∆rG = ∆rG0 + RTln( Q/ Q0), where:

=C D

A B

n nDC

n nA B

C CQC C

1

Q0 is always 1:1 bar, 1 molar, or X (molefraction) = 1 (pure liquid or solid)

except for the biochemical standard state where C(H+ ) = 10-7 M

Calculate ∆rG for any reaction nA A + nBBnCC +nDD at any concentrations:

361 Lec18Mon 1oct18

This whole term =the change in-T ∆S due to concentrations;= Lechatelier part: tells which way it goesActual ∆G

Standard state value:(a constant) all concentrations =1(get from Table A5)

( ) ( )( ) ( )

1= =C D

A B

n n0 0C D0

n n0 0A B

C CQ

C C

Actual conc.(variable)

All C0 = 1(constant)

∆G = ∆G0 + RTln(Q/Q0)what are the WORDS?“∆G is ∆G when all concentrations are 1 and ideal, plus a correction for actual concentrations, RTln(Q/Q0) which is a correction to -T∆S

At equilibrium: ∆G =0; so ∆G0 = -RTln(Qeq/Q0)which is the same as ∆G0 =-RT2.3log (Qeq/Q0)

∆G0 = -2.3RTlog (Qeq/Q0) At 298 K: ∆G0 = - (2.3)(8.3145)(298)log (Qeq/Q0)

-5700 log (Qeq/Q0) in J-5.7 log (Qeq/Q0) in kJ

Qeq/Q0 = K/1 = K , the equilibrium constantTherefore: ∆G0 = -5.7log (K) (at 298K)

2

361 Lec 18Mon, 1oct18

∆G0 = -2.3RTlog (Qeq/Q0) = -5.7log (K)

3

∆G0

+17.1+11.4+ 5.7

0.0- 5.7

- 11.4- 17.1

K = ? =10-∆G0/5.7

K=10-∆G0/5.7 at 298K and IDEAL

10-3 = 0.00110-2 = 0.0110-1 = 0.110-0 = 1.010+1 = 1010+2 = 10010+3 = 1000

K=10-∆G0/2.3RT solves for equilibrium concentrations only if ideal! ∆G = ∆G0 + RTln(Q/Q0) solves for the correct ∆G only if ideal!

4

Only ACTIVITIES instead of concentrations gives the exact ∆G

What exactly are activities?

K=10-∆G0/2.3RT solves for exact equilibrium concentrations only if ideal!

∆G = ∆G0 + RTln(Q/Q0) solves for the exact ∆G only if ideal!

5

Activity = a = γCwhere C = concentrationand γ = activity coefficient

activity coefficients come from experimentalmeasurements of the equilibrium concentrations

ormeasurements of the reversible work provided by

the reaction.

Next: THE REAL Q

6

THE REAL Q:

7

ideal =C D

A B

n nDC

n nA B

C CQC C

( ) ( )( ) ( )

a aa a

γ γ

γ γ= =

C DC D

A B A B

n nn n

C C D D C DREAL n n n n

A BA A B B

C CQ

C C

( ) ( )( ) ( )

0 lnG G RTγ γ

γ γ

∆ = ∆ +

C D

A B

n nC C D D

n nA A B B

C C

C C

If activity coefs =1 then we have IDEAL behavior.What happens to ∆G if the activity coef. of the products are less than 1?More negative, more spontaneous= shifts to right, What happens to ∆G if the activity coef. of the products are greater than 1?More positive, LESS spontaneous= shifts to LEFT

Activity coefficient of ions in solution:is about ion atmospheres:

crystal: perfect correlation

Real solution: good correlation

strong negative terms weak positive terms

Net negative ∆G for interaction.Will activity coef. be greater or less than 1?

9From G. Barrow, Physical Chemistry, 3rd Ed. 1973, page 693

10

γ=0.5 is like saying the concentrations of H+ andA- are less. LeChateliersays this will shift the equilibrium to the right.

Electrostatic attractionmakes ∆G more negative;The reaction goes furtherright than expected fromideal behavior. Actualconcentration is 0.2M foreach ion.

11

!!!??? Really?

Importance of the “Biochemical Standard State”

This H+is buriedin the Pi

12

ATP + H2O ADP + H+ + Pi

2

i

H O00

[ADP][H ][P ][ATP][X ]

G G ln 9 ln1

QRT RTQ

+

∆ = ∆ + = + +

2

7i

H O0'7 70'

[ADP][H ][P ] (1)1x10 )[ATP][X ] (1)(1)G G ln ln ln 31 /

(1)1x10 ) (1)1x10 )(1)(1) (1)(1)

QRT RT RT kJ molQ

+ −

− −∆ = ∆ + = + = = −-31

That is the same as finding the biological standard ∆G = ∆G0’

What does LeChatelier say if we are at pH 7 and other activities = 1 ?

317 −=−+=−×++= 9.399)(7.59

Let’s SEPARATE the H+ from the Pi

7

0' 00

[1]1x10 [1][1][1]G G ln 9 2.3 log 31 /

1QRT RT kJ molQ

−

∆ = ∆ + = + = −

Using the biological standard, ∆G0’, find ∆G if at pH 7 if all otheractivities = 1

Reaction becomes more spontaneous (equil. shifts RIGHT)

2

70' 7i

H O

[ADP][H ][P ] (1)(1 10 )(1)Q 1 10[ATP][X ] (1)(1)

+ −−×

= = = ×

13

The famed “high energy phosphate bond” yields littlemore free energy than a hydrogen bond!

What is K when ∆G0’ = -31 kJ/mol ?

K = 10 ∆G0’/5.7 = 10-31/5.7= 2.7 x 105

i.e., enough but not too much.

14

LeChatlier:adding heat (raising temperature) shifts

equilibrium (changes K) in the direction thatuses the heat (lowers temperature).

Endothermic (∆H = +) absorbs heat; rxn will therefore use heat and lower the T, i.e.,will shift to the right (K increases with rising T)

Exothermic (∆H = negative) gives off heat and raise the T; therefore shift to the left (K decreases with rising T)

How does Kequil depend on temperature?

15

Saying the same thing with ∆G0

You will do this for theLab experiment on the dissolving of butanol in water.

divide by -RT and call it T2

do same for T1