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

The picture we keep coming back to: $Δ G^{\circ}$ says where the valley bottom sits, and $Δ G$ (the local slope) says which way a ball rolls right now.

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

Level 1 — Recognition

(Can you pick the right form and plug in without slipping on units or signs?)

Exercise 1.1. State, without calculation, the sign of $Δ G^{\circ}$ for each: (a) $K = 5 \times 1 0^{6}$ , (b) $K = 1$ , (c) $K = 3 \times 1 0^{- 4}$ .

Recall Solution 1.1

Use $Δ G^{\circ} = - R T ln K$ . The sign of $Δ G^{\circ}$ is the opposite of the sign of $ln K$ .

(a) $K > 1 \Rightarrow ln K > 0 \Rightarrow Δ G^{\circ} < 0$ (products favoured).
(b) $K = 1 \Rightarrow ln K = 0 \Rightarrow Δ G^{\circ} = 0$ (balanced).
(c) $K < 1 \Rightarrow ln K < 0 \Rightarrow Δ G^{\circ} > 0$ (reactants favoured). Mnemonic: "Negative G, greater than 1."

Exercise 1.2. A reaction at $298 K$ has $Δ G^{\circ} = - 11.5 kJ mol^{- 1}$ . Find $K$ .

Recall Solution 1.2

What: rearrange to $ln K = - Δ G^{\circ} / (R T)$ . Why SI: convert kJ → J so $R$ matches. $ln K = - \frac{- 11500}{8.314 \times 298} = \frac{11500}{2477.6} = 4.641$ $K = e^{4.641} \approx 1.04 \times 1 0^{2}$ Sign check: $Δ G^{\circ} < 0 \Rightarrow K > 1$ . ✔

Exercise 1.3. For a reaction $K = 0.0450$ at $298 K$ . Find $Δ G^{\circ}$ in kJ mol⁻¹.

Recall Solution 1.3

$Δ G^{\circ} = - R T ln K = - (2477.6) ln (0.0450) = - (2477.6) (- 3.101) = + 7683 J mol^{- 1}$ $\approx + 7.68 kJ mol^{- 1}$ $K < 1 \Rightarrow Δ G^{\circ} > 0$ . ✔

Level 2 — Application

(Plug into the full machinery, including $Q$ .)

Exercise 2.1. For $N_{2} O_{4} ⇌ 2 N O_{2}$ , $Δ G^{\circ} = + 4.74 kJ mol^{- 1}$ at $298 K$ . A vessel holds $P_{N_{2} O_{4}} = 2.0 bar$ and $P_{N O_{2}} = 0.50 bar$ (standard pressure $= 1 bar$ ). Compute $Δ G$ and state the direction.

Recall Solution 2.1

Step 1 — build $Q$ from activities $a_{i} = P_{i} / P^{\circ} = P_{i} / (1 bar)$ : $Q = \frac{a _{N O_{2}}^{2}}{a _{N_{2} O_{4}}} = \frac{( 0.50 ) ^{2}}{2.0} = \frac{0.25}{2.0} = 0.125$ Step 2 — plug into $Δ G = Δ G^{\circ} + R T ln Q$ (convert $Δ G^{\circ}$ to J): $Δ G = 4740 + (2477.6) ln (0.125) = 4740 + (2477.6) (- 2.079)$ $= 4740 - 5152 = - 412 J mol^{- 1}$ Step 3 — read the sign: $Δ G < 0 \Rightarrow$ reaction runs forward (makes more $N O_{2}$ ). Why: even though $Δ G^{\circ} > 0$ , here $Q = 0.125$ is below $K$ (from Ex. 2 of the parent, $K = 0.148$ ), so the ball is on the reactant side of the valley and rolls toward products.

Exercise 2.2. Same reaction/ $Δ G^{\circ}$ . What value of $Q$ makes $Δ G = 0$ ? Confirm it equals $K$ .

Recall Solution 2.2

Set $Δ G = 0$ : $0 = Δ G^{\circ} + R T ln Q \Rightarrow ln Q = - Δ G^{\circ} / R T$ . $ln Q = - \frac{4740}{2477.6} = - 1.913 \Rightarrow Q = e^{- 1.913} = 0.148$ This is exactly $K$ . Why: $Δ G = 0$ is defined as the equilibrium condition, and at equilibrium $Q = K$ by definition. So $Δ G = 0$ and $Q = K$ are the same statement.

Exercise 2.3. At $298 K$ , $K_{w}$ for water autoionisation is $1.0 \times 1 0^{- 14}$ . Find $Δ G^{\circ}$ .

Recall Solution 2.3

$Δ G^{\circ} = - R T ln K_{w} = - (2477.6) ln (1.0 \times 1 0^{- 14})$ $= - (2477.6) (- 32.24) = + 79880 J mol^{- 1} \approx + 79.9 kJ mol^{- 1}$ A very small $K$ ⇒ a large positive $Δ G^{\circ}$ : pure water is overwhelmingly un-ionised. ✔

Level 3 — Analysis

(Compare, invert, and reason about behaviour.)

Exercise 3.1. Reaction A has $Δ G_{A}^{\circ} = - 20 kJ mol^{- 1}$ ; reaction B has $Δ G_{B}^{\circ} = - 40 kJ mol^{- 1}$ (both $298 K$ ). By what factor is $K_{B}$ larger than $K_{A}$ ?

Recall Solution 3.1

$ln K = - Δ G^{\circ} / R T$ , so $ln \frac{K _{B}}{K _{A}} = ln K_{B} - ln K_{A} = \frac{- Δ G _{B}^{\circ} + Δ G _{A}^{\circ}}{R T} = \frac{- ( - 40000 ) + ( - 20000 )}{2477.6}$ $= \frac{20000}{2477.6} = 8.073 \Rightarrow \frac{K _{B}}{K _{A}} = e^{8.073} \approx 3.2 \times 1 0^{3}$ Insight: doubling the depth of the valley did not double $K$ — it multiplied it by ~3000. Because $K$ depends exponentially on $Δ G^{\circ}$ , small energy differences produce enormous ratio changes.

Exercise 3.2. A reaction has $Δ G^{\circ} = - 30 kJ mol^{- 1}$ and is currently held at $Q = 1 \times 1 0^{8}$ . Is it at equilibrium? If not, which direction, and what is $Δ G$ ?

Recall Solution 3.2

First find $K$ : $ln K = 30000/2477.6 = 12.11 \Rightarrow K = e^{12.11} = 1.82 \times 1 0^{5}$ . Now $Q = 1 0^{8} > K$ , so products are over-supplied — expect reverse motion. Confirm with $Δ G$ : $Δ G = Δ G^{\circ} + R T ln Q = - 30000 + (2477.6) ln (1 0^{8})$ $= - 30000 + (2477.6) (18.42) = - 30000 + 45640 = + 15640 J mol^{- 1}$ $Δ G > 0 \Rightarrow$ forward reaction is uphill ⇒ it runs backward (consumes product). Consistent with $Q > K$ . ✔

Exercise 3.3. For a reaction at $298 K$ , adding up the standard values gives $Δ H^{\circ} = + 52.0 kJ mol^{- 1}$ and $Δ S^{\circ} = + 176 J K^{- 1} mol^{- 1}$ . Find $Δ G^{\circ}$ and then $K$ .

Recall Solution 3.3

Step 1 — get $Δ G^{\circ}$ from $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ (this is where Gibbs Free Energy G = H - TS feeds in). Convert $Δ S^{\circ}$ term to kJ: $T Δ S^{\circ} = 298 \times 176 = 52448 J = 52.448 kJ$ . $Δ G^{\circ} = 52.0 - 52.448 = - 0.448 kJ mol^{- 1} = - 448 J mol^{- 1}$ Step 2 — get $K$ : $ln K = - \frac{- 448}{2477.6} = 0.1808 \Rightarrow K = e^{0.1808} \approx 1.20$ $K$ just above 1: the reaction is nearly balanced because $Δ H^{\circ}$ and $T Δ S^{\circ}$ almost cancel at this $T$ .

Level 4 — Synthesis

(Combine $Δ G^{\circ} = - R T ln K$ with a second relation.)

Exercise 4.1. A reaction has $K_{1} = 4.0 \times 1 0^{- 2}$ at $T_{1} = 298 K$ and $K_{2} = 4.0 \times 1 0^{- 1}$ at $T_{2} = 350 K$ . Assuming $Δ H^{\circ}$ constant, use the two-point Van't Hoff Equation $ln \frac{K _{2}}{K _{1}} = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ to find $Δ H^{\circ}$ .

Recall Solution 4.1

Step 1 — left side: $ln (K_{2} / K_{1}) = ln (0.40/0.040) = ln 10 = 2.3026$ . Step 2 — the temperature bracket: $\frac{1}{T _{2}} - \frac{1}{T _{1}} = \frac{1}{350} - \frac{1}{298} = 0.0028571 - 0.0033557 = - 4.986 \times 1 0^{- 4} K^{- 1}$ Step 3 — solve for $Δ H^{\circ}$ : $2.3026 = - \frac{Δ H ^{\circ}}{8.314} \times (- 4.986 \times 1 0^{- 4})$ $Δ H^{\circ} = \frac{2.3026 \times 8.314}{4.986 \times 1 0 ^{- 4}} = \frac{19.15}{4.986 \times 1 0 ^{- 4}} \approx 3.84 \times 1 0^{4} J mol^{- 1} \approx + 38.4 kJ mol^{- 1}$ Why positive? $K$ rose when $T$ rose, and only an endothermic ( $Δ H^{\circ} > 0$ ) reaction has $K$ increasing with temperature — exactly Le Chatelier's Principle applied to heat.

Exercise 4.2. For that same reaction, find $Δ G^{\circ}$ and $Δ S^{\circ}$ at $298 K$ .

Recall Solution 4.2

$Δ G^{\circ}$ at $298 K$ from $K_{1}$ : $Δ G^{\circ} = - R T ln K_{1} = - (2477.6) ln (0.040) = - (2477.6) (- 3.219) = + 7976 J mol^{- 1} \approx + 7.98 kJ mol^{- 1}$ $Δ S^{\circ}$ from $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} \Rightarrow Δ S^{\circ} = (Δ H^{\circ} - Δ G^{\circ}) / T$ : $Δ S^{\circ} = \frac{38400 - 7976}{298} = \frac{30424}{298} \approx 102 J K^{- 1} mol^{- 1}$ Positive $Δ S^{\circ}$ : the products are more disordered — consistent with a reaction driven forward by heating.

Level 5 — Mastery

(A multi-step chain; everything must click together.)

Exercise 5.1. For $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ at $298 K$ , $Δ G^{\circ} = - 140.0 kJ mol^{- 1}$ . (a) Find $K$ . (b) A reactor holds $P_{S O_{2}} = 0.10 bar$ , $P_{O_{2}} = 0.20 bar$ , $P_{S O_{3}} = 2.0 bar$ . Find $Q$ and $Δ G$ ; state the direction. (c) Interpret whether the mixture is far from or near equilibrium.

Recall Solution 5.1

(a) $ln K = - Δ G^{\circ} / R T = 140000/2477.6 = 56.50 \Rightarrow K = e^{56.50} \approx 3.5 \times 1 0^{24}$ . Enormous — $S O_{3}$ is heavily favoured. (b) Build $Q$ (all activities $= P_{i} /1 bar$ ): $Q = \frac{a _{S O_{3}}^{2}}{a _{S O_{2}}^{2} a _{O_{2}}} = \frac{( 2.0 ) ^{2}}{( 0.10 ) ^{2} ( 0.20 )} = \frac{4.0}{0.0020} = 2000$ Then $Δ G = Δ G^{\circ} + R T ln Q = - 140000 + (2477.6) ln (2000)$ $= - 140000 + (2477.6) (7.601) = - 140000 + 18832 = - 121168 J mol^{- 1} \approx - 121 kJ mol^{- 1}$ $Δ G ≪ 0 \Rightarrow$ strongly forward (more $S O_{3}$ ). (c) Compare $Q = 2000$ with $K = 3.5 \times 1 0^{24}$ : $Q / K \approx 6 \times 1 0^{- 22}$ , i.e. $Q$ is twenty-two orders of magnitude below $K$ . The mixture is extremely far from equilibrium on the reactant side, which is exactly why $Δ G$ is so large and negative. The tiny $18.8 kJ$ push from the $R T ln Q$ term barely dents the $- 140 kJ$ standard drop.

Exercise 5.2. For the same reaction, at what total picture would $Δ G = 0$ ? Show that the required $Q$ equals $K$ , and explain in one sentence what that means physically.

Recall Solution 5.2

Set $Δ G = 0$ : $ln Q = - Δ G^{\circ} / R T = 56.50 \Rightarrow Q = e^{56.50} = 3.5 \times 1 0^{24} = K$ . Physical meaning: $Δ G = 0$ is the equilibrium condition — it is the flat bottom of the free-energy valley where the reaction has no further tendency to shift, and there the reaction quotient has climbed all the way up to $K$ .

Wrap-up recall

Recall One-line summary of every level

Which relation decides the direction of a real mixture? ::: $Δ G = Δ G^{\circ} + R T ln Q$ — the sign of $Δ G$ (equivalently, $Q$ vs $K$ ). Which relation gives the position of equilibrium? ::: $Δ G^{\circ} = - R T ln K$ . Why does small $ΔΔ G^{\circ}$ cause huge $Δ K$ ? ::: because $K = e^{- Δ G^{\circ} / R T}$ is exponential in $Δ G^{\circ}$ .

Connections

Gibbs Free Energy G = H - TS — source of $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ (Ex. 3.3, 4.2).
Reaction Quotient Q and Equilibrium — building $Q$ from activities (Ex. 2.1, 5.1).
Van't Hoff Equation — temperature dependence of $K$ (Ex. 4.1).
Le Chatelier's Principle — sign of $Δ H^{\circ}$ vs how $K$ moves with $T$ .
Spontaneity and the Second Law — why $Δ G < 0$ means "goes."
Chemical Potential μ — the $μ = μ^{\circ} + R T ln a$ backbone under every $Q$ .