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how far? Entropy CHAPTER 2:

Chapter 2 - Entropy

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how far? Entropy

CHAPTER 2:

•Thermodynamics is the study of heat energy changes and other related quantities which occur during physical and chemical changes.

•A spontaneous change, once begun, continues on its own with no further outside assistance.

•Thermodynamics is able to tell us whether a change is spontaneous or not - if we consider all of the possible factors which affect spontaneity.

•The heat energy content change is called the enthalpy change of a reaction.

•Exothermic reactions, where ΔH is negative, occur more frequently than endothermic ones and so appear to be more favoured.

Introduction

•Entropy is a:• measure of the ways of sharing energy packets

between particles • measure of the way the particles can arrange

themselves in space.

•Entropy is often called a measure of the randomness of a system and it is also a measure of the probability of a particular arrangement occurring.

Entropy ΔS

Entropy may be defined as : ‘The trend in a chemical reaction is from order to disorder’

‘order’ means few arrangements of energy

‘disorder’ means many arrangements of energy

For the states of matter : Solids (least entropy) < liquids < gases

EntropyIt’s just a measure of disorder

Units of entropy

Symbol S Units JK-1mol-1 Other thermodynamic units are KJ K-1mol-1 Entropy increases with temperature Simple molecules have lower entropies

than complicated molecules.

2nd Law of Thermodynamics

In a spontaneous change (reaction) the total entropy change (entropy of the universe) must increase.

Total entropy means that of the system and surroundings.

DSuniverse = DSsystem + DSsurroundings

Entropy can explain why some reactions occur, even though the enthalpy change is not favourable (i.e endothermic reactions)

Entropy and your bedroom

Low Entropy

High Entropy

No matter what we say or try, the entropy of the universe is increasing!

The molecules in the solid are in a much more restricted or ordered state so they have low entropy.

Entropy increases when melting.

Entropy decreases when freezing

A box containing air molecules, divided in two compartments with a hole in the dividing wall. Since there are many more molecules in the left compartment, there will also be many more molecules crossing the divide from left to right.

The second law of thermodynamics states that in an isolated system the entropy will increase. In the statistical definition of entropy according to Boltzmann, this means that the system will evolve to its most probable state.

Entropy, Marble Example

Suppose you have a bag with 50 red marbles and 50 green marbles

You draw a marble, record its color, return it to the bag, and draw another

Continue until four marbles have been drawn

What are the possible results and what are their probabilities?

Entropy, Marble Example, Results

The most ordered are the least likely The most disorder is the most likely

EntropY CHANGE OF SYSTEM

Entropy is a state function; therefore, the entropy change for a chemical reaction can be calculated as follows:

Sreactiono Sprod .

o Sreact .o

Example

Balance the following reaction and determine DS°rxn.

Fe(s) + H2O(g) Fe2O3(s) + H2(g) 2Fe(s) + 3H2O(g) Fe2O3(s) + 3H2(g)

DS°rxn = (S°(Fe2O3(s)) + 3S°H2(g)) -(2S°Fe(s) + 3S°H2O(g))

DS°rxn = -141.5 J/K

ENTROPY CHANGE OF SURROUNDING

Consider a reaction driven by heat flow from the surroundings at constant P. Exothermic Process: DSsurr = heat/T

Endothermic Process: DSsurr = -heat/T

• Heat transferred = qP,surr = - qP,system= -DHsys

Ssurr H sys

T

Example

For the following reaction at 298 K:

Sb4O6(s) + 6C(s) 4Sb(s) + 6CO2(g) DH = +778 kJ

What is DSsurr?DSsurr = -DH/T = -778 kJ/298K

= -2.6 kJ/K

2nd Law OF THERMODYNAMIC (cont.)

Three possibilities: IfDSuniv > 0…..process is spontaneous If DSuniv < 0…..process is spontaneous

in opposite direction. If DSuniv = 0….equilibrium

Here’s the catch: We need to know DS for both the system and surroundings to predict if a reaction will be spontaneous!

• Chemical reaction are favoured if they are accompanied by an increase in entropy

•Many endothermic reactions proceed spontaneously under normal conditions because there is an increase in entropy.

•Some exothermic reactions do not proceed spontaneously because there is a decrease in entropy.

Big Example

Is the following reaction spontaneous at 298 K?

2Fe(s) + 3H2O(g) Fe2O3(s) + 3H2(g)

DS°rxn = DS°system = -141.5 J/K

DS°surr = -DH°sys/T = -DH°rxn/T

(Is DS°univ > 0?)

DH°rxn= DH°f(Fe2O3(s)) + 3DH°f(H2(g)) - 2DH°f(Fe (s)) - 3 DH°f(H2O(g))

Big Example (cont.)

DS°surr = -DH°sys/T = +331.5 J/K

DH° sys = -824.2 – 3(-241.8) = -98.8 kJ

DS°univ = DS°sys + DS°surr

= -141.5 J/K + 331.5 J/K = +190 J/K

DS°univ > 0 ; therefore, reaction is spontaneous

Entropy and Phase Changes

Phase Change: Reaction in which a substance goes from one phase of state to another.

Example:

H2O(l) H2O(g) @ 373 K

• Phase changes are equilibrium processes such that:

DSuniv = 0

S and Phase Changes (cont.)

H2O(l) H2O(g) @ 373 K

• Now, DS°sys = S°(H2O(g)) - S°(H2O(l)) = 195.9 J/K - 86.6 J/K = 109.1 J/K

• And, DS°surr = -DHsys/T = -40.7 kJ/373 K = -109.1 J/K

• Therefore, DSuniv = DSsys + DSsurr = 0

Example

Determine the temperature at which liquid bromine boils:

Br2(l) Br2(g)

• Now, DS°rxn = S°(Br2 (g)) - S°(Br2(l)) = 245.38 J/K - 152.23 J/K = 93.2 J/K

Example (cont.)

• Now, DS°surr = - DS°sys = -93.2 J/K = -DHsys/T

• Therefore, calculate DHsys and solve for T!

• Now, DH°rxn = DH°f(Br2(g)) - DH°f(Br2(l)) = 30.91 kJ - 0 = 30.91 kJ

• Such that, -93.2 J/K = -30.91kJ/T Tboiling = 331.6 K

(standard state)

0

3rd LAW OF THERMODYNAMICS

The third law: The entropy of a perfect crystal at 0K is zero.

The third law provides the reference state for use in calculating absolute entropies.

What is a Perfect Crystal?

Perfect crystal at 0 K Crystal deforms at T > 0 K

Calculations of ΔH and ΔS can be used to probe the “driving force” or the most important factor in deciding whether a particular reaction is spontaneous or not

ΔH is exothermic and ΔS is positive, then the reaction will be spontaneous

Other combinations of ΔH and ΔS are tricky to deal with.

What happens when one potential driving force is favorable and the other is not?

Gibbs Free Energy , g

Gibbs free energy (G)

The change in the Gibbs free energy of the system that occurs during a reaction is equal to the change in the enthalpy of the system minus the change in the product of the temperature times the entropy of the system.

If the data are collected under standard-state conditions, the result is the standard-state free energy of reaction

ΔG = ΔH - TΔS

If ΔH is negative (exothermic) and ΔS is positive then we can guarantee that ΔG is negative and the reaction is spontaneous at any temperature

ΔG = ΔH - TΔS

Critical temperature Because the ΔG depends on TΔS, the effect of entropy becomes

more important at higher temperatures.

If ΔH is -ve and ΔS is +ve, the rxn will be spontaneous at all T.

If ΔH is +ve and ΔS is -ve, the rxn will not be spontaneous at any T.

If ΔH and ΔS are both +ve, the rxn will be spontaneous only above a certain T.

If ΔH and ΔS are both -ve, the rxn will be spontaneous only below a certain T.

The critical T is the T at which ΔG=0, so ΔH = TΔS.

T = ΔH/ ΔS

Dissolving ionic compounds

Ions are strongly attracted to water, since water is a polar molecule and so cations are attracted to the O atoms in water and anions are attracted to H atom.

The energy released when a gaseous ions is dissolved in water is known as the hydration enthalpy of the ion:

Mx+ (g) →Mx+ (aq)

Not all ionic compounds are soluble, however, since the ions in the solid state are also attracted to each other (lattice energy).

The energy required to break up an ionic lattice is known as the lattice dissociation enthalpy.

The energy change when 1 mole of an ionic compound dissolves in excess water is known as the enthalpy of solution.

The more exothermic the enthalpy of the solution, the more likely the compound to dissolve.

Na+ (g) + Cl-

(g)

NaCl (s)

Na+ (aq) + Cl-

(aq)

E.g:

The hydration enthalpy depends on the charge and size of the ions: the larger the charge and the smaller the size, the larger the hydration enthalpy.

Ion Na+ K+ Mg2+ Ca2+ Cl- Br-

Hydration energy/kJ mol-1

-406 -322 -1920

-1650

-364 -337

Solubility of ionic compounds in water thus varies widely, and depends on the balance between the hydration energy of the ions and the lattice energy of the compound. Most ionic compounds are at least partially soluble in water.

The solubility of the ionic compounds can be calculated using the lattice energy of the compound and the hydration energy of the ions according to the following Hess’ Law cycle:

ΔH soln = Σ(ΔH hyd) - ΔH lattice formation

OR ΔHsoln = Σ(ΔH hyd) + ΔH lattice dissociation

Example for sodium chloride:lattice dissociation enthalpy of NaCl =

+780kJmol-1

ΔH hyd of Na+ = -406 kJ mol-1

ΔH hyd of Cl- = - 364 kJ mol-1

thus, ΔH soln of NaCl = 780 – 406 – 364 = +10

kJ mol-1

The greater the lattice energy, the harder it is to separate the ions and the lower the solubility.

The greater the hydration energy, the greater the energy released when the ions dissolve in water and the greater the solubility.

A REAL QUESTION!

(a) Explain in terms of ΔGΘ, why a reaction for which both ΔHΘ and positive is sometimes spontaneous and sometimes not. [4]

(b) Consider the following reaction.

N 2 (g) + 3H 2 (g) → 2NH 3(g)

(i) Using the average bond enthalpy values in Table 10 of the Data Booklet, calculate the standard enthalpy change for this reaction. [3]

(ii) The absolute entropy values, S, at 300 K for are 193, 131 and 192 J K−1 mol−1 respectively for N 2(g), H 2(g) and NH3 (g). Calculate for ΔSΘ the reaction and explain the sign of. ΔS Θ [2]

(iii) Calculate ΔGΘ for the reaction at 300 K. [1]

QUESTION CONTINUED

(iv) If the ammonia were produced as a liquid and not as a gas, state and explain the effect this would have on the value of ΔHΘ for the reaction. [5]

(c) Define the term standard enthalpy of formation, and write the equation for the standard enthalpy of formation of ethanol. [2]

(d) Bond enthalpies are tabulated as average bond enthalpies. Explain what this term means. [4]

(e) Enthalpies of reactions, for example combustion, can be calculated using average bond enthalpies or enthalpies of formation. The two methods give closer results for cyclohexane than they do for benzene. Explain this difference. [1]