KF lecture

7/27/2019 KF lecture

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1

Lecture 10: The Meaning of ΔG

• Reading: Zumdahl 10.10-10.13

• Outline

– A descriptive example of ΔG – Temperature Dependence of K

– van’t Hoff Relation for K and T.

• Problems:

– Z10.6, Z10.71 (Demo), Z10.72, Z10.74

– Z10.77-79, Z10.81, Z10.90 (qualitative)



2

A Descriptive Example

• Consider the following gas-phase

“reaction”:

OClO (g) ClOO (g)

• Now, ΔGrxn = ΔG°rxn +RTln(Q)= -RTln(K) + RTln(Q)

= RTln(Q/K)



3

Review

OClO (g) ClOO (g)

Using ΔGrxn = RTln(Q/K)

Qualitatively:

If Q/K < 1 ΔGrxn < 0

If Q/K > 1 ΔGrxn > 0

OClO (g) ClOO (g)

OClO (g) ClOO (g)

If Q/K = 1 ΔGrxn = 0 equilibrium

Both Forms Exist in equimolar amounts



4

Sample Calculations

• K = 0.005

• Reaction (as written) is Not Spontaneous in

standard state.

ΔGrxn = RTln(Q/K)

= RTln(.001/.005) = RTln(.2) = -1.6RT < 0

Q = PClOO/POClO = .001

ΔG°rxn = -RTln(K) = +13.1 kJ/mol

• Let’s start the rxn off with 5 atm of OClO and

0.005 atm of ClOO. (Lots of Reactants.)

ΔGrxn < 0, reaction is proceeding towards

products, at fixed T and fixed Total Pressureof 5.005 atm.



5

Finding Equilibrium

• Where is equilibrium? Back to Chem 142 and

ICE table for OClO (g) ClOO (g)

x = 0.02

POClO = 5 - .02 = 4.98 atm

( )

PClOO = 0.025 atm

( )

( )

0.005

(5 )

@ 0.005

ClOO

P

OClO

P

xP

Q x P x

EQ Q x K

+

= = −

= =

Initial Pressure StoicCoeffs

Change New Conc

P(OClO)=5 1(-) -x 5-x

P(ClOO)=0.005 1(+) +x 0.005+x

For this reaction the total pressure will not change as moles of reactant go to product.



6

Change initial conditions

ΔGrxn = RTln(Q/K) Q = PClOO/POClO = 10

• Let’s start the rxn off with 5 atm. of ClOO and

0.5 atm of OClO, total pressure now 5.5 atm.

= RTln(10/0.005)

= RTln(2000) >> 0

ΔGrxn > 0, reaction is proceeding towards

reactants (i.e. backwards) with about 8kJ/mole

of driving energy.



7

Temperature Dependence of K

• K cannot depend on pressure (it is define for 1 Atm of eachspecies)

• However it can depend on temperature (here’s how)

• We have two definitions for ΔG°

• Rearranging:

Plot of ln(K) vs 1/T is a straight line, the slope gives the enthalpy and

the intercept (at infinite T) gives the entropy for the reaction. Weassume Enthalpy and Entropy are independent of Temperature.

lno o o

rxn rxn rxnG RT K H T S Δ = − = Δ − Δ

1ln

o o

rxn rxn H S

K R T R

y m x b

Δ Δ= − ⋅ +

= +



8

Plot lnK(T)

• If we measure K as a function of T, we candetermine ΔH° by determining the slope of theline

ln(K

)=

−Δ H °

R

1

T

⎝ ⎜

⎠⎟+

ΔS °

R

slope

intercept

Work Problems 10.72 and 10.74



9

T Dependence of K

• Once we know the Temperature dependence of K,we can predict K at another temperature:

2 1

2 1

2 12 1

2

1 2 1

1 1ln( ) ln( )

1 1

ln( ) ln( )

1 1ln

H S H S K Subtct K R T R R T R

H

K K R T T

K H

K R T T

⎛ ⎞ ⎛ ⎞−Δ ° Δ ° −Δ ° Δ °= + = +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

⎧ ⎫⎛ ⎞ ⎛ ⎞−Δ ° ⎪ ⎪

− = −⎨ ⎬⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎩ ⎭

⎛ ⎞ ⎧ ⎫−Δ °⎛ ⎞= −⎨ ⎬⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎩ ⎭

The van’t Hoff equation.



10

K(T) For a Specific Reaction

• For the following reaction:

CO(g) + 2H2(g) CH3OH(l)

What is K at T=T2=340 K?

• First, what is K eq when T =T1= 298 K?

( ) 5

1 1

30 ln 2.48ln

29ln 11.7

2.48

1.2 10

o

rxnG RT K K

K

K K T

Δ = − = − = −

= =

= = ⋅

30o

rxnkJ G

molΔ = −



11

Calculate K(T)

To use the van’t Hoff Eq., we need Δ

H°

CO(g) + 2H2(g) CH3OH(l)

ΔHf °(CO(g)) = -110. kJ/mol

ΔHf °(H2(g)) = 0

ΔHf °(CH3OH(l)) = -240 kJ/mol

ΔH°rxn = ΣΔH°f (products) - ΣΔH° f (reactants)

= ΔH°f (CH3OH(l)) - ΔH°f (CO(g))= -239 kJ - (-110.5 kJ)

= -130 kJ

130o

rxnkJ H

molΔ = −

K(T)



12

K(T)

• With ΔH°, we’re ready for the van’t Hoff Eq.

( )

{ }

2 1

1 2 1 1 2

32

1

3 5 3

2 1

1 1ln 1

130 2981 52 1 .88 6.32.48 340

exp 6.3 1.8 10

1.8 10 1.2 10 1.8 10 200

o

rxn H K T H

K R T T RT T

K

K

K K

−

− −

⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ−Δ °= − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠

− ⎛ ⎞= − = − − = −⎜ ⎟⎝ ⎠

⎛ ⎞= − = ⋅⎜ ⎟

⎝ ⎠= ⋅ ⋅ = ⋅ ⋅ ⋅ =

Why is K reduced (a lot!)?Reaction is Exothermic!

LeChatelier: Increase T, Shift Eq. To React.

K eq (T) will then decrease with increasing T;

Entropy is not part of the reason for the Change in K with T



13

The Entropy for this reaction

• Compute the Entropy change for the reaction

• To save problems with dimensions (becauseenergies are in kJ and entropy is in J/K)

3 3130 30 10010 10 330298 300

o oo o o o rxn rxnrxn rxn rxn rxn

orxn

H GG H T S S

T

J S mol K

Δ − ΔΔ = Δ − Δ ⇒ Δ =

− +Δ = ⋅ = − ⋅ = − −

@ 25 2.48

130 30 402.48

o o

o rxn rxnrxn

kJ T C RT mol

H GS R R R RT

= ° =

Δ − Δ − +⎛ ⎞Δ = ⋅ = =⎜ ⎟⎝ ⎠

This is a better way to report entropy; all units are in R; uses the

ratio of kJ in energies; less prone to mistakes.

G l C l i



14

General Conclusions

• K and ΔGo

are independent functions of T.• How K changes with T depends only on the sign

of the Enthalpy

• How ΔGo changes with T depends only on the

sign of the Entropy.

• Principles of LeChatelier are born out by van’tHoff equation.

• Now know where the equilibrium constants came

from and why and how they depend on thereaction energy (viz. Enthalpy)



15

Gibbs Energy and Work (at fixed P and T)

• Preparation to talk about batteries (next chapter)

• From Calorimetry: to actually get heat (or an

enthalpy change) you actually have to change the number of

moles of something (X).

• Now to get work you need to consume some amount of

reactant (X) :

• This shows that to do work you need a (chemical) potential

times a (chemical) distance. And the work you get is

proportional to the amount of stuff you use up.

• The “useful” work (text) is non-PdV work, and is electrical

work; like a battery

• Calling the work useful and not useful (or PdV) work, whichthe book does, is not useful. (Sec 10.12-13).

rxn H H X Δ = Δ ⋅

max

electrical

rxnw G G X = Δ = Δ ⋅

ΔG d W k (Fi d T d P)



16

ΔG and Work (Fixed T and P)

• Consider a reversible reaction, and work is the sum of thePdV part and extra work. Then

) ( )

( ) x x

rev rev rev rev

x nonPdV

rev rev

G H TS E PV T S

q P V w P V q w

G w w

Δ = Δ − Δ = Δ + Δ − Δ

= − Δ + + Δ − =

Δ = =

We previously showed that the reversible work is the

maximum work (as a negative number) you can getout of a system. Now see Gibbs energy is the same as

the maximum non-PdV work (i.e. electrical) we can

get from a chemical reaction when run at constant Tand P.

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