Advanced Topics Advanced Topics in Chemistryin Chemistry

ThermochemistryThermochemistry

ObjectivesObjectives1· Define and apply the terms standard 1· Define and apply the terms standard state, standard enthalpy change of state, standard enthalpy change of formation, and standard enthalpy formation, and standard enthalpy change of combustionchange of combustion

2· Determine the enthalpy change of a 2· Determine the enthalpy change of a reaction using standard enthalpy reaction using standard enthalpy changes of formation and combustionchanges of formation and combustion

1/20131/2013Dr. DuraDr. Dura 22

Heat is the transfer of thermal energy between two bodies that are at different temperatures.

Standard Enthalpy Changes in Chemical Reactions

Temperature is a measure of the thermal energy.

900C400C

greater thermal energy

Temperature = Thermal Energy

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Thermochemistry is the study of heat change in chemical reactions.

The system is the specific part of the universe that is of interest in the study.

open

mass & energyExchange:

closed

energy

isolated

nothing1/20131/2013Dr. DuraDr. Dura 44

Exothermic process is any process that gives off heat – transfers thermal energy from the system to the surroundings.

Endothermic process is any process in which heat has to be supplied to the system from the surroundings.

2H2 (g) + O2 (g) 2H2O (l) + energy

H2O (g) H2O (l) + energy

energy + 2HgO (s) 2Hg (l) + O2 (g)

energy + H2O (s) H2O (l)

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Thermodynamics is the scientific study of the interconversion of heat and other kinds of energy.

State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.

Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.

energy, pressure, volume, temperature

E = Efinal - Einitial

P = Pfinal - Pinitial

V = Vfinal - Vinitial

T = Tfinal - Tinitial

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First law of thermodynamics – energy can be converted from one form to another, but cannot be created or destroyed.

Esystem + Esurroundings = 0

or

Esystem = -Esurroundings

C3H8 + 5O2 3CO2 + 4H2O

Exothermic chemical reaction!

Chemical energy lost by combustion = Energy gained by the surroundingssystem surroundings 1/20131/2013Dr. DuraDr. Dura 77

Another form of the first law for Esystem

E = q + w

E is the change in internal energy of a system

q is the heat exchange between the system and the surroundings

w is the work done on (or by) the system

w = -PV when a gas expands against a constant external pressure

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Work Done On the Systemw = F x d

w = -P V

P x V = x d3 = F x d = wFd2

V > 0

-PV < 0

wsys < 0

Work is not a state function!

w = wfinal - winitial

initial final 1/20131/2013Dr. DuraDr. Dura 99

A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at constant temperature. What is the work done in joules if the gas expands (a) against a vacuum and (b) against a constant pressure of 3.7 atm?

w = -P V

(a) V = 5.4 L – 1.6 L = 3.8 L P = 0 atm

W = -0 atm x 3.8 L = 0 L•atm = 0 joules

(b) V = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm

w = -3.7 atm x 3.8 L = -14.1 L•atm

w = -14.1 L•atm x 101.3 J1L•atm

= -1430 J

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Enthalpy and the First Law of Thermodynamics

E = q + w

E = H - PV

H = E + PV

q = H and w = -PV

At constant pressure:

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

H = H (products) – H (reactants)

H = heat given off or absorbed during a reaction at constant pressure

Hproducts < Hreactants

H < 0Hproducts > Hreactants

H > 0 1/20131/2013Dr. DuraDr. Dura 1212

Thermochemical Equations

H2O (s) H2O (l) H = 6.01 kJ

Is H negative or positive?

System absorbs heat

Endothermic

H > 0

6.01 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm.

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Thermochemical Equations

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H = -890.4 kJ

Is H negative or positive?

System gives off heat

Exothermic

H < 0

890.4 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm.

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H2O (s) H2O (l) H = 6.01 kJ

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

Thermochemical Equations

• If you reverse a reaction, the sign of H changes

H2O (l) H2O (s) H = -6.01 kJ

• If you multiply both sides of the equation by a factor n, then H must change by the same factor n.

2H2O (s) 2H2O (l) H = 2 x 6.01 = 12.0 kJ

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H2O (s) H2O (l) H = 6.01 kJ

• The physical states of all reactants and products must be specified in thermochemical equations.

Thermochemical Equations

H2O (l) H2O (g) H = 44.0 kJ

How much heat is evolved when 266 g of white phosphorus (P4) burn in air?

P4 (s) + 5O2 (g) P4O10 (s) H = -3013 kJ

266 g P4

1 mol P4

123.9 g P4

x3013 kJ1 mol P4

x = 6470 kJ

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A Comparison of H and E

2Na (s) + 2H2O (l) 2NaOH (aq) + H2 (g) H = -367.5 kJ/mol

E = H - PV At 25 0C, 1 mole H2 = 24.5 L at 1 atm

PV = 1 atm x 24.5 L = 2.5 kJ

E = -367.5 kJ/mol – 2.5 kJ/mol = -370.0 kJ/mol

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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.

C = m x s

Heat (q) absorbed or released:

q = m x s x t

q = C x t

t = tfinal - tinitial

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Chemistry in Action:

Fuel Values of Foods and Other Substances

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) H = -2801 kJ/mol

1 cal = 4.184 J

1 Cal = 1000 cal = 4184 J

Substance Hcombustion (kJ/g)

Apple -2

Beef -8

Beer -1.5

Gasoline -341/20131/2013Dr. DuraDr. Dura 1919

Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?

Establish an arbitrary scale with the standard enthalpy of formation (H0) as a reference point for all enthalpy expressions.

f

Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.

f

The standard enthalpy of formation of any element in its most stable form is zero.

H0 (O2) = 0f

H0 (O3) = 142 kJ/molf

H0 (C, graphite) = 0f

H0 (C, diamond) = 1.90 kJ/molf1/20131/2013Dr. DuraDr. Dura 2020

The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atm.

rxn

aA + bB cC + dD

H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]

H0rxn nH0 (products)f= mH0 (reactants)f-

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

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Calculate the standard enthalpy of formation of CS2 (l) given that:

C(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJrxn

S(rhombic) + O2 (g) SO2 (g) H0 = -296.1 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn

1. Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)

2. Add the given rxns so that the result is the desired rxn.

rxnC(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJ

2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -296.1x2 kJrxn

CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)

H0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn 1/20131/2013Dr. DuraDr. Dura

2222

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

H0rxn nH0 (products)f= mH0 (reactants)f-

H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]

H0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ

-5946 kJ2 mol

= - 2973 kJ/mol C6H6

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Sample Problem

Given the following thermochemical equations:

B2O3 (s) + 3 H2O (g) → 3 O2 (g) + B2H6 (g) (ΔH = 2035 kJ/mol)H2O (l) → H2O (g) (ΔH = 44 kJ/mol)H2 (g) + (1/2) O2 (g) → H2O (l) (ΔH = -286 kJ/mol)2 B (s) + 3 H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol)

Determine the ΔHf of:

2 B (s) + (3/2) O2 (g) → B2O3 (s)

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After the multiplication and reversing of the equations (and their enthalpy changes), the result is:

B2H6 (g) + 3 O2 (g) → B2O3 (s) + 3 H2O (g) (ΔH = -2035 kJ/mol)3 H2O (g) → 3 H2O (l) (ΔH = -132 kJ/mol)3 H2O (l) → 3 H2 (g) + (3/2) O2 (g) (ΔH = 858 kJ/mol)2 B (s) + 3 H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol)

Adding these equations and canceling out the common terms on both sides, we get2 B (s) + (3/2) O2 (g) → B2O3 (s) (ΔH = -1273 kJ/mol)

ObjectivesObjectives3· Define and apply the terms lattice enthalpy and 3· Define and apply the terms lattice enthalpy and electron affinityelectron affinity

4. Explain how the relative sizes and the charges of 4. Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic ions affect the lattice enthalpies of different ionic compoundscompounds

5. Construct a Born-Haber cycle for group 1 and group 5. Construct a Born-Haber cycle for group 1 and group 2 oxides and chlorides, and use it to calculate an 2 oxides and chlorides, and use it to calculate an enthalpy change (9.1.12.A.1, 9.1.12.B.1)enthalpy change (9.1.12.A.1, 9.1.12.B.1)

6· Discuss the difference between theoretical and 6· Discuss the difference between theoretical and experimental lattice enthalpy values of ionic experimental lattice enthalpy values of ionic compounds in terms of their covalent charactercompounds in terms of their covalent character

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Experimental lattice energies cannot be Experimental lattice energies cannot be determined directly. An energy cycle based on determined directly. An energy cycle based on Hess’s Law, known as the Born-Haber cycle is Hess’s Law, known as the Born-Haber cycle is used. The Key Concepts in Born-Haber Cycle used. The Key Concepts in Born-Haber Cycle are:are:

Heat of formation Δ HHeat of formation Δ Hff

Ionization Energy I.E.Ionization Energy I.E.

Electron AffinityElectron Affinity E.A. E.A.

Atomization (Sublimation) s Atomization (Sublimation) s g g

Bond enthalpy B.E.Bond enthalpy B.E.

Lattice Energy – is the enthalpy change that occurs Lattice Energy – is the enthalpy change that occurs when one mole of a solid ionic compound is separated when one mole of a solid ionic compound is separated into ions under standard conditions. Example: NaCl(s) into ions under standard conditions. Example: NaCl(s) Na+ (g) + Cl -(g) (endothermic). Na+ (g) + Cl -(g) (endothermic).

Note that in forming ionic compounds, the oppositely Note that in forming ionic compounds, the oppositely charged gaseous ions come together to form an ionic charged gaseous ions come together to form an ionic lattice – this is a very exothermic process as there is a lattice – this is a very exothermic process as there is a strong attraction between the ionsi.e. Na+ (g) + Cl -strong attraction between the ionsi.e. Na+ (g) + Cl -(g) (g) NaCl(s)NaCl(s) 1/20131/2013Dr. DuraDr. Dura 2727

Na(s) + ½ Cl2(g) NaCl(s)

Na(g) Cl(g)

Hsub ½ BDE

Na+(g) Cl-(g)

IE EA

Hf

Lattice E. = 786.5 kJ/mol

+

Born-Haber Cycle for NaCl

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Construct the Born-Haber cycle for KF and using appropriate data, calculate the electron affinity of fluorine.

Thermodynamic dataK(s) → K(g) ΔH°a = 90 kJ mol-1K(g) → K+(g) + e- IE = 419 kJ mol-1

½ F2(g) → F(g) ΔH°a = 79.5 kJ mol-1K(s) + F(g) → KF(s) ΔH°f = -569 kJ mol-1K+(g) + F-(g) → KF(s) ΔH°lattice = -821 kJ mol-1

Choose the correct answer:a) EA = 336.5 kJ mol-1b) EA = -336.5 kJ mol-1c) EA = -840.5 kJ mol-1

Answer: b

Do NOW

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Draw the Born-Haber cycle for potassium sulphide K2S.

Thermodynamic dataPotassium K Sulphur S Potassium Sulphide K2SIEI = 419 kJ mol-1 ΔH°a= 279 kJ mol-1 ΔH°f = -257 kJ mol-1ΔH°a = 78 kJ mol-1 EAI= -199.5 kJ mol-1 ΔH°lattice = -1979 kJ mol

EAII - 648.5 kJ mol-1

Practice problem

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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 0C and ice melts above 0 0C

• 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

spontaneous

nonspontaneous

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Does a decrease in enthalpy mean a reaction proceeds spontaneously?

CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) H0 = -890.4 kJ

H+ (aq) + OH- (aq) H2O (l) H0 = -56.2 kJ

H2O (s) H2O (l) H0 = 6.01 kJ

NH4NO3 (s) NH4+(aq) + NO3

- (aq) H0 = 25 kJH2O

Spontaneous reactions

ObjectivesObjectives7. State and explain the factors that 7. State and explain the factors that increase entropy in a systemincrease entropy in a system8· Predict whether the entropy change 8· Predict whether the entropy change for a given reaction or process is for a given reaction or process is positive or negative·positive or negative·

9. Calculate the standard entropy 9. Calculate the standard entropy change for a reaction using standard change for a reaction using standard entropy values (5.2.12.D.2, 9.1.12.A.1)entropy values (5.2.12.D.2, 9.1.12.A.1)

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Entropy (S) is a measure of the randomness or disorder of a system.

order SdisorderS

S = Sf - Si

If the change from initial to final results in an increase in randomness

Sf > Si S > 0

For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than

gas stateSsolid < Sliquid << Sgas

H2O (s) H2O (l) S > 0

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Processes that lead to an

increase in entropy (S > 0)

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How does the entropy of a system change for each of the following processes?

(a) Condensing water vapor

Randomness decreases Entropy decreases (S < 0)

(b) Forming sucrose crystals from a supersaturated solution

Randomness decreases Entropy decreases (S < 0)

(c) Heating hydrogen gas from 600C to 800C

Randomness increases Entropy increases (S > 0)

(d) Subliming dry ice

Randomness increases Entropy increases (S > 0)

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Entropy

State functions are properties that are determined by the state of the system, regardless of how that condition was

achieved.

Potential energy of hiker 1 and hiker 2 is the same even though they took

different paths.

energy, enthalpy, pressure, volume, temperature, entropy

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First Law of Thermodynamics

Energy can be converted from one form to another but energy cannot be created or destroyed.

Second Law of Thermodynamics

The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium

process.

Suniv = Ssys + Ssurr > 0Spontaneous process:

Suniv = Ssys + Ssurr = 0Equilibrium process:

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Entropy Changes in the System (Ssys)

aA + bB cC + dD

S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]

S0rxn nS0(products)= mS0(reactants)-

The standard entropy of reaction (S0 ) is the entropy change for a reaction carried out at 1 atm and 250C.

rxn

What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) 2CO2 (g)

S0(CO) = 197.9 J/K•molS0(O2) = 205.0 J/K•mol

S0(CO2) = 213.6 J/K•mol

S0rxn = 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]

S0rxn= 427.2 – [395.8 + 205.0] = -173.6 J/K•mol

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Entropy Changes in the System (Ssys)When gases are produced (or consumed)

• If a reaction produces more gas molecules than it consumes, S0 > 0.

• If the total number of gas molecules diminishes, S0 < 0.

• If there is no net change in the total number of gas molecules, then S0 may be positive or negative BUT S0 will be a small number.

What is the sign of the entropy change for the following reaction?

2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down,

S is negative.

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Third Law of Thermodynamics

The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.

S = k ln W

W = 1 where W is the number of

microstates. There is only one way to

arrange the atoms or molecules to form a perfect

crystal.

S = 0

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Objectives10.Predict whether a reaction or process will be spontaneous using the sign of ΔG11· Calculate ΔG for a reaction using the equation with enthalpy, temperature, and entropy and by using values of free energy change of formation, ΔGf (9.1.12.A.1) 12. Predict the effect of a change in temperature on the spontaneity of a reaction using standard entropy and enthalpy changes and the equation

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Suniv = Ssys + Ssurr > 0Spontaneous process:

Suniv = Ssys + Ssurr = 0Equilibrium process:

Gibbs Free Energy

For a constant-temperature process:

G = Hsys -TSsysGibbs free energy (G)

G < 0 The reaction is spontaneous in the forward direction.

G > 0 The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction.

G = 0 The reaction is at equilibrium.

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aA + bB cC + dD

G0rxn dG0 (D)fcG0 (C)f= [ + ] - bG0 (B)faG0 (A)f[ + ]

G0rxn nG0 (products)f= mG0 (reactants)f-

The standard free-energy of reaction (G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.

rxn

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.

f

.G0 of any element in its stable form is zero

f

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2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

G0rxn nG0 (products)f= mG0 (reactants)f-

What is the standard free-energy change for the following reaction at 25 0C?

G0rxn 6G0 (H2O)f12G0 (CO2)f= [ + ] - 2G0 (C6H6)f[ ]

G0rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ

Is the reaction spontaneous at 25 0C?

G0 = -6405 kJ < 0

spontaneous

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G = H - TS