DG 1

1. Chlorine trifluoride is a toxic, intensely reactive gas. It was used in World War II to make incendiary bombs. It reacts with ammonia and forms nitrogen, chlorine, and hydrogen fluoride gases. When two moles of chlorine trifluoride reacts, 1196 kJ of heat is evolved.– Write the thermochemical equations for the reaction– What is the ∆Ho

f for ClF3?

2. Consider the combustion of one mole of acetylene, C2H2, which yields carbon dioxide and liquid water. Using the given below,

• ΔHf° (kJ/mol): C2H2(g) = +226.7, CO2(g) = -393.5, • H2O(l) = -285.8• (a) Calculate ΔHrxn

• (b) Calculate ΔE at 25°C

3. A styrofoam coffee cup serves as an inexpensive calorimeter for measurements that do not require high accuracy. To 25.0 mL of water in such a cup at 24.33°C, a 1.00 g portion of KCl(s) was added. It dissolved completely after a short period of gentle stirring (using the thermometer). The minimum temperature reached was 22.12°C. Estimate ΔH° for the heat of dissolution of KCl in kJ/mol. You may assume that the solution has the same heat capacity as the water (4.18 J/g K) and that the heat capacity of the calorimeter itself is negligible.[AW (g/mol): Cl 35.45, K 39.10]

Chemical Thermodynamics 2: Spontaneity and Free Energy

• The second law of thermodynamics states, “In spontaneous changes the universe tends towards a state of greater disorder.”

• Spontaneous processes have two requirements:1. The free energy change of the system must be negative. 2. The entropy of universe must increase.

2nd Law of Thermodynamics

Entropy, S• Entropy is a measure of the disorder or randomness

of a system.

• As with H, entropies have been measured and tabulated in Appendix K as So

298.

• When:– S > 0 disorder increases (which favors

spontaneity).– S < 0 disorder decreases (does not favor

spontaneity).

Entropy, S

• From the Second Law of Thermodynamics, for a spontaneous process to occur: 0S S S gssurroundinsystemuniverse

In general for a substance in its three states of matter:

Sgas > Sliquid > Ssolid

Entropy, S

3rd Law of Thermodynamics

• “The entropy of a pure, perfect, crystalline solid at 0 K is zero.”

• This law permits us to measure the absolute values of the entropy for substances.– To get the actual value of S, cool a substance to 0 K, or as

close as possible, then measure the entropy increase as the substance heats from 0 to higher temperatures.

– Notice that Appendix K has values of S not S. THEY ARE ABSOLUTE VALUES

Entropy Values• Standard states similar to enthalpy standard

states conditons

• However, all substances at 25oC has NONZERO ENTROPY values.

• The reference state is the ABSOLUTE ZERO temperature, 0 K.

• Unit: J/K-mole

• Entropy Values

Substance Entropy, So , J/K-mole

N2(g) 191.61

Fe(s) 27.28

Br(l) 152.23

Cu2+ (aq) -99.6

Cs1+(aq) 133.05

CO32-

(aq) -56.9

H+(aq) 0

Entropy, S

• Entropy changes for reactions can be determined similarly to H for reactions.

oreactants

n

oproducts

n

o298 SnSnS

Entropy, S

• Example 1: Calculate the entropy change for the following reaction at 25oC. Use values in appendix K.

4(g)22(g) ONNO 2

K molkJ

K molJ

K molJ

K molJ

oNO

oON

oreactants

n

oproducts

n

orxn

0.1758-or 8.175

)0.240(2)2.304(

S2S

Sn S nS

2(g)4(g)2

Entropy, S• Example 2: Calculate So

298 for the reaction below. Use values in appendix K.

g2g2g NOONNO 3

K mol

kJ K mol

J

K molJ

0NO

0NO

0ON

0298

0.1724-or 4.172

210.43 - 240.0 7.219

S 3SSSgg2g2

What is Spontaneity?• Spontaneous changes happen without any

continuing outside influences. – A spontaneous change has a natural direction (you have to

consider the conditions of the system).

Gibb’s Free Energy

• The change in the Gibbs Free Energy, G, is a reliable indicator of spontaneity of a physical process or chemical reaction.– G does not tell us how quickly the process occurs. It will

only tell if a reaction is spontaneous or non spontaneous

• G is the Energy released to the surroundings (after the spontaneous process of the system) to do useful work!

Gibb’s Free Energy: Sign Convention

+

–

Free Energy 0

∆G < 0, spontaneousUNSTABLE

∆G = 0, system equilibrium, STABLE SYSTEM

∆G > 0, nonspontaneous

Free energy released to the surroundings

Energy must be applied from the surroundings

Gibb’s Free Energy and Reversibility

• If a process is spontaneous i.e. one with highly negative ∆G, it will IRREVERSIBLY proceed to give-off energy (that’s why its called FREE ENERGY)

• Consider rusting of iron (Fe)

O2 + 4 e- + 2 H2O → 4 OH-

Fe → Fe2+ + 2 e−

4 Fe2+ + O2 → 4 Fe3+ + 2 O2−

2e- + O2- + 2H2O → 4OH-

Fe3+ + 3 H2O Fe(OH)⇌ 3 + 3 H+

Fe(OH)3 FeO(OH) + H⇌ 2O 2FeO(OH) Fe⇌ 2O3 + H2O

Gibb’s Free Energy and Reversibility

Free EnergykJ/mole

Reaction pathway

Fe(s) , O2(g) , H2O(l)

Fe2O3(s)

∆G

Surroundings receives free energy

RUST FORMATION wastes important materials and ENERGY!

Gibb’s Free Energy and Reversibility

• Once an egg is broken, you need a lot of external energy to put it together again!

Not Spontaneous!

WORK!

Free Energy Change, G, and Spontaneity

• Changes in free energy obey the same type of relationship we have described for enthalpy, H, and entropy, S, changes.Standard states also in 25oC and 1 atmElements in their standard forms also has ZERO

free energy change of formation (∆Gfo= 0, also H+

(aq))All other properties is similar to Enthalpy and

Internal energy

0reactants

n

0products

n

0298 GnGnG

Calculation of Free Energy Change Example: Calculate Go

298 for the decomposition of CaCO3(s).

)(2(g)3(s) CaO CO CaCO s

molkJ

molkJ

oCaCO f

oOCa f

oCO f

orxn

4.130

)}8.1128()]0.604()4.394{[(

]G[]GG[G(s)3)(2(g)

s

Calculation of Free Energy Change

Example: Calculate Go298 for the oxidation of glucose,

C6H12O6(s).)(22(g)2(g)6(s) 126 O6H 6CO 6O OHC

molkJ

molkJ

oO f

oHHC f

oOH f

oCO f

orxn

2.2881

)]}0(6)0.909[()]3.237(6)4.394(6{[

]G6G[]G6G6[G2(g)(s)6126)(22(g)

Very Spontaneous! But why are we still alive? We should have been liquefied! :B

Activation energy prevents many spontaneous reactions from happening immediately.

Calculation of Free Energy Change

Free Energy Calculations

Consider the formation of ice- spontaneous only at low temp (≤ 0o C), but is

an ordered process (entropy decreases)Free Energy, Disorder, and Enthalpy

G = H - T S (at constant T & P)

∆G = ∆H – T∆SBecause 0 ≤ H ≥ 0 and 0 ≤ S ≥ 0, there are four

possibilities for G.

H S G Reaction spontaneity – + – Spontaneous at all T’s. – – Temp dependent Spont at low Temp. + – Temp dependent Spont at high Temp. + – + NON Spont at all T’s.

Calculations at Equilibrium

• Example: Use thermodynamic data to estimate the normal boiling point of water.

S

H T and ST H Thus

0. G process mequilibriuan is thisBecause

OHOH (g)2)(2

Calculations at Equilibrium• Assumptions

Enthalpy of Vaporization at 25oC is the same as Enthalpy of Vaporization at the boiling point

Entropy of Vaporization at 25oC is the same as Entropy of Vaporization at the boiling point

Calculations at Equilibrium

Calculations at Equilibrium

T =HS

H

S

.0 kJ0.1188

K

370 K - 273 K = 97 C

o

o kJK

o

44

370

% error =

370 - 373 K

K error

% error of less than 1%!!373

100% 0 80% .

Coupled Reactions• First step in utilizing glucose (in all organisms)

Non spontaneous

Coupled Reaction

Spontaneous, yeah!