Thermodynamics

Thermochemistry II

Chapter 19

"the study of possibilities."

Chemical thermodynamics is concerned with energy relationships in chemical reactions.

Three things to consider...

1. changes in heat content in a system...enthalpy (∆H).

2. randomness or disorder in the reaction...entropy (∆S).

3. the work obtainable from a thermodynamic system...this

is called free energy or Gibbs free energy (∆G).

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Josiah Gibbs is known for saying that free energy was "the study of possibilities."

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Spontaneous Processes

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New name: thermodynamically favored

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Spontaneity --thermodynamically favored

spontaneous process: a process that is capable of proceeding in a given direction without needing to be driven by an outside source of energy.

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Any process that occurs without outside intervention is a spontaneous process.

spontaneity examples

• Drop an egg on the ground and it will break.

A process that is spontaneous in one direction is not spontaneous in the

opposite direction.

Dropping a broken egg on the ground will not make it spontaneously

come back together.

• Water melts spontaneously above 0º C, but not below zero.

Temperature affects the spontaneity of a process…(s)←→(l)

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spontaneity examples

In order for a process to go back to what it once was, work must be done to it.

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•The path taken back to the original state is exactly the reverse of the forward process.

Quick Facts:

• Chemical systems in equilibrium are reversible. However, energy must be put in!
• Completely reversible processes are too slow to be attained in practice, and therefore any spontaneous process is irreversible!

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POP QUIZ: spontaneous or not?

1. A piece of metal heated to 150^oC is plunged into water at 40^oC?

A:

2. An exothermic reaction? H₂ + O₂ → H₂O ∆H=-572KJ

A:

3. CO₂ breaking down into its elements? CO₂ (g) → C (s) + 2O(g)

A:

4. Dissolving NaCl in water? NaCl (s) → Na⁺ (aq) + Cl^- (aq)

A:

5. The freezing of water at 5^oC to form ice?

A:

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Enthalpy

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Enthalpy (∆H)

enthalpy: measure of energy in a thermodynamic system.

It includes the internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure.

Change in heat always goes HOT→COLD

This can be measured using various quantities: a. temperature change-q=m•∆T•C, b. heats for formation-∆H_f, and moles

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furthermore...

Increasing entropy usually tends to be spontaneous!

Entropy (∆S): the goal of many chemical processes is to proceed to a state of minimal energy.

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A prime example is the universe!

Always expanding!

WHY?

Lower energy!

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ENTROPY

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For example:

Entropy of Dissolving--

•more disorder

•NaCl → Na⁺ + Cl^-

Entropy (S): a measure of the disorder of a system.

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ENTROPY (∆S)

∆S_universe = ∆S_system + ∆S_surroundings > 0

This is known as the second law of thermodynamics.

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ENTROPY (∆S)

In general, any irreversible process results in an overall increase in entropy...the entropy of the “universe” always increases.

This is known as the second law of thermodynamics.

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The first law of thermodynamics is essentially the law of conservation of energy.

ENTROPY (∆S) FAST FACTS

Entropy is a state function…(the path doesn’t matter, only the initial and final states of the system make any difference.)

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If ∆ S > 0 (+) the randomness increases,

if ∆ S < 0 (-) the order increases.

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Calculating entropy.

Ar gas vaporizes at -185.7^oC with an enthalpy of 6.52 KJ/mol. What is the entropy of the phase change?

∆S_sys = q_(rev)/T

∆S_sys = 6.52KJ/mol

(-185.7^oC+273)

∆S_sys = _6520J/mol_____

87.5 K

∆S_sys = 74.5 J/K-mol

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Common unit for entropy:

J/K-mol

ENTROPY (∆S) examples

1) Consider the melting of ice (a change of state):

- When it melts (SOLID→LIQUID), the molecules have more freedom to move increasing degrees of freedom.

- The molecules are more randomly distributed and have more entropy.

2) Burning of wood (a chemical reaction):

- When the single piece of wood burns, it releases numerous gases and leaves solid carbon (C) behind.

- The molecules are more random than at the start, therefore more entropy!

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ENTROPY (∆S) FAST FACTS

Entropy is a state function…(the path doesn’t matter, only the initial and final states of the system make any difference.)

∆S = ∑nS_products – ∑nS_reactants

Just like enthalpy (∆H)!

If ∆ S > 0 (+) the randomness increases,

if ∆ S < 0 (-) the order increases.

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ENTROPY (∆S)

again...∆S = ∑nS_final – ∑nS_initial

If ∆ S > 0 (+) the randomness increases,

if ∆ S < 0 (-) the order increases.

ex. Calculate the entropy change for the following reaction at 25^oC.

2 SO₂ (g) + O₂ (g) → 2 SO₃ (g)

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Std. Entropy Values: SO₂ (g) = 248.5 J/K-mol

O₂ (g) = 205.0 J/mol-K

SO₃ (g) = 256.2 J/mol-K

A: -189.6J/K

again...∆S = ∑nS_final – ∑nS_initial.

2 SO₂ (g) + O₂ (g) → 2 SO₃ (g)

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Std. Entropy Values: SO₂ (g) = 248.5 J/K-mol

O₂ (g) = 205.0 J/mol-K

SO₃ (g) = 256.2 J/mol-K

A: -189.6J/K

THERMODYNAMICS

British scientist and author C.P. Snow stated the laws of thermodynamics most simply:

• You cannot win (that is, you cannot get something for nothing because matter and energy are conserved).
• You cannot break even (useful energy degenerates into non-useable, disorganized energy, because there is always an increase in disorder).

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POP QUIZ: Predicting Entropy

1. H₂O (l) → H₂O (g)

A: +∆S

2. Ag⁺ (aq) + Cl^- (aq) → AgCl (s)

A: -∆S

3. 1 mol H₂(g) or 1 mol SO₂(g): which has a greater S?

A: SO₂ will have greater S

4. N₂ (g) + O₂ (g) → 2 NO (g)

A: 0...minimal or no difference

5. 4 Fe (s) + 3 O₂ (g) → 2 Fe₂O₃ (s)

A: -∆S

Fe in no.5 has fewer "microstates" because it is a solid.

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FREE ENERGY

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Gibbs Free Energy:

Predicting Spontaneity

Generally, exothermic reactions tend to be spontaneous, AND thermodynamically favored reactions increase entropy of the universe.

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But what’s the deal with thermodynamically favored endothermic reactions?

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Or how about a thermodynamically favored exothermic reaction that decreases entropy?

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Gibbs Free Energy:

Predicting Spontaneity

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Both variables, enthalpy and entropy, drive the spontaneity (or thermodynamically favorability) of chemical reactions.

How do we balance S and H to predict whether a reaction is going to be spontaneous?

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Gibbs Free Energy:

Predicting Spontaneity

free energy:

∆G = ∆H - T∆S

Josiah Gibbs is known for saying that free energy was "the study of possibilities."

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Josiah Gibbs

1839-1903

American

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Gibbs Free Energy:

Predicting Spontaneity

Gibbs Free Energy:

G = H – TS

For a process occurring at constant temperature…

∆G = ∆H–T∆S

If ∆G is (–), the reaction is thermodynamically favored in the forward direction.

If ∆G is (+), the reaction is thermodynamically favored in the reverse direction.

If ∆G = 0 the reaction is at equilibrium.

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Under standard conditions: ∆G^o = ∆H^o–T∆S^o

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Gibbs Free Energy:

Predicting Spontaneity

Gibbs Free Energy:

∆G^o = ∆H^o–T∆S^o

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p.825

FREE ENERGY CHANGE–

HYGGE

relax!

FREE ENERGY CHANGE–

HYGGE

relax!

Calculating FREE ENERGY CHANGE

ex. Calculate the standard free energy change from the formation of NO (g) from N₂ (g) and O₂ (g) at 298K.

N₂ (g) + O₂ (g) → 2 NO (g)

Given that ∆H = 180.7KJ and ∆S = 24.7 J/K. Is the reaction spontaneous under these circumstances?

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∆G^o = ∆H^o–T∆S^o

= 180.7KJ - 298K•24.7J / K

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∆S–19.23 (like 24)

The normal boiling point of methanol (CH₃OH) is 65.0^oC, and it’s molar enthalpy is 72.0kJ/mol.

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• When CH₃OH (l) boils at its normal boiling point, does its entropy increase or decrease?
• Calculate the value of ∆S when 10.0 grams of CH₃OH (l) is vaporized at 65.0^oC.

∆S_sys = q_(rev)/T

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∆S–19.23 (like 24)

The normal boiling point of methanol (CH₃OH) is 65.0^oC, and it’s molar enthalpy is 72.0kJ/mol.

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• When CH₃OH (l) → CH₃OH (g)

entropy increases

• CH₃OH (l)--10.0 grams of CH₃OH (l) is vaporized at 65.0^oC.

A: ~67.0 J/K

∆S_sys = q / T

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thermodynamics

A general conclusion:

Things don’t just happen...they are composed of the right conditions!

Gibbs FREE E!

19.53

For a reaction ∆H=-35kJ and ∆S=-86 J/K.

• Is the reaction EXO or ENDO?
• Does the reaction lead to an increase or decrease in the disorder of the system?
• Calculate ∆G for the reaction at 298K.
• Is the reaction thermodynamically favorable under these conditions?

∆G = ∆H - T∆S

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Gibbs FREE E!

19.53

For a reaction ∆H=-35kJ and ∆S=-86 J/K.

• Is the reaction EXO or ENDO?

EXOTHERMIC

• Does the reaction lead to an increase or decrease in the disorder of the system?

DECREASE in DISORDER (increase of order)

• Calculate ∆G for the reaction at 298K.

A: ~ -10.0 kJ

• Is the reaction thermodynamically favorable under these conditions?

∆G = ∆H - T∆S

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Duluth/Lake Superior Thermodynamics

rxn in a bag

CaCl₂ (s) → Ca⁺²(aq) + 2Cl^-(aq)

NaHCO₃ (s) → Na⁺(aq) + HCO₃^-(aq) + HOH(l)

CaCl₂ (s) → Ca⁺²(aq) + 2Cl^-(aq)

NaHCO₃ (s) → Na⁺(aq) + H₂O (l) + CO₂(g) + OH^- (aq)

Ca⁺²(aq) + HCO₃^-(aq) → CaCO₃(s) + H⁺(aq)

+ H₂CO₃^-2(aq) + OH^-(aq)

+ H₂O (l) + CO₂(g) + OH^- (aq)

rxn in a bag

Conclusion:

Things don’t just happen...they are composed of the right conditions!

rxn in a bag

CaCl₂ (s) → Ca⁺²(aq) + 2Cl^-(aq)

NaHCO₃ (s) → Na⁺(aq) + HCO₃^-(aq) + HOH(l)

CaCl₂ (s) → Ca⁺²(aq) + 2Cl^-(aq)

NaHCO₃ (s) → Na⁺(aq) + H₂O (l) + CO₂(g) + OH^- (aq)

Ca⁺²(aq) + HCO₃^-(aq) → CaCO₃(s) + H⁺(aq)

Identify the following…

Where is the heat in the above reactions? (add a “q” to where it occurred)

Where is the acid? Based upon the color change of the phenol red…

Where is the base? Based upon the color change of the phenol red...

“Add” the reactions as you would using Hess’s Law.

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+ H₂CO₃^-2(aq) + OH^-(aq)

+ H₂O (l) + CO₂(g) + OH^- (aq)

Standard Free Energies (G) of Formation

∆G^o = Σn∆G^o_products − Σn∆G^o_reactants

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State of Matter Standard State

solid  → pure solid

liquid → pure liquid

gas  → 1 atm pressure

solution →   1 M concentration

elements  → defined as zero in its standard state

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Free Energy: try this.

p.823:

Predict if ∆G^o at 298K for the combustion of propane is more negative or less negative than H^o=-2220KJ.

C₃H₈ (g) + 5 O₂ (g) → 3 CO₂ (g) + 4 H₂O (l)

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What if the reaction were in the reverse?

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Gibbs Free Energy:

Predicting Spontaneity

Gibbs Free Energy:

∆G^o = ∆H^o–T∆S^o

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Calculating FREE ENERGY CHANGE

ex. Calculate the standard free energy change from the formation of NO (g) from N₂ (g) and O₂ (g) at 298K.

N₂ (g) + O₂ (g) → 2 NO (g)

Given that ∆H = 180.7KJ and ∆S = 24.7 J/K. Is the reaction spontaneous under these circumstances?

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∆G^o = ∆H^o–T∆S^o

= 180.7KJ - 298K•24.7J / K

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Gibbs Free Energy:

Predicting Spontaneity

Gibbs Free Energy:

G = H – TS

Under standard conditions: ∆G^o = ∆H^o–T∆S^o

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WINT-O-GREEN LIFESAVERS

EMISSION

EXCITATION

Emission spectrum of Wint-O-Green Lifesavers

Lightning in your mouth!

• Upon chomping on a Wint-o-green LifeSaver, a sugar molecule is broken. Triboluminescence occurs when these sugar molecules are crushed, forcing some electrons out of their atomic fields. These free electrons bump into nitrogen molecules in the air.
• The nitrogen electrons are excited and emit UV radiation.
• The UV radiation in turn excites the electrons in methyl salicylate (the flavoring of wintergreen) and therefore, causes emission of visible light!

thermo-favored examples

In order for a process to go back to what it once was, work must be done to it.

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•The path taken back to the original state is exactly the reverse of the forward process.

In order for a process to go back to what it once was, work must be done to it.

•The path taken back to the original state is exactly the reverse of the forward process. But this can be slooooowwwwww.

Entropy Changes in Chemical Reactions

Absolute entropy can be determined from complicated measurements/experiments.

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However, we can easily calculate it out...using prepared data.

For a chemical reaction:

S^o = ΣnS^o_products − ΣnS^o_reactants

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Standard molar entropy, S: entropy of a substance in its

standard state…(this is similar in concept to H.)

Units: J/mol-K… (NOTE: units of H are in kJ/mol.)

Standard molar entropies of elements are not zero!

Entropy Changes in Chemical Reactions

Consider the reaction:

N_2(g) + 3H_2(g) → 2NH_3(g)

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S^o = ΣnS^o_products − ΣnS^o_reactants

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S^o = [2S^o(NH₃)] – [S^o(N₂) + 3S^o(H₂)]

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The values for standard entropies are in Appendix C in the textbook. PLUG & CHUG!

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Standard Free Energy Changes

We can tabulate standard free energies of formation, G^o_f under standard states.

The standard free-energy change for a process is given by:

∆Gº = Σ ∆Gº_{f (products)} - Σ ∆Gº_{f (reactants)}

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--Units of G are in kJ/mol

NOTE: ∆G does not give us information on the speed of a reaction!

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ENTROPY (∆S)

again...∆S = ∑nS_final – ∑nS_initial

If ∆ S > 0 (+) the randomness increases,

if ∆ S < 0 (-) the order increases.

ex. Calculate the entropy change for the following reaction at 25^oC.

2 SO₂ (g) + O₂ (g) → 2 SO₃ (g)

​

Std. Entropy Values: SO₂ (g) = 248.5 J/K-mol

O₂ (g) = 205.0 J/mol-K

SO₃ (g) = 256.2 J/mol-K

A: -189.6J/K

Relating Entropy to

Heat Transfer & Temperature

Suppose a system changes reversibly between state 1 and state 2…

ex: A phase change occurs at constant T with the reversible addition of heat such as freezing/melting equilibrium for water at 0º C.

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Relating Entropy to

Heat Transfer & Temperature

Then, the change in entropy at constant temperature is given by: ∆S_sys = q_(rev)/T

The subscript “rev” reminds us that the path between states is reversible.

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POP QUIZ: spontaneous or not?

1. A piece of metal heated to 150^oC is plunged into water at 40^oC?

A:

2. An exothermic reaction? H₂ + O₂ → H₂O ∆H=-572KJ

A:

3. CO₂ breaking down into its elements? CO₂ (g) → C (s) + 2O(g)

A:

4. Dissolving NaCl in water? NaCl (s) → Na⁺ (aq) + Cl^- (aq)

A:

5. The freezing of water at 5^oC and 1 atm pressure to form ice?

A:

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Free Energy and the Equilibrium Constant

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Free Energy and K_eq

It is useful to determine whether substances under any conditions will react.

Here’s a equation to determine if a reaction will occur:

∆G = ∆Gº + RT lnQ

AS NOTE:

• Q is the reaction quotient which is similar to K_eq but it is not at standard conditions.
• “R” is the ideal gas constant, 8.314 J/mole·K, and “T” is the temperature in Kelvin.
• When a reaction is at standard conditions, Q = 1, so… ln Q = 0.

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Free Energy and K_eq

At equilibrium, Q = K_eq and G = 0, so…

∆Gº = − RT lnK_eq

From the above we can conclude the following:

If G < 0, then K > 1…the reaction will go forward.

If G = 0, then K = 1…the reaction is at equilibrium.

If G > 0, then K < 1…the reaction will go in reverse.

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Using Gibb's Masterpiece: ∆G^o = ∆H^o–T∆S^o

a. Is a reaction spontaneous if T∆S^o > 0 and ∆H^o < 0?

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b. What is the temperature that NaCl reversibly melts, that is, when the melting occurs so slowly that the solid phase is always in equilibrium with the liquid phase? The enthalpy of melting is 30.3kJ/mol, and the entropy change upon melting is 28.2J/mol-K.

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Using Gibb's Masterpiece: ∆G^o = ∆H^o–T∆S^o

∆G = ∆Gº + RT lnQ

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ex. Does the following reaction occur spontaneously at 25^oC? ∆G^o_rxn= +173.1 KJ

N₂ (g) + O₂ (g) → 2 NO (g)

P (atm) 1.0 1.0 4.0

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Using Gibb's Masterpiece: ∆G^o = ∆H^o–T∆S^o

Here’s the equation rearranged to solve for K_eq:

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K_eq = e^–∆Go/RT

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Be sure ∆Gº is in units of kJ/mol, and the values in the equilibrium expression are in atm or moles/L.

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Finally, if you want a nonspontaneous reaction to proceed, you have to supply energy to the system to “drive” it forward. Nature does this often.

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Using Gibb's Masterpiece: ∆G^o = ∆H^o–T∆S^o

ex. The following reaction is the oxidation of glucose at standard conditions.

C₆H₁₂O₆ (s) + 6 O₂ (g) → 6 CO₂ (g) + 6 H₂O (l)

a. Calculate ∆G^o. ...sum of prod - sum of react...

∆G^o= -2882.9KJ

b. Is the oxidation of glucose spontaneous at standard conditions?

c. Calculate the equilibrium constant for the reaction.

∆Gº = − RT lnK_{eq OR} K_eq = e^–∆Go/RT

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A: A HUGE #,

e¹¹⁶⁰ = 1 x 10⁵⁰⁵.

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EXTRA! EXTRA!

The human body can generate 10-100 millivolts of electricity!

Parkinson’s disease sends continual electrical impulses between the brain and nervous system...tiring the muscles.

My dad had his device turned on last week. The device better regulates the electrical impulses between the brain and nervous system.

spontaneity examples

In order for a process to go back to what it once was, work must be done to it.

•The path taken back to the original state is exactly the reverse of the forward process.

Quick Facts:

• Chemical systems in equilibrium are reversible.
• Completely reversible processes are too slow to be attained in practice.

ENTROPY—level of disorder!

“Let the entropy begin!” ∆S: (+) is more disorder (-) is more order

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1. Which has greater entropy?

a. NH₃ (l) or NH₃ (g)

b. Mg crystal at 0K or 200K?

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2. Does S go up? Does S go down? Stay the same?

a. Na₂SO₄ (s) 2Na⁺¹ (aq) + SO4^-2 (aq)

b. 2H₂ (g) + O₂ (g) 2H₂O (g)

ex.:

a. Na₂SO₄ (s) --> 2 Na (aq) + SO₄ (aq)

b. 2H₂ (g) + O₂ (g) --> 2H₂O (g)

c. Which has more order? NH₃ (l) or NH₃ (g)

d. Which has more order? SO₃ or Cl₂

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Entropy Changes in Chemical Reactions

Consider the reaction:

Calculate the entropy of

2SO₂ (g) + O₂ (g) → 2SO₃ (g)

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S^o = ΣnS^o_products − ΣnS^o_reactants

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The values for standard entropies are in Appendix C in the textbook. PLUG & CHUG!

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Entropy & the 2^nd Law of Thermodynamics

2nd Law of Thermodynamics: In any spontaneous process, the entropy of the universe increases.

The change in entropy of the universe is the sum of the change in entropy of the system and the change in entropy of the surroundings.

∆S_univ = ∆ S_sys + ∆ S_surr

For a reversible process:

∆ S_univ = ∆ S_sys + ∆ S_surr = 0

For a spontaneous process (i.e., irreversible):

∆ S_univ = ∆ S_sys + ∆ S_surr > 0

Therefore, entropy is not conserved… ∆ S_univ is continually increasing.

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Entropy & the 3^rd Law of Thermodynamics

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In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules. Therefore, this is a state of perfect order.

Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero.

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more Thermodynamics

LAWS

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1st Law of Thermodynamics:

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2nd Law of Thermodynamics: Is the entropy of the universe increasing?

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 ∆S_universe = ∆S_system + ∆S_surroundings

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3rd Law of Thermodynamics: A perfect crystal's entropy is 0 when at ___K

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Gibbs Free Energy:

Predicting Spontaneity

Gibbs Free Energy:

∆G^o = ∆H^o–T∆S^o

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•If ∆H < 0 and ∆S > 0, then ∆G is always negative…(spontaneous.)

• If ∆H > 0 and ∆S < 0, then ∆G is always positive…(nonspontaneous)
• If ∆H < 0 and ∆S < 0, then ∆G is negative at low temperatures.
• If ∆H > 0 and ∆S > 0, then ∆G is negative at high temperatures.

NOTE: Even though a reaction has a negative ∆G it may occur too slowly to be observed.

p.823:

Calculate G^o at 298K for the combustion of methane.

CH₄ (g) + 2 O₂ (g) --> CO₂ (g) + 2 H₂O (g)

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What about the reverse reaction?

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Chapter 5 Exam:

5,6 !! Limiting Reactant.

2,8 !! Reading carefully/thought.

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CORRECTIONS:

• Question #
• Why is it incorrect?
• Corrected answer. Use words. Or numbers & words!

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