ΔG° and equilibrium constant - ΔG° = −RT ln K: Visual walkthrough

Before any algebra, meet the cast in plain words.

We build all five now, one picture at a time.

Step 1 — Free energy is a valley; equilibrium is its bottom

WHAT. Draw the free energy $G$ of the whole mixture on the vertical axis, and the extent-of-reaction dial $ξ$ (0 = pure reactants, 1 = pure products) on the horizontal axis. The curve dips to a lowest point somewhere in between — see the red minimum in the figure.

WHY this picture first. A physical system always drifts downhill in free energy (this is Spontaneity and the Second Law). So the mixture rolls, like a ball, toward the lowest point of the curve. Wherever that lowest point sits is equilibrium — the reaction stops changing there. Everything else on this page is just: where exactly is the bottom, and what number describes it?

PICTURE. The ball (red) starts high on the left and rolls to the valley floor.

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

Step 2 — Why one gas's free energy grows with the logarithm of its pressure

WHAT. Before we can describe the valley, we need the free energy of a single substance. Take one mole of ideal gas in a piston and slowly squeeze it. Track how its free energy $G$ climbs as the pressure $P$ rises. The result is a logarithm curve (red), not a straight line.

WHY a logarithm, and why not something simpler? Two facts collide:

At fixed temperature, compressing costs free energy at a rate $d G = V d P$ (from Gibbs Free Energy G = H - TS: $d G = V d P - S d T$ with $d T = 0$ ).
For an ideal gas the volume is $V = R T / P$ — a smaller box when pressure is high.

So each extra bit of pressure costs $d G = \frac{R T}{P} d P$ . The cost per unit pressure shrinks as $P$ grows (the $1/ P$ ). Adding up shrinking costs is exactly what produces a logarithm — that is why $ln$ appears here and not, say, a plain multiple of $P$ . We choose the integral (the tool for "add up a continuously changing rate") to sum these costs.

$free energy now G (P) = value at the reference P^{\circ} = 1 bar G^{\circ} + energy scale: R = 8.314, T in K R T how many e -folds above reference ln \frac{P}{P ^{\circ}}$

PICTURE. Free energy vs pressure: the red logarithm passes through the reference point where $P = P^{\circ}$ and $ln (P / P^{\circ}) = 0$ .

Step 3 — Add up the substances: assembling $Δ G$ for the reaction

WHAT. A reaction $a A + b B ⇌ c C + d D$ destroys reactants and builds products. Its free-energy change is (free energy of products) − (free energy of reactants), each counted with its coefficient. Picture four stacked bars — two products going in, two reactants coming out.

WHY. The slope of the valley (Step 1) is exactly this sum. Each substance contributes its own $μ = μ^{\circ} + R T ln a$ from Step 2, weighted by how many moles the equation uses.

$Δ G = products, built (c μ_{C} + d μ_{D}) - reactants, spent (a μ_{A} + b μ_{B})$

PICTURE. Product bars point up (gained), reactant bars point down (lost); the net height is $Δ G$ .

Step 4 — Split each bar into "standard part" + "log part"

WHAT. Substitute $μ_{i} = μ_{i}^{\circ} + R T ln a_{i}$ into every bar and sort the pieces into two piles: the plain $μ^{\circ}$ pieces, and the $R T ln a$ pieces.

WHY. The $μ^{\circ}$ pieces don't depend on how much of anything is present — they are pure constants. Pooling them gives one fixed number we name $Δ G^{\circ}$ . The remaining log pieces are the only part that changes as the mixture changes.

$Δ G = Δ G^{\circ} (the fixed standard number) (c μ_{C}^{\circ} + d μ_{D}^{\circ} - a μ_{A}^{\circ} - b μ_{B}^{\circ}) + R T (c ln a_{C} + d ln a_{D} - a ln a_{A} - b ln a_{B})$

PICTURE. The tower of bars separates into a solid black block ( $Δ G^{\circ}$ , unchanging) and a red stack of log terms (the part that breathes with composition).

Step 5 — Collapse the logs into one ratio $Q$

WHAT. Use two schoolbook log rules to fuse the four separate log terms into a single logarithm of one fraction.

WHY these two rules, and no others?

$n ln x = ln x^{n}$ turns each coefficient into an exponent, so $c ln a_{C}$ becomes $ln a_{C}^{c}$ .
$ln x + ln y = ln (x y)$ and $ln x - ln y = ln (x / y)$ turn sums into products and subtractions into division.

Together they say: coefficients go up as powers, products go on top, reactants go on the bottom.

$R T (c ln a_{C} + d ln a_{D} - a ln a_{A} - b ln a_{B}) = R T ln Q (reaction quotient) \frac{a _{C}^{c} a _{D}^{d}}{a _{A}^{a} a _{B}^{b}}$

So the whole story compresses to one clean line — the master equation:

$Δ G = Δ G^{\circ} + R T ln Q$

Term by term: $Δ G$ is the current slope of the valley, $Δ G^{\circ}$ is the fixed standard block, and $R T ln Q$ is the sliding correction set by how much of each species is present right now (this is the Reaction Quotient Q and Equilibrium $Q$ ).

PICTURE. The red stack of four logs funnels into a single fraction: products stacked on top, reactants on the bottom.

Step 6 — Stand at the valley bottom: $Δ G^{\circ} = - R T ln K$

WHAT. Return to the valley of Step 1. At the very bottom the ground is flat, so the slope is zero: $Δ G = 0$ . And "the composition at the bottom" is by definition the equilibrium composition, so its quotient $Q$ earns a special name — the equilibrium constant $K$ .

WHY. These two facts ( $Δ G = 0$ and $Q = K$ ) are the only extra ingredients needed. Drop them into the master equation from Step 5:

$0 = Δ G^{\circ} + R T ln K ⟹ Δ G^{\circ} = - R T ln K$

Read it: $Δ G^{\circ}$ (how deep and which side the valley bottom sits) equals minus- $R T$ times $ln K$ (how lopsided the settling ratio is).

PICTURE. Zoom on the valley floor: the tangent line is dead flat (slope $= Δ G = 0$ ), and the horizontal position of the bottom is labelled $Q = K$ .

Step 7 — Edge cases: the shapes the valley can take

WHAT. The valley in Step 1 was a generic bowl. Now sweep through the three shapes it can have and one thing it can never be.

WHY. The contract: the reader must never meet a scenario we didn't draw. There are exactly three positions for the bottom, plus the degenerate "no bottom" that thermodynamics forbids.

Bottom near $ξ = 1$ (right): valley leans product-side, $Δ G^{\circ}$ big negative, $K ≫ 1$ . Reaction nearly completes.
Bottom near $ξ = 0$ (left): valley leans reactant-side, $Δ G^{\circ}$ big positive, $K ≪ 1$ . Barely any product — but note it is still non-zero: even a tiny valley on the right has some forward pull if you start with $Q$ below $K$ .
Bottom in the middle: $Δ G^{\circ} \approx 0$ , $K \approx 1$ . Reactants and products roughly balanced.
Degenerate — a straight downhill ramp (no bottom): this would need $K$ to be infinite or the log to blow up. Real mixing always bends the line back up (the $R T ln Q$ term shoots to $- \infty$ as products run out and $+ \infty$ as reactants run out), so a genuine minimum always exists. A reaction can never fully reach $ξ = 0$ or exactly $ξ = 1$ — the logarithm forbids it. That is why $K$ is finite and non-zero.

PICTURE. Three coloured valley shapes side by side (bottom left / centre / right), with the forbidden straight ramp crossed out.

The one-picture summary

Everything on one canvas. The valley of $G$ vs $ξ$ , with: the red rolling ball; three slope markers (left $Δ G < 0$ , bottom $Δ G = 0$ , right $Δ G > 0$ ); the master equation $Δ G = Δ G^{\circ} + R T ln Q$ printed along the curve; and the bottom labelled $Q = K$ feeding into the boxed result $Δ G^{\circ} = - R T ln K$ .

Recall Feynman retelling — the whole walkthrough in plain words

Picture a marble in a bowl. The bowl is "free energy" and the sideways position is "how far the reaction has run." The marble always rolls downhill and settles at the lowest point — that's equilibrium. To draw the bowl we first learned how squeezing one gas costs free energy: because a squeezed gas has less room, each extra push costs a little less than the last, and adding up shrinking pushes gives a logarithm — that's the $R T ln$ . We stacked up the logarithms of every product (on top) and every reactant (on the bottom) into one fraction $Q$ , and split off all the fixed bits into a single constant $Δ G^{\circ}$ . That gave the master line $Δ G = Δ G^{\circ} + R T ln Q$ : the marble's local steepness equals the fixed block plus a sliding log term. Finally we walked to the bottom of the bowl, where the floor is flat (steepness zero) and the fraction has its settled value $K$ . Setting the steepness to zero there rearranges into $Δ G^{\circ} = - R T ln K$ : how deep and which way the bowl leans tells you exactly how lopsided the final mixture is.

Recall Quick self-test

Why must $Δ G = 0$ at the valley bottom? ::: The bottom is a minimum of $G$ , so its slope — which is $Δ G$ — is zero. Where does the logarithm physically come from? ::: From $d G = V d P = (R T / P) d P$ ; integrating $1/ P$ gives $ln (P / P^{\circ})$ . Why is $K$ dimensionless? ::: It is built from activities $a = P / P^{\circ}$ , which are ratios and carry no units. Can a valley leaning toward reactants still run forward? ::: Yes — if the mixture starts with $Q < K$ , the local slope $Δ G$ is negative and it rolls to the bottom.

Connections

Gibbs Free Energy G = H - TS — supplies $d G = V d P - S d T$ used in Step 2.
Chemical Potential μ — the $μ = μ^{\circ} + R T ln a$ backbone of Steps 2–4.
Reaction Quotient Q and Equilibrium — the $Q$ assembled in Step 5.
Spontaneity and the Second Law — why the marble rolls downhill (Step 1).
Van't Hoff Equation — what happens to the valley when $T$ changes.
Le Chatelier's Principle — the qualitative shadow of $Q$ vs $K$ .

valley bottom leans toward	$Δ G^{\circ}$	$ln K$	$K$
products (deep on the right)	$< 0$	$> 0$	$> 1$
dead centre	$= 0$	$= 0$	$= 1$
reactants (deep on the left)	$> 0$	$< 0$	$< 1$