CHE 116 No. 1 Chapter Nineteen Copyright © Tyna L. Heise 2001 - 2002 All Rights Reserved

CHE 116

No. No. 11

Chapter Nineteen

Copyright © Tyna L. Heise 2001 - 2002Copyright © Tyna L. Heise 2001 - 2002

All Rights ReservedAll Rights Reserved

CHE 116

No. No. 22

Spontaneous Processes

Understanding and designing chemical Understanding and designing chemical reactions:reactions:

How rapidly does the reaction proceed?How rapidly does the reaction proceed?

- reaction rates- reaction rates

- controlled by a factor related to - controlled by a factor related to energy energy

- the lower the activation energy, - the lower the activation energy, the faster the reaction proceeds the faster the reaction proceeds

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No. No. 33


How far toward completion will the How far toward completion will the reaction go? reaction go?

- equilibrium constants- equilibrium constants

- depends on rates of forward and - depends on rates of forward and reverse reactions reverse reactions

- equilibrium should also be - equilibrium should also be dependent on energy in some way dependent on energy in some way due to dependence on reaction due to dependence on reaction rates rates

CHE 116

No. No. 44


Chemical thermodynamics is the relationship Chemical thermodynamics is the relationship between equilibrium and energy.between equilibrium and energy.

First Law of Thermodynamics: for a First Law of Thermodynamics: for a reaction that occurs at constant reaction that occurs at constant pressure, the enthalpy change equals pressure, the enthalpy change equals the heat transferred between the system the heat transferred between the system and its surroundingsand its surroundings

Energy is conserved!!Energy is conserved!!

* Enthalpy is important in helping us * Enthalpy is important in helping us determine if a reaction will proceed!determine if a reaction will proceed!

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No. No. 55


Spontaneous Processes:Spontaneous Processes:energy is neither created nor destroyed energy is neither created nor destroyed in any process, energy can only be in any process, energy can only be transferred or converted from one form to transferred or converted from one form to anotheranother

E = q + wE = q + wa spontaneous process occurs without a spontaneous process occurs without any outside intervention as energy is any outside intervention as energy is conservedconserved

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No. No. 66


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No. No. 77


Spontaneous Processes:Spontaneous Processes:temperature is going to effect the temperature is going to effect the spontaneity of a processspontaneity of a process

if discussing a phase change, at the if discussing a phase change, at the temperature of the phase change, the temperature of the phase change, the phases compete for spontaniety, and phases compete for spontaniety, and neither is said to win over the otherneither is said to win over the other

CHE 116

No. No. 88


Sample exercise: Under 1 atm pressure, Sample exercise: Under 1 atm pressure, COCO22(s) sublimes at -78°C. Is the (s) sublimes at -78°C. Is the transformation of COtransformation of CO22(s) to CO(s) to CO22(g) a (g) a spontaneous process at -100°C?spontaneous process at -100°C?

CHE 116

No. No. 99


Sample exercise: Under 1 atm pressure, Sample exercise: Under 1 atm pressure, COCO22(s) sublimes at -78°C. Is the (s) sublimes at -78°C. Is the transformation of COtransformation of CO22(s) to CO(s) to CO22(g) a (g) a spontaneous process at -100°C?spontaneous process at -100°C?

*No, if the temp had been higher it *No, if the temp had been higher it would change phase spontaneously, but would change phase spontaneously, but lower than the sublimation point favors the lower than the sublimation point favors the reverse, so the solid remains a solid.reverse, so the solid remains a solid.

CHE 116

No. No. 1010


Reversible and Irreversible Processes:Reversible and Irreversible Processes:State functions: define a state and do not State functions: define a state and do not depend upon the pathwaydepend upon the pathway

temperaturetemperatureinternal energyinternal energyenthalpyenthalpy

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No. No. 1111


Reversible and Irreversible Processes:Reversible and Irreversible Processes:Reversible processes is a unique way for Reversible processes is a unique way for a system to change its state, than go back a system to change its state, than go back to its state by following the exact same to its state by following the exact same path but in the opposite directionpath but in the opposite direction

phase changes at constant tempphase changes at constant temponly one specific value of only one specific value of qq (heat) (heat)system in equilibriumsystem in equilibrium

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No. No. 1212


Reversible and Irreversible Processes:Reversible and Irreversible Processes:Irreversible processes cannot be simply Irreversible processes cannot be simply restored to their original state using the restored to their original state using the same path, it may be forced to go back, same path, it may be forced to go back, but by a different pathwaybut by a different pathway

phase changes at different tempsphase changes at different tempstwo q values need to be established two q values need to be established qqforwardforward and and qqreversereverse

any spontaneous reactionany spontaneous reaction

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No. No. 1313


Thermodynamics can tell usThermodynamics can tell usdirection of reactiondirection of reactionextent of reactionextent of reaction

NOTNOTspeed of reactionspeed of reaction

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No. No. 1414

Entropy and the 2nd Law

Spontaniety depends upon two factorsSpontaniety depends upon two factors

Enthalpy (Enthalpy (H): heat of reactionH): heat of reactionexothermic normally spontaneousexothermic normally spontaneousendothermic normally NOT spontaneousendothermic normally NOT spontaneous

Entropy (Entropy (S): disorder of the systemS): disorder of the systemnatural law indicates reactions go in the natural law indicates reactions go in the direction that leads to more disorderdirection that leads to more disorder

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No. No. 1515


The Spontaneous Expansion of a Gas:The Spontaneous Expansion of a Gas:

When the stopcock is opened, When the stopcock is opened, the gas will spontaneously the gas will spontaneously flow to fill the empty half, flow to fill the empty half,

but but it WILL NOT flow backward it WILL NOT flow backward without work being done on without work being done on the system.the system.

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No. No. 1616


The Spontaneous Expansion of a Gas:The Spontaneous Expansion of a Gas:

Gas expands because of the tendency for the Gas expands because of the tendency for the molecules to ‘spread out’ among the different molecules to ‘spread out’ among the different arrangements that they can take.arrangements that they can take.

CHE 116

No. No. 1717


Entropy: measurement of randomness or Entropy: measurement of randomness or chaoschaosmelting icemelting icedissolving saltsdissolving salts

The more disordered or random a system, The more disordered or random a system, the larger its entropythe larger its entropy

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No. No. 1818


Sample exercise: Indicate whether each of Sample exercise: Indicate whether each of the following reactions produces an increase the following reactions produces an increase or decrease in the entropy of the system:or decrease in the entropy of the system:

a) COa) CO22(s) (s) CO CO22(g)(g)

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No. No. 1919



a) COa) CO22(s) (s) CO CO22(g)(g)

Entropy increasesEntropy increases

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No. No. 2020



b) CaO(s) + COb) CaO(s) + CO22(g) (g) CaCO CaCO33(s)(s)

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No. No. 2121



b) CaO(s) + COb) CaO(s) + CO22(g) (g) CaCO CaCO33(s)(s)

Entropy decreasesEntropy decreases

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No. No. 2222


Entropy: measurement of randomness or Entropy: measurement of randomness or chaoschaosstate functionstate functionfor a process that occurs at constant for a process that occurs at constant temperature, the entropy change is temperature, the entropy change is dependent on the heat transferred during the dependent on the heat transferred during the reverse of the reaction (reverse of the reaction (qqrevrev))

S = S = qqrevrev/T/T

CHE 116

No. No. 2323


Sample exercise: The normal boiling point of Sample exercise: The normal boiling point of ethanol, Cethanol, C22HH55OH, is 78.3°C, and its molar OH, is 78.3°C, and its molar enthalpy of vaporization is 38.56 kJ/mol. enthalpy of vaporization is 38.56 kJ/mol. What is the change in entropy when 25.8 g of What is the change in entropy when 25.8 g of CC22HH55OH(g) at 1 atm pressure condenses to OH(g) at 1 atm pressure condenses to liquid at the normal boiling point?liquid at the normal boiling point?

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No. No. 2424



S = S = qqrevrev/T/T

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No. No. 2525



S = S = qqrevrev/T /T qqrevrev = = vapvap

= 38.56kJ/mol= 38.56kJ/mol

T = 351.45 KT = 351.45 K

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No. No. 2626



S = S = qqrevrev/T /T 38.56kJ 1000 J 1 mol 38.56kJ 1000 J 1 mol mol 1 kJ 46 g mol 1 kJ 46 g

T = 351.45 KT = 351.45 K

CHE 116

No. No. 2727



S = S = qqrevrev/T /T 838.26 J/g * 25.8 g = 838.26 J/g * 25.8 g =

-21627 J-21627 J

T = 351.45 KT = 351.45 K

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No. No. 2828



S = S = qqrevrev/T/T == -21627 J/351.45 K-21627 J/351.45 K

= -61.5 J/K= -61.5 J/K

CHE 116

No. No. 2929


Second law of Thermodynamics: In any Second law of Thermodynamics: In any reversible process, reversible process, SSunivuniv = 0= 0. In any . In any irreversible reaction, irreversible reaction, SSunivuniv >0>0..

SSuniv univ = = SSsys sys + + SSsurrsurr

InIn an isolated system, just the entropy of the an isolated system, just the entropy of the system is considered.system is considered.

CHE 116

No. No. 3030

Molecular InterpretationOn the microscopic level, the number of gas On the microscopic level, the number of gas molecules can be directly related to the molecules can be directly related to the amount of entropy in a system.amount of entropy in a system.

The more gas molecules present, the The more gas molecules present, the higher the entropy valuehigher the entropy valuea phase change that increases the a phase change that increases the number of gas molecules would increase number of gas molecules would increase entropyentropya phase change that decreases the a phase change that decreases the number of gas molecules would decrease number of gas molecules would decrease entropyentropy

CHE 116

No. No. 3131

Molecular InterpretationThree moles of gas combine to form two Three moles of gas combine to form two moles of gas, thus decreasing the number of moles of gas, thus decreasing the number of molecules.molecules.

CHE 116

No. No. 3232

Molecular InterpretationDegrees of freedomDegrees of freedom

creating new bonds decreases the creating new bonds decreases the freedom of movement atoms may have freedom of movement atoms may have had.had.

3 degrees3 degreesmotion in one direction, motion in one direction, translational movementtranslational movementvibrational movementvibrational movementspinning, rotational movementspinning, rotational movement

CHE 116

No. No. 3333

Molecular InterpretationFigure 19.12Figure 19.12

CHE 116

No. No. 3434

Molecular InterpretationSample exercise: Choose the substance with Sample exercise: Choose the substance with the greatest entropy in each case:the greatest entropy in each case:

1 mol of H1 mol of H22(g) at STP or 1 mol of H(g) at STP or 1 mol of H22(g) at (g) at 100°C and 0.5 atm.100°C and 0.5 atm.

CHE 116

No. No. 3535


1 mol of H1 mol of H22O(s) at 0°C or 1 mol of HO(s) at 0°C or 1 mol of H22O(l) at O(l) at 25°C.25°C.

CHE 116

No. No. 3636


1 mol of H1 mol of H22(g) at STP or 1 mol of SO(g) at STP or 1 mol of SO22(g) at (g) at STP.STP.

CHE 116

No. No. 3737


1 mol of N1 mol of N22OO44(g) at STP or 2 mol of NO(g) at STP or 2 mol of NO22(g) at (g) at STP.STP.

CHE 116

No. No. 3838

Molecular InterpretationSample exercise: Predict whether is Sample exercise: Predict whether is S is S is positive or negative in each of the following positive or negative in each of the following processes:processes:

HCl(g) + NHHCl(g) + NH33(g) (g) NH NH44Cl(s)Cl(s)

CHE 116

No. No. 3939


2SO2SO22(g) + O(g) + O22(g) (g) 2SO 2SO33(s)(s)

CHE 116

No. No. 4040


cooling of nitrogen gas from 20°C cooling of nitrogen gas from 20°C to -50°Cto -50°C

CHE 116

No. No. 4141

Calculation of Entropy ChangesEntropy Calculations:Entropy Calculations:

no easy method for measuring entropyno easy method for measuring entropyexperimental measurements on the experimental measurements on the variation of heat capacity with variation of heat capacity with temperature can give a value known as temperature can give a value known as absolute entropyabsolute entropyzero point of reference for perfect zero point of reference for perfect crystalline solidscrystalline solidstabulated as molar quantities, J/mol-Ktabulated as molar quantities, J/mol-K

CHE 116

No. No. 4242

Calculation of Entropy ChangesEntropy differs from enthalpyEntropy differs from enthalpy

standard molar entropies are not 0standard molar entropies are not 0entropies of gases are greater than those entropies of gases are greater than those of liquids and solidsof liquids and solidsentropies increase with molar massentropies increase with molar massentropies increase with number of atoms entropies increase with number of atoms in formulain formulaS° = nS(products) - mS(reactants)S° = nS(products) - mS(reactants)

CHE 116

No. No. 4343

Gibbs Free Energy

Spontaneity involves both Spontaneity involves both thermodynamic concepts: entropy and thermodynamic concepts: entropy and enthalpyenthalpy

G = G = H – TH – TSS

G = Gibbs Free EnergyG = Gibbs Free Energy

H = EnthalpyH = Enthalpy

T = Temperature (K)T = Temperature (K)

S = EntropyS = Entropy

CHE 116

No. No. 4444

Gibbs Free Energy

Gibbs Free EnegryGibbs Free Enegry If If G is negative, the reaction is spontaneous G is negative, the reaction is spontaneous

and proceeds in the forward directionand proceeds in the forward direction If If G is zero, the reaction is at equilibriumG is zero, the reaction is at equilibrium If If G is positive, the reaction is G is positive, the reaction is

nonspontaneous and proceeds in the reverse nonspontaneous and proceeds in the reverse directiondirection

CHE 116

No. No. 4545

Gibbs Free Energy

Gibbs Free EnegryGibbs Free Enegry State functionState function Table 19.3Table 19.3

CHE 116

No. No. 4646

Gibbs Free Energy

Sample exercise: By using data from Appendix Sample exercise: By using data from Appendix C, Calculate C, Calculate GG at 298 K for the combustion at 298 K for the combustion of methane:of methane:

CHCH44(g) + 2O(g) + 2O22(g) (g) COCO22(g) + 2H(g) + 2H22O(g)O(g)

CHE 116

No. No. 4747

Gibbs Free Energy



-50.8 0 -394.4 2(-228.57)-50.8 0 -394.4 2(-228.57)

CHE 116

No. No. 4848

Gibbs Free Energy



-50.8 0 -394.4 2(-228.57)-50.8 0 -394.4 2(-228.57)

GG = products - reactants = products - reactants

CHE 116

No. No. 4949

Gibbs Free Energy



-50.8 0 -394.4 2(-228.57)-50.8 0 -394.4 2(-228.57)

GG = products – reactants = products – reactants

(-394.4 + 2(-228.57)) – (-50.8 + 0) (-394.4 + 2(-228.57)) – (-50.8 + 0)

CHE 116

No. No. 5050

Gibbs Free Energy



-50.8 0 -394.4 2(-228.57)-50.8 0 -394.4 2(-228.57)

GG = products – reactants = products – reactants

(-394.4 + 2(-228.57)) – (-50.8 + 0) (-394.4 + 2(-228.57)) – (-50.8 + 0)

-800.7-800.7

CHE 116

No. No. 5151

Gibbs Free Energy

Sample exercise: Consider the combustion of Sample exercise: Consider the combustion of propane to form COpropane to form CO22(g) and H(g) and H22O(g) at 298K. O(g) at 298K. Would you expect Would you expect GG to be more negative or to be more negative or less negative than less negative than HH? ?

CHE 116

No. No. 5252

Gibbs Free Energy

Sample exercise: Consider the combustion of Sample exercise: Consider the combustion of propane to form COpropane to form CO22(g) and H(g) and H22O(g) at 298K. O(g) at 298K. Would you expect Would you expect GG to be more negative or to be more negative or less negative than less negative than HH? ?

more negative, using Gibbs Free Enegry more negative, using Gibbs Free Enegry formula, more moles of gas being produced formula, more moles of gas being produced would be increasing entropy, +would be increasing entropy, +S, S,

CHE 116

No. No. 5353

Free Energy and Temperature

Table 19.4Table 19.4

CHE 116

No. No. 5454


Sample exercise: Using standard enthalpies of Sample exercise: Using standard enthalpies of formation and standard entropies in Appendix formation and standard entropies in Appendix C, calculate C, calculate HH and and SS at 298 K for the at 298 K for the following reaction:following reaction:

2SO2SO22(g) + O(g) + O22(g) (g) 2SO 2SO33(g)(g)

CHE 116

No. No. 5555




2(-296.9) 0 2(-395.2)2(-296.9) 0 2(-395.2)

H = 2(-395.2) - 2(-296.9) = -196.6 kJH = 2(-395.2) - 2(-296.9) = -196.6 kJ

CHE 116

No. No. 5656




2(248.5) 205.0 2(256.2)2(248.5) 205.0 2(256.2)

S = 2(256.2) – (2(248.5)+205.0) = -189.6 JS = 2(256.2) – (2(248.5)+205.0) = -189.6 J

CHE 116

No. No. 5757


Sample exercise: Using the values obtained, Sample exercise: Using the values obtained, estimate estimate GG at 400 K at 400 K


CHE 116

No. No. 5858


Sample exercise: Using the values obtained, Sample exercise: Using the values obtained, estimate estimate GG at 400 K at 400 K


= -196.6 kJ – 400(-0.1896 kJ)= -196.6 kJ – 400(-0.1896 kJ)

= -120.8= -120.8

CHE 116

No. No. 5959

Free Energy and the Equilibrium Constant

2 other important ways free energy is a 2 other important ways free energy is a powerful toolpowerful toolTabulate free energy under nonstandard Tabulate free energy under nonstandard

conditionsconditionsDirectly relate free energy to equilibrium Directly relate free energy to equilibrium

constantsconstants

CHE 116

No. No. 6060


Tabulate free energy under Tabulate free energy under nonstandard conditionsnonstandard conditions

G = G = G° + RTlnQG° + RTlnQ

R = 8.314 J/mol-KR = 8.314 J/mol-K

T = temperature KT = temperature K

Q = reaction quotientQ = reaction quotient

CHE 116

No. No. 6161


Sample exercise: Calculate Sample exercise: Calculate G at 298 K G at 298 K for the reaction of nitrogen and hydrogen for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture to form ammonia if the reaction mixture consists of 0.50 atm Nconsists of 0.50 atm N22, 0.75 atm H, 0.75 atm H22, and , and 2.0 atm NH2.0 atm NH3.3.


CHE 116

No. No. 6262


Sample exercise: Calculate Sample exercise: Calculate G at 298 K G at 298 K for the reaction of nitrogen and hydrogen for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture to form ammonia if the reaction mixture consists of 0.50 atm Nconsists of 0.50 atm N22, 0.75 atm H, 0.75 atm H22, and , and 2.0 atm NH2.0 atm NH3.3.


2(-16.6) + 2(-16.6) + 0.008314(298)ln(2.00.008314(298)ln(2.022/0.50(0.75/0.50(0.7533))))

-26.0 kJ-26.0 kJ

CHE 116

No. No. 6363


At equilibrium, At equilibrium, G is equal to 0 so…G is equal to 0 so…

G° = -RTlnKG° = -RTlnKeqeq

G° negative: K > 1G° negative: K > 1

G° zero : K = 1G° zero : K = 1

G° positive: K < 1G° positive: K < 1

KKeqeq = = ee--G°/RTG°/RT

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CHE 116 No. 1 Chapter Nineteen Copyright © Tyna L. Heise 2001 - 2002 All Rights Reserved