Useful Information

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ΔE =q+ w

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ΔE =nCvΔT

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ΔH =nCpΔT

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w=−PextΔV

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wrev=−nRTlnVfinalVinitial

⎛

⎝ ⎜

⎞

⎠ ⎟

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ΔS=dqrevT∫

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S=kln Ω( )

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Ω= A!ai!

i∏

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ΔS =nRlnVfinalVinitial

⎛

⎝ ⎜

⎞

⎠ ⎟

€

ΔS=nCP lnTfinalTinitial

⎛

⎝ ⎜

⎞

⎠ ⎟

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ΔS=nCv lnTfinalTinitial

⎛

⎝ ⎜

⎞

⎠ ⎟

€

ΔG=ΔG°+RT (lnQ) and

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K=Q ( )at equilibrium

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aA+bB ⏐ → ⏐ ← ⏐ ⏐ cC+ dD

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K =C[ ]

c D[ ]d

A[ ]a B[ ]

b

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ΔH°rxn= cΔH°fproducts∑ − cΔH°f

reactants∑

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ΔS°rxn= cS°products∑ − cS°

reactants∑

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ΔG°rxn= cΔG°fproducts∑ − cΔG°f

reactants∑

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(lnK)=−ΔH°R

1T ⎛ ⎝ ⎜

⎞ ⎠ ⎟+ΔS°R

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lnK2

K1

⎛

⎝ ⎜

⎞

⎠ ⎟=−ΔH°R

1T2

−1T1

⎛

⎝ ⎜

⎞

⎠ ⎟

Lecture 10: ΔG, Q, and K

• Reading: Zumdahl 10.10, 10.11

• Outline– Relating ΔG to Q– Relating ΔG to K– The temperature dependence of K

Relating ΔG to Q

• Recall from Lecture 6: ΔS = R ln (Ωfinal/Ωinitial)

• For the expansion of a gas⏐⏐⏐ Ωfinal Volume

Relating ΔG to Q (cont.)

• Given this relationship ΔS = R ln (Vfinal/Vinitial)

€

ΔS = R ln

nRTPfinal

nRTPinitial

⎛

⎝

⎜ ⎜ ⎜

⎞

⎠

⎟ ⎟ ⎟= R ln

PinitialPfinal

⎛

⎝ ⎜

⎞

⎠ ⎟= −R ln

PfinalPinitial

⎛

⎝ ⎜

⎞

⎠ ⎟

Relating ΔG to Q (cont.)

• This equation tells us what the change in entropy will be for a change in concentration away from standard state.

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ΔSfi = ΔSfi°− R lnPfinalPinitial

⎛

⎝ ⎜

⎞

⎠ ⎟

Entropy change for processoccurring under standardconditions

Additional term for changein concentration.

(1 atm, 298 K) (P ≠ 1 atm)

Relating ΔG to Q (cont.)

• How does this relate to ΔG?

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ΔGrxn = ΔH rxn − TΔSrxn

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ΔGrxn = ΔH orxn − TΔSorxn + RT ln

PfinalPinitial

⎛

⎝ ⎜

⎞

⎠ ⎟

€

ΔGrxn = ΔGrxno + RT ln

PfinalPinitial

⎛

⎝ ⎜

⎞

⎠ ⎟

Relating ΔG to Q (cont.)

€

ΔGrxn = ΔGrxno + RT ln

PfinalPinitial

⎛

⎝ ⎜

⎞

⎠ ⎟

€

ΔGrxn = ΔGrxno + RT ln Q( )

• Generalizing to a multicomponent reaction:

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Q =

C[ ]

C[ ]reference

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

coeff

prod .

∏

C[ ]

C[ ]reference

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

coeff

react .

∏

• Where

An Example

• Determine ΔGrxn at 298 K for:

C2H4(g) + H2O(l) C2H5OH(l)

where PC2H4 = 0.5 atm (others at standard state)

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ΔGrxn = ΔGrxno + RT ln Q( )

ΔG°rxn = -6 kJ/mol (from Lecture 9)

An Example (cont.)C2H4(g) + H2O(l) C2H5OH(l)

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ΔGrxn = ΔGrxno + RT ln Q( )

ΔGrxn = -6 kJ/mol + (8.314 J/mol.K)(298K)ln(2)

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Q =1PC2H4

1atm

⎛

⎝ ⎜

⎞

⎠ ⎟

=1

0.5atm

1atm

⎛

⎝ ⎜

⎞

⎠ ⎟= 2

= -4.3 kJ/mol

ΔG and K

• The Reaction Quotient (Q) corresponds to a situation where the concentrations of reactants and products are not those at equilibrium.

• At equilibrium, we have K.

• What is the relationship between ΔG and K?

ΔG and K (cont.)

• At equilibrium, ΔGrxn = 0

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ΔGrxn = ΔGrxno + RT ln Q( )

0 K

0 = ΔG°rxn +RTln(K)

ΔG°rxn = -RTln(K)

ΔG and K (cont.)• Let’s look at the interaction between ΔG° and K

ΔG°rxn = -RTln(K)

If ΔG° < 0

then > 1

Products are favored over reactants

reactants

products

ΔGrxn

ΔG and K (cont.)• Let’s look at the interaction between ΔG° and K

ΔG°rxn = -RTln(K)

If ΔG° = 0

then = 1

Products and reactants are equally favored

reactants

products

ΔGrxn = 0

ΔG and K (cont.)• Let’s look at the interaction between ΔG° and K

ΔG°rxn = -RTln(K)

If ΔG° > 0

then < 1

Reactants are favored over products

reactantsΔGrxn

An Example

• For the following reaction at 298 K: HBrO(aq) + H2O(l) BrO-(aq) + H3O+(aq)

Ka = 2.3 x 10-9 What is ΔG°rxn?

ΔG°rxn = -RTln(K)= -RTln(2.3 x 10-9)

= 49.3 kJ/mol

An Example (cont.)

• What is ΔGrxn when pH = 5, [BrO-] = 0.1 M, and [HBrO] = 0.2 M ?

HBrO(aq) + H2O(l) BrO-

(aq) + H3O+(aq)

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Q =H 3O

+[ ] BrO

−[ ]

HBrO[ ]=

10−5( ) 0.1( )

(0.2)= 5x10−6

An Example (cont.)• Then:

€

ΔGrxn = ΔGrxno + RT ln Q( )

= 49.3 kJ/mol + (8.314 J/mol.K)(298 K)ln(5 x 10-6)

= 19.1 kJ/mol

ΔGrxn < ΔG°rxn “shifting” reaction towards products

Temperature Dependence of K

• We now have two definitions for ΔG°

ΔG°rxn = -RTln(K)= ΔH° - TΔS°

• Rearranging (dividing by -RT)

y = m x + b

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ln(K ) =−ΔH°

R

1

T

⎛

⎝ ⎜

⎞

⎠ ⎟+

ΔS°

R

• Plot of ln(K) vs 1/T is a straight line

T Dependence of K (cont.)

• If we measure K as a function of T, we can determine ΔH° by determining the slope of the line

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ln(K ) =−ΔH°

R

1

T

⎛

⎝ ⎜

⎞

⎠ ⎟+

ΔS°

R

slope

intercept

T Dependence of K (cont.)

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

€

ln(K2 ) =−ΔH°

R

1

T2

⎛

⎝ ⎜

⎞

⎠ ⎟+

ΔS°

R-

€

ln(K1) =−ΔH°

R

1

T1

⎛

⎝ ⎜

⎞

⎠ ⎟+

ΔS°

R

€

lnK2

K1

⎛

⎝ ⎜

⎞

⎠ ⎟=

−ΔH°

R

1

T2

−1

T1

⎛

⎝ ⎜

⎞

⎠ ⎟

the van’t Hoff equation.

An Example• For the following reaction:

CO(g) + 2H2(g) CH3OH(l) ΔG° = -29 kJ/mol

What is K at 340 K?

• First, what is Keq when T = 298 K?

ΔG°rxn = -RTln(K)= -29 kJ/mol

ln(K298) = (-29 kJ/mol)

-(8.314 J/mol.K)(298K) = 11.7

K298 = 1.2 x 105

An Example (cont.)• Next, to use the van’t Hoff Eq., we need ΔH°

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

ΔHf°(CO(g)) = -110.5 kJ/mol

ΔHf°(H2(g)) = 0ΔHf°(CH3OH(l)) = -239 kJ/mol

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

= ΔH°f(CH3OH(l)) - ΔH°f(CO(g)) = -239 kJ - (-110.5 kJ) = -128.5 kJ

An Example (cont.)• With ΔH°, we’re ready for the van’t Hoff Eq.

€

lnK2

K1

⎛

⎝ ⎜

⎞

⎠ ⎟=

−ΔH°

R

1

T2

−1

T1

⎛

⎝ ⎜

⎞

⎠ ⎟

K340 = 2.0 x 102

Why is K reduced?

Reaction is Exothermic.

Increase T, Shift Eq. To React.

Keq will then decrease

€

lnK340

1.2x105

⎛

⎝ ⎜

⎞

⎠ ⎟=

−(−128500 J)

(8.314 J /K)

1

340K−

1

298K

⎛

⎝ ⎜

⎞

⎠ ⎟

€

lnK340

1.2x105

⎛

⎝ ⎜

⎞

⎠ ⎟= −6.4