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Thermochemistry. Chapters 6 and11. TWO Trends in Nature. ____________  Disorder   ______ energy ____ energy . 2H 2 ( g ) + O 2 ( g ) 2H 2 O ( l ) + energy. H 2 O ( g ) H 2 O ( l ) + energy.

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thermochemistry

Thermochemistry

Chapters 6 and11

two trends in nature
TWO Trends in Nature
  • ____________ Disorder

 

  • ______ energy ____ energy

slide3

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

H2O (g) H2O (l) + energy

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

energy + H2O (s) H2O (l)

___________ 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.

6.2

slide4

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

DH = H (products) – H (reactants)

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

Hproducts ? Hreactants

Hproducts ? Hreactants

DH < 0

DH > 0

6.4

slide5

H2O (s) H2O (l)

DH = 6.01 kJ

Thermochemical Equations

Is DH negative or positive?

System absorbs heat

_____thermic

DH ? 0

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

6.4

slide6

DH = -890.4 kJ

CH4(g) + 2O2(g) CO2(g) + 2H2O (l)

Thermochemical Equations

Is DH negative or positive?

System gives off heat

_____thermic

DH ? 0

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

6.4

slide7

2H2O (s) 2H2O (l)

H2O (s) H2O (l)

H2O (l) H2O (s)

DH = -6.01 kJ

DH = 6.01 kJ/mol ΔH = 6.01 kJ

DH = 2 mol x 6.01 kJ/mol= 12.0 kJ

Thermochemical Equations

  • The stoichiometric _____________always refer to the number of _______ of a substance
  • If you ________ a reaction, the sign of DH changes
  • If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.

6.4

slide8

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

x

H2O (l) H2O (g)

H2O (s) H2O (l)

3013 kJ

1 mol P4

x

DH = 44.0 kJ

DH = 6.01 kJ

1 mol P4

123.9 g P4

Thermochemical Equations

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

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

= 6470 kJ

266 g P4

6.4

slide9

DH0 (O2) = 0

DH0 (O3) = 142 kJ/mol

DH0 (C, graphite) = 0

DH0 (C, diamond) = 1.90 kJ/mol

f

f

f

f

Standard enthalpy of formation (DH0) 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 ______.

6.6

slide11

The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.

rxn

aA + bB cC + dD

-

[

+

]

[

+

]

=

DH0

DH0

rxn

rxn

DH0 (_______)

dDH0 (D)

aDH0 (A)

bDH0 (B)

cDH0 (C)

f

f

f

f

f

-

DH0 (_______)

S

S

=

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 _______of steps.

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

6.6

slide12

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

-

S

S

=

DH0

DH0

DH0

-

[

]

[

+

]

=

rxn

rxn

rxn

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

=

12DH0 (CO2)

2DH0 (C6H6)

f

f

= - ________ kJ/mol C6H6

6DH0 (H2O)

-6535 kJ

f

2 mol

DH0 (reactants)

DH0 (products)

f

f

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.

6.6

slide13

C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ

rxn

S(rhombic) + O2(g) SO2(g)DH0 = -296.1 kJ

rxn

CS2(l) + 3O2(g) CO2(g) + 2SO2(g)DH0 = -1072 kJ

rxn

2S(rhombic) + 2O2(g) 2SO2(g)DH0 = -296.1x2 kJ

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

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

rxn

rxn

C(graphite) + O2(g) CO2(g)DH0 = -393.5 kJ

+

CO2(g) + 2SO2(g) CS2(l) + 3O2(g)DH0 = +1072 kJ

rxn

DH0 = -393.5 + (2x-296.1) + 1072 =_____ kJ

rxn

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

1. Write the enthalpy of formation reaction for CS2

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

6.6

slide14

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

Chemistry in Action:

Fuel Values of Foods and Other Substances

1 cal = ______ J

1 ____ = 1000 cal = 4184 J

slide15

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

DHsoln = Hsoln - Hcomponents

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

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

6.7

slide16

The Solution Process for NaCl

DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol

6.7

slide17

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.

__________ process:

DSuniv = DSsys + DSsurr > 0

__________process:

DSuniv = DSsys + DSsurr = 0

18.4

slide18

The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C.

rxn

aS0(A)

bS0(B)

-

[

+

]

cS0(C)

dS0(D)

[

+

]

=

aA + bB cC + dD

-

S

S0(reactants)

S

S0(products)

=

DS0

DS0

DS0

DS0

rxn

rxn

rxn

rxn

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

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

= 427.2 – [395.8 + 205.0] = ?????J/K•mol

Entropy Changes in the System (DSsys)

S0(CO) = 197.9 J/K•mol

S0(CO2) = 213.6 J/K•mol

S0(O2) = 205.0 J/K•mol

18.4

slide19

What is the sign of the entropy change for the following reaction? 2Zn (s) + O2(g) 2ZnO (s)

Entropy Changes in the System (DSsys)

When gases are produced (or consumed)

  • If a reaction produces more gas molecules than it consumes, DS0? 0.
  • If the total number of gas molecules diminishes, DS0 ? 0.
  • If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a _______number.

The total number of gas molecules goes down, DS is ______.

18.4

slide20

nonspontaneous

spontaneous

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

18.2

slide21

Gibbs Free Energy

Spontaneous process:

DSuniv = DSsys + DSsurr > 0

Equilibrium process:

DSuniv = DSsys + DSsurr = 0

For a constant-temperature process:

Gibbs free energy (G)

DG = DHsys -TDSsys

DG ? 0 The reaction is spontaneous in the forward direction.

DG ? 0 The reaction is nonspontaneous as written. The

reaction is spontaneous in the reverse direction.

DG ? 0 The reaction is at equilibrium.

18.5

slide23

The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.

rxn

aA + bB cC + dD

-

[

+

]

[

+

]

=

-

DG0 (?)

S

S

=

f

Standard free energy of formation (DG0) is the free-energy change that occurs when _____of the compound is formed from its elements in their standard states.

DG0

DG0

rxn

rxn

f

DG0 of any element in its stable form is ______.

f

dDG0 (D)

DG0 (?)

aDG0 (A)

bDG0 (B)

cDG0 (C)

f

f

f

f

f

18.5

slide24

-

DG0 (reactants)

S

S

=

f

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

DG0

DG0

DG0

-

[

]

[

+

]

=

rxn

rxn

rxn

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

=

Is the reaction spontaneous at 25 0C?

12DG0 (CO2)

2DG0 (C6H6)

f

f

6DG0 (H2O)

f

DG0 (products)

f

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

DG0 = -_____kJ

?0

____________

18.5

slide26

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

slide27

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

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

q = msDt

= 869 g x 0.444 J/g •0C x –890C

= ______ J

6.5

slide28

Constant-Pressure Calorimetry

qsys = qwater + qcal + qrxn

qsys = 0

qrxn = - (qwater + qcal)

qwater = msDt

qcal = CcalDt

Reaction at Constant P

DH = qrxn

No heat enters or leaves!

6.5

slide30

Phase Changes

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

slide31

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)?

Where’s Waldo?

Carbon Dioxide

slide33

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

slide34

H2O (s) H2O (g)

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

___________

___________

DHsub = DH??? + DH????

( Hess’s Law)

11.8

sample problem
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 problem1
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