ΔG° and equilibrium constant - ΔG° = −RT ln K: Question bank

Definition Every symbol used on this page, in plain words

$R$ = the gas constant, $R = 8.314 J K^{- 1} mol^{- 1}$ . It is the fixed conversion factor that turns "temperature" into "energy per mole" — it is what makes $R T$ an energy.
$T$ = the absolute temperature in kelvin (K). Always positive; $298 K$ is room temperature.
$R T$ = a bundle of thermal energy per mole (about $2.48 kJ mol^{- 1}$ at $298 K$ ). It sets the scale of every free-energy term below.
$P^{\circ}$ = the standard pressure, fixed at $P^{\circ} = 1 bar$ . It is only a reference peg so that ratios like $P / P^{\circ}$ are pure numbers.
$a_{i}$ = the activity of species $i$ — an "effective, dimensionless amount." For an ideal gas $a_{i} = P_{i} / P^{\circ}$ ; for a dilute solute $a_{i} = c_{i} / c^{\circ}$ (with $c^{\circ} = 1 mol L^{- 1}$ ); for a pure solid or liquid $a_{i} = 1$ . Because it is a ratio, activity carries no units, which is exactly why $ln a_{i}$ and $ln K$ are even allowed to be written.
$Δ G$ — the free-energy "slope" of the reaction right now, at the mixture you actually have.
$Δ G^{\circ}$ — the fixed value of that slope in the imaginary case where every species sits at activity $1$ (the standard state). Constant at a given $T$ .
$Q$ vs $K$ — $Q = \frac{products (in activities)}{reactants (in activities)}$ now; $K$ is that same ratio frozen at equilibrium.

True or false — justify

Δ G^{\circ}

changes as a reaction proceeds toward equilibrium.

False.

Δ G^{\circ}

is fixed at a given

T

(it is built only from the

μ^{\circ}

standard values); it is

Δ G

and

Q

that change during the reaction — visually,

Δ G

is the local slope of the Figure-1 valley, while

Δ G^{\circ}

is a single fixed number.

Δ G^{\circ} = 0

then the reaction is at equilibrium the moment you mix reactants.

False.

Δ G^{\circ} = 0

only forces

K = 1

; equilibrium requires

Q = K

, so a mixture with

Q \neq = 1

still has

Δ G \neq = 0

and must react.

A reaction with a large positive

Δ G^{\circ}

can never produce any product.

False. It produces product until

Q

climbs to

K

; a large positive

Δ G^{\circ}

just makes

K

tiny, so the amount of product is small, not zero.

K

has units of pressure for a gas-phase reaction.

False.

K

is built from activities

P / P^{\circ}

, which are pure numbers, so

K

is dimensionless — that is the only way

ln K

can even be written.

At equilibrium

Δ G = 0

, so the reaction has stopped completely.

False.

Δ G = 0

means no net drive; forward and reverse still happen at equal rates. It is a dynamic balance, not a frozen one — the ball rests at the valley floor but still jiggles.

Doubling

T

while keeping

Δ G^{\circ}

fixed always increases

K

False (and the premise is shaky). At fixed

Δ G^{\circ} < 0

ln K = - Δ G^{\circ} / R T

shrinks in magnitude, pushing

K

toward 1 — but really

Δ G^{\circ}

itself is

T

-dependent, so use the Van't Hoff Equation.

Δ G < 0

the reaction is spontaneous, so

Δ G^{\circ}

must also be negative.

False. Example:

Δ G^{\circ} > 0

but

Q < K

gives

R T ln Q

negative enough to make

Δ G < 0

. Sign of

Δ G

and sign of

Δ G^{\circ}

are independent.

The relation

Δ G^{\circ} = - R T ln K

is a definition, not a derived result.

False. It is derived by imposing equilibrium (

Δ G = 0

Q = K

) on the general law

Δ G = Δ G^{\circ} + R T ln Q

(see the "Why questions" derivation of that general law).

K > 1

guarantees the reaction is fast.

False.

K

and

Δ G^{\circ}

are thermodynamic (where it ends up), not kinetic (how quickly). A hugely favourable reaction can be immeasurably slow without a catalyst.

All of the activity relations above (

a = P / P^{\circ}

a = c / c^{\circ}

) hold exactly for any real system.

False. Those are the ideal forms. Real gases and concentrated solutions deviate; then

a = γ (P / P^{\circ})

with an activity coefficient

γ

that is

1

only in the ideal limit. Everything on this page assumes ideal behaviour (

γ = 1

Why questions

Why does

Δ G = Δ G^{\circ} + R T ln Q

hold in the first place?

Each species contributes

μ_{i} = μ_{i}^{\circ} + R T ln a_{i}

; summing with stoichiometric signs

ν_{i}

gives

Δ G = \sum ν_{i} μ_{i}^{\circ} + R T \sum ν_{i} ln a_{i}

. The first sum is

Δ G^{\circ}

; the log terms collapse (using

ln x^{n} = n ln x

and

ln x + ln y = ln x y

) into a single

R T ln Q

Why does the free-energy contribution of an ideal gas grow as

R T ln (P / P^{\circ})

and not linearly with

P

For one mole at constant

T

, molar free energy obeys

d G = V d P

; assuming ideal-gas behaviour

V = R T / P

, so

d G = (R T / P) d P

. Integrating

1/ P

from

P^{\circ}

P

gives the logarithm

R T ln (P / P^{\circ})

. Squeezing already-dense gas costs proportionally less, which is exactly the logarithmic shape of Figure 2.

Why is it

Δ G^{\circ}

(standard) and not

Δ G

that equals

- R T ln K

At equilibrium

Δ G = 0

, so

0 = Δ G^{\circ} + R T ln K

. The constant left standing is

Δ G^{\circ}

; it is the fixed reference that

K

measures.

Why must

Δ G = 0

exactly at equilibrium?

Look at Figure 1: equilibrium is the minimum of the

G

-vs-extent valley, and the slope of any curve at its minimum is zero. That slope is

Δ G

Why does the parent call

Δ G^{\circ}

"not the driving force at every moment"?

The moment-to-moment driving force is

Δ G = Δ G^{\circ} + R T ln Q

, which changes with composition (the changing slope in Figure 1).

Δ G^{\circ}

is only the value at the single special composition where all activities equal 1.

Why is

K

dimensionless even for

K_{p}

that "looks like" pressures?

Each pressure enters as an activity

P / P^{\circ}

with

P^{\circ} = 1 bar

, so the units cancel. Without this,

ln K

would be mathematically undefined.

Why can two reactions with the same

Δ G^{\circ}

end at different product amounts?

They can't at equilibrium — same

Δ G^{\circ}

means same

K

, hence the same equilibrium ratio. Different starting amounts change how far along you travel, but not

K

itself.

Why does a more negative

Δ G^{\circ}

give a larger

K

K = e^{- Δ G^{\circ} / R T}

; making

Δ G^{\circ}

more negative makes the exponent more positive, and the exponential blows up. In Figure 1 this is a deeper "products valley" ⇒ more product.

The single sentence that resolves most traps above is ::: direction is set by $Δ G$ (compare $Q$ with $K$ ); $Δ G^{\circ}$ only fixes where equilibrium sits (the value of $K$ ), not whether a given mixture moves forward.

Connections

Reaction Quotient Q and Equilibrium — the $Q$ -vs- $K$ comparison behind most traps here.
Gibbs Free Energy G = H - TS — source of the " $Δ G^{\circ} > 0$ yet $Δ H^{\circ} < 0$ " edge case.
Chemical Potential μ — the $μ = μ^{\circ} + R T ln a$ origin of the log term.
Van't Hoff Equation — the honest $T$ -dependence the "double $T$ " traps warn about.
Spontaneity and the Second Law — why $Δ G < 0$ , not $Δ G^{\circ} < 0$ , means "goes."
Le Chatelier's Principle — qualitative companion to the shift arguments.

Question bank — ΔG° and equilibrium constant - ΔG° = −RT ln K

True or false — justify

Spot the error

Why questions

Edge cases

Connections