Spontaneous Change - Part 2

York University CHEM 1001 3.0 Spontaneous Change - 37

Spontaneous Change - Part 2

Reading: from Petrucci, Harwood and Herring (8th edition):

Required for Part 2: Sections 20-5 and 20-6.

Recommended for Part 2: Sections 7-7 and 7-8.

Examples for Part 2: 20-5 through 20-8.

Problem Set for Part 2:

Chapter 20 questions: 43, 45, 49, 51, 60, 63, 68, 75


Chemical Reactions

The changes of enthalpy, entropy, and free energy duringa chemical reaction determine:

C the direction in which the reaction proceeds

C the position of equilibrium

C dependence on temperature

)H, )S, and )G can be determined from quantitiestabulated for individual reactants and products.The procedures are very similar for all three.


Enthalpy Change During Reaction

Example: N2(g) + 3H2(g) W 2NH3(g)

Reacting one mole of N2 with enough H2 to form NH3 at298.15 K and one bar pressure releases 92.22 kJ of energyto the surroundings as heat.

For the surroundings: )Hsur = 92.22 kJ

For the system (reactants and products): )H = -92.22 kJ

)H depends on how much material reacts.

Forming one mole of NH3: )H = -46.11 kJ

Reacting one mole of H2: )H = -30.74 kJ


Measuring the Amount of ReactionA mole of reaction is the amount of reaction that wouldproduce one mole of a product with a stoichiometriccoefficient of one. (just like reaction rate laws)

Examples:

C 1/2 N2(g) + 3/2 H2(g) W NH3(g)

1.00 mole of reaction produces 1.00 mole NH3

C N2(g) + 3H2(g) W 2NH3(g)

consumes 1.00 mol N2, produces 2.00 mol NH3

C 3O2(g) W 2O3(g)

consumes 3.00 mol O2, produces 2.00 mol O3


Enthalpy of ReactionThe enthalpy of reaction, )rH, is the change in enthalpyper mole of reaction. It depends on how the reaction iswritten.

Examples: (at 298.15 K)

C 1/2 N2(g) + 3/2 H2(g) W NH3(g) )rH = -46.11 kJ mol-1

Producing 2 moles of NH3 requires 2 moles of reaction.So )H = (-46.11 kJ mol-1) (2 mol) = -92.22 kJ.

C N2(g) + 3H2(g) W 2NH3(g) )rH = -92.22 kJ mol-1

Producing 2 moles of NH3 requires 1 mole of reaction.So )H = (-92.22 kJ mol-1) (1 mol) = -92.22 kJ.


Enthalpy of Reaction - continued

Consequences of the fact that enthalpy is a state function:

C Reversing a reaction changes the sign of )rH.

N2(g) + 3H2(g) W 2NH3(g) )rH = -92.22 kJ mol-1

2NH3(g) W N2(g) + 3H2(g) )rH = 92.22 kJ mol-1

C Hess's Law: )H for an overall process is the sum of the)H values for the individual steps.

N2(g) + 2H2(g) W N2H4(g) )rH = 95.40 kJ mol-1

N2H4(g) + H2(g) W 2NH3(g) )rH = -187.62 kJ mol-1

N2(g) + 3H2(g) W 2NH3(g) )rH = -92.22 kJ mol-1


Standard States

C Quantities such as H, S, and G depend on temperature,pressure, and composition.

C In tabulating these quantities for individual chemicalspecies, some set of conditions must be chosen.

C Data are usually tabulated at a temperature of 298.15 Kand a pressure of one bar (1×105 Pa . 1 atm).

C For a given substance, data are given for one or morestandard states.

C Standard states are chosen for ease of calculation. Theymay be hypothetical.


Common Standard States

The most commonly used standard states (all at one bar):

C the pure solid

C the pure liquid

C the ideal gas at a partial pressure of one bar

C the ideal solution at a concentration of one molal

We will use 1 mol L-1 . 1 molal for aqueous solutions.

We will treat solutions as being approximately ideal.

Standard states may be at the temperature of interest.


Standard Enthalpy of Reaction

The standard enthalpy of reaction, )rH°, is the changein enthalpy per mole of reaction when all reactants andproducts are in their standard states.

What if species are not in standard states?

C Enthalpies depend only weakly on pressure.

C For ideal solutions, enthalpies per mole are independentof concentration.

C Unless a system is strongly non-ideal or pressures arevery high, )rH . )rH°.


Standard Enthalpy of FormationThe standard enthalpy of formation, )fH°, of a substanceis the )rH°, per mole of the species, for its formationfrom elements in their reference states.

The reference state of an element is the standard state of theelement in its most stable form at a pressure of one bar.

Reference states: He(g), Cl2(g), Br2(l), I2(s), C(graphite)

Enthalpies of formation: (at 298.15 K)

H2(g) + ½O2(g) W H2O(g) )fH° = -241.8 kJ mol-1

H2(g) + ½O2(g) W H2O(l) )fH° = -285.8 kJ mol-1

For an element in its reference state, )fH° = 0.


Using Enthalpies of FormationExample: Calculate the standard enthalpy of reaction for

CH4(g) + 2O2(g) W CO2(g) + 2H2O(l)

Solution: Use Hess's Law.

(a) C(s,gr) + O2(g) W CO2(g) )fH° = -393.5 kJ mol-1

(b) H2(g) + ½O2(g) W H2O(l) )fH° = -285.8 kJ mol-1

(c) C(s,gr) + 2H2(g) W CH4(g) )fH° = -74.8 kJ mol-1

Reaction of interest equals (a) + 2×(b) - (c).

In general,

)rH° = 3<p)fH°(products) - 3<r)fH°(reactants)

)rH° = -393.5 - 2(-285.8) + 74.8 kJ mol-1 = -890.3 kJ mol-1


Standard Entropy of ReactionThe standard entropy of reaction, )rS°, is the changein entropy per mole of reaction when all reactants andproducts are in their standard states.

)rS° is calculated from standard absolute molar entropies.

)rS° = 3<pSm°(products) - 3<rSm°(reactants)

Example: Find )rS° for the reaction

C(s,gr) + 2H2(g) W CH4(g)

Sm°(C) = 5.74 J K-1 mol-1 Sm°(H2) = 130.7 J K-1 mol-1

Sm°(CH4) = 186.3 J K-1 mol-1

Y )rS° = -80.8 J K-1 mol-1


Entropy of Reaction

The entropy change per mole of reaction, )rS, dependsweakly on total pressure but strongly on concentration.

C For liquids or solids, higher pressure changes structure.Entropy per mole is usually somewhat lower.

C For gases, higher concentration (partial pressure) meansless room to move around. Entropy per mole is lower.

C For species in solution, higher concentration means lessroom to move around. Entropy per mole is lower.

Y If reactants and/or products are not in standard states,)rS can be very different from )rS°.


Effect of Concentration on SS = k lnW

For ideal gases and ideal solutions:

C energy per mole is independent of concentration

C for each molecule, the number of microstates, W1, isproportional to the volume:

W1 = "V

C for N molecules, W equals the product of all the W1

W = W1N = ("V)N

C In terms of concentration, C / N/V,

W = ("N/C)N


Effect of Concentration - continuedS = k lnW

For ideal gases and ideal solutions, W = ("N/C)N. So

S = k ln("N/C)N = kN ln("N/C)

For one mole, N = NA. So

Sm = kNA ln("NA/C) = Rln("NA) - Rln(C)

For a change in concentration from Ci to Cf,

)Sm = Sm,f - Sm,i = Rln(Ci/Cf)

Note that we can replace Ci and Cf with molarconcentrations or, for gases, partial pressures.


Relation between Sm and Sm° - GasesFor a change in concentration (ideal), )Sm = Rln(Ci/Cf).

For an ideal gas, )Sm = Rln(Pi/Pf). Let

Pi = 1 bar, Pf = P, )Sm = Sm - Sm°

then Sm = Sm° - RlnP

where P is the partial pressure of the gas, in bar.

Example: For O2(g) at 298.15 K, Sm° = 205.1 J K-1 mol-1.Find Sm at a partial pressure of 0.100 bar.

Sm = 205.1 J K-1 mol-1 - (8.314 J K-1 mol-1) ln(0.100)

Sm = 224.2 J K-1 mol-1


Relation between Sm and Sm° - Solutions

For a solute in an ideal solution, )Sm = Rln(Ci/Cf). Let

Ci = 1 M, Cf = [X] )Sm = Sm - Sm°

then Sm = Sm° - Rln[X]

where [X] is the concentration of the solute, in M.

Example: For Na+(aq) at 298.15 K, Sm° = 59.0 J K-1 mol-1.Find Sm when [Na+] = 2.0×10-3 M.

Sm = 59.0 J K-1 mol-1 - (8.314 J K-1 mol-1) ln(0.0020)

Sm = 110.7 J K-1 mol-1


Standard Free Energy of Reaction

The standard Gibbs free energy of reaction, )rG°, isthe change in free energy per mole of reaction when allreactants and products are in their standard states.

Can be calculated two different ways:

C )rG° = )rH° - T)rS°

C )rG° = 3<p)fG°(products) - 3<r)fG°(reactants)

where the )fG° are standard free energies of formationfrom elements in their reference states.

For elements in their reference states, )fG° = 0.


Standard Free Energy of Reaction - example

CH4(g) + 2O2(g) W CO2(g) + 2H2O(l)

Data at 298.15 K

substance )fH° (kJ/mol) )fG° (kJ/mol) Sm° (J/K/mol) CH4(g) -74.81 -50.72 186.3 O2(g) 0 0 205.1 CO2(g) -393.5 -394.4 213.7 H2O(l) -285.8 -237.1 69.91

)rH° = -890.3 kJ mol-1 )rS° = -243.0 J K-1 mol-1

At T = 298.15 K: )rG° = )rH° - T)rS° = -817.9 kJ mol-1

Using tabulated )fG°: )rG° = -817.9 kJ mol-1


Thermochemical Calculations - Summary C )rH, )rS, and )rG are per mole of reaction. They obey

Hess's Law.

C Data are tabulated for standard states of individual species.

C )fH° and )fG° are for the formation of one mole of aspecies from elements in their reference states.

C Sm° are absolute molar entropies.

C )rH°, )rS°, and )rG° for a reaction may be computed from)fH°, Sm°, and )fG° for the reactants and products.

C )rG° = )rH° - T)rS°

C For non-standard conditions, )rH . )rH° but )rS � )rS° and)rG � )rS°.


Free Energy of Reaction

)rG = )rH - T)rS

C )rG is per mole of reaction.

C Unless a solution is strongly non-ideal, )rH . )rH°.

C S depends strongly on concentrations.

The free energy of reaction, )rG, depends on theconcentrations of reactants and products.

To determine this, we need to determine how G depends onconcentration for individual species.


Concentration Dependence of G and )rG

For an ideal system: Hm=Hm°, so Gm - Gm° = -T(Sm-Sm°)

Ideal gas: Gm - Gm° = RTlnP (P in bar)

Ideal solute: Gm - Gm° = RTln[X] ([X] in mol L-1)

Solid or liquid: Gm - Gm° = 0

Applied to reactions:

3O2(g) W 2O3(g)

)rG = 2)fG(O3) - 3)fG(O2)

)rG = 2)fG°(O3) + 2RTlnPO3 - 3)fG°(O2) - 3RTlnPO2

)rG = )rG° + RTln(P2O3/P

3O2) = )rG° + RTlnQ


General Relation of )rG to )rG°

The activity, a, of a chemical species is defined so that

Gm - Gm° / RTlna

For an ideal gas: a = P = partial pressure, in bar

For an ideal solute: a = [X] = concentration in mol kg-1

For a pure solid or liquid: a = 1

For a chemical reaction:

)rG = )rG° + RTlnQ

where Q is the reaction quotient using activities.


Approach to Equilibrium

For a reaction at constanttemperature and pressure:

C if )rG < 0, reaction isspontaneous to the right,Q increases

C if )rG = 0, reaction is atequilibrium

C if )rG > 0, reaction isspontaneous to the left,Q decreases


The Equilibrium Constant

The equilibrium constant, Keq, is the reaction quotient atequilibrium.

)rG = )rG° + RTlnQ

At equilibrium, )rG = 0 and Q = Keq. Therefore

)rG° = -RTlnKeq

or Keq = exp(-)rG°/RT)

This enables us to calculate equilibrium constants fromtabulated data for the individual species in the reaction.


Calculating Equilibrium Constants

Given that at 298.15 K, )fG°(NO2) = 51.31 kJ mol-1 and)fG°(N2O4) = 97.89 kJ mol-1, calculate equilibriumconstants for the following reactions at 298.15 K:

(a) ½N2(g) + O2(g) W NO2(g)

)rG° = )fG°(NO2) = 51.31 kJ mol-1

Keq = exp(-(51.31×103 J/mol)/(8.314 J/K/mol)(298.15K))Keq = 1.02×10-9

(b) 2NO2(g) W N2O4(g)

)rG° = )fG°(N2O4) - 2)fG°(NO2) = -4.73 kJ mol-1

Keq = 6.74


Effect of How the Reaction is Written

NO2(g) W ½N2O4(g)

)rG° = ½(-4.73 kJ mol-1) Keq,1 = 2.6 = (PN2O4)½ (PNO2)

-1

2NO2(g) W N2O4(g)

)rG° = -4.73 kJ mol-1 Keq,2 = 6.74 = (PN2O4) (PNO2)-2

C Note that Keq,2 = (Keq,1)2.

C Doubling a reaction doubles )rG° and squares Keq.

C Writing the reaction either way gives the sameequilibrium partial pressures of NO2 and N2O4.


Sequences of Reactions

C Standard free energy changes are added together.

C Equilibrium constants are multiplied together.

Example: Data at 298.15 K.

)rG° (kJ mol-1) Keq

N2(g) + 2O2(g) W 2NO2(g) 2(51.31) (1.02×10-9)2

2NO2(g) W N2O4(g) -4.73 6.74

N2(g) + 2O2(g) W N2O4(g) 97.89 7.0×10-18


Ions in Solution

Experimental fact: We can not make a solution with ions ofonly one charge.

Consequence: One type of ion must be given arbitraryvalues for )fH°, )fG°, and Sm°.

Convention: The hydrogen (hydronium) ion, H+, is assigned)fH° = 0, )fG° = 0, and Sm° = 0 at all temperatures.

One consequence: ions can have "absolute" entropies < 0.


Ions in Solution - continued

Example: HF(g) W H+(aq) + F-(aq)

Data at 298.15 Ksubstance )fH° (kJ/mol) )fG° (kJ/mol) Sm° (J/K/mol) HF(g) -271.1 -273.2 173.8 F-(aq) -332.6 -278.8 -13.8

)rH° = -332.6 + 0 - (-271.1) kJ mol-1 = -61.5 kJ mol-1

)rG° = -278.8 + 0 - (-273.2) kJ mol-1 = -5.6 kJ mol-1

)rS° = -13.8 + 0 - 173.8 J mol-1 K-1 = -187.6 J mol-1 K-1

Keq = [H+][F-] / PHF = exp(-)rG°/RT) = 9.6 at 298.25 K


The Equilibrium Constant - Summary C )rG depends on the activities of reactants and products.

C Activities may approximated with concentrations (insolution) or partial pressures (gases).

C )rG = )rG° + RTlnQ

C The sign of )rG determines the direction of spontaneity.

C At equilibrium, Q = Keq.

C )rG° = -RTlnKeq

C For sequences of reactions, the )rG° are added and theKeq are multiplied.

C H+ is assigned )fH° = 0, )fG° = 0, and Sm° = 0.

Spontaneous Change - Part 2

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