Second law of Thermodynamics

- 2

• If an irreversible process occurs in a closed system, the entropy S of the system always increase; it never decreases.

• There are two equivalent ways to define the change in entropy of a system:

(1) In terms of the system’s temperature and the energy it gains or loses as heat

(2) By counting the ways in which the atoms or molecules that make up the system can be arranged.

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• To predict spontaneity we need to know the energy change and the entropy.

• Entropy (S) is a measure of the randomness or disorder of a system.

• The greater the disorder, the greater the entropy.

• Nature tends to the greatest entropy.

• Ssolid < Sliquid < Sgas

Entropy & Chemical Reactions

• The decomposition of N2O4 (O2N–NO2) is also accompanied by an increase in randomness.

• Whenever molecules break apart, randomness increases.

• Consider the gas phase reaction:

3A2 + 6B = 6AB

•)a (Write a balanced equation for the reaction.

) b (Predict the sign of S for the reaction.

Standard Molar Entropies

• Standard Molar Entropy, S°: The entropy of 1 mol of the pure substance at 1 atm pressure and a specified temperature, usually 25°C.

• Standard molar entropies are absolute entropies measured against an absolute reference point.

• Standard entropy of reaction:

•S°= ΣnS°(products) – ΣnS°(reactants)

Standard Molar Entropies

CpdS° (J/K.mol)CpdS° (J/K.mol)

C2H2200.8CaCO392.9

NH3192.3CaO39.7

CO197.6CH3COOH160

CO2213.6H2O69.9

C2H4219.5C diamond2.4

H2130.6C graphite3.6

CH4186.2Fe27.3

QUIZ!!!!!!

• Calculate the standard entropy of reaction at 25°C for the synthesis of ammonia:

• N2(g) + 3 H2(g) = 2 NH3(g)

• • Calculate the standard entropy of reaction at 25°C for the decomposition of calcium carbonate:

• CaCO3(s) = CaO(s) + CO2(g)

Gibbs Free Energy, Gibbs Free Energy, GG

Multiply through by -TMultiply through by -T

-T∆S-T∆Sunivuniv = ∆H = ∆Hsyssys - T∆S - T∆Ssyssys

--T∆ST∆Sunivuniv = change in Gibbs free = change in Gibbs free

energy for the system = ∆Genergy for the system = ∆Gsystemsystem

Under Under standard conditionsstandard conditions — —

∆∆GGoosyssys = ∆H = ∆Hoo

syssys - T∆S - T∆Soosyssys

Suniv = Hsys

T + Ssys

∆∆SSunivuniv = ∆S = ∆Ssurrsurr + ∆S + ∆Ssyssys

∆∆GGoo = ∆H = ∆Hoo - T∆S - T∆Soo

Gibbs Gibbs free energyfree energy change = total energy change for change = total energy change for system - energy lost in disordering the systemsystem - energy lost in disordering the system

If reaction isIf reaction is

- exothermic (negative ∆ Hexothermic (negative ∆ Hoo) ) (energy dispersed) (energy dispersed) and and entropy increases (positive ∆Sentropy increases (positive ∆Soo) ) matter dispersed) matter dispersed)

- thenthen ∆G ∆Goo must bemust be NEGATIVE NEGATIVE

•reaction is spontaneous (and product-favored).

∆∆GGoo = ∆H = ∆Hoo - T∆S - T∆Soo

Gibbs free energy change = total energy Gibbs free energy change = total energy

change for system change for system

energy lost in disordering the systemenergy lost in disordering the system

If reaction isIf reaction is : :

•• endothermic (positive ∆Hendothermic (positive ∆Hoo) and entropy ) and entropy decreases (negative ∆Sdecreases (negative ∆Soo))

•• then then ∆G∆Goo must be must be POSITIVEPOSITIVE reaction is reaction is not spontaneousnot spontaneous (and is (and is reactant-reactant-favoredfavored).).

Gibbs Free Energy, GGibbs Free Energy, G

∆∆GGoo = ∆H = ∆Hoo - T∆S - T∆Soo

∆∆HHoo ∆S∆Soo ∆G∆Goo ReactionReaction

exo(–)exo(–) increase(+)increase(+) –– Prod-favoredProd-favored

endo(+)endo(+) ecrease(-)ecrease(-) ++ React-favoredReact-favored

exo(–)exo(–) decrease(-)decrease(-) ?? T dependentT dependent

endo(+)endo(+) increase(+)increase(+) ?? T dependentT dependent

Calculating ∆GCalculating ∆Goorxnrxn

Is the dissolution of ammonium nitrate Is the dissolution of ammonium nitrate product-favored? product-favored?

If so, is it enthalpy- or entropy-drivenIf so, is it enthalpy- or entropy-driven??

NHNH44NONO33(s) + heat ---> NH(s) + heat ---> NH44NONO33(aq)(aq)

• Practice Problem • Calculate ∆H° and ∆S°   for the following reaction:

• NH4NO3(s)  + H2O(l)  =  NH4+ (aq) + NO3

- (aq)• Use the results of this calculation to determine the value of ∆Go for this

reaction at 25o C, and explain why NH4NO3 spontaneously dissolves in water at room temperature.

• Solution• Using a standard-state enthalpy of formation and absolute entropy data

table, we find the following information:• Compound              Hfo(kJ/mol)         S°(J/mol-K)• NH4NO3(s)                 -365.56             151.08• NH4

+ (aq)                 -132.51               113.4• NO3

- (aq)                 -205.0                 146.4• This reaction is endothermic, and the enthalpy of reaction is therefore

unfavorable:• ∆Hf

o(products) - ∆Hfo(reactants)

•         = [1 mol NH4+ x 132.51 kJ/mol + 1 mol NO3

-   x -205.0 kJ/mol] - [1 mol NH4NO3 x -365.56 kJ/mol]         = 28.05 kJ

• The reaction leads to a significant increase in the disorder of the system, however, and is therefore favored by the entropy of reaction:

• ∆So = So (products) – So (reactants)•         = [1 mol NH4

+ x 113.4 J/mol-K + 1 mol NO3-   x 146.4 J/mol-K] - [1 mol

NH4NO3 x 151.08 J/mol-K]         = 108.7 J/K

• To decide whether NH4NO3 should dissolve in water at 25o C we have to compare the ∆Ho and T∆So to see which is larger.  Before we can do this, we have to convert the temperature from oC to kelvin:

• TK = 25o C + 273.15 = 298.15 K• We also have to recognize that the units of  ∆Ho for this

reaction are kilojoules and the units of  So are joules per kelvin.  At some point in this calculation, we therefore have to convert these quantites to a consistent set of untis.  Perhaps the easiest way of doing this is to convert  Ho to joules.  We then multiply the entropy term by the absolute temperature and subtract this quantity from the enthalpy term:

• ∆Go = ∆Ho - T∆So         = 28,050 J - (298.15 K x 108.7 J/K)•    = 28,050 J - 32,410 J         = -4360

J• At 25 oC, the standard-state free energy for this reaction is

negative because the entropy term at this temperature is larger that the enthalpy term:

• ∆Go = -4.4 kJ• The reaction is therefore spontaneous at room temp.

Helmholtz Free Energy

Gibbs Free Energy

• The increase in entropy principle can be summarized as follows:

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processreversibleeprocessIrrevrsibl

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