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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!