General Chemistry revisited - jila.colorado.edujila.colorado.edu/~wcl/Chem4521/images/Second Law Thermo 8 4521... · General Chemistry revisited A(g) ... the effects of temperature

General Chemistry revisitedA(g) + B(g) C(g) + D(g)

We said that G = H –TSh H H(C) + H(D) H(A) H(B)where, eg, H = fH(C) + fH(D) - fH(A) - fH(B)

G < 0 implied spontaneous to rightG > 0 implied spontaneous to left

In a very unconnected way, we discussed the equilibrium

A(g) + B(g) C(g) + D(g)with equilibrium constant C D

eqA B

P PKP P

We talked about the reaction proceeding in both directions to establish equilibrium.

A B

ie, proceeding both when ΔG < 0 and when ΔG > 0 !!!

What’s going on??

General Chemistry revisitedA(g) + B(g) C(g) + D(g)

We said that ΔG = ΔH –T ΔSWe said that ΔG ΔH T ΔSwhere, eg, ΔH = ΔfH(C) + ΔfH(D) - ΔfH(A) - ΔfH(B)ΔG < 0 implied spontaneous to rightΔG > 0 impli d sp nt n us t l ftΔG > 0 implied spontaneous to left

The description above basically treats A B C and D as pure componentsSo where is equilibrium? What’s wrong?

The description above basically treats A,B, C and D as pure componentsthat never mix. That is to say, as though the composition is constant.This has to be made more sophisticated if we are to couple this conceptwith that of equilibrium.with that of equilibrium.

We now know about chemical potentials and molar Gibbs energies,so that we can follow G as the composition changes, gaining a muchmore sophisticated viewpoint

We now do just that!

more sophisticated viewpoint.

The importance of free energy of mixingfree energy of mixing

G(A)No Mixing

A B

A BG = (1-nA)GB –nAGA < 0

With Mixing

P P

G(B)

APA, PB

P ln lno B

rA

PG RT K RTP

Gibbs Energy Minimum

The composition that makes the chemical potentials equalis the equilibrium composition.

ConcepTest 12 NO (g) N O (g) 300K

A. For the above reaction, find K°C if K°P = 6.9 bar–1

2 NO2 (g) N2O4 (g) 300K

RT=0 083 bar dm3 K-1 mol-1 x 300 K = 24 9 bar dm3 mol-1

A 172 dm3/molK°C = K°P/(RT) Δn

RT=0.083 bar dm K mol x 300 K = 24.9 bar dm mol

C 0.28 mol/dm3A 172 dm /molB 0.28 dm3/mol

C 0.28 mol/dmD 172 mol/dm3

ConcepTest 12 NO (g) N O (g) 300K

A. For the above reaction, find K°C if K°P = 6.9 bar–1

2 NO2 (g) N2O4 (g) 300K

RT=0 083 bar dm3 K-1 mol-1 x 300 K = 24 9 bar dm3 mol-1

A 172 dm3/molK°C = K°P/(RT)Δn Here Δn = -1

RT=0.083 bar dm K mol x 300 K = 24.9 bar dm mol

C 0.28 mol/dm3A 172 dm /molB 0.28 dm3/mol

C 0.28 mol/dmD 172 mol/dm3

Sample ProblemIf 1 00 mol NO2 is injected into a 1 dm3 vessel at 300 K

K°C = 172 dm3/mol and K°P = 6.9 bar–1

If 1.00 mol NO2 is injected into a 1 dm vessel at 300 K,find [N2O4] at equilibrium.

2 NO2 (g) N2O4 (g) 300KAt equilibrium, x (1-x)/2 moles of each

Assuming gases are ideal, then PNO2 nNO2

and PN2O4 nN2O4

We use K°C but could have used K°P1 x mol

2344 1 orx xSimplifying33

22

3

2172 dmdmmol molx

dm

2

344 1 or344 1 0

x xx x

Simplifying,

Answer: an additional constraint: 0< x < 1

Using my quadratic equation solver, I find x = 0.05248 and x =-0.0554Two solutions!!Are both right? x= xNO2 = 0.05248 mol/dm3 NO2

xN2O4 = (1-x)/2 = 0.474 mol/dm3 N2O4

Are both right?Are both wrong?Did I make a mistake?

Effects of Changes on EquilibriaEff f g EqThe Le Chatelier Principle results from consideration of the effects of temperature pressure or quantity the effects of temperature, pressure or quantity changes on an equilibrium constant. E.g.,

2 NO2 (g) N2O4 (g)The reaction shifts to the left when the # moles of N2O4 is increased

2 NO2 (g) N2O4 (g)

of N2O4 is increased.The reaction shifts to the right when the total pressure is increased.

We now consider in a bit more detail pressure and temperature effects on equilibria.p q(Reasons that we might do this include having tests forequilibrium, and to have ways to measure H or S.)

Effects of Pressure on EquilibriaEff f Eq

G° = –RT ln K°PWe know PΔG is defined at P = P = 1 bar, and so it is independent of pressure. (ideal gases assumed)

For the gas-phase reaction aA + bB yY + zZ

y zY Zo

P b

P PK

Obviously, KP must also be pressure independent. P a b

A B

KP P

y P p p

There are three ways that one could imagine changing the pressure:a) Add an inert gas to the reaction mixtureb) Change the volume of the reaction mixtureb) Change the volume of the reaction mixturec) Change the quantity of one of the components (eg, A,B, Y or Z in above)

Take a brief look at each of them.

Effect of adding an ideal inert gas, M

aA + bB yY + zZ

G° = –RT ln K°P

y zY Zo

P a b

P PK

P PAdd M

If there is no change in volume then the partial pressures of each of the ideal

A BP P

mM+ aA + bB yY + zZ+ mM

Add gas M

y z mY Z Mo P P P

K

If there is no change in volume, then the partial pressures of each of the ideal gas components remains unchanged by the addition of M. No change

in any partialpressures

Y Z MoP a b m

A B My z

KP P P

P P

pressures.K is unchanged!

However, M might If the reaction weretaking place in an

Y Za b

A B

P P

P P

, gwell affect therate at which theprocess occurs.

isolated system, M wouldalmost surely also affectthe final temperature

Change the volumeg m

Changing the volume can have an effect if the number of moles of gas on the left and right sides of the equation are different.

N2O4 2 NO2

N(1 ) 2N N i l

Let be the fraction of thedimer that has dissociated.

N(1-) 2N N in moles

Total number of moles of gas = N(1 - ) + 2N = N(1 + )g ( ) ( )We now write the mole fractions of each component:

1 1 2 2N N

2 4 2

1 1 2 21 1 1 1N O NO

N NN N

Change the volume (2)g m ( )

We use the mole fractions to obtain the partial ppressures of each component:

21P P P P

2 4 21 1N O NOP P P P

Use these to write the pressure equilibrium constantUse these to write the pressure equilibrium constant

2

2 2 2

2

42 11 11

NOP

N O

PK P P

P P

2 4

1 11N OP P

Solve for : 4

PKK P

The degree of dissociation decreaseswhen the pressure increases. 4PK P p

Another manifestation of the Le Chatelier Principle

Change quantity of one componentg q y f mp

2 NO ( ) N O ( )2 NO2 (g) N2O4 (g)

2 4N OPK 2 4

2

2PNO

KP

Th ill hif i di i f h h The system will shift in a direction away from the component that was increased.This is a method that is often used to determine whether or not equilibrium has actually been attainedequilibrium has actually been attained.(Rather than simply having an equilibration time >> observation time)

Effect of Temperature on EquilibriumEff f mp Eq m

G° = –RT ln K°PWe know

0 0G H

PWe couple this ΔG relationship with the Gibbs-Helmholtz equation derived in Chapter 3,

2 (3.169)P

G HT T T

to obtain an expression that gives the temperature dependence of K°Pp g p p P

0 0

2

ln the van't Hoff equationPd K H

dT RT

0ld

0 0

2

ln1Since , we can rewrite, giving 1 /

Pd KdT HdT RT d T

This last form shows that a plot of ln KP vs 1/T has slope -H/R

Effect of Temperature on Equilibrium Eff f mp Eq mThis last form shows that a plot of ln KP vs 1/T has slope -H/R,if H does not depend strongly on temperature.if H does not depend strongly on temperature.

Note that the positive slopemeans that H is negative, an

H H h lm g ,exothermic reaction. Here H has a large

T dependence.

Can the slope be negative? What would a negative slope imply?

The Y – intercept of this line is also interesting!

Effect of Temperature on Equilibrium

The Y – intercept is also interesting!W k th t

Eff f mp Eq m

We know that

0 0 0G

G° = –RT ln K°P

So

The y-axis is located at

0 0 00 1ln P

G H T SKRT R T RT

1 0The y axis is located at

So the value of the intercept is S/R

0T

S/RThese plots can be a very useful wayto determine basic thermochemical S/Rproperties of molecules and reactions.

Sample Problem Consider the reaction

4 NH3 (g) + 3 O2 (g) 2 N2 (g) + 6 H2O(g) with the dimensionless equilibrium constant Kp

o = 2.84 104 at 298 K. A reaction flask is filled to a pressure of 1.0 bar with gases at the following partial pressures: P(NH3) = 0.10 bar, P(O2) = 0.10 bar, P(H2O) = 0.60 bar, and P(N2) = 0.20 bar. The gases are well mixed. You may further assume that the total pressure of the reaction flask is maintained constant, the gases behave ideally, and standard state corresponds to partial pressures of 1 bar of all gases.Compared with the initial composition will the flask contain fewer reactants or Compared with the initial composition, will the flask contain fewer reactants or fewer products when it reaches equilibrium?

6 26 240.60 0.20

1 87 10H O NP PQ

2 2

3 2

44 3 4 3 1.87 10

0.10 0.10H O N

NH O

QP P

Q < K so by Le Chatelier’s Principle products will be produced Q < K, so by Le Chatelier s Principle, products will be produced as the reaction proceeds to equilibrium.Therefore, fewer reactants will be present at equilibrium.

Sample Problem Consider the reaction 4 NH3 (g) + 3 O2 (g) 2 N2 (g) + 6 H2O(g) with the dimensionless equilibrium constant Kp

o = 2.84 104 at 298 K. A reaction flask is filled to a pressure of 1.0 bar with gases at the following partial pressures: P(NH3) = 0.10 bar, P(O2) = 0.10 bar, P(H2O) = 0.60 bar, and P(N2) = 0.20 bar. The gases are well mixed. You may further assume that the total pressure of the reaction flask is maintained constant, the gases behave ideally, and standard statepressure of the reaction flask is maintained constant, the gases behave ideally, and standard state corresponds to partial pressures of 1 bar of all gases.

Calculate G for this reaction for the concentrations and temperature given above. What does this value imply about the spontaneity of this point in the reaction?What does this value imply about the spontaneity of this point in the reaction?

4ln 0.0083 298 ln 2.84 10 25.4o op

kJ kJG RT K Kmol K mol

4

ln

25 4 0 0083 298 ln 1 87 10

oG G RT QkJ kJ K

25.4 0.0083 298 ln 1.87 10

1.04

Kmol mol KkJ

l

mol

Since G < 0, this reaction is spontaneousat the given concentrations and temperature

Sample Problem Consider the reaction 4 NH3 (g) + 3 O2 (g) 2 N2 (g) + 6 H2O(g) with the dimensionless equilibrium constant Kp

o = 2.84 104 at 298 K. A reaction flask is filled to a pressure of 1.0 bar with gases at the following partial pressures: P(NH3) = 0.10 bar, P(O2) = 0.10 bar, P(H2O) = 0.60 bar, and P(N2) = 0.20 bar. The gases are well mixed. You may further assume that the total pressure of the reaction flask is maintained constant, the gases behave ideally, and standard statepressure of the reaction flask is maintained constant, the gases behave ideally, and standard state corresponds to partial pressures of 1 bar of all gases.

What will be the value of G at equilibrium? Gi t f j tifi tiGive one sentence of justification.

G = 0 at equilibriumG = 0 at equilibriumAt equilibrium, the reaction is neither spontaneous (G < 0)nor non-spontaneous (G > 0).

Sample Problem (2) Use the thermochemical data below to estimate the boiling temperature of hydrazoic acid, HN3.Thermodynamic data (Standard state: 298 K, 1 bar)Hf

oHN3 (liquid) = 264.0 kJ mol-1 Sfo HN3 (liquid) = 140.6 J K-1 mol-1

Hfo HN3 (gas) = 294.0 kJ mol-1 Sf

o HN3 (gas) = 238.97 J K-1 mol-1

Density: (HN3 (liquid)) = 1.09 kg dm-3Density: (HN3 (liquid)) 1.09 kg dm

3 3 The vaporization reactionThe boiling temperature is when the above Rxn is in equilibrium at 1 bar. HN HN g

g p q

0 at boiling temperaturerxno o

rxn rxn

GH T S

0o

o o rxnrxn rxn o

HH T S TS

rxnS

3 3 294.0 264.0 30.0o o orxn f f

kJ kJH H HN g H HNmol mol

30 0 kJ

3 3 238.97 140.6 98.4o o orxn f f

J JS S HN g S HNmolK molK

30.0305

0.0984boil

molT KkJ Kmol

Why is this only an estimate?