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

This page assumes you know nothing about the symbols in the parent note. We will earn every letter — $G$, $\mu$, $a$, $Q$, $K$, $R$, $T$, $\ln$, $\nu$ — one at a time, each with a picture, each explaining why the topic cannot live without it.

0. The absolute starting point: a reaction that can go both ways

Write a reaction with a double arrow: $aA+bB\rightleftharpoons cC+dD$

• $A,B$ = the reactants (what you start with).
• $C,D$ = the products (what you make).
• The double arrow $\rightleftharpoons$ means the reaction runs forward and backward at once; it can stop partway, with some of everything left over.
• $a,b,c,d$ = stoichiometric numbers — how many molecules of each take part. In "$2H_2+O_2\rightleftharpoons 2H_{2}O$" they are $2,1,2$.

1. $G$ — free energy, the "height" the reaction rolls downhill in

Look at the figure. The horizontal axis is the extent of reaction — a slider from "all reactants" on the left to "all products" on the right. The vertical axis is $G$, the free energy, drawn as a valley.

• The bottom of the valley is where $G$ is smallest — the mixture has nowhere lower to roll. That point is equilibrium.
• On the left slope the ball rolls right (reaction goes forward); on the right slope it rolls left (goes backward).

Why the topic needs $G$: without a "height," there is no notion of "downhill," and no way to say which direction a reaction runs. $G$ is the landscape everything else lives on.

2. $\Delta$ and $\Delta G$ — the slope, not the height

The symbol $\Delta$ (Greek capital delta) means "change in" — a difference, final minus initial.

• $\Delta G$ for a reaction = free energy of products minus free energy of reactants, measured for a tiny step of reaction. Geometrically it is the slope of the valley at your current position.

$\Delta G<0\Rightarrow \text{forward},\Delta G=0\Rightarrow \text{stuck (equilibrium)},\Delta G>0\Rightarrow \text{backward}$

Why the topic needs $\Delta G$: it is the compass. The sign of $\Delta G$ tells you which way, right now, the reaction wants to go. See Spontaneity and the Second Law for why "downhill" is the natural direction.

3. The little circle: $G^{\circ }$ and the "standard state"

The superscript ${}^{\circ }$ (a small circle, read "standard" or "naught") is a fixed reference setting, like sea level on a map.

So:

• $G^{\circ }$ = free energy of a species when it sits exactly at that reference.
• $\Delta G^{\circ }$ = the slope of the valley measured at the standard reference composition — a single fixed number for a given reaction at a given temperature.

4. $a$ — activity: "how crowded" a species is

Before we can measure free energy at any composition, we need a number for "how much of a species is present." That number is the activity $a$.

Because we divide by the reference, activity is a pure ratio — it has no units. A gas at 1 bar has $a=1$; at 3 bar, $a=3$; at the reference itself, $a=1$.

Why the topic needs $a$: it is the dial that says how far from the standard state you are — and (next section) free energy depends on it logarithmically. See Chemical Potential μ for the full story.

5. $R$, $T$, and $\ln$ — the machinery of the log law

Three symbols always travel together in this topic. Meet them:

• $T$ = absolute temperature in kelvin (K). Zero kelvin is the coldest possible; room temperature is about $298\text{ K}$.
• $R$ = the gas constant, $R=8.314{\text{J K}}^{-1}{\text{mol}}^{-1}$. It is the fixed exchange rate between "temperature" and "energy per mole." The product $RT$ has units of energy per mole — exactly the units of $\Delta G$.
• $\ln$ = the natural logarithm. $\ln x$ answers the question "e to what power gives $x$?" Its partner, $e^y$, undoes it: $\ln (e^y)=y$ and $e^{\ln x}=x$.

Look at the curve of $\ln x$ in the figure. Three facts you will use constantly:

Why the topic needs them: the master equations are literally built out of $RT\ln (\text{activity})$. Without these three you cannot even write $\Delta G^{\circ }=-RT\ln K$.

6. $\mu$ — chemical potential, free energy per mole of one species

The Greek letter $\mu$ ("mu") is the free energy carried by one mole of a single species at its current activity:

$\mu ={\mu }^{\circ }+RT\ln a$

Read it in plain words: (free energy per mole now) = (free energy per mole at the reference) + (a correction $RT\ln a$ for being more or less crowded than the reference).

• If $a>1$ (more crowded than reference): $\ln a>0$, so $\mu$ is raised.
• If $a=1$ (at reference): $\ln a=0$, so $\mu ={\mu }^{\circ }$.
• If $a<1$ (dilute): $\ln a<0$, so $\mu$ is lowered.

Why the topic needs $\mu$: the reaction's total slope $\Delta G$ is just the $\mu$'s of the products minus the $\mu$'s of the reactants. $\mu$ is the brick; $\Delta G$ is the wall. Full detail in Chemical Potential μ.

7. $\nu$ — stoichiometric number as a signed weight

To add up the $\mu$'s correctly we use $\nu$ (Greek "nu"), the stoichiometric number with a sign:

• $\nu$ is positive for products, negative for reactants, and equal in size to the coefficient ($a,b,c,d$).

Then the reaction slope is one clean sum: $\Delta G={\sum }_i{\nu }_i{\mu }_i=(c{\mu }_C+d{\mu }_D)-(a{\mu }_A+b{\mu }_B)$

Why the topic needs $\nu$: it lets "products minus reactants" be written as a single sum, which is exactly the step the parent note uses to unpack $\Delta G$.

8. $Q$ and $K$ — the reaction quotient and its equilibrium value

When you substitute ${\mu }_i={\mu }_i^{\circ }+RT\ln a_i$ into that sum, every $\ln$ collects (using $\ln x^n=n\ln x$ and $\ln x+\ln y=\ln xy$) into one ratio:

$Q=\frac{a_C^c{a}_D^d}{a_A^a{a}_B^b}(\text{products on top, each raised to its coefficient})$

Both are built from activities, so both are dimensionless — that is the only reason $\ln Q$ and $\ln K$ are even allowed. Follow the figure: as the ball rolls toward the valley bottom, $Q$ climbs (or falls) until it lands on $K$.

Why the topic needs $Q$ and $K$: $Q$ tells the general law where you are; $K$ marks the destination. See Reaction Quotient Q and Equilibrium.

9. Putting the bricks together — the two master equations

Now every symbol is earned, and the parent note's boxes read like sentences:

Reading the sign is now automatic (this is what $\ln$ from §5 buys you):

For temperature effects on $K$, this same $\Delta G^{\circ }$ feeds straight into the Van't Hoff Equation; the qualitative "push-back" picture is Le Chatelier's Principle.

Prerequisite map

Equipment checklist

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