . . .

Microstates and Entropy Microstate - a specific microscopic configuration of a thermodynamic system that the system may occupy with a certain probability in the course of its thermal fluctuations. In this description, microstates appear as different possible ways the system can achieve a partic-ular macrostate.

W = # of microstates

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https://en.wikipedia.org/wiki/Thermodynamic_system

https://en.wikipedia.org/wiki/Thermal_fluctuations

https://en.wikipedia.org/wiki/Thermal_fluctuations


Changes in Entropy

A system with fewer microstates (smaller W) among which to spread its energy (small dispersal) will have smaller entropy.

A system with more microstates (larger W) among which to spread its energy (large dispersal) will have larger entropy.

Processes that Lead to an INCREASE in Entropy

Example18.1 Practice Exercise 12: How does the entropy of the system change for each of the following processes? (a) Condensing water vapor

(b) Forming sucrose crystals from a supersaturated solution

(c) Heating hydrogen gas from 60oC to 80oC

(d) Subliming dry ice

Section 18.3 Review of Concept: For which of the following physical changes is ΔS positive? (a) Condensing ether vapor

(b) Melting iron

(c) Subliming solid iodine

(d) Freezing benzene

Section 18.4: The Second Law of Thermodynamics Objectives:

Identify & define: standard entropy of reaction. State & explain the 2nd Law of Thermodynamics. State & explain the 3rd Law of Thermodynamics. Relate ΔSuniverse for spontaneous processes to processes at equilibrium. Calculate ΔSreaction. Mathematically relate ΔSsurroundings to ΔHsystem and temperature.


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The Second Law of Thermodynamics: • The state of entropy of the entire universe, as an isolated system, will always increase over time. • The changes in the entropy in the universe can never be negative. • The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. A nonspontaneous

process can be made to occur, but at the expense of an increase in entropy elsewhere in the universe.

Entropy Changes in the System (ΔSsys) The standard entropy of reaction (ΔSsys) is the entropy change for a reaction carried out at 1 atm and 25oC. Consider the following reaction: a A + b B → c C + d D ΔSorxn = [c So(C) + d So(D)] − [a So(A) + b So(B)] or ΔSorxn = Σn So(products) − Σm So(reactants)

Example18.2Practice Exercise 13: Calculate the standard enthalpy change for the following reactions at 25oC. (a) 2 CO(g) + O2(g) → 2 CO2(g)

(b) 3 O2(g) → 2 O3(g)

(c) 2 NaHCO(s) → Na2CO3(s) + H2O(l) + CO2(g)

Factors that influence the amount of entropy that a system has in a particular state: 1. A liquid has a higher entropy than the solid from which it is formed.

2. A gas has a higher entropy than the liquid from which it is formed.

3. Increasing the temperature of a substance increases its entropy.

4. System progresses from fewer to more moles of a gas

5. System progresses from simpler to more complex molecules

6. System progresses from shorter to longer molecules

7. System progresses from ionic solids with STRONG attractions to ionic solids with weaker attractive forces

8. System progresses from independent solute and solvent particles to solutions (with hydrated particles)

Example18.3Practice Exercise 14: Qualitatively, discuss the sign of the entropy change expected for each of the following processes: (a) I2(s) → 2 I(g)

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https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Thermodynamics/Fundamentals_of_Thermodynamics/A_System_and_Its_Surroundings#Closed_System


(b) 2 Zn(s) + O2(g) → 2 ZnO(s)

(c) N2(g) + O2(g) → 2 NO(g)

Section 18.4 Review of Concepts: Consider the gas-phase reaction of A2 (dark) and B2 (light) to form AB3. (a) What is the balanced equation for the reaction?

(b) What is the sign of ΔS for the reaction? Why?

Entropy changes in the surroundings Consider the change in the entropy of the surroundings for an exothermic process and an endothermic process. Explain why the changes is so . . .

x x

Third Law of Thermodynamic Third Law of Thermodynamics: A completely ordered pure crystalline solid has an entropy of 0 at 0K.

Question: The standard enthalpy of formation for elements in their stan-dard states is zero. Yet the entropy of elements in their standard states is NOT zero. Why is this so?

Changes in entropy values S = k ln W where W = number of microstates

When T = 0K, there is only 1 microstate so S = k ln(1) = 0

But as T increases, W also increases, and S is a positive value.

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Section 18.5: Gibbs Free Energy Objectives:

Identify, Identify, define, and explain: Gibbs free energy, free energy, standard free energy change, standard free energy of formation. Express Gibbs free energy in terms of H, T, & S. Predict if a reaction is spontaneous, spontaneous in reverse, or in equilibrium, from the sign of ∆G. Calculate ∆G from data. Predict the direction of a spontaneous reaction given ∆H, ∆S, and T. Calculate ∆S for phase changes, given ∆Hfusion or ∆Hvaporization and melting/boiling temperatures.


Gibbs Free Energy = ΔG The Gibbs free energy, expressed in terms of enthalpy and entropy, refers only to the system, and can be used to predict spontaneity.

The Gibbs free energy change is equal to the maximum possible work that can be obtained from a process. Any process that occurs spontaneously can be utilized to perform useful work. The Gibbs free energy of a system will decrease (ΔG < 0) in a spontaneous process because, as work is done by the system, the system loses its free energy. The Gibbs free energy of a system will increase (ΔG > 0) in a nonspontaneous process because, to make the nonspontaneous process occur, free energy must be absorbed from the sur-roundings, and the system gains energy.

If ΔG < 0 the process is spontaneous in the forward direction. If ΔG > 0 the process is nonspontaneous in the forward direction, but would be spontaneous in the reverse direction (ΔG < 0). If ΔG = 0 the process is in equilibrium.

Calculating Gibbs Free Energy ΔGsys = ΔHsys − T ΔSsys

for a process at constant temperature and pressure.

The standard free-energy of reaction (∆Go) is the free-energy change for a reaction when it occurs under standard-state condi-tions.

Consider the following reaction: a A + b B → c C + d D ΔGorxn = [c ΔGo(C) + d ΔGo(D)] − [a ΔGo(A) + b ΔGo(B)] or ΔGorxn = Σn ΔGo(products) − Σm ΔGo(reactants)

Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states.

ΔGo of any element in its standard state (most stable form) at standard conditions is ZERO.

Example18.4 Practice Exercise 15 Calculate the standard free-energy for the following reactions at 25oC. (a) H2(g) + Br2(l) → 2 HBr(g)

(b) 2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(l)

Section 18.5 Review of Concept #1: (a) Under what circumstances will an endothermic reaction occur spontaneously?

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(b) Explain why, in many reactions in which both the reactant and product species are in the solution phase, ΔH often gives a good hint about the spontaneity of the reaction at room temperature?

Temperature and Free Energy Changes Two factors that govern a process:

If both enthalpy and entropy are favorable for a process, what happens?

If both enthalpy and entropy are unfavorable for a process, what happens?

If one is favorable and one is unfavorable, what happens?

From the equation ΔG = ΔH − T ΔS, it is evident that temperature plays a role influences the spontaneity of a process.

In terms of ENTHALPY: spontaneous/favorable: nonspontaneous/unfavorable:

In terms of ENTROPY: spontaneous/favorable: nonspontaneous/unfavorable:

How Does Temperature Relate to Spontaneity?

Practice Exercise 16: For a given reaction, ΔH = −8.84 kJ/mol, ΔS= −14.3 J/mol-K, and T = 300.oC. (a) Calculate the ΔG at 300.oC.

Since ΔG is ______________________, the change ______________________ take place spontaneously at 300.oC.

Situation ΔH and ΔS Signs TEMPERATRURE/Comment

1 favorable ΔH = _____

favorable ΔS = _____

2 unfavorable ΔH = _____

unfavorable ΔS = _____

3 favorable ΔH = _____

unfavorable ΔS = _____

4 unfavorable ΔH = _____

favorable ΔS = _____

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(b) Use the above values for the reaction occurring at a temperature of 400.oC. Calculate ΔG.

Since ΔG is ______________________, the change ______________________ take place spontaneously at 400.oC.

(c) At what temperature is this process in equilibrium?

Practice Exercise 17: Is the following reaction spontaneous under standard conditions? 4 KClO3(s) → 3 KClO4(s) + KCl(s)

KClO3(s) ΔHfo = −397.7 kJ/mol ΔSfo = 143.1 J/mol-K KClO4(s) ΔHfo = −432.8 kJ/mol ΔSfo = 151.0 J/mol-K KCl(s) ΔHfo = −436.7 kJ/mol ΔSfo = 82.6 J/mol-K

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Practice Exercise 18: The molar heats of fusion and vaporization of argon are 1.3 kJ/mol and 6.3 kJ/mol, respectively, and argon’s melting and boiling points are -190oC and -186oC, respectively. Calculate the energy changes for fusion and vaporization.

Section 18.5 Review of Concept #2: Consider the sublimation of iodine (I2) at 45oC in a closed flask, shown to the right. If the enthalpy of sublimation is 62.4 kJ/mol, what is the ∆S for sublimation?

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honors chemistry unit 12: thermodynamics - notes ch 6, 9 ... · unit 12: thermodynamics - notes...

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