Gibbs free energy ΔG = ΔH − TΔS; spontaneity criteria: Exercises

Constants used throughout (write these on your scratch paper): $R = 8.314 J mol^{- 1} K^{- 1}, F = 96485 C mol^{- 1}, 0^{\circ} C = 273.15 K$

Figure — Gibbs free energy ΔG = ΔH − TΔS; spontaneity criteria

Definition Two symbols you will meet in Levels 4 and 5

Standard Gibbs free energy $Δ G^{\circ}$ — the value of $Δ G$ when every reactant and product is in its standard state (gases at $1 bar$ , solutes at $1 mol/L$ , pure solids/liquids). It is a fixed number for a given reaction at a given temperature, independent of how much has actually reacted. The plain $Δ G$ (no circle) is the value for the current mixture.
Reaction quotient $Q$ — the same ratio of products-over-reactants that defines the equilibrium constant, but evaluated at the current, not-yet-equilibrium composition. For $a A + b B \to c C + d D$ , $Q = \frac{a _{C}^{c} a _{D}^{d}}{a _{A}^{a} a _{B}^{b}} .$ Here each $a_{X}$ is the activity — an "effective concentration." In practice a dissolved species uses its concentration $[X]$ (relative to $1 mol/L$ ) and a gas uses its partial pressure (relative to $1 bar$ ). Crucially, pure solids, pure liquids, and the solvent have activity $= 1$ and are OMITTED from $Q$ (they don't appear as $[]$ terms), because their "concentration" doesn't change as the reaction proceeds. So for $CaCO_{3} (s) \to CaO (s) + CO_{2} (g)$ , $Q = p_{CO_{2}}$ only. At equilibrium $Q = K$ . The two $Δ G$ 's are linked by $Δ G = Δ G^{\circ} + R T ln Q .$ When $Q$ is small (few products yet), $ln Q$ is very negative and $Δ G$ can be negative even if $Δ G^{\circ} > 0$ — see Chemical Equilibrium and K.

Level 1 — Recognition

(Can you read the sign rule and the four-case table?)

Recall Solution L1.1

The rule (from the parent note) is: $Δ G < 0 \Rightarrow$ spontaneous forward. Here $Δ G = - 47 kJ < 0$ , so the forward reaction is spontaneous. Mnemonic: Go Down = Go.

Recall Solution L1.2

Read $Δ G = Δ H - T Δ S$ term by term, and locate each on the quadrant map. (a) $Δ H$ negative, $- T Δ S$ negative → $Δ G$ always negative → always spontaneous — this is the red top-left box of Figure 1. (b) $Δ H$ positive, $- T Δ S$ positive → $Δ G$ always positive → never spontaneous — the bottom-right box. (c) $Δ H$ negative (helps), $- T Δ S$ positive (hurts). Enthalpy wins only when $T$ is small → spontaneous only at low $T$ — the bottom-left box (thermometer reads "cold"). (d) $Δ H$ positive (hurts), $- T Δ S$ negative (helps). Entropy wins only when $T$ is big → spontaneous only at high $T$ — the top-right box (thermometer reads "hot"). Cold loves H, Hot loves S.

Recall Solution L1.3

At the boiling point the two phases coexist — neither wins — so the process is at equilibrium and $Δ G = 0.$ On Figure 2 this is precisely where the red line crosses the zero-axis. (That is why $T_{b o i l} = Δ H /Δ S$ ; see L2 below.)

Level 2 — Application

(Plug numbers into $Δ G = Δ H - T Δ S$ correctly.)

Recall Solution L2.1

Convert $Δ S$ to kJ: $150.0 J/K = 0.1500 kJ/K$ . $Δ G = 120.0 - (298) (0.1500) = 120.0 - 44.70 = + 75.30 kJ .$ $Δ G > 0$ → not spontaneous at 298 K. (Endothermic + entropy-increasing → the top-right "hot only" box of Figure 1; 298 K isn't hot enough yet.)

Recall Solution L2.2

Spontaneity switches on when $Δ G = 0$ , i.e. $Δ H = T Δ S$ : $T = \frac{Δ H}{Δ S} = \frac{120000 J}{150.0 J/K} = 800.0 K .$ Above $800 K$ , $T Δ S > Δ H$ so $Δ G < 0$ → spontaneous. In the language of Figure 2: this reaction's line starts at intercept $+ 120 kJ$ (positive, since $Δ H > 0$ ) and tilts down with steepness $Δ S = 150 J/K$ (because slope $= - Δ S < 0$ ). It reaches the zero-axis at $T = 800 K$ — the crossover. Left of that point the line is above zero (non-spontaneous); right of it, below zero (spontaneous).

Recall Solution L2.3

Coexistence means equilibrium, $Δ G = 0$ : $T = \frac{Δ H}{Δ S} = \frac{6010 J/mol}{22.0 J/mol\cdotK} = 273.2 K = 0.05^{\circ} C .$ That is the melting point of ice (spot-on with $273.15 K$ ). See Phase Transitions.

Level 3 — Analysis

(Reason about behaviour, not just plug-and-chug.)

Recall Solution L3.1

Compare to $Δ G = Δ H - T Δ S = Δ H + (- Δ S) T$ — the same $y = c + m x$ shape as Figure 2. The intercept is $Δ H = + 40.0 kJ$ → endothermic ( $Δ H > 0$ ). The slope is $- Δ S = - 0.100 kJ/K$ , so $Δ S = + 0.100 kJ/K = + 100 J/K$ → entropy increases. (Recall the derivative box: slope $= d (Δ G) / d T = - Δ S$ , which is why reading the coefficient of $T$ gives $- Δ S$ .) Crossover: $Δ G = 0 \Rightarrow T = 40.0/0.100 = 400 K$ . Since the line falls, it is spontaneous above $400 K$ . This is the top-right (case-d) box of Figure 1 — Hot loves S.

Recall Solution L3.2

From the derivation in the parent note, $Δ S_{s u r r} = - Δ H / T$ (see Entropy and ΔS): $Δ S_{s u r r} = \frac{- ( - 88000 J )}{298 K} = + 295.3 J/K .$ Positive because the system dumps heat into the surroundings, spreading energy there and raising their entropy. Exothermic reactions "pay" the universe in surroundings-entropy.

Recall Solution L3.3

This is case (c) — the bottom-left "cold only" box of Figure 1: both negative → spontaneous only at low $T$ . It flips at $Δ G = 0$ : $T = \frac{Δ H}{Δ S} = \frac{- 92400 J}{- 198.3 J/K} = 465.96 K \approx 466 K .$ On a Figure-2 line this reaction rises (slope $= - Δ S > 0$ , since $Δ S < 0$ ) from a negative intercept, crossing zero at $\approx 466 K$ . Above that the reaction is non-spontaneous — which is exactly why industry can't just crank the temperature to speed it up.

Level 4 — Synthesis

(Combine Gibbs energy with equilibrium and electrochemistry.)

Recall Solution L4.1

Use $Δ G^{\circ} = - R T ln K$ (see Chemical Equilibrium and K), so $ln K = - Δ G^{\circ} / (R T)$ : $ln K = \frac{28000 J}{( 8.314 ) ( 298 )} = \frac{28000}{2477.6} = 11.30 \Rightarrow K = e^{11.30} \approx 8.1 \times 1 0^{4} .$ Large negative $Δ G^{\circ}$ → $K ≫ 1$ → products strongly favoured. Consistent sign logic.

Recall Solution L4.2

$Δ G^{\circ} = - R T ln K = - (8.314) (298) ln (2.5 \times 1 0^{- 3}) .$ $ln (2.5 \times 1 0^{- 3}) = - 5.9915$ , so $Δ G^{\circ} = - (2477.6) (- 5.9915) = + 14847 J \approx + 14.8 kJ .$ $Δ G^{\circ} > 0$ → under standard conditions the reaction is non-spontaneous (reactants favoured, $K < 1$ ). Note the actual $Δ G$ could still be negative if the reaction quotient $Q$ (defined at the top of the page) is small — few products present makes $R T ln Q$ strongly negative.

Recall Solution L4.3

Use $Δ G^{\circ} = - n F E^{\circ}$ (see Electrochemistry ΔG = −nFE): $Δ G^{\circ} = - (2) (96485) (1.10) = - 212267 J \approx - 212.3 kJ .$ $Δ G^{\circ} < 0$ → spontaneous cell reaction. A positive cell voltage always means a negative $Δ G^{\circ}$ (the minus sign links them).

Level 5 — Mastery

(Multi-step problems that stitch everything together.)

Recall Solution L5.1

(a) $T^{*} = Δ H /Δ S = 55000/140.0 = 392.9 K$ . (b) At 298 K: $Δ G = 55000 - (298) (140.0) = 55000 - 41720 = + 13280 J = + 13.28 kJ$ . Then $ln K = - Δ G / (R T) = - 13280/ (8.314 \times 298) = - 5.360$ , so $K = e^{- 5.360} = 4.7 \times 1 0^{- 3}$ . (c) At 500 K: $Δ G = 55000 - (500) (140.0) = 55000 - 70000 = - 15000 J = - 15.0 kJ$ . Then $ln K = 15000/ (8.314 \times 500) = 3.609$ , so $K = e^{3.609} = 36.9$ . (d) Below $T^{*}$ , $Δ G > 0$ and $K < 1$ (reactants favoured); crossing $T^{*}$ flips $Δ G$ negative and pushes $K$ above 1 — the reaction "turns on." This is exactly the shape of Figure 2: the line falls through zero at $\approx 393 K$ , and the further below zero $Δ G$ goes, the larger $K$ grows.

Recall Solution L5.2

(a) $Δ G^{\circ} = - n F E^{\circ} = - (2) (96485) (0.34) = - 65610 J \approx - 65.6 kJ$ . (b) $ln K = - Δ G^{\circ} / (R T) = 65610/ (8.314 \times 298) = 26.49$ , so $K = e^{26.49} \approx 3.2 \times 1 0^{11}$ . (c) Yes — all three say "products/forward strongly favoured": $E^{\circ} > 0 \Leftrightarrow Δ G^{\circ} < 0 \Leftrightarrow K > 1$ . They are three languages for the same fact, linked by $Δ G^{\circ} = - n F E^{\circ} = - R T ln K$ .

Recall Solution L5.3

Both expressions give the same $Δ G^{\circ}$ , so set them equal: $Δ H - T Δ S = - R T ln K .$ Solve for $T$ : bring the $T$ -terms together, $Δ H = T Δ S - R T ln K = T (Δ S - R ln K)$ , hence $T = \frac{Δ H}{Δ S - R l n K} .$ Numbers: $R ln K = 8.314 \times ln 10 = 8.314 \times 2.3026 = 19.14 J/K$ . $T = \frac{80000}{160.0 - 19.14} = \frac{80000}{140.86} = 568.0 K .$ At about $568 K$ the equilibrium constant reaches 10. Sanity check against Figure 2: the plain crossover (where $K = 1$ ) is $T^{*} = 80000/160 = 500 K$ ; to make $K$ climb from 1 up to 10 you must push $T$ past that point, and indeed $568 K > 500 K$ . The formula $T = Δ H / (Δ S - R ln K)$ reduces to the familiar $T^{*} = Δ H /Δ S$ exactly when $ln K = 0$ , i.e. $K = 1$ — the crossover is just the special case.

Recall One-line self-test before you leave

Given any two of ${E^{\circ}, Δ G^{\circ}, K}$ , can you get the third at $298 K$ ? ::: Yes — chain them through $Δ G^{\circ} = - n F E^{\circ} = - R T ln K$ .

Connections

Parent topic (Hinglish) — the theory these drills exercise.
Second Law of Thermodynamics — origin of the $Δ G < 0$ rule.
Enthalpy and ΔH and Entropy and ΔS — the two terms you keep balancing.
Chemical Equilibrium and K — L4/L5 use $Δ G^{\circ} = - R T ln K$ and $Q$ .
Electrochemistry ΔG = −nFE — L4.3, L5.2 use the cell link.
Phase Transitions — L2.3 melting point.