9.5 - Free Energy and Equilibrium

Overview

Topic 9.3 used

\Delta G^\circ

to decide whether a process is thermodynamically favored. This topic ties that same number to the equilibrium constant

K

from Unit 7. A negative

\Delta G^\circ

goes with

K > 1

(products favored at equilibrium), and a positive

\Delta G^\circ

goes with

K < 1

(reactants favored). One equation,

\Delta G^\circ = -RT\ln K

, takes you from either one to the other, and a quick comparison with

RT

lets you estimate

K

without a calculator.

What "thermodynamically favored" means at equilibrium

\Delta G^\circ

describes a reaction under standard conditions: gases at 1 atm, solutes at 1 M, and pure solids and liquids (Topic 9.3). Under those conditions every term in the reaction quotient equals 1, so

Q = 1

. Now compare that

Q

with

K

, the way you did in Topic 7.3:

If $K > 1$ , then $Q < K$ . Starting from standard conditions, the reaction runs forward and makes more products before it reaches equilibrium. Products are favored. This is what $\Delta G^\circ < 0$ means.

If $K < 1$ , then $Q > K$ . The reaction runs in reverse, and reactants are favored at equilibrium. This is what $\Delta G^\circ > 0$ means.

If $K = 1$ , the standard mixture is already at equilibrium, and $\Delta G^\circ = 0$ .

So thermodynamically favored (

\Delta G^\circ < 0

) means products are favored at equilibrium (

K > 1

) under standard conditions. It doesn't mean the reaction goes to completion, and it says nothing about speed (Topic 9.4). A reaction with

\Delta G^\circ > 0

still makes some products; it just reaches equilibrium with mostly reactants.

$\Delta G^\circ$	$K$	At equilibrium
Negative ( $< 0$ )	Greater than 1	Products favored
Zero	Equal to 1	Neither side favored
Positive ( $> 0$ )	Less than 1	Reactants favored

The equation linking ΔG° and K

This equation is on the AP reference sheet:

\Delta G^\circ = -RT\ln K

Solving for

K

(divide both sides by

-RT

, then raise

e

to each side) gives the other form:

K = e^{-\Delta G^\circ/RT}

$R$ is the gas constant, $8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ . Its energy unit is joules, so $\Delta G^\circ$ has to be in $\mathrm{J/mol_{rxn}}$ , not kJ. Forgetting to convert is the most common mistake on this topic.

$T$ is the temperature in kelvins. At 25 °C, $T = 298$ K.

$K$ is the same equilibrium constant you wrote in Topic 7.3, with gas pressures in atm and solute concentrations in M. It has no units. For a reaction of gases, this makes it $K_p$ .

$\Delta G^\circ$ belongs to the equation as written, just like $K$ . Double the coefficients and $\Delta G^\circ$ doubles while $K$ is squared. Reverse the reaction and $\Delta G^\circ$ changes sign while $K$ becomes $1/K$ (Topic 7.6). The exponential form makes both rules agree.

Because

K

depends exponentially on

\Delta G^\circ

, a small change in

\Delta G^\circ

makes a big change in

K

. Keep extra digits in the exponent until the end, and report

K

to about two significant figures.

Example 1: Finding K from ΔG°

Colorless dinitrogen tetroxide breaks apart into brown nitrogen dioxide:

\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}

For

\mathrm{N_2O_4(g)}

\Delta G_f^\circ = +97.9

kJ/mol, and for

\mathrm{NO_2(g)}

\Delta G_f^\circ = +51.3

kJ/mol. Find

\Delta G^\circ

and

K

at 25 °C. Which side is favored?

Step 1:

\Delta G^\circ

from free energies of formation (Topic 9.3):

\Delta G^\circ = 2(51.3) - 97.9 = +4.7\ \mathrm{kJ/mol_{rxn}}

Step 2: Convert to joules (

+4700\ \mathrm{J/mol_{rxn}}

) and substitute:

K = e^{-\Delta G^\circ/RT} = e^{-(4700)/[(8.314)(298)]} = e^{-1.90} = 0.15

Answer:

K = 0.15

\Delta G^\circ

is positive, so

K < 1

and

\mathrm{N_2O_4}

is favored. But

\Delta G^\circ

is small, less than twice

RT

, so

K

isn't far from 1: both gases are present in real amounts at equilibrium. That's why a sealed tube of these gases at room temperature is light brown, not colorless and not dark brown. (This

K

uses pressures in atm, so it's

K_p

. The

K_c = 0.360

that Unit 7 uses for this reaction is a different constant: it uses concentrations, and it's for a higher temperature.)

Estimating K without a calculator

The AP exam often asks you to reason about

K

instead of calculating it. The trick is to compare

\Delta G^\circ

with

RT

. At 298 K,

RT = (8.314\ \mathrm{J\,mol^{-1}\,K^{-1}})(298\ \mathrm{K}) = 2478\ \mathrm{J/mol} \approx 2.5\ \mathrm{kJ/mol}

$\Delta G^\circ$ near zero (a few kJ/mol, about the size of $RT$ ): the exponent $-\Delta G^\circ/RT$ is small, so $K$ is close to 1. Reactants and products are both present in significant amounts at equilibrium.

$\Delta G^\circ$ much larger than $RT$ (tens of kJ/mol or more, positive): $K$ is much less than 1. Reactants are strongly favored.

$\Delta G^\circ$ much more negative than $-RT$ : $K$ is much greater than 1. Products are strongly favored.

A handy rule at 25 °C: every 5.7 kJ/mol of

\Delta G^\circ

changes

K

by a factor of 10, because

2.303RT = 5.7

kJ/mol. So

\Delta G^\circ = -57\ \mathrm{kJ/mol_{rxn}}

gives

K \approx 10^{10}

, and

\Delta G^\circ = +57\ \mathrm{kJ/mol_{rxn}}

gives

K \approx 10^{-10}

On a log scale the relationship is a straight line through

\Delta G^\circ = 0

K = 1

. Notice how fast

K

changes: going from

\Delta G^\circ = 0

\Delta G^\circ = -40\ \mathrm{kJ/mol_{rxn}}

takes

K

from 1 to about

10^{7}

Example 2: Estimating K (no calculator)

At 25 °C, three reactions have these values of

\Delta G^\circ

. Estimate

K

for each, and say which side is favored. (a)

+45\ \mathrm{kJ/mol_{rxn}}

(b)

-2\ \mathrm{kJ/mol_{rxn}}

(c)

-120\ \mathrm{kJ/mol_{rxn}}

(a) $45/5.7 \approx 8$ , and $\Delta G^\circ$ is positive, so $K \approx 10^{-8}$ . Reactants are strongly favored.

(b) 2 kJ/mol is smaller than $RT$ (2.5 kJ/mol), so $K$ is close to 1, a little above 1 because $\Delta G^\circ$ is negative. (A calculator gives $K = 2.2$ .) Products are only slightly favored.

(c) $120/5.7 \approx 21$ , and $\Delta G^\circ$ is negative, so $K \approx 10^{21}$ . Products are so strongly favored that the reaction essentially goes to completion.

Example 3: ΔG° from K

Acetic acid, the acid in vinegar, has

K_a = 1.8 \times 10^{-5}

at 25 °C:

\mathrm{CH_3COOH(aq) + H_2O(l) \rightleftharpoons H_3O^{+}(aq) + CH_3COO^{-}(aq)}

Find

\Delta G^\circ

for this ionization and explain what its sign means.

\Delta G^\circ = -RT\ln K = -(8.314)(298)\ln(1.8 \times 10^{-5}) = -(2478)(-10.93) = +2.71 \times 10^{4}\ \mathrm{J/mol_{rxn}}

Answer:

\Delta G^\circ = +27.1\ \mathrm{kJ/mol_{rxn}}

. It's positive, so ionization is not thermodynamically favored: at equilibrium, almost all of the acetic acid is still un-ionized molecules. That's what "weak acid" means (Topic 8.3). Any

K

less than 1 goes with a positive

\Delta G^\circ

Temperature, ΔG°, and K

Temperature shows up twice.

\Delta G^\circ

itself changes with temperature, since

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

(Topic 9.3), and

T

also sits in the exponent. For AP problems, assume

\Delta H^\circ

and

\Delta S^\circ

stay about the same as the temperature changes. Then:

Calculate $\Delta G^\circ$ at the new temperature from $\Delta H^\circ - T\Delta S^\circ$ .

Use the new $\Delta G^\circ$ and the new $T$ in $K = e^{-\Delta G^\circ/RT}$ .

Whether

K

goes up or down with temperature depends on the sign of

\Delta H^\circ

, just as Le Châtelier's principle predicts (Topic 7.9):

K

increases with temperature for an endothermic reaction and decreases for an exothermic one. Substituting

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

shows why:

\ln K = -\frac{\Delta G^\circ}{RT} = -\frac{\Delta H^\circ}{RT} + \frac{\Delta S^\circ}{R}

Only the first term depends on

T

. (You don't need this form on the exam; it just explains the rule.)

Example 4: Limestone in a lime kiln (AP-style)

Lime (

\mathrm{CaO}

) is made by heating limestone (

\mathrm{CaCO_3}

\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}

\Delta H^\circ = +178.3\ \mathrm{kJ/mol_{rxn}}

and

\Delta S^\circ = +160.6\ \mathrm{J/(K{\cdot}mol_{rxn})}

. Topic 9.3 found

\Delta G^\circ = +130.4\ \mathrm{kJ/mol_{rxn}}

at 298 K and showed that

\Delta G^\circ

changes sign near 1110 K. (a) Calculate

K

at 298 K. (b) Calculate

\Delta G^\circ

and

K

at 1200 K, assuming

\Delta H^\circ

and

\Delta S^\circ

don't change. (c) Write the expression for

K

, and use it to explain why limestone is stable at room temperature but breaks down in a kiln.

(a) Convert

\Delta G^\circ

to joules and substitute:

K = e^{-(130{,}400)/[(8.314)(298)]} = e^{-52.63} = 1.4 \times 10^{-23}

(b) First

\Delta G^\circ

at 1200 K, with

\Delta S^\circ

in kJ:

\Delta G^\circ = 178.3 - (1200)(0.1606) = 178.3 - 192.7 = -14.4\ \mathrm{kJ/mol_{rxn}}

K = e^{-(-14{,}400)/[(8.314)(1200)]} = e^{1.443} = 4.2

(c)

\mathrm{CaCO_3}

and

\mathrm{CaO}

are pure solids, so they're left out of

K

(Topic 7.3):

K = P_{\mathrm{CO_2}}

At 298 K, the equilibrium pressure of

\mathrm{CO_2}

above limestone is about

10^{-23}

atm. Essentially nothing decomposes, which is why limestone buildings last for centuries. At 1200 K,

K \approx 4.2

, so

\mathrm{CO_2}

keeps forming until its pressure reaches about 4 atm. A kiln is open to the air, where the

\mathrm{CO_2}

pressure is far below that, so

Q

stays less than

K

and the limestone keeps breaking down.

The graph puts the example together.

\Delta G^\circ

falls in a straight line as

T

rises (its slope is

-\Delta S^\circ

and its intercept is

\Delta H^\circ

) and crosses zero, where

K = 1

, at about 1110 K (837 °C). The reaction is endothermic, so

K

grows with temperature, from

10^{-23}

to about 4.

Common mistakes

Mixing kJ and J. $R = 8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$ , so convert $\Delta G^\circ$ to joules before you divide by $RT$ . Otherwise the exponent is off by a factor of 1000.

Using °C. $T$ must be in kelvins.

Dropping the negative sign in $\Delta G^\circ = -RT\ln K$ . Quick check: $K > 1$ must give a negative $\Delta G^\circ$ .

Reading $\Delta G^\circ > 0$ as "no reaction." It means $K < 1$ . Some products still form.

Reading $\Delta G^\circ < 0$ as "fast." $\Delta G^\circ$ tells you where equilibrium lies, not how quickly the reaction gets there (Topic 9.4).

Saying $K$ always increases with temperature. It increases only for endothermic reactions.

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