Thermodynamics
Gibbs Free Energy & Feasibility
Reference Data Tables (Click to Open)
Table 1: Reaction Entropies ($\Delta S$)
| Reaction | $\Delta S$ / J K-1 mol-1 |
|---|---|
| $\frac{1}{2}N_2(g) + \frac{3}{2}H_2(g) \rightarrow NH_3(g)$ | -98.9 |
| $4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g)$ | 181.0 |
| $P_2O_5(s) + 3H_2O(l) \rightarrow 2H_3PO_4(aq)$ | 270.4 |
| $C_2H_6(g) + 3.5O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$ | -285 |
| $(CH_3CO)_2O(g) \rightarrow H_2C=C=O(g) + CH_3COOH(g)$ | 111.84 |
Table 2: Reaction Enthalpies ($\Delta H$)
| Reaction | $\Delta H$ / kJ mol-1 |
|---|---|
| $\Delta H_f$ of $NH_3$ | -46.1 |
| $\Delta H_r$ of $4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g)$ | -905.2 |
| $\Delta H_r$ of $P_2O_5(s) + 3H_2O(l) \rightarrow 2H_3PO_4(aq)$ | -96.2 |
| $\Delta H_c$ of $C_2H_6$ | -1561 |
| $\Delta H_r$ of $(CH_3CO)_2O(g) \rightarrow H_2C=C=O(g) + CH_3COOH(g)$ | 83.4 |
Table 3: Formation, Combustion & Vaporisation Data
| Substance / Reaction | $\Delta H$ / kJ mol-1 | Substance / Reaction | $\Delta H$ / kJ mol-1 |
|---|---|---|---|
| $\Delta H_c$ $C_2H_4(l)$ | -1411 | $\Delta H_f$ CO | 110.5 |
| $\Delta H_c$ $C_4H_{10}(g)$ | -2877.3 | $\Delta H_f$ $CO_2$ | -393.5 |
| $\Delta H_c$ Ethane | -1560 | $\Delta H_f$ $H_2O$ | -285.8 |
| $\Delta H_c$ Ethene | -1411.20 | $\Delta H_f$ Iron (III) oxide | -825.50 |
| $\Delta H_c$ Hydrogen | -285.8 | $\Delta H_f$ $MgCO_3$ | -1111 |
| $\Delta H_f$ butan-1-ol | -328 | $\Delta H_f$ MgO | -601.6 |
| $\Delta H_f$ $C_2H_4$ | -84.0 | $\Delta H_f$ $O_3$ | 142.7 |
| $\Delta H_f$ $C_2H_5OH$ | -269.3 | $\Delta H_f$ Phosphorus acid | -1284 |
| $\Delta H_f$ $C_4H_{10}$ | -124.7 | $\Delta H_f$ $PCl_3$ | -306.0 |
| $\Delta H_f$ $CaCO_3$ | -1206.9 | $\Delta H_{vap}$ $C_2H_5OH(l)$ | 38.60 |
| $\Delta H_f$ CaO | -635 | $\Delta H_{vap}$ $C_6H_{14}(l)$ | 28.90 |
| $\Delta H_f$ $CH_3OH$ | -205 | $\Delta H_{vap}$ $H_2O(l)$ | 40.70 |
| $\Delta H_f$ chlorobutane | -188.2 |
Table 4: Entropies of Substances ($S$)
| Substance | S / J K-1 mol-1 | Substance | S / J K-1 mol-1 |
|---|---|---|---|
| 1-Chlorobutane (l) | 375.6 | Iron (III) oxide (s) | 87.40 |
| Butan-1-ol (l) | 225.7 | Iron (s) | 27.28 |
| Butane (g) | 310.0 | Methanol (l) | 127.2 |
| Calcium carbonate (s) | 92.90 | Oxygen (g) | 205.2 |
| Calcium oxide (s) | 39.70 | Oxygen difluoride (g) | 247.5 |
| Carbon Dioxide (g) | 213.8 | Ozone (g) | 238.9 |
| Carbon monoxide (g) | 197.7 | Phosphorus acid (s) | 110.5 |
| Ethane (g) | 229.5 | Phosphorus (III) chloride (l) | 124.6 |
| Ethanol (g) | 160.7 | Sulfur hexafluoride (g) | 291.5 |
| Ethene (g) | 219.3 | Sulfur oxyfluoride (g) | 283.6 |
| Fluorine (g) | 202.8 | Sulfur trioxide (g) | 256.8 |
| Hexane (g) | 388.8 | Tridecane (l) | 522.9 |
| Hexane (l) | 296.0 | Undecane (l) | 464.0 |
| Hydrogen (g) | 130.7 | Water (g) | 188.8 |
| Water (l) | 69.95 |
Table 5: Bond Enthalpies
| Bond | $\Delta H$ / kJ mol-1 |
|---|---|
| C-C | 348 |
| C=O | 740 |
| C-H | 412 |
| O=O | 498 |
| O-H | 464 |
| S=O | 495 |
| S-F | 327 |
1. Calculate $\Delta G$ of the following reactions and state whether they are feasible given the temperature, $\Delta H$ and $\Delta S$. (Use Tables 1 & 2).
a) ½N₂(g) + 1.5H₂(g) → NH₃(g) at 200°C
Step 1: Convert Units
Temperature: $200^\circ C + 273 = 473 \text{ K}$
$\Delta S$: $-98.9 \text{ J K}^{-1} \text{mol}^{-1} = -0.0989 \text{ kJ K}^{-1} \text{mol}^{-1}$
$\Delta H$: $-46.1 \text{ kJ mol}^{-1}$
Step 2: Calculate $\Delta G$
$\Delta G = \Delta H – T\Delta S$
$\Delta G = -46.1 – (473 \times -0.0989)$
$\Delta G = -46.1 – (-46.78) = +0.68 \text{ kJ mol}^{-1}$
Reaction Not Feasible (since $\Delta G > 0$)
b) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g) at 298 K
Step 1: Identify Data
$\Delta H = -905.2 \text{ kJ mol}^{-1}$
$\Delta S = 181.0 \text{ J K}^{-1} \text{mol}^{-1} = 0.181 \text{ kJ K}^{-1} \text{mol}^{-1}$
Step 2: Calculate $\Delta G$
$\Delta G = -905.2 – (298 \times 0.181)$
$\Delta G = -905.2 – 53.94 = -959.1 \text{ kJ mol}^{-1}$
Reaction Feasible (Exothermic & Entropy Increase)
c) P₄O₁₀(s) + 6H₂O(l) → 4H₃PO₄(aq) at 25°C
Note: The stoichiometry in the question is double the reaction provided in Tables 1 & 2. We must double the data values.
Step 1: Adjust Data for Stoichiometry
$\Delta H = 2 \times -96.2 = -192.4 \text{ kJ mol}^{-1}$
$\Delta S = 2 \times 270.4 = 540.8 \text{ J K}^{-1} \text{mol}^{-1} = 0.5408 \text{ kJ K}^{-1} \text{mol}^{-1}$
$T = 298 \text{ K}$
Step 2: Calculate $\Delta G$
$\Delta G = -192.4 – (298 \times 0.5408)$
$\Delta G = -192.4 – 161.2 = -353.6 \text{ kJ mol}^{-1}$
Reaction Feasible
d) C₂H₆(g) + 3.5O₂(g) → 2CO₂(g) + 3H₂O(l) at 298 K
Step 1: Identify Data
$\Delta H = -1561 \text{ kJ mol}^{-1}$
$\Delta S = -285 \text{ J K}^{-1} \text{mol}^{-1} = -0.285 \text{ kJ K}^{-1} \text{mol}^{-1}$
Step 2: Calculate $\Delta G$
$\Delta G = -1561 – (298 \times -0.285)$
$\Delta G = -1561 + 84.9 = -1476 \text{ kJ mol}^{-1}$
Reaction Feasible
e) (CH₃CO)₂O(g) → H₂C=C=O(g) + CH₃COOH(g) at 100°C
Step 1: Identify Data & Convert Units
$\Delta H = 83.4 \text{ kJ mol}^{-1}$
$\Delta S = 111.84 \text{ J K}^{-1} \text{mol}^{-1} = 0.11184 \text{ kJ K}^{-1} \text{mol}^{-1}$
$T = 100 + 273 = 373 \text{ K}$
Step 2: Calculate $\Delta G$
$\Delta G = 83.4 – (373 \times 0.11184)$
$\Delta G = 83.4 – 41.7 = +41.7 \text{ kJ mol}^{-1}$
Reaction Not Feasible
2. Calculate $\Delta G$ of the following reactions and state whether they are feasible given the temperature, enthalpy and entropy data given in the tables.
a) CaCO₃(s) → CaO(s) + CO₂(g) at 298 K
Step 1: Calculate $\Delta H$
$\Delta H = [\Delta H_f(CaO) + \Delta H_f(CO_2)] – \Delta H_f(CaCO_3)$
$\Delta H = [-635 + (-393.5)] – [-1206.9] = \mathbf{+178.4 \text{ kJ mol}^{-1}}$
Step 2: Calculate $\Delta S$
$\Delta S = [S(CaO) + S(CO_2)] – S(CaCO_3)$
$\Delta S = [39.7 + 213.8] – 92.9 = \mathbf{+160.6 \text{ J K}^{-1} \text{mol}^{-1}}$
Step 3: Calculate $\Delta G$
$\Delta G = 178.4 – (298 \times 0.1606) = 178.4 – 47.9 = \mathbf{+130.5 \text{ kJ mol}^{-1}}$
Reaction Not Feasible
b) C₄H₁₀(g) + 4.5O₂(g) → 4CO(g) + 5H₂O(l) at 80°C
Step 1: Calculate $\Delta H$
$\Delta H = [4(\Delta H_f CO) + 5(\Delta H_f H_2O)] – [\Delta H_f C_4H_{10}]$
$\Delta H = [4(110.5) + 5(-285.8)] – [-124.7]$
$\Delta H = [442 – 1429] + 124.7 = \mathbf{-862.3 \text{ kJ mol}^{-1}}$
Step 2: Calculate $\Delta S$
$\Delta S = [4(S_{CO}) + 5(S_{H2O})] – [S_{Butane} + 4.5(S_{O2})]$
$\Delta S = [4(197.7) + 5(69.95)] – [310.0 + 4.5(205.2)]$
$\Delta S = 1140.6 – 1233.4 = \mathbf{-92.8 \text{ J K}^{-1} \text{mol}^{-1}}$
Step 3: Calculate $\Delta G$
$T = 80 + 273 = 353 \text{ K}$
$\Delta G = -862.3 – (353 \times -0.0928) = -862.3 + 32.8 = \mathbf{-829.5 \text{ kJ mol}^{-1}}$
Reaction Feasible
c) SF₆(g) + 2SO₃(g) → 3SO₂F₂(g) at 25°C
Step 1: Calculate $\Delta H$
Using Bond Enthalpies (Break – Make):
- Break: 6(S-F) + 2(3 S=O) = 6(327) + 6(495) = 4932
- Make: 3(2 S=O + 2 S-F) = 6(495) + 6(327) = 4932
$\Delta H = 4932 – 4932 = \mathbf{0 \text{ kJ mol}^{-1}}$
Step 2: Calculate $\Delta S$
$\Delta S = [3(283.6)] – [291.5 + 2(256.8)]$
$\Delta S = 850.8 – 805.1 = \mathbf{+45.7 \text{ J K}^{-1} \text{mol}^{-1}}$
Step 3: Calculate $\Delta G$
$\Delta G = 0 – (298 \times 0.0457) = \mathbf{-13.6 \text{ kJ mol}^{-1}}$
Reaction Feasible
d) C₂H₄(g) + H₂O(g) → C₂H₅OH(g) at 150°C
Step 1: Calculate $\Delta H$ (Gas Phase)
$\Delta H_f(g) = \Delta H_f(l) + \Delta H_{vap}$
- $\Delta H_f$ Ethanol(g) = $-269.3 + 38.6 = -230.7$
- $\Delta H_f$ Water(g) = $-285.8 + 40.7 = -245.1$
$\Delta H_{rxn} = [-230.7] – [-84.0 + (-245.1)] = -230.7 – (-329.1) = \mathbf{+98.4 \text{ kJ mol}^{-1}}$
Step 2: Calculate $\Delta S$
$\Delta S = 160.7 – (219.3 + 188.8) = \mathbf{-247.4 \text{ J K}^{-1} \text{mol}^{-1}}$
Step 3: Calculate $\Delta G$
$T = 150 + 273 = 423 \text{ K}$
$\Delta G = 98.4 – (423 \times -0.2474) = 98.4 + 104.7 = \mathbf{+203.1 \text{ kJ mol}^{-1}}$
Reaction Not Feasible
e) C₆H₁₄(g) → C₄H₁₀(g) + C₂H₄(g) at 400°C
Step 1: Calculate $\Delta H$ (Bond Enthalpies)
Net Change: Break 2 C-C. Form 1 C=C.
$\Delta H \approx [2(348)] – [612] = 696 – 612 = \mathbf{+84 \text{ kJ mol}^{-1}}$
Step 2: Calculate $\Delta S$
$\Delta S = [310.0 + 219.3] – 388.8 = \mathbf{+140.5 \text{ J K}^{-1} \text{mol}^{-1}}$
Step 3: Calculate $\Delta G$
$T = 400 + 273 = 673 \text{ K}$
$\Delta G = 84 – (673 \times 0.1405) = 84 – 94.6 = \mathbf{-10.6 \text{ kJ mol}^{-1}}$
Reaction Feasible
3. Calculate the range of temperatures at which the following reactions are feasible.
a) 2FeO(s) + O₂(g) → Fe₂O₃(s)
Data
$\Delta S = -172.3 \text{ J K}^{-1}$ (Negative)
$\Delta H = -109.8 \text{ kJ mol}^{-1}$ (Exothermic)
Calculation
Since $\Delta S$ is negative, feasible at low temperatures.
$T < \frac{\Delta H}{\Delta S} = \frac{-109.8}{-0.1723}$
$T < 637 \text{ K}$
b) 3CH₃CH₂CH₂CH₂OH(l) + PCl₃(l) → 3CH₃CH₂CH₂CH₂Cl(l) + H₃PO₃(s)
Step 1: Calculate $\Delta H$
$\Delta H = [3(-188.2) + (-1284)] – [3(-328) + (-306)]$
$\Delta H = -1848.6 – (-1290) = \mathbf{-558.6 \text{ kJ}}$
Step 2: Calculate $\Delta S$
$\Delta S = [3(375.6) + 110.5] – [3(225.7) + 124.6]$
$\Delta S = 1237.3 – 801.7 = \mathbf{+435.6 \text{ J K}^{-1}}$
Feasible at ALL temperatures (since $\Delta H < 0, \Delta S > 0$)
c) C₂H₄(g) + H₂(g) → C₂H₆(g)
Step 1: Calculate $\Delta H$ (Combustion Data)
$\Delta H_c(\text{Reactants}) – \Delta H_c(\text{Products})$
$\Delta H = [-1411.2 + (-285.8)] – [-1560]$
$\Delta H = -1697 + 1560 = \mathbf{-137 \text{ kJ}}$
Step 2: Calculate $\Delta S$
$\Delta S = 229.5 – (219.3 + 130.7) = \mathbf{-120.5 \text{ J K}^{-1}}$
Step 3: Calculate Range
Both negative, so feasible at lower temperatures.
$T < \frac{-137}{-0.1205}$
$T < 1137 \text{ K}$
d) Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(g)
Step 1: Calculate $\Delta H$
$\Delta H = [2(0) + 3(-245.1)] – [-825.5]$
$\Delta H = -735.3 + 825.5 = \mathbf{+90.2 \text{ kJ}}$
Step 2: Calculate $\Delta S$
$\Delta S = [2(27.3) + 3(188.8)] – [87.4 + 3(130.7)]$
$\Delta S = 621 – 479.5 = \mathbf{+141.5 \text{ J K}^{-1}}$
Step 3: Calculate Range
Both positive, feasible at high temperatures.
$T > \frac{90.2}{0.1415}$
$T > 637.5 \text{ K}$
e) 2O₃(g) → 3O₂(g)
Step 1: Calculate $\Delta H$
$\Delta H = 3(0) – 2(142.7) = \mathbf{-285.4 \text{ kJ}}$
Step 2: Calculate $\Delta S$
$\Delta S = 3(205.2) – 2(238.9) = \mathbf{+137.8 \text{ J K}^{-1}}$
Feasible at ALL temperatures (since $\Delta H < 0, \Delta S > 0$)
4. Calculate the following data based on the information in the table.
a) ΔH of C₁₃H₂₈(l) → C₂H₄(g) + C₁₁H₂₄(l)
Step 1: Calculate $\Delta S$
$\Delta S = [219.3 + 464.0] – 522.9 = \mathbf{+160.4 \text{ J K}^{-1}}$
Step 2: Calculate $\Delta H$
Given $\Delta G = -81.3$ at 298 K
$\Delta H = \Delta G + T\Delta S$
$\Delta H = -81.3 + (298 \times 0.1604) = -81.3 + 47.8$
$\Delta H = -33.5 \text{ kJ mol}^{-1}$
b) ΔS of MgCO₃(s) → MgO(s) + CO₂(g)
Step 1: Calculate $\Delta H$
$\Delta H = [-601.6 + (-393.5)] – [-1111] = \mathbf{+115.9 \text{ kJ mol}^{-1}}$
Step 2: Calculate $\Delta S$
$\Delta S = \frac{\Delta H – \Delta G}{T}$
$\Delta S = \frac{115.9 – (-143.6)}{298} = \frac{259.5}{298} = \mathbf{0.871 \text{ kJ K}^{-1}}$
$\Delta S = 871 \text{ J K}^{-1} \text{mol}^{-1}$
5. Calculate the mean bond enthalpy of the C-H bond in ethane (C₂H₆). Use the enthalpy of combustion of ethane and the enthalpy of vaporization of water from Table 3. Use Table 5 for other bond enthalpies.
Step 1: Calculate Gas Phase Reaction Enthalpy
$\Delta H_c(\text{Ethane}) = -1560 \text{ kJ mol}^{-1}$ (to liquid water).
Convert to gas: $\Delta H_r = \Delta H_c + 3 \times \Delta H_{vap}(H_2O)$
$\Delta H_r = -1560 + 3(40.7) = \mathbf{-1437.9 \text{ kJ mol}^{-1}}$
Step 2: Set up Bond Enthalpy Equation
Reaction: $C_2H_6(g) + 3.5 O_2(g) \rightarrow 2 CO_2(g) + 3 H_2O(g)$
- Break: 1(C-C) + 6(C-H) + 3.5(O=O)
- Make: 4(C=O) + 6(O-H)
Step 3: Substitute & Solve for E(C-H)
Break = $348 + 6x + 3.5(498) = 2091 + 6x$
Make = $4(740) + 6(464) = 5744$
$\Delta H_r = \text{Break} – \text{Make}$
$-1437.9 = (2091 + 6x) – 5744$
$-1437.9 = 6x – 3653$
$6x = 2215.1$
C-H Bond Enthalpy = 369.2 kJ mol-1