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# What makes entropy a state function? Description

##### Description: These notes cover classes from February 2-4.
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CHEM 1112

## What makes entropy a state function?

Dr. Martín Zysmilich

Class Notes for February 2-4

Entropy (February 2)

∙ Units of enthalpy: kJ/mol

Units of entropy: J/K∙mol

∙ 3rd Law of Thermodynamics: the entropy of a perfect crystal of any substance at 0 K is equal to 0 o All molecules in the system have a fixed position; only one possible arrangement o Also, all molecules are in the lowest energy state possible: no translational or rotational  energy (no kinetic energy) and lowest possible vibrational energy

o Increased temperature increases entropy; therefore, entropy has an absolute value because it can be compared to the absolute value of 0 at 0 K

▪ Enthalpy does not have to be expressed as ΔS like enthalpy must be expressed  as ΔH

## What are the trends in entropy?

∙ Entropy at 1 atm = standard entropy (S°)

o Entropy is a state function, so Hess’s Law can be applied to entropy as well as enthalpy o Molecular complexity increases entropy (more ways of movement and more ways to  store energy in more complex molecules)

∙ Trends in Entropy

o S°gas > S°liquid > S°solid

o S° increases with molar mass

o S° increases with molecular complexity

∙ ΔS° (change in entropy) for a chemical reaction:

aA + bB → cC + dD

ΔS° = enthalpy of products – enthalpy of reactants

## What is the process for computing for the second law of thermodynamics?

We also discuss several other topics like What are the negative effects of teratogens?
If you want to learn more check out How is the navier-stokes equation applied?

= (cS°C + dS°D) – (aS°A + bS°B)

∙ SAMPLE PROBLEM:

Calculate the enthalpy change for the reaction of formation of water  a) as a gas.

H2 (g) + ½O2 (g) → H2O(g)

ΔS° = S°products – S°reactants

= S°H2O (g) – (S°H2 (g) + ½S°O2 (g))

Given:

S°H2 (g) = 130.6 J/K∙mol S°O2 (g) = 205.0 J/K∙mol S°H2O (g) = 188.8 J/K∙mol S°H2O (l) = 69.9 J/K∙mol

= 188.8 J/K∙mol – (130.6 J/K∙mol + ½(205.0 J/K∙mol)) = -44.3 J/K∙mol  b) as a liquid.

H2 (g) + ½O2 (g) → H2O(l)

ΔS° = S°products – S°reactants

= S°H2O (l) – (S°H2 (g) + ½S°O2 (g))

= 69.9 J/K∙mol – (130.6 J/K∙mol + ½(205.0 J/K∙mol)) = -163.2 J/K∙mol ∙ Predicting the Sign of ΔS°sys Don't forget about the age old question of Which agreement officially ended the thirty years war?

o Processes that lead to increased entropy:

▪ Melting

▪ Vaporization

▪ Sublimation

▪ Temperature increase

▪ Reaction resulting in a greater number of gas molecules

∙ SAMPLE PROBLEM:

Determine the sign of ΔS°sys:

a) decomposition of KClO3 (s) to give K2O(s), Cl2 (g), and O2 (g)

KClO3 (s)→ K2O(s) + Cl2 (g) + O2 (g)

creation of gas molecules: ΔS°sys > 0

b) condensation of water vapor on a cold surface

H2O(g) → H2O(l) Don't forget about the age old question of Why do we have to have consideration in a contract?

condensation goes to a more ordered system: ΔS°sys < 0

c) reaction of NH3 (g) and HCl(g) to give NH4Cl(s)

NH3 (g) + HCl(g) → NH4Cl(s)

less gas molecules in products: ΔS°sys < 0

∙ Change in entropy of a system is not enough to predict spontaneity of a process: condensation  of water has negative entropy but we know it is spontaneous If you want to learn more check out What are the two pathways of photosynthesis?

o Spontaneity depends on ΔS of the universe, which in total increases in entropy o Universe = system + surroundings If you want to learn more check out Enumerate the types of unemployed people.

o Therefore, the entropy of both the system and the surroundings are needed to  determine the spontaneity of a process

∙ Calculating ΔSsurr

o Δ���������� = −Δ��������

��

o If the system is losing heat (exothermic), the surroundings absorb that heat  (endothermic) and vice versa, which is the reason for the negative sign

o At low temperatures, increase in temperature greatly increases the energy of the  system

o At higher temperature, adding energy doesn’t change the energy of the system quite as  much because the distribution of energy among molecules is more broad at higher  temperatures

∙ 2nd Law of Thermodynamics: ΔSuniv > 0 for spontaneous processes

ΔSuniv = ΔSsys + ΔSsurr

ΔSuniv = ΔSsys –Δ��������

��

–TΔSuniv = –TΔSsys + ΔHsys

ΔG

ΔG = ΔH – TΔS

∙ ΔG = Gibbs Free Energy (state function)

o ΔG < 0 for a spontaneous process

o ΔG > 0 for a nonspontaneous process

o ΔG = 0 for an equilibrium process

∙ Under standard conditions:

ΔG° = ΔH° – TΔS°

∙ How temperature affects the sign of ΔG:

ΔH

ΔS

Spontaneity

+

+

Spontaneous at high temperatures

-

-

Spontaneous at low temperatures

+

-

Never spontaneous at any temperature

-

+

Always spontaneous at any temperature

∙ Calculating ΔG°rxn:

ΔG°rxn = Σ(ΔG°products) – Σ(ΔG°reactants)

∙ SAMPLE PROBLEM:

Is this reaction spontaneous at 200°C under standard conditions? If yes, at what temperature  does it become nonspontaneous? If no, at what temperature does it become spontaneous?

CaCO3 (s) → CaO(s) + CO2 (g)Given:

CO2 (g)

CaO (s)

CaCO3 (s)

ΔHf°(kJ/mol)

-393.5

-635.6

-1206.9

S° (J/K-mol)

213.6

39.8

92.9

200°C = 473 K

ΔG°rxn = (ΔH° – TΔS°products) – (ΔH° – TΔS°reactants)

= [(-393.5 kJ/mol – 635.6 kJ/mol) – 473 K (213.6 J/K-mol + 39.8 J/K-mol)]   – [-1206.9 kJ/mol – 473 K (92.9 J/K-mol)]

= (-1029 kJ/mol – 119.9 kJ/mol) – (-1207 kJ/mol – 43.94 kJ/mol)  = -1149 kJ/mol + 1251 kJ/mol = 101.9 kJ/mol

ΔG°rxn > 0, therefore the process is nonspontaneous at 200°C

It changes to spontaneous at ΔG°rxn = ΔH° – TΔS° = 0

ΔH° = TΔS°

T = ΔH°

ΔS°

T = (−393.5 kJ/mol – 635.6 kJ/mol) – (−1206.9 kJ/mol)

(39.8J/K−mol +213.6J/K−mol)−92.9J/K−mol = 1108 K = 835°C

The class on February 4 was a TA-led review session for the exam.

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