Thermochemistry

Chapters 6 and11

TWO Trends in Nature

• ____________ Disorder

• ______ energy ____ energy

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

________ process is any process in which heat has to be _____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 _______flow into or out of a system in a process that occurs at ______ 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

_____thermic

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

_____thermic

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 _____________always refer to the number of _______ of a substance

Thermochemical Equations

• If you ________ 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 ________ at a pressure of 1 atm.

f

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

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 (_______)f= H0 (_______)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 _______of steps.

(Enthalpy is a ______ 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

= - ________ 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 =_____ 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 = ______ J

1 ____ = 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

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 > 0__________ process:

Suniv = Ssys + Ssurr = 0__________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] = ?????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 _______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 ______.

Spontaneous Physical and Chemical Processes

• A waterfall runs downhill

• A lump of sugar dissolves in a cup of coffee

• 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 (?)f= G0 (?)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 _____of the compound is formed from its elements in their standard states.

f

G0 of any element in its stable form is ______.

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 ] = -_______ kJ

Is the reaction spontaneous at 25 0C?

G0 = -_____kJ?0

____________

18.5

Recap: Signs of Thermodynamic ValuesRecap: Signs of Thermodynamic Values

Negative Positive

Enthalpy (ΔH) ____thermic _____thermic

Entropy (ΔS) _____ disorder

____ disorder

Gibbs Free Energy (ΔG)

_________ ____________

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.

Heat (q) absorbed or released:

q =_______

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 = ______ 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 __________________is equal to the external pressure.

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

11.8

Phase Changes

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

The critical pressure (Pc) is the minimum pressure that must be applied to bring about ________ at the _______ 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

____

___

11.8__

____

_

H2O (s) H2O (l)

The_______ point of a solid or the ________ point of a liquid is the temperature at which the solid and liquid phases coexist in ________

____

____

___

11.8

____

____

___

H2O (s) H2O (g)

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

Hsub = H??? + H????

( Hess’s Law)

Molar heat of fusion (Hfus) is the energy required to ____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 _____ J/g deg C x 8 deg C = ___ J = ___ kJ

Step 2: Convert the solid to liquid ΔH fusion

Q = 2.0 mol x ______ kJ/mol = ____ kJ

Step 3: Heat the liquid Q=mcΔT

Q = 36g x 4.184 J/g deg C x 100 deg C = _____ J = ___ 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 ________kJ/mol = _____ kJ

Step 5: Heat the gas Q=mcΔT

Q = 36 g x ______ J/g deg C x 20 deg C = _______ J = ___ kJ

Now, add all the steps together

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