ΔG° and equilibrium constant - ΔG° = −RT ln K: Worked examples

We lean on just two boxed facts, both derived in the parent: $Δ G = Δ G^{\circ} + R T ln Q and Δ G^{\circ} = - R T ln K .$

Throughout, $R = 8.314 J K^{- 1} mol^{- 1}$ (SI — so $Δ G$ must be in joules, not kilojoules), and $T$ is in kelvin. $ln$ is natural log; $lo g$ is base-10, and they connect by $ln x = 2.303 lo g x$ .

The scenario matrix

Every question on this topic falls into one of these cells. The examples below are labelled with the cell they cover.

Cell	Case class	What is special about it
A	$Δ G^{\circ} < 0$	products favoured, $K > 1$ — find $K$
B	$Δ G^{\circ} > 0$	reactants favoured, $K < 1$ — find $Δ G^{\circ}$
C	$Δ G^{\circ} = 0$ (degenerate)	perfectly balanced, $K = 1$
D	$Q \neq = K$ , non-standard mixture	$Δ G^{\circ} > 0$ but reaction still goes forward
E	Limiting extreme: $T \to \infty$ and $T \to 0$	how $K$ drifts to 1 or blows up
F	Van't Hoff twist: $K$ at a second temperature	uses $Δ H^{\circ}, Δ S^{\circ}$
G	Real-world word problem	ammonia-type industrial equilibrium
H	Exam twist: units / base-10 log trap	catches the kJ-vs-J and $2.303$ slips

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

Look at the figure: the single curve $K = e^{- Δ G^{\circ} / R T}$ passes through the point $(0, 1)$ (that is Cell C), rises steeply for negative $Δ G^{\circ}$ (Cell A) and sinks toward zero for positive $Δ G^{\circ}$ (Cell B). Every example below is a point on this curve — or a shift along it (temperature) or off it (non-standard $Q$ ).

The examples

Worked example Example 1 —

Cell A: negative $Δ G^{\circ}$ , find $K$ A reaction has $Δ G^{\circ} = - 34.0 kJ mol^{- 1}$ at $T = 298 K$ . Find $K$ .

Forecast: $Δ G^{\circ}$ is negative, so should $K$ be bigger than 1 or smaller than 1? Guess a rough size before reading on.

Convert to joules. $Δ G^{\circ} = - 34000 J mol^{- 1}$ . Why this step? $R$ is in joules; mixing kJ with J is the single most common wrong answer (Cell H).
Rearrange the master equation. From $Δ G^{\circ} = - R T ln K$ , $ln K = - \frac{Δ G ^{\circ}}{R T} = - \frac{- 34000}{8.314 \times 298} = \frac{34000}{2477.6} = 13.72.$ Why this step? We want $K$ , so we isolate $ln K$ first — undoing multiplication before undoing the log.
Undo the log. $K = e^{13.72} \approx 9.1 \times 1 0^{5} .$ Why this step? $e^{(\cdot)}$ is the inverse of $ln$ — it answers "which number has this natural log?"

Verify: $Δ G^{\circ} < 0 \Rightarrow ln K > 0 \Rightarrow K > 1$ . ✔ A large $K \sim 1 0^{6}$ says products dominate — consistent with a strongly negative $Δ G^{\circ}$ . Units: $K$ is dimensionless. ✔

Worked example Example 4 —

Cell D: non-standard mixture, $Δ G^{\circ} > 0$ but still goes forward Take Example 2's reaction ( $Δ G^{\circ} = + 4735 J mol^{- 1}$ , so $K_{p} = 0.148$ ). Right now the mixture has $Q = 0.010$ . Which direction does it move, forward or backward?

Forecast: $Δ G^{\circ}$ is positive. A naive reader says "can't go forward." Is that right? Compare $Q$ and $K$ first.

Compare $Q$ with $K$ . $Q = 0.010 < K = 0.148.$ Why this step? Direction is decided by $Δ G$ (the actual slope), and $Δ G$ 's sign is set by whether $Q$ is below or above $K$ — not by $Δ G^{\circ}$ alone.
Compute $R T ln Q$ . $ln (0.010) = - 4.605$ , so $R T ln Q = 2477.6 \times (- 4.605) = - 11409 J mol^{- 1} .$ Why this step? This is the correction term that turns the standard number into the actual number.
Add them. $Δ G = Δ G^{\circ} + R T ln Q = 4735 + (- 11409) = - 6674 J mol^{- 1} .$ Why this step? $Δ G < 0$ ⇒ forward. The starved-of-product mixture climbs toward equilibrium by making more $NO_{2}$ .

Verify: $Q < K$ should always give $Δ G < 0$ . ✔ It does. Lesson: $Δ G^{\circ} > 0$ does not forbid forward reaction — see the parent's first mistake callout. ✔

Worked example Example 5 —

Cell E (limiting behaviour): $T \to \infty$ and $T \to 0$ A reaction has a fixed $Δ G^{\circ} = - 20.0 kJ mol^{- 1}$ (treat it as constant here). What happens to $K$ as $T$ becomes very large, and as $T$ becomes very small?

Forecast: Does raising temperature push $K$ toward 1 or away from 1? Sketch $ln K = - Δ G^{\circ} / (R T)$ against $T$ in your head first.

Write $K$ as a function of $T$ . $ln K (T) = - \frac{Δ G ^{\circ}}{R T} = \frac{20000}{8.314 T} = \frac{2405.6}{T} .$ Why this step? Isolating $T$ in the denominator makes the two limits obvious.
Let $T \to \infty$ . $ln K \to 0$ , so $K \to 1$ . Why this step? A huge denominator crushes the fraction to zero — the "no preference" balanced state. Physically, thermal energy $R T$ swamps the fixed free-energy gap, so the system stops caring which side it's on.
Let $T \to 0^{+}$ . $ln K \to + \infty$ , so $K \to \infty$ . Why this step? A tiny denominator makes the fraction enormous — at very low $T$ the negative $Δ G^{\circ}$ dominates totally and products win overwhelmingly.
Check a concrete value, $T = 298 K$ : $ln K = 2405.6/298 = 8.072$ , so $K = e^{8.072} \approx 3.2 \times 1 0^{3} .$

Verify: matches the recall prompt in the parent: raising $T$ (at fixed $Δ G^{\circ} < 0$ ) shrinks $∣ ln K ∣$ , moving $K$ toward 1. ✔ Both limits sit on the correct side of 1 for a negative $Δ G^{\circ}$ . ✔ (Caveat: real $Δ G^{\circ}$ drifts with $T$ — Example 6 handles that properly.)

Worked example Example 6 —

Cell F (Van't Hoff twist): $K$ at a second temperature A reaction has $Δ H^{\circ} = + 40.0 kJ mol^{- 1}$ and $Δ S^{\circ} = + 100 J K^{- 1} mol^{- 1}$ (both taken as $T$ -independent). Find $K$ at $T_{1} = 298 K$ and at $T_{2} = 500 K$ .

Forecast: This reaction is endothermic ( $Δ H^{\circ} > 0$ ). By Le Chatelier's Principle, heating should favour products — so should $K$ grow or shrink as we go from 298 K to 500 K?

Build $Δ G^{\circ}$ from $Δ H^{\circ} - T Δ S^{\circ}$ . $Δ G^{\circ} (298) = 40000 - 298 (100) = 40000 - 29800 = + 10200 J mol^{- 1} .$ Why this step? $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ is the Gibbs Free Energy G = H - TS definition — it is where the temperature dependence lives.
Convert to $K$ at 298 K. $ln K_{1} = - \frac{10200}{2477.6} = - 4.117 \Rightarrow K_{1} = e^{- 4.117} \approx 1.63 \times 1 0^{- 2} .$
Repeat at 500 K. $Δ G^{\circ} (500) = 40000 - 500 (100) = 40000 - 50000 = - 10000 J mol^{- 1} .$ $ln K_{2} = - \frac{- 10000}{8.314 \times 500} = \frac{10000}{4157} = 2.406 \Rightarrow K_{2} = e^{2.406} \approx 11.1.$ Why this step? Same recipe, new $T$ — notice $Δ G^{\circ}$ actually flipped sign, crossing the Cell-C point $Δ G^{\circ} = 0$ at $T = Δ H^{\circ} /Δ S^{\circ} = 400 K$ .

Verify: $K$ jumped from $0.016$ to $11.1$ — it grew with temperature, exactly what an endothermic reaction should do (Le Chatelier). ✔ This is the content of the Van't Hoff Equation: $K$ rises with $T$ when $Δ H^{\circ} > 0$ . ✔

Worked example Example 7 —

Cell G (real-world word problem): ammonia synthesis At $298 K$ the Haber process $N_{2} + 3 H_{2} ⇌ 2 NH_{3}$ has $K_{p} \approx 6.0 \times 1 0^{5}$ . A chemist wants the standard free energy change. Find $Δ G^{\circ}$ , and comment on why the plant still runs hot despite this favourable number.

Forecast: $K$ is huge ( $≫ 1$ ). Positive or negative $Δ G^{\circ}$ ? Roughly how many kJ?

Take the log. $ln (6.0 \times 1 0^{5}) = ln 6.0 + 5 ln 10 = 1.792 + 11.513 = 13.305.$ Why this step? Splitting into mantissa + power of ten keeps the arithmetic honest for very large $K$ .
Apply the formula. $Δ G^{\circ} = - R T ln K = - (2477.6) (13.305) = - 3.297 \times 1 0^{4} J mol^{- 1} \approx - 33.0 kJ mol^{- 1} .$
Interpret the plant. $Δ G^{\circ}$ is strongly negative ⇒ thermodynamics loves ammonia at 298 K. But at 298 K the rate is hopeless (a kinetics problem, not a thermodynamics one), so the plant heats up — which by Example 6's logic actually lowers $K$ for this exothermic reaction, a trade-off patched by high pressure and a catalyst. Why this step? $Δ G^{\circ}$ tells you the destination (how far), never the speed (how fast) — a classic exam distinction.

Verify: $K > 1 \Rightarrow Δ G^{\circ} < 0$ . ✔ Magnitude $\sim 33 kJ$ is the standard textbook value for $NH_{3}$ formation ( $2 \times (- 16.4) kJ mol^{- 1}$ ). ✔

Worked example Example 8 —

Cell H (exam twist): the base-10 / units trap A question gives $K = 1000$ at $298 K$ and asks for $Δ G^{\circ}$ using the base-10 form $Δ G^{\circ} = - 2.303 R T lo g K$ . A student answers $- 5.7 J mol^{- 1}$ . Find the correct value and name the trap.

Forecast: $K = 1000 = 1 0^{3}$ , so $lo g K = 3$ exactly. Should the answer be single-digit joules, or thousands of joules?

Read the log correctly. $lo g_{10} (1000) = 3.$ Why this step? Using base-10 log is fine — but you must carry the $2.303$ factor, and the student's tiny answer hints they used $lo g$ where the formula wanted $ln$ , or dropped a factor.
Plug into the base-10 form. $Δ G^{\circ} = - 2.303 R T lo g K = - 2.303 (8.314) (298) (3) .$
Compute. $2.303 \times 8.314 \times 298 = 5706$ , times $3$ gives $17118$ , so $Δ G^{\circ} = - 17118 J mol^{- 1} \approx - 17.1 kJ mol^{- 1} .$ Why this step? Cross-check with the natural-log form: $ln 1000 = 6.908$ , and $- 2477.6 \times 6.908 = - 17116 J mol^{- 1}$ — same answer. ✔ The two forms agree because $2.303 lo g K = ln K$ .

Verify: The student's $- 5.7 J$ was off by exactly $\times 3000$ — they used $- 2.303 R T lo g K$ with $lo g K$ but wrote it as $lo g K /1000$ (dropped both the $\times 3$ and the kJ→J conversion). Correct answer $\approx - 17.1 kJ mol^{- 1}$ . ✔ $K > 1 \Rightarrow Δ G^{\circ} < 0$ . ✔

Coverage check

Recall Did we hit every cell?

A → Ex 1 (find $K$ , $Δ G^{\circ} < 0$ ) · B → Ex 2 ( $Δ G^{\circ} > 0$ from $K < 1$ ) · C → Ex 3 (degenerate $Δ G^{\circ} = 0$ ) · D → Ex 4 (non-standard $Q < K$ ) · E → Ex 5 (limits $T \to \infty, 0$ ) · F → Ex 6 (second temperature, Van't Hoff) · G → Ex 7 (Haber word problem) · H → Ex 8 (units / base-10 trap). Every row of the matrix is worked. ✔

Sign forecast before you check

K > 1

then

Δ G^{\circ} < 0

; if

K < 1

then

Δ G^{\circ} > 0

; if

K = 1

then

Δ G^{\circ} = 0

Which quantity decides real-mixture direction

Δ G = Δ G^{\circ} + R T ln Q

, i.e. the sign of

Δ G

set by

Q

K

— not

Δ G^{\circ}

Endothermic reaction, raise

T

: does

K

rise or fall

Rises (Le Chatelier / Van't Hoff), because

Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}

becomes more negative.

Connections

Parent topic (Hinglish) — the derivation these examples exercise.
Reaction Quotient Q and Equilibrium — the machinery of Cell D.
Gibbs Free Energy G = H - TS — used to build $Δ G^{\circ}$ in Cell F.
Van't Hoff Equation — the temperature story of Cells E and F.
Le Chatelier's Principle — the qualitative check in Cells F and G.
Spontaneity and the Second Law — why the sign of $Δ G$ means "goes."
Chemical Potential μ — the $μ = μ^{\circ} + R T ln a$ root of everything.