Chapter 19

John A. SchreifelsChemistry 212

Chapter 19-1

8–1

Chapter 19

Thermodynamics and Equilibrium


Chapter 19-2

8–2

Overview

• First Law of Thermodynamics• Spontaneous Processes and Entropy

– Entropy and the Second Law of Thermodynamics– Standard Entropies and the Third Law of Thermodynamics

• Free Energy Concept– Free Energy and Spontaneity– Interpretation of Free Energy

• Free Energy and Equilibrium Constants– Relating G° to the Equilibrium Constant– Change of Free Energy with Temperature


Chapter 19-3

8–3

First Law of Thermodynamics

• First law can be written as:E = q + w

where q = heat involved in the process and w = work done by or to the system.

• Work can be electrical or pressure –volume • Work = force acting over some distance: w = d x F (referenced

to the system).– During reactions often there is an expansion of gases against some

pressure where pressure is equal to the force per unit area:

or .

• Work is obtained by substitution: – w = d x F = d x (PxA) or – w = PV.

• The first law can be restated as E = q PV.

A

FP PxAF


Chapter 19-4

8–4

Energy and Enthalpy

• From the first law: q = E + PV.

• With no change in volume the equation simplifies to qV = E.

• At constant pressure: qP = E + PV.

• There are times when both volume and pressure can change; the heat involved in the reaction is then a more complicated function of E.

• Enthalpy: the heat output at constant pressure. H = E + PV.• In general, H = E + PV + VP. • At constant pressure, a change in enthalpy is given by:

H = E + PV = qP.

• Normally, H and E are fairly close to each other in magnitude. In the combustion of propane (see book), E = 2043 kJ, H = 2041 kJ and w = PV = 2kJ.


Chapter 19-5

8–5

Spontaneous Processes

• Spontaneous process: one that occurs without any continuous external influence; it may not occur rapidly.E.g. combustion of hydrogen with oxygen spontaneous; decomposition of water never occurs spontaneously.

H2(g) + O2(g) ½H2O(l)E.g.2 Weak acid

HF(aq) + H2O(aq) F(aq) + H3O+(aq) Ka = 3.52x104

E.g. 3 Solubility of solids.

AgCl(s) Ag+(aq) + Cl(aq) Ksp = 1.8x1010

• Moves to equilibrium amounts of reactants and product.• Rate of spontaneous change related to ratio of rate constants

for forward and reverse reactions. • Extent of reaction related to energy difference between forward

and reverse reactions (H).• Goal: Determine criteria necessary to predict spontaneity.


Chapter 19-6

8–6

Enthalpy, Entropy, and Spontaneous Processes: Review.

• Negative value for enthalpy thought to be criterion for spontaneity since most reactions with negative H are spontaneous, but exceptions exist:

• The reverse of the spontaneous reaction in not spontaneous. Reactions can only be spontaneous in one direction.

• Qc compared to Kc indicates the direction of the reaction (will be related to the thermodynamic quantity soon).

E.g. determine the direction of reaction when 0.100M acetic acid (Ka = 1.75x105), 0.100 M sodium acetate, and 0.100 M HCl are mixed together.

Solution: LeChatelier’s principle says reaction shifted to the left. What about when the reaction is reversed?

NH4Cl(s) )aq(Cl)aq(NH4OH2 Ho = +14.8 kJ

H2O(l) H2O(g) Ho = +40.7 kJ

OH2 KCl(s) K+(aq) + Cl (aq) Ho = +17.2 kJ


Chapter 19-7

8–7

ENTROPY

• Entropy: thermodynamic quantity that is measure of randomness of system.

• All changes tend to increase in randomness (second law of thermodynamics).Stotal = Sfinal Sinitial

Stotal =Ssys + Ssurr 0. Ssys is sometimes negative, but is more than offset by Ssurr. S is positive for a transition from ordered to less ordered

system (such as with melting and vaporization) and negative for the reverse. E.g. determine the sign of S for the following changes (what about the reverse of these?):

N2(g) + 3H2(g) 2NH3(g)CaCO3(s) Ca(s) + CO2(g)H2(g) + O2(g) H2O(l)H2O(l) H2O(s)


Chapter 19-8

8–8

Entropy and the Second Law

• Second Law: all systems tend towards an increase in the randomness of the system. In terms of entropy, the total entropy of the universe always increases. Stotal =Ssys + Ssurr 0. Srev = Sfwd.–

S of phase change determined from heat of phase change. S = H/T.

E.g. Determine S of melting and vaporization for water. Hfus = 6.0kJ/mol and Hvap = 40.8 kJ/mol.

• Generally, Sfus < Svap. Entropy of s < l < g.• Since S = H/T, we can determine temperature of phase

change (mp or bp) if H and S of phase change are known.E.g. Determine boiling point of chlorine if Hvap = +6.41kJ/mol & Svap =+37.3 J/Kmol.

T

HSsurr


Chapter 19-9

8–9

Entropy and Probability

• Randomness is related to the probability of finding a molecule in a particular microstate; E.g. determine the probability of having perfectly order system (HH) when:

• Two microstates (heads, H, and tails,T) of same exact energy (same likelihood of existing).

• Two particles. Then the possible states are: HH, TT, TH, HT. Probability, P, of perfectly order system 25% (PHH = 22*100%).E.g.2 determine PHHH for 3 particles.

• Possible states: HHH, TTT, HHT, HTH, THH, HTT,THT, TTH. (PHHH = 23*100%).

• Ludwig Boltzmann = statistical approach to entropy, S, given by S = klnW where k = Boltzmann's const. = R/N = 1.38x1023J/K.

• W = total # of ways that all atoms in a sample can be arranged and still have same energy.

E.g. Determine molar entropy of CO (2 states) and HCl (1 state) molecules in a crystal.


Chapter 19-10

8–10

Entropy and Probability

• Third Law = perfectly ordered crystal has zero entropy.

• entropy change when molecules filling a container after expansion is related probability of this expansion and is given by the relationship (constant T and n):

E.g. determine the entropy change that occurs when a gas expands to twice its initial volume.

E.g.2 twice it’s initial pressure.

final

initial

initial

final

P

PlnR

V

VlnRS


Chapter 19-11

8–11

Entropy and Temperature

• At absolute zero there is no molecular motion and the S of a perfectly order system is zero.

• Entropy expressed in absolute quantities.• Entropy increases with temperature.• Entropy increases dramatically as the melting and boiling points.• Standard Molar Entropies, So,& So.• Standard molar entropy, So: entropy of 1 mol of a compound at 1 atm

and 25°C. • Entropy of a compound is the total entropy. Not relative to the

reference state.• As with Ho it is convenient to present all entropies at reference set of

conditions.• As with Ho, So of the reaction can be determined. For general

reaction aA +bB + cC + ... mM + nN + oO + ...So = (mSM + nSN + ...) (aSA + bSB + ...).

E.g. determine So of the reaction using data in the table from the text.:NH3(g) 3/2 H2(g) + ½ N2(g)


Chapter 19-12

8–12

GIBB'S FREE ENERGY

• Spontaneity can also be expressed without Ssurr. • Recall: Stotal = Ssys + Ssurr 0 and

Ssurr = H/T.– Combining: Stotal = Ssys H/T 0 or TStotal = G = H TSsys 0

• Gibb's Free Energy, G and in general terms is G = H TS. Criterion for spontaneity is G 0. Always found to be true.

• If H and S are known, spontaneity of reaction can be determined.• Possible signs of enthalpy and entropy:

E.g. determine Go for the melting of water. Ho = 6.03 kJ/mol; So = 22.1 J/K at 10°C.

H S G Spontaneous? or + At low T (enthalpy driven). + Always. + + Never. + + or + At high T (entropy driven.


Chapter 19-13

8–13

G – Temperature Dependence

• Spontaneity: Reaction becomes spontaneous when G goes from + to . We use G = 0 to tell us when reaction just becomes spontaneous or 0 = H TS or T = H/S.

E.g. determine the temperature at which the synthesis of HI(g) becomes spontaneous.

Ho = +52.96 kJ and So = +166.4 J/mol

H2(g) + I2(g) 2HI(g)


Chapter 19-14

8–14

Standard Free Energy Changes for Reactions

Free Energy is a state function and an extensive property so that its value is dictated by the number of moles appearing in the equation.

½ N2(g) + 3/2 H2(g) NH3(g) Go = 16.5 kJ

N2(g) + 3 H2(g) 2 NH3(g) Go = 33.0 kJ


Chapter 19-15

8–15

Standard free energies of formation, GfO.

Go = 0 for elements in most stable form (as with enthalpy).• Free Energies are listed in tables and can be used to determine

G of reaction and spontaneity.• For general reaction aA +bB + cC + ... mM + nN + oO + ...

E.g. Determine G for

4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g).• As with enthalpy reactions can be coupled to determine G for

unknown reaction. • Determine G: Fe2O3(s) + 3/2C(g) 3/2CO2 + 2Fe(s). • Given:

– Fe2O3(s) 2Fe(s) + 3/2O2(g) Go = +742 kJ – 3/2 C(gr) +3/2O2(g) 3/2CO2(g) Go = 592 kJ.

oC,f

oB,f

oA,f

oO,f

oN,f

oM,f

orxn GcGbGaGoGnGmG


Chapter 19-16

8–16

EQUILIBRIUM CONSTANTS AND G

• Equilibrium constant for a reaction aA + bB +... mM + nN + ... is defined as

• Tells how far to right reaction proceeds. – Large value mostly products.– Small value mostly reactants.

• At equilibrium this equation must always be obeyed no matter what relative amount of reactant and was started with.

• Reaction quotient, Q, defined in same way as K but refers to some initial condition that might not be at equilibrium.

• These two are related to G by the equation:• G = RTln(Q/K).

When Q = K equilibrium exists and G = 0.

...][B][A

...][N][M =K

ba

nm


Chapter 19-17

8–17

EQUILIBRIUM CONSTANTS AND G°

• K and Q often separated by rewriting:G = Go + RTlnQ and Go = RTlnK.

• From initial conditions and Go, the free energy of the reaction can be determined.E.g. determine G at 25°C for the reaction: H2O(l) H+ (aq) + OH(aq) if the concentration of [H+] = 1.00x105 M; [OH] = 1.00x105 M

• K can be determined under standard conditions if G is known.E.g. Determine K for 3NO(g) N2O(g) + NO2(g)E.g.2 Determine K for NH4NO3(s) N2O(g) + 2H2O(g).

Go can be determined if K is known.

E.g. determine Go for the solubility of AgCl(s) if its Ksp = 1.8x1010;E.g. 2 determine Go for ionization of HF in water. Ka = 3.5x104.

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