Thermochemistry
This presentation is the property of its rightful owner.
Sponsored Links
1 / 42

Thermochemistry PowerPoint PPT Presentation


  • 50 Views
  • Uploaded on
  • Presentation posted in: General

Thermochemistry. Chapters 6 and 18. TWO Trends in Nature. Order  Disorder   High energy  Low energy . 2H 2 ( g ) + O 2 ( g ) 2H 2 O ( l ) + energy. H 2 O ( g ) H 2 O ( l ) + energy.

Download Presentation

Thermochemistry

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript


Thermochemistry

Thermochemistry

Chapters 6 and 18


Two trends in nature

TWO Trends in Nature

  • Order  Disorder

     

  • High energy  Low energy


Thermochemistry

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

H2O (g) H2O (l) + energy

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

energy + H2O (s) H2O (l)

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.

6.2


Thermochemistry

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


Thermochemistry

H2O (s) H2O (l)

DH = 6.01 kJ

Thermochemical Equations

Is DH negative or positive?

System absorbs heat

Endothermic

DH > 0

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

6.4


Thermochemistry

DH = -890.4 kJ

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

Thermochemical Equations

Is DH negative or positive?

System gives off heat

Exothermic

DH < 0

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

6.4


Thermochemistry

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 coefficients always refer to the number of moles of a substance

  • If you reverse 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


Thermochemistry

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


Thermochemistry

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 elements at a pressure of 1 atm.

f

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

6.6


Thermochemistry

6.6


Thermochemistry

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 (products)

dDH0 (D)

aDH0 (A)

bDH0 (B)

cDH0 (C)

f

f

f

f

f

-

DH0 (reactants)

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 series of steps.

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

6.6


Thermochemistry

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

= - 3267 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


Thermochemistry

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 = 86.3 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


Thermochemistry

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

Chemistry in Action:

Fuel Values of Foods and Other Substances

1 cal = 4.184 J

1 Cal = 1000 cal = 4184 J


Thermochemistry

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


Thermochemistry

The Solution Process for NaCl

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

6.7


Energy diagrams

Energy Diagrams

Exothermic

Endothermic

  • Activation energy (Ea) for the forward reaction

  • Activation energy (Ea) for the reverse reaction

  • (c) Delta H


Thermochemistry

S

order

disorder

S

H2O (s) H2O (l)

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

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

DS > 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

DS > 0

18.3


Thermochemistry

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.

Spontaneous process:

DSuniv = DSsys + DSsurr > 0

Equilibrium process:

DSuniv = DSsys + DSsurr = 0

18.4


Thermochemistry

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] = -173.6 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


Thermochemistry

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 small number.

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

18.4


Thermochemistry

nonspontaneous

spontaneous

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

18.2


Thermochemistry

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


Thermochemistry

DG = DH - TDS

18.5


Thermochemistry

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 (reactants)

S

S

=

f

Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole 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 zero.

f

dDG0 (D)

DG0 (products)

aDG0 (A)

bDG0 (B)

cDG0 (C)

f

f

f

f

f

18.5


Thermochemistry

-

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 ] = -6405 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 = -6405 kJ

< 0

spontaneous

18.5


Recap signs of thermodynamic values

Recap: Signs of Thermodynamic Values


Thermochemistry

Gibbs Free Energy and Chemical Equilibrium

DG = DG0 + 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

Q = K

DG = 0

0 = DG0 + RT lnK

DG0 = -RT lnK

18.6


Thermochemistry

DG0 = -RT lnK

18.6


Thermochemistry

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 = msDt

q = CDt

Dt = tfinal - tinitial

6.5


Thermochemistry

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

= -34,000 J

6.5


Thermochemistry

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


Thermochemistry

6.5


Thermochemistry

Phase Changes

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


Thermochemistry

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)?

Where’s Waldo?

Carbon Dioxide


Thermochemistry

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

Freezing

Melting

11.8


Thermochemistry

H2O (s) H2O (g)

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

Sublimation

Deposition

DHsub = DHfus + DHvap

( Hess’s Law)

11.8


Thermochemistry

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

11.8


Thermochemistry

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 iceQ=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 liquidΔH fusion

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

Step 3: Heat the liquidQ=mcΔT

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

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

Step 5: Heat the gasQ=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


  • Login