Thermochemistry Chapters 6 and 18 TWO Trends in Nature Order Disorder High energy Low energy.

Thermochemistry

Chapters 6 and 18

TWO Trends in Nature

• Order Disorder

• High energy Low energy

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)

6.2

energy + H2O (s) H2O (l)

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 6.4

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.

6.4

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.

6.4

H2O (s) H2O (l) H = 6.01 kJ/mol ΔH = 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 mol x 6.01 kJ/mol = 12.0 kJ

6.4

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

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

Thermochemical Equations

6.4

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) Hreaction = -3013 kJ

266 g P4

1 mol P4

123.9 g P4

x3013 kJ1 mol P4

x = 6470 kJ

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/molf

6.6

6.6

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 H0 (products)

f= H0 (reactants)f

-

6.6

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.)

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 H0 (products)f= H0 (reactants)f-

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

H0rxn = [ 12 × -393.5 + 6 × -285.8 ] – [ 2 × 49.04 ] = -6535 kJ

-6535 kJ2 mol

= - 3267 kJ/mol C6H6

6.6

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 kJrxn6.6

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

The enthalpy of solution (Hsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.

Hsoln = Hsoln - Hcomponents

6.7

Which substance(s) could be used for melting ice?

Which substance(s) could be used for a cold pack?

The Solution Process for NaCl

Hsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol6.7

Energy Diagrams

Exothermic Endothermic

(a) Activation energy (Ea) for the forward reaction

(b) Activation energy (Ea) for the reverse reaction

(c) Delta H

50 kJ/mol 300 kJ/mol

150 kJ/mol 100 kJ/mol

-100 kJ/mol +200 kJ/mol

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

order SdisorderS

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

S > 0

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

Ssolid < Sliquid << Sgas

H2O (s) H2O (l) S > 018.3

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:

18.4

Entropy Changes in the System (Ssys)

aA + bB cC + dD

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

S0rxn S0(products)= S0(reactants)-

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

rxn

18.4

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

Entropy Changes in the System (Ssys)

18.4

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.

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

18.2

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.

18.5

G = H - TS

18.5

18.5

aA + bB cC + dD

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

G0rxn G0 (products)f= G0 (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

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

G0rxn G0 (products)f= G0 (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

18.5

Recap: Signs of Thermodynamic ValuesRecap: Signs of Thermodynamic Values

Negative Positive

Enthalpy (ΔH) Exothermic Endothermic

Entropy (ΔS) Less disorder More disorder

Gibbs Free Energy (ΔG)

Spontaneous Not spontaneous

Gibbs Free Energy and Chemical Equilibrium

G = G0 + RT lnQ

R is the gas constant (8.314 J/K•mol)

T is the absolute temperature (K)

Q is the reaction quotient

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

18.6

G0 = RT lnK

18.6

The specific heat (s) [most books use lower case c] 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 = ms

Heat (q) absorbed or released:

q = mst

q = Ct

t = tfinal - tinitial

6.5

How much heat is given off when an 869 g iron bar cools from 940C to 50C?

s of Fe = 0.444 J/g • 0C

t = tfinal – tinitial = 50C – 940C = -890C

q = mst = 869 g x 0.444 J/g • 0C x –890C = -34,000 J

6.5

Constant-Pressure Calorimetry

No heat enters or leaves!

qsys = qwater + qcal + qrxn

qsys = 0

qrxn = - (qwater + qcal)

qwater = mst

qcal = Ccalt

6.5

Reaction at Constant PH = qrxn

6.5

The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure.

The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm.

11.8

Phase Changes

The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure.

The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature.

11.8

Where’s Waldo?Can you find…

The Triple Point?

Critical pressure?

Critical temperature?

Where fusion occurs?

Where vaporization occurs?

Melting point (at 1 atm)?

Boiling point(at 6 atm)?

Carbon Dioxide

Mel

ting

11.8F

reez

ing

H2O (s) H2O (l)

The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium

Sub

limat

ion

11.8

Dep

ositi

on

H2O (s) H2O (g)

Molar heat of sublimation (Hsub) is the energy required to sublime 1 mole of a solid.

Hsub = Hfus + Hvap

( Hess’s Law)

Molar heat of fusion (Hfus) is the energy required to melt 1 mole of a solid substance.

11.8

11.8

Sample Problem• How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C?

Step 1: Heat the ice Q=mcΔT

Q = 36 g x 2.06 J/g deg C x 8 deg C = 593.28 J = 0.59 kJ

Step 2: Convert the solid to liquid ΔH fusion

Q = 2.0 mol x 6.01 kJ/mol = 12 kJ

Step 3: Heat the liquid Q=mcΔT

Q = 36g x 4.184 J/g deg C x 100 deg C = 15063 J = 15 kJ

Sample Problem• How much heat is required to change 36 g of

H2O from -8 deg C to 120 deg C?

Step 4: Convert the liquid to gas ΔH vaporization

Q = 2.0 mol x 44.01 kJ/mol = 88 kJ

Step 5: Heat the gas Q=mcΔT

Q = 36 g x 2.02 J/g deg C x 20 deg C = 1454.4 J = 1.5 kJ

Now, add all the steps together

0.59 kJ + 12 kJ + 15 kJ + 88 kJ + 1.5 kJ = 118 kJ