Thermodynamics

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# Thermodynamics - PowerPoint PPT Presentation

Thermodynamics. Chapter 19. 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. energy + 2HgO ( s ) 2Hg ( l ) + O 2 ( g ).

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### Thermodynamics

Chapter 19

TWO Trends in Nature
• Order  Disorder

 

• High energy  Low energy

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.

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

H2O (s) H2O (l)

DH = 6.01 kJ

Thermochemical Equations

Is DH negative or positive?

System absorbs heat

Endothermic

DH > 0

Positive enthalpy does not favor spontaneity (favors the reverse reaction of forming reactants)

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

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

Negative enthalpy favors spontaneous reactions (favors the forward reaction of making products)

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

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.

x

H2O (l) H2O (g)

H2O (s) H2O (l)

3013 kJ

1 mol P4

x

DH = 6.01 kJ

DH = 44.0 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

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.

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

dDH0 (D)

cDH0 (C)

bDH0 (B)

f

f

f

f

-

DH0 (reactants)

S

DH0 (products)

S

=

f

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.)

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.

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.

Chemistry in Action:

Fuel Values of Foods and Other Substances

1 cal = 4.184 J

1 Cal = 1000 cal = 4184 J

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?

The Solution Process for NaCl

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

First Law of Thermodynamics

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

Energy can be transferred between a system and the surroundings or can be converted from one form to another, but the total energy remains constant. It can be expressed mathematically as DE = q + w

:

A process that occurs of its own accord without any ongoing outside interventions is said to be spontaneous.

Processes that are spontaneous in one direction are nonspontaneous in the opposite direction.

Just because a process is spontaneous does not mean that it will occur at an observable rate. It could occur very rapidly, but it could also occur very slowly.

Thermodynamics can tell us the direction and extent of a reaction but tells us nothing about the speed of the reaction.

During the 1870s Bertholet, a famous chemist of that era, suggested that the direction of spontaneous changes in chemical systems was also determined by the loss of energy. He proposed that all spontaneous chemical and physical changes were exothermic.

Although the majority of spontaneous reactions are exothermic, there are some spontaneous endothermic reactions as well.

S

order

disorder

S

Entropy (S) is a measure of the randomness or disorder of a system or the degrees of freedom. Entropy is a state function.

2nd Law of Thermodynamics states that the entropy of the universe is increasing. Entropy is not conserved.

3rd Law of Thermodynamics states that the entropy of a pure crystalline substance at absolute zero is zero.

+DS (positive entropy) favors spontaneous reactions – favors the forward reaction making products.

-DS (negative entropy) does not favor spontaneity – favors the reverse reaction of forming reactants.

Rules for determining entropy

• Increasing entropy from solid to liquid to gas. This one has the highest priority.
• Compounds have more entropy than elements (comparing same phase of each).
• If elements are all in the same phase, then the element with the larger molar mass has more entropy.
• The higher the temperature, the more positive the entropy (more disorder)
• All elements and compounds have an entropy value (none are zero)
• In a chemical reaction, if all the reactants and products are in the gas phase, then the side with the greater number of moles favors more entropy
• Units are typically in J/mole*K

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

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

Gibbs Free Energy

Ultimately determines whether a reaction will be spontaneous or not. It is a balancing act of enthalpy, entropy, and temperature. Gibbs Free Energy is named after Willard Gibbs, a Yale professor who came up with a simple way to solve for whether a reaction will be spontaneous.

Units are typically kJ/mol

Free energy is a state function

-DG The reaction is spontaneous in the forward direction.

+DG The reaction is nonspontaneous as written. The

reaction is spontaneous in the reverse direction.

DG = 0 The reaction is at equilibrium.

DG = DH – TDS

• - - + Always spontaneous
• - - - Spontaneous at low temps
• - + + Spontaneous at high temps
• + + - Never spontaneous

This equation can be used with numerical values and without numerical values to determine whether a reaction is spontaneous or not.

If a reaction is at a transition state, such as melting point, freezing point, vaporization point, sublimation point, boiling point, … the reaction is said to be at equilibrium.

At equilibrium G = 0, so 0 = H - T S

therefore H/ S = T

This scenario allows us to solve for the actual transition temperature point of a substance.

At what temperature is the following process spontaneous at 1 atm? Br2(l)  Br2(g) H = 31 kJ/mol S = 93 J/mol*K

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)

bDG0 (B)

cDG0 (C)

f

f

f

f

f

18.5

-

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

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

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

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

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

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

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

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

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

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