Enthalpy Changes Definitions Enthalpy (H H H
An enthalpy change (Δ H \Delta H Δ H
Sign Convention Type Sign Energy Flow Exothermic Δ H < 0 \Delta H \lt 0 Δ H < 0 System releases heat to surroundings Endothermic Δ H > 0 \Delta H \gt 0 Δ H > 0 System absorbs heat from surroundings
Standard Conditions Standard enthalpy changes are measured under standard conditions:
Pressure: 100 k P a 100\mathrm{ kPa} 100 kPa Concentration: 1 m o l / L 1\mathrm{ mol/L} 1 mol/L Temperature: 298 K 298\mathrm{ K} 298 K 25 ° C 25\degree\mathrm{C} 25° C All substances in their standard states Types of Enthalpy Change Type Symbol Definition Standard enthalpy of formation Δ H f ∘ \Delta H_f^\circ Δ H f ∘ Enthalpy change when 1 mol of compound forms from its elements in standard states Standard enthalpy of combustion Δ H c ∘ \Delta H_c^\circ Δ H c ∘ Enthalpy change when 1 mol of substance burns completely in oxygen Standard enthalpy of neutralisation Δ H n e u t ∘ \Delta H_{\mathrm{neut}}^\circ Δ H neut ∘ Enthalpy change when 1 mol of water forms from acid-base reaction Standard enthalpy of atomisation Δ H a t ∘ \Delta H_{\mathrm{at}}^\circ Δ H at ∘ Enthalpy change to form 1 mol of gaseous atoms from element in standard state
Δ H f ∘ \Delta H_f^\circ Δ H f ∘ Δ H f ∘ \Delta H_f^\circ Δ H f ∘ 2 _2 2 Δ H f ∘ \Delta H_f^\circ Δ H f ∘
Calorimetry Principle Calorimetry measures the heat exchanged during a chemical or physical process.
Specific Heat Capacity The amount of energy required to raise the temperature of 1 g 1\mathrm{ g} 1 g 1 ° C 1\degree\mathrm{C} 1° C
Q = m c Δ T Q = mc\Delta T Q = m c Δ T Where:
q q q m m m c c c ° \degree ° Δ T \Delta T Δ T ° \degree ° Substance c c c ° \degree ° Water 4.18 Ice 2.09 Aluminium 0.90 Copper 0.39 Iron 0.45
Measuring Enthalpy of Reaction For a reaction in solution:
Δ H = − m c Δ T n \Delta H = -\frac{mc\Delta T}{n} Δ H = − n m c Δ T The negative sign accounts for the convention: heat lost by the reaction is gained by the solution.
Assumptions in Calorimetry No heat loss to the surroundings (use a calorimeter). The calorimeter itself has negligible heat capacity. The solution has the same density and specific heat capacity as water. Example
50.0 m L 50.0\mathrm{ mL} 50.0 mL 1.0 M 1.0\mathrm{ M} 1.0 M 50.0 m L 50.0\mathrm{ mL} 50.0 mL 1.0 M 1.0\mathrm{ M} 1.0 M 21.0 ° C 21.0\degree\mathrm{C} 21.0° C 27.5 ° C 27.5\degree\mathrm{C} 27.5° C
M = 100.0 g ( a s s u m i n g d e n s i t y o f w a t e r ) M = 100.0\mathrm{ g} \mathrm{ (assuming density of water)} M = 100.0 g ( assumingdensityofwater ) Q = m c Δ T = 100.0 × 4.18 × 6.5 = 2717 J = 2.717 k J Q = mc\Delta T = 100.0 \times 4.18 \times 6.5 = 2717\mathrm{ J} = 2.717\mathrm{ kJ} Q = m c Δ T = 100.0 × 4.18 × 6.5 = 2717 J = 2.717 kJ N ( H 2 O ) = 0.050 m o l ( l i m i t e d b y t h e r e a g e n t v o l u m e s ) N(\mathrm{H}_2\mathrm{O}) = 0.050\mathrm{ mol} \mathrm{ (limited by the reagent volumes)} N ( H 2 O ) = 0.050 mol ( limitedbythereagentvolumes ) Δ H = − 2.717 0.050 = − 54.3 k J / m o l \Delta H = -\frac{2.717}{0.050} = -54.3\mathrm{ kJ/mol} Δ H = − 0.050 2.717 = − 54.3 kJ/mol Bomb Calorimetry Used for combustion reactions. The calorimeter constant C C C
Q r e a c t i o n = − ( m c Δ T + C Δ T ) Q_{\mathrm{reaction}} = -(mc\Delta T + C\Delta T) Q reaction = − ( m c Δ T + C Δ T ) Hess”s Law Statement The total enthalpy change for a reaction is independent of the route taken. Only on the Initial and final states.
Δ H t o t a l = Δ H 1 + Δ H 2 + ⋯ \Delta H_{\mathrm{total}} = \Delta H_1 + \Delta H_2 + \cdots Δ H total = Δ H 1 + Δ H 2 + ⋯ Using Enthalpy Cycles To find an unknown enthalpy change, construct a cycle with known enthalpy changes and apply Hess’s Law.
Δ H r ∘ = ∑ Δ H f ∘ ( p r o d u c t s ) − ∑ Δ H f ∘ ( r e a c t a n t s ) \Delta H_r^\circ = \sum \Delta H_f^\circ(\mathrm{products}) - \sum \Delta H_f^\circ(\mathrm{reactants}) Δ H r ∘ = ∑ Δ H f ∘ ( products ) − ∑ Δ H f ∘ ( reactants ) Example
Calculate Δ H r ∘ \Delta H_r^\circ Δ H r ∘ 4 _4 4 2 _2 2 → \to → 2 _2 2 2 _2 2
Given:
Δ H f ∘ \Delta H_f^\circ Δ H f ∘ 4 _4 4 − 74.8 k J / m o l -74.8\mathrm{ kJ/mol} − 74.8 kJ/mol Δ H f ∘ \Delta H_f^\circ Δ H f ∘ 2 _2 2 − 393.5 k J / m o l -393.5\mathrm{ kJ/mol} − 393.5 kJ/mol Δ H f ∘ \Delta H_f^\circ Δ H f ∘ 2 _2 2 − 285.8 k J / m o l -285.8\mathrm{ kJ/mol} − 285.8 kJ/mol Δ H r ∘ = [ ( − 393.5 ) + 2 ( − 285.8 ) ] − [ ( − 74.8 ) + 2 ( 0 ) ] \Delta H_r^\circ = [(-393.5) + 2(-285.8)] - [(-74.8) + 2(0)] Δ H r ∘ = [( − 393.5 ) + 2 ( − 285.8 )] − [( − 74.8 ) + 2 ( 0 )] = ( − 393.5 − 571.6 ) − ( − 74.8 ) = − 965.1 + 74.8 = − 890.3 k J / m o l = (-393.5 - 571.6) - (-74.8) = -965.1 + 74.8 = -890.3\mathrm{ kJ/mol} = ( − 393.5 − 571.6 ) − ( − 74.8 ) = − 965.1 + 74.8 = − 890.3 kJ/mol Using Standard Enthalpies of Combustion Δ H r ∘ = ∑ Δ H c ∘ ( r e a c t a n t s ) − ∑ Δ H c ∘ ( p r o d u c t s ) \Delta H_r^\circ = \sum \Delta H_c^\circ(\mathrm{reactants}) - \sum \Delta H_c^\circ(\mathrm{products}) Δ H r ∘ = ∑ Δ H c ∘ ( reactants ) − ∑ Δ H c ∘ ( products ) Note the reversed order compared to formation enthalpies.
Bond Enthalpies Definition The bond enthalpy (bond dissociation energy) is the enthalpy change when one mole of covalent Bonds in the gas phase is broken.
Average Bond Enthalpies Since bond enthalpies vary slightly depending on the molecular environment, tables give average Values.
Bond Average Enthalpy (kJ/mol) C—C 347 C=C 612 C≡ \equiv ≡ 838 C—H 413 C—O 358 C=O 743 O—H 463 O=O 495 H—H 436 N≡ \equiv ≡ 945 N—H 391
Calculating Enthalpy Change from Bond Enthalpies Δ H = ∑ ( b o n d s b r o k e n ) − ∑ ( b o n d s f o r m e d ) \Delta H = \sum (\mathrm{bonds broken}) - \sum (\mathrm{bonds formed}) Δ H = ∑ ( bondsbroken ) − ∑ ( bondsformed ) Bonds broken (positive — energy absorbed) and bonds formed (negative — energy released).
Example
Calculate the enthalpy change for: CH4 _4 4 2 _2 2 → \to → 2 _2 2 2 _2 2
Bonds broken : 4(C—H) + 2(O=O) = 4 ( 413 ) + 2 ( 495 ) = 1652 + 990 = 2642 k J / m o l = 4(413) + 2(495) = 1652 + 990 = 2642\mathrm{ kJ/mol} = 4 ( 413 ) + 2 ( 495 ) = 1652 + 990 = 2642 kJ/mol
Bonds formed : 2(C=O) + 4(O—H) = 2 ( 743 ) + 4 ( 463 ) = 1486 + 1852 = 3338 k J / m o l = 2(743) + 4(463) = 1486 + 1852 = 3338\mathrm{ kJ/mol} = 2 ( 743 ) + 4 ( 463 ) = 1486 + 1852 = 3338 kJ/mol
Δ H = 2642 − 3338 = − 696 k J / m o l \Delta H = 2642 - 3338 = -696\mathrm{ kJ/mol} Δ H = 2642 − 3338 = − 696 kJ/mol
Bond enthalpy calculations give approximate values because average bond enthalpies are used. Values From Hess’s law with formation data are more accurate. Bond enthalpy calculations only apply to Gases.
Born-Haber Cycles Purpose Born-Haber cycles determine the lattice energy of an ionic compound using a thermodynamic cycle.
Lattice Energy The lattice energy Δ H l a t t \Delta H_{\mathrm{latt}} Δ H latt
Steps in a Born-Haber Cycle For an ionic compound MX:
Δ H f ∘ \Delta H_f^\circ Δ H f ∘ Δ H a t ( M ) \Delta H_{\mathrm{at}}(M) Δ H at ( M ) → \to → Δ H a t ( X 2 ) \Delta H_{\mathrm{at}}(X_2) Δ H at ( X 2 ) 1 2 \frac{1}{2} 2 1 2 _2 2 → \to → IE1 _1 1 2 _2 2 EA1 _1 1 Δ H l a t t \Delta H_{\mathrm{latt}} Δ H latt Applying Hess’s Law Δ H f ∘ = Δ H a t ( M ) + 1 2 Δ H a t ( X 2 ) + I E + E A + Δ H l a t t \Delta H_f^\circ = \Delta H_{\mathrm{at}}(M) + \frac{1}{2}\Delta H_{\mathrm{at}}(X_2) + \mathrm{IE} + \mathrm{EA} + \Delta H_{\mathrm{latt}} Δ H f ∘ = Δ H at ( M ) + 2 1 Δ H at ( X 2 ) + IE + EA + Δ H latt Example
Calculate the lattice energy of NaCl.
Given:
Δ H f ∘ \Delta H_f^\circ Δ H f ∘ − 411 k J / m o l -411\mathrm{ kJ/mol} − 411 kJ/mol Δ H a t \Delta H_{\mathrm{at}} Δ H at + 108 k J / m o l +108\mathrm{ kJ/mol} + 108 kJ/mol 1 2 Δ H a t \frac{1}{2}\Delta H_{\mathrm{at}} 2 1 Δ H at 2 _2 2 + 122 k J / m o l +122\mathrm{ kJ/mol} + 122 kJ/mol IE1 _1 1 + 496 k J / m o l +496\mathrm{ kJ/mol} + 496 kJ/mol EA1 _1 1 − 349 k J / m o l -349\mathrm{ kJ/mol} − 349 kJ/mol Δ H l a t t = Δ H f ∘ − Δ H a t ( N a ) − 1 2 Δ H a t ( C l 2 ) − I E 1 − E A 1 \Delta H_{\mathrm{latt}} = \Delta H_f^\circ - \Delta H_{\mathrm{at}}(\mathrm{Na}) - \frac{1}{2}\Delta H_{\mathrm{at}}(\mathrm{Cl}_2) - \mathrm{IE}_1 - \mathrm{EA}_1 Δ H latt = Δ H f ∘ − Δ H at ( Na ) − 2 1 Δ H at ( Cl 2 ) − IE 1 − EA 1 = − 411 − 108 − 122 − 496 − ( − 349 ) = − 411 − 108 − 122 − 496 + 349 = − 788 k J / m o l = -411 - 108 - 122 - 496 - (-349) = -411 - 108 - 122 - 496 + 349 = -788\mathrm{ kJ/mol} = − 411 − 108 − 122 − 496 − ( − 349 ) = − 411 − 108 − 122 − 496 + 349 = − 788 kJ/mol Factors Affecting Lattice Energy Factor Effect Larger ionic charges Higher lattice energy Smaller ionic radii Higher lattice energy Higher charge density Higher lattice energy
Entropy Definition Entropy (S S S
Factors Affecting Entropy Factor Effect on Entropy More particles Higher entropy Higher temperature Higher entropy Gas → \to → → \to → Decreasing entropy Dissolving solid in solvent Increasing entropy
Standard Entropy Change Δ S ∘ = ∑ S ∘ ( p r o d u c t s ) − ∑ S ∘ ( r e a c t a n t s ) \Delta S^\circ = \sum S^\circ(\mathrm{products}) - \sum S^\circ(\mathrm{reactants}) Δ S ∘ = ∑ S ∘ ( products ) − ∑ S ∘ ( reactants ) Example
Calculate Δ S ∘ \Delta S^\circ Δ S ∘ 3 _3 3 → \to → 2 _2 2
Given: S ∘ S^\circ S ∘ 3 _3 3 92.9 J / ( m o l ⋅ K ) 92.9\mathrm{ J/(mol}\cdot\mathrm{K)} 92.9 J/ ( mol ⋅ K ) S ∘ S^\circ S ∘ 39.7 J / ( m o l ⋅ K ) 39.7\mathrm{ J/(mol}\cdot\mathrm{K)} 39.7 J/ ( mol ⋅ K ) S ∘ S^\circ S ∘ 2 _2 2 213.7 J / ( m o l ⋅ K ) 213.7\mathrm{ J/(mol}\cdot\mathrm{K)} 213.7 J/ ( mol ⋅ K )
Δ S ∘ = ( 39.7 + 213.7 ) − 92.9 = 253.4 − 92.9 = 160.5 J / ( m o l ⋅ K ) \Delta S^\circ = (39.7 + 213.7) - 92.9 = 253.4 - 92.9 = 160.5\mathrm{ J/(mol}\cdot\mathrm{K)} Δ S ∘ = ( 39.7 + 213.7 ) − 92.9 = 253.4 − 92.9 = 160.5 J/ ( mol ⋅ K ) The positive Δ S \Delta S Δ S
Gibbs Free Energy Definition Gibbs free energy combines enthalpy and entropy to predict spontaneity:
Δ G = Δ H − T Δ S \Delta G = \Delta H - T\Delta S Δ G = Δ H − T Δ S Where T T T
Spontaneity Δ G \Delta G Δ G Spontaneity Δ G < 0 \Delta G \lt 0 Δ G < 0 Spontaneous (thermodynamically favourable) Δ G = 0 \Delta G = 0 Δ G = 0 At equilibrium Δ G > 0 \Delta G \gt 0 Δ G > 0 Non-spontaneous
Standard Gibbs Free Energy Change Δ G ∘ = ∑ Δ G f ∘ ( p r o d u c t s ) − ∑ Δ G f ∘ ( r e a c t a n t s ) \Delta G^\circ = \sum \Delta G_f^\circ(\mathrm{products}) - \sum \Delta G_f^\circ(\mathrm{reactants}) Δ G ∘ = ∑ Δ G f ∘ ( products ) − ∑ Δ G f ∘ ( reactants ) Relationship to Equilibrium Constant Δ G ∘ = − R T ln K \Delta G^\circ = -RT\ln K Δ G ∘ = − R T ln K Where R = 8.314 J / ( m o l ⋅ K ) R = 8.314\mathrm{ J/(mol}\cdot\mathrm{K)} R = 8.314 J/ ( mol ⋅ K ) K K K
Temperature Dependence A reaction that is non-spontaneous at low temperature may become spontaneous at high temperature if Δ S > 0 \Delta S \gt 0 Δ S > 0
Δ H \Delta H Δ H Δ S \Delta S Δ S Spontaneous when Negative Positive Always spontaneous Negative Negative Low temperature Positive Positive High temperature Positive Negative Never spontaneous
Example
For the reaction: CaCO3 _3 3 → \to → 2 _2 2
Δ H = + 178 k J / m o l \Delta H = +178\mathrm{ kJ/mol} Δ H = + 178 kJ/mol Δ S = + 160.5 J / ( m o l ⋅ K ) \Delta S = +160.5\mathrm{ J/(mol}\cdot\mathrm{K)} Δ S = + 160.5 J/ ( mol ⋅ K )
Find the temperature at which the reaction becomes spontaneous.
Δ G = 0 w h e n T = Δ H Δ S = 178000 160.5 = 1109 K \Delta G = 0 \mathrm{ when } T = \frac{\Delta H}{\Delta S} = \frac{178000}{160.5} = 1109\mathrm{ K} Δ G = 0 when T = Δ S Δ H = 160.5 178000 = 1109 K The reaction is spontaneous above 1109 K 1109\mathrm{ K} 1109 K
IB Exam-Style Questions Question 1 (Paper 1 style) Using bond enthalpies, calculate Δ H \Delta H Δ H 2 _2 2 2 _2 2 → \to →
Bonds broken: H—H (436 436 436 242 242 242 = 678 k J / m o l = 678\mathrm{ kJ/mol} = 678 kJ/mol
Bonds formed: 2 × 2 \times 2 × 431 431 431 = 862 k J / m o l = 862\mathrm{ kJ/mol} = 862 kJ/mol
Δ H = 678 − 862 = − 184 k J / m o l \Delta H = 678 - 862 = -184\mathrm{ kJ/mol} Δ H = 678 − 862 = − 184 kJ/mol Question 2 (Paper 2 style) 25.0 c m 3 25.0\mathrm{ cm}^3 25.0 cm 3 1.0 M 1.0\mathrm{ M} 1.0 M 25.0 c m 3 25.0\mathrm{ cm}^3 25.0 cm 3 1.0 M 1.0\mathrm{ M} 1.0 M 20.0 ° C 20.0\degree\mathrm{C} 20.0° C 26.5 ° C 26.5\degree\mathrm{C} 26.5° C
(a) Calculate the enthalpy change of neutralisation per mole of water formed.
Q = 50.0 × 4.18 × 6.5 = 1359 J Q = 50.0 \times 4.18 \times 6.5 = 1359\mathrm{ J} Q = 50.0 × 4.18 × 6.5 = 1359 J N = 0.025 × 1.0 = 0.025 m o l N = 0.025 \times 1.0 = 0.025\mathrm{ mol} N = 0.025 × 1.0 = 0.025 mol Δ H = − 1.359 0.025 = − 54.4 k J / m o l \Delta H = -\frac{1.359}{0.025} = -54.4\mathrm{ kJ/mol} Δ H = − 0.025 1.359 = − 54.4 kJ/mol (b) Explain why the experimental value differs from the theoretical value of − 57.1 k J / m o l -57.1\mathrm{ kJ/mol} − 57.1 kJ/mol
Heat loss to the surroundings, calorimeter absorbs some heat, incomplete reaction, or the assumption That the solution has the same properties as pure water.
Question 3 (Paper 2 style) Given the following data, calculate the lattice energy of MgO:
Δ H f ∘ \Delta H_f^\circ Δ H f ∘ − 602 k J / m o l -602\mathrm{ kJ/mol} − 602 kJ/mol Δ H a t \Delta H_{\mathrm{at}} Δ H at + 148 k J / m o l +148\mathrm{ kJ/mol} + 148 kJ/mol 1 2 Δ H a t \frac{1}{2}\Delta H_{\mathrm{at}} 2 1 Δ H at 2 _2 2 + 249 k J / m o l +249\mathrm{ kJ/mol} + 249 kJ/mol IE1 _1 1 2 _2 2 + 2188 k J / m o l +2188\mathrm{ kJ/mol} + 2188 kJ/mol EA1 _1 1 2 _2 2 + 603 k J / m o l +603\mathrm{ kJ/mol} + 603 kJ/mol Δ H l a t t = − 602 − 148 − 249 − 2188 − 603 = − 3790 k J / m o l \Delta H_{\mathrm{latt}} = -602 - 148 - 249 - 2188 - 603 = -3790\mathrm{ kJ/mol} Δ H latt = − 602 − 148 − 249 − 2188 − 603 = − 3790 kJ/mol Question 4 (Paper 1 style) For which reaction is Δ S \Delta S Δ S
A. 2H2 _2 2 2 _2 2 → \to → 2 _2 2 4 _4 4 → \to → 3 _3 3 2 _2 2 → \to → 2 _2 2 2 _2 2 2 _2 2 → \to → 3 _3 3
Answer: B — a solid produces two gases, increasing disorder.
Summary Formula Expression Heat energy q = m c Δ T q = mc\Delta T q = m c Δ T Enthalpy from formation Δ H r = ∑ Δ H f ( p r o d u c t s ) − ∑ Δ H f ( r e a c t a n t s ) \Delta H_r = \sum \Delta H_f(\mathrm{products}) - \sum \Delta H_f(\mathrm{reactants}) Δ H r = ∑ Δ H f ( products ) − ∑ Δ H f ( reactants ) From bond enthalpies Δ H = ∑ ( b o n d s b r o k e n ) − ∑ ( b o n d s f o r m e d ) \Delta H = \sum(\mathrm{bonds broken}) - \sum(\mathrm{bonds formed}) Δ H = ∑ ( bondsbroken ) − ∑ ( bondsformed ) Entropy change Δ S = ∑ S ( p r o d u c t s ) − ∑ S ( r e a c t a n t s ) \Delta S = \sum S(\mathrm{products}) - \sum S(\mathrm{reactants}) Δ S = ∑ S ( products ) − ∑ S ( reactants ) Gibbs free energy Δ G = Δ H − T Δ S \Delta G = \Delta H - T\Delta S Δ G = Δ H − T Δ S Equilibrium relation Δ G ∘ = − R T ln K \Delta G^\circ = -RT\ln K Δ G ∘ = − R T ln K
For Hess’s law questions, draw the energy cycle . For calorimetry, always account for the Total mass of the solution. For Gibbs free energy, pay attention to units — Δ H \Delta H Δ H Δ S \Delta S Δ S ⋅ \cdot ⋅
Thermochemistry Extended Enthalpy of Solution The enthalpy change when one mole of solute dissolves in a solvent to form an infinitely dilute Solution:
Δ H s o l = Δ H l a t t i c e + Δ H h y d r a t i o n \Delta H_{\mathrm{sol}} = \Delta H_{\mathrm{lattice}} + \Delta H_{\mathrm{hydration}} Δ H sol = Δ H lattice + Δ H hydration If Δ H s o l > 0 \Delta H_{\mathrm{sol}} \gt 0 Δ H sol > 0 4 _4 4 3 _3 3 If Δ H s o l < 0 \Delta H_{\mathrm{sol}} \lt 0 Δ H sol < 0 Enthalpy of Hydration The enthalpy change when gaseous ions are surrounded by water molecules:
Δ H h y d = Δ H a t + Δ H E A + o t h e r t e r m s \Delta H_{\mathrm{hyd}} = \Delta H_{\mathrm{at}} + \Delta H_{\mathrm{EA}} + \mathrm{other terms} Δ H hyd = Δ H at + Δ H EA + otherterms Trends in Lattice Energy Trend Effect on Lattice Energy Higher ionic charge Increases Smaller ionic radius Increases Higher charge density Increases
Comparison Higher Lattice Energy NaCl vs NaBr NaCl (Br− ^- − NaCl vs MgCl2 _2 2 MgCl2 _2 2 2 + ^{2+} 2 + NaCl vs Na2 _2 2 Na2 _2 2 2 − ^{2-} 2 −
Using Born-Haber Cycles to Compare Compounds Example
Explain why the lattice energy of MgO (− 3791 k J / m o l -3791\mathrm{ kJ/mol} − 3791 kJ/mol − 788 k J / m o l -788\mathrm{ kJ/mol} − 788 kJ/mol
MgO has Mg2 + ^{2+} 2 + 2 − ^{2-} 2 − + ^+ + − ^- − 2 + ^{2+} 2 + + ^+ +
Entropy: Extended Analysis Predicting Entropy Changes Process Δ S \Delta S Δ S Reason Solid → \to → Positive More disorder Liquid → \to → Positive Much more disorder Dissolving ionic solid in water positive Ions become dispersed Gas → \to → Negative Much less disorder Decreasing volume of gas Negative Fewer microstates Increasing temperature Positive More molecular motion
Calculating Δ S \Delta S Δ S Example
Calculate Δ S ∘ \Delta S^\circ Δ S ∘ 2 _2 2 2 _2 2 → \to → 2 _2 2
S ∘ S^\circ S ∘ 2 _2 2 = 131 J / ( m o l ⋅ K ) = 131\mathrm{ J/(mol}\cdot\mathrm{K)} = 131 J/ ( mol ⋅ K ) S ∘ S^\circ S ∘ 2 _2 2 = 205 J / ( m o l ⋅ K ) = 205\mathrm{ J/(mol}\cdot\mathrm{K)} = 205 J/ ( mol ⋅ K ) S ∘ S^\circ S ∘ 2 _2 2 = 70 J / ( m o l ⋅ K ) = 70\mathrm{ J/(mol}\cdot\mathrm{K)} = 70 J/ ( mol ⋅ K )
Δ S ∘ = 2 ( 70 ) − [ 2 ( 131 ) + 205 ] = 140 − 467 = − 327 J / ( m o l ⋅ K ) \Delta S^\circ = 2(70) - [2(131) + 205] = 140 - 467 = -327\mathrm{ J/(mol}\cdot\mathrm{K)} Δ S ∘ = 2 ( 70 ) − [ 2 ( 131 ) + 205 ] = 140 − 467 = − 327 J/ ( mol ⋅ K ) The large negative Δ S \Delta S Δ S
Gibbs Free Energy: Extended Calculating Δ G \Delta G Δ G Δ G f ∘ \Delta G_f^\circ Δ G f ∘ Δ G r ∘ = ∑ Δ G f ∘ ( p r o d u c t s ) − ∑ Δ G f ∘ ( r e a c t a n t s ) \Delta G_r^\circ = \sum \Delta G_f^\circ(\mathrm{products}) - \sum \Delta G_f^\circ(\mathrm{reactants}) Δ G r ∘ = ∑ Δ G f ∘ ( products ) − ∑ Δ G f ∘ ( reactants ) Example
Calculate Δ G ∘ \Delta G^\circ Δ G ∘ 2 _2 2 → \to → 298 K 298\mathrm{ K} 298 K
Given: Δ G f ∘ \Delta G_f^\circ Δ G f ∘ 2 _2 2 = − 394 k J / m o l = -394\mathrm{ kJ/mol} = − 394 kJ/mol Δ G f ∘ \Delta G_f^\circ Δ G f ∘ = − 137 k J / m o l = -137\mathrm{ kJ/mol} = − 137 kJ/mol
Δ G ∘ = 2 ( − 137 ) − ( − 394 ) = − 274 + 394 = + 120 k J / m o l \Delta G^\circ = 2(-137) - (-394) = -274 + 394 = +120\mathrm{ kJ/mol} Δ G ∘ = 2 ( − 137 ) − ( − 394 ) = − 274 + 394 = + 120 kJ/mol The reaction is not spontaneous at 298 K 298\mathrm{ K} 298 K Δ S > 0 \Delta S \gt 0 Δ S > 0
Using Δ G \Delta G Δ G K = e − Δ G ∘ / R T K = e^{-\Delta G^\circ/RT} K = e − Δ G ∘ / R T Example
For a reaction with Δ G ∘ = − 5.4 k J / m o l \Delta G^\circ = -5.4\mathrm{ kJ/mol} Δ G ∘ = − 5.4 kJ/mol 298 K 298\mathrm{ K} 298 K
K = e − ( − 5400 ) / ( 8.314 × 298 ) = e 2.18 = 8.85 K = e^{-(-5400)/(8.314 \times 298)} = e^{2.18} = 8.85 K = e − ( − 5400 ) / ( 8.314 × 298 ) = e 2.18 = 8.85 Since K > 1 K \gt 1 K > 1
Additional IB Exam-Style Questions Question 5 (Paper 2 style) The enthalpy of combustion of methanol is − 726 k J / m o l -726\mathrm{ kJ/mol} − 726 kJ/mol 200 g 200\mathrm{ g} 200 g 20.0 ° C 20.0\degree\mathrm{C} 20.0° C 65.0 ° C 65.0\degree\mathrm{C} 65.0° C 1.50 g 1.50\mathrm{ g} 1.50 g
(a) Calculate the experimental enthalpy of combustion.
Q = 200 × 4.18 × 45 = 37620 J = 37.62 k J Q = 200 \times 4.18 \times 45 = 37620\mathrm{ J} = 37.62\mathrm{ kJ} Q = 200 × 4.18 × 45 = 37620 J = 37.62 kJ N ( C H 3 O H ) = 1.50 32.04 = 0.0468 m o l N(\mathrm{CH}_3\mathrm{OH}) = \frac{1.50}{32.04} = 0.0468\mathrm{ mol} N ( CH 3 OH ) = 32.04 1.50 = 0.0468 mol Δ H c = − 37.62 0.0468 = − 804 k J / m o l \Delta H_c = -\frac{37.62}{0.0468} = -804\mathrm{ kJ/mol} Δ H c = − 0.0468 37.62 = − 804 kJ/mol (b) Calculate the percentage error compared to the literature value.
P e r c e n t a g e e r r o r = ∣ − 804 − ( − 726 ) ∣ 726 × 100 % = 78 726 × 100 % = 10.7 % \mathrm{Percentage error} = \frac{|-804 - (-726)|}{726} \times 100\% = \frac{78}{726} \times 100\% = 10.7\% Percentageerror = 726 ∣ − 804 − ( − 726 ) ∣ × 100% = 726 78 × 100% = 10.7% (c) Explain two sources of error.
Heat loss to the surroundings; incomplete combustion of methanol; not all heat transferred to the Water; the calorimeter absorbs some heat.
Question 6 (Paper 1 style) For the reaction: N2 _2 2 4 _4 4 ⇌ \rightleftharpoons ⇌ 2 _2 2 Δ H = + 57 k J / m o l \Delta H = +57\mathrm{ kJ/mol} Δ H = + 57 kJ/mol Δ S = + 176 J / ( m o l ⋅ K ) \Delta S = +176\mathrm{ J/(mol}\cdot\mathrm{K)} Δ S = + 176 J/ ( mol ⋅ K )
At what temperature does the reaction become spontaneous?
Δ G = 0 w h e n T = Δ H Δ S = 57000 176 = 324 K \Delta G = 0 \mathrm{ when } T = \frac{\Delta H}{\Delta S} = \frac{57000}{176} = 324\mathrm{ K} Δ G = 0 when T = Δ S Δ H = 176 57000 = 324 K The reaction is spontaneous above 324 K 324\mathrm{ K} 324 K 51 ° C 51\degree\mathrm{C} 51° C
Question 7 (Paper 2 style) Using the data below, calculate the lattice energy of CaF2 _2 2
Δ H f ∘ \Delta H_f^\circ Δ H f ∘ 2 _2 2 = − 1220 k J / m o l = -1220\mathrm{ kJ/mol} = − 1220 kJ/mol Δ H a t \Delta H_{\mathrm{at}} Δ H at = + 178 k J / m o l = +178\mathrm{ kJ/mol} = + 178 kJ/mol 1 2 Δ H a t \frac{1}{2}\Delta H_{\mathrm{at}} 2 1 Δ H at 2 _2 2 = + 79 k J / m o l = +79\mathrm{ kJ/mol} = + 79 kJ/mol IE1 _1 1 2 _2 2 = + 1735 k J / m o l = +1735\mathrm{ kJ/mol} = + 1735 kJ/mol EA1 _1 1 = − 328 k J / m o l = -328\mathrm{ kJ/mol} = − 328 kJ/mol Δ H f ∘ = Δ H a t ( C a ) + 2 [ 1 2 Δ H a t ( F 2 ) + E A ( F ) ] + I E 1 + I E 2 + Δ H l a t t \Delta H_f^\circ = \Delta H_{\mathrm{at}}(\mathrm{Ca}) + 2\left[\frac{1}{2}\Delta H_{\mathrm{at}}(\mathrm{F}_2) + \mathrm{EA}(\mathrm{F})\right] + \mathrm{IE}_1 + \mathrm{IE}_2 + \Delta H_{\mathrm{latt}} Δ H f ∘ = Δ H at ( Ca ) + 2 [ 2 1 Δ H at ( F 2 ) + EA ( F ) ] + IE 1 + IE 2 + Δ H latt − 1220 = 178 + 2 ( 79 − 328 ) + 1735 + Δ H l a t t -1220 = 178 + 2(79 - 328) + 1735 + \Delta H_{\mathrm{latt}} − 1220 = 178 + 2 ( 79 − 328 ) + 1735 + Δ H latt − 1220 = 178 + 2 ( − 249 ) + 1735 + Δ H l a t t -1220 = 178 + 2(-249) + 1735 + \Delta H_{\mathrm{latt}} − 1220 = 178 + 2 ( − 249 ) + 1735 + Δ H latt − 1220 = 178 − 498 + 1735 + Δ H l a t t = 1415 + Δ H l a t t -1220 = 178 - 498 + 1735 + \Delta H_{\mathrm{latt}} = 1415 + \Delta H_{\mathrm{latt}} − 1220 = 178 − 498 + 1735 + Δ H latt = 1415 + Δ H latt Δ H l a t t = − 1220 − 1415 = − 2635 k J / m o l \Delta H_{\mathrm{latt}} = -1220 - 1415 = -2635\mathrm{ kJ/mol} Δ H latt = − 1220 − 1415 = − 2635 kJ/mol Thermochemistry: Extended Topics Enthalpy of Atomisation of Elements The enthalpy of atomisation is the enthalpy change to form one mole of gaseous atoms from the Element in its standard state under standard conditions.
Element Standard State Δ H a t \Delta H_{\mathrm{at}} Δ H at Na Solid + 108 +108 + 108 Mg Solid + 148 +148 + 148 Al Solid + 330 +330 + 330 Cl2 _2 2 Gas + 122 +122 + 122 H2 _2 2 Gas + 218 +218 + 218 C (graphite) Solid + 717 +717 + 717 O2 _2 2 Gas + 249 +249 + 249
Electron Affinity Trends Electron affinity is the enthalpy change when one mole of gaseous atoms gains one electron.
Trend Effect Across a period (left to right) Generally becomes more negative Down a group Generally becomes less negative Noble gases Positive (unfavourable) Group 17 Most negative (most favourable)
Ionisation Energy Trends Trend Effect Across a period Increases (nuclear charge increases, same shielding) Down a group Decreases (atomic radius increases) Large jumps Occur when removing electron from new shell
Successive Ionisation Energies Each successive ionisation energy is larger than the previous one. Large jumps indicate removal from A new shell.
Example
The first four ionisation energies of aluminium are (in kJ/mol): 578, 1817, 2745, 11578.
The large jump between the 3rd and 4th IE indicates that the 4th electron is being removed from a New (inner) shell. This confirms Al has 3 valence electrons.
Calculating Enthalpy Changes from Calorimetry Data When using a calorimeter, account for the heat absorbed by the calorimeter itself:
Q r e a c t i o n = − ( m s o l u t i o n c s o l u t i o n Δ T + C c a l o r i m e t e r Δ T ) Q_{\mathrm{reaction}} = -(m_{\mathrm{solution}} c_{\mathrm{solution}} \Delta T + C_{\mathrm{calorimeter}} \Delta T) Q reaction = − ( m solution c solution Δ T + C calorimeter Δ T ) Example
50 c m 3 50\mathrm{ cm}^3 50 cm 3 1.0 M 1.0\mathrm{ M} 1.0 M 50 c m 3 50\mathrm{ cm}^3 50 cm 3 1.0 M 1.0\mathrm{ M} 1.0 M 15 J / K 15\mathrm{ J/K} 15 J/K 20.0 ° C 20.0\degree\mathrm{C} 20.0° C 26.8 ° C 26.8\degree\mathrm{C} 26.8° C
Q t o t a l = ( 100 ) ( 4.18 ) ( 6.8 ) + 15 ( 6.8 ) = 2842.4 + 102 = 2944.4 J Q_{\mathrm{total}} = (100)(4.18)(6.8) + 15(6.8) = 2842.4 + 102 = 2944.4\mathrm{ J} Q total = ( 100 ) ( 4.18 ) ( 6.8 ) + 15 ( 6.8 ) = 2842.4 + 102 = 2944.4 J N ( H 2 O ) = 0.050 m o l N(\mathrm{H}_2\mathrm{O}) = 0.050\mathrm{ mol} N ( H 2 O ) = 0.050 mol Δ H = − 2944.4 0.050 = − 58888 J / m o l = − 58.9 k J / m o l \Delta H = -\frac{2944.4}{0.050} = -58888\mathrm{ J/mol} = -58.9\mathrm{ kJ/mol} Δ H = − 0.050 2944.4 = − 58888 J/mol = − 58.9 kJ/mol Additional IB Exam-Style Questions Question 8 (Paper 1 style) Which process has a positive entropy change?
A. C a 2 + ( a q ) + C O 3 2 − ( a q ) → C a C O 3 ( s ) \mathrm{Ca}^{2+}(aq) + \mathrm{CO}_3^{2-}(aq) \to \mathrm{CaCO}_3(s) Ca 2 + ( a q ) + CO 3 2 − ( a q ) → CaCO 3 ( s ) N H 4 C l ( s ) → N H 3 ( g ) + H C l ( g ) \mathrm{NH}_4\mathrm{Cl}(s) \to \mathrm{NH}_3(g) + \mathrm{HCl}(g) NH 4 Cl ( s ) → NH 3 ( g ) + HCl ( g ) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( l ) 2\mathrm{H}_2(g) + \mathrm{O}_2(g) \to 2\mathrm{H}_2\mathrm{O}(l) 2 H 2 ( g ) + O 2 ( g ) → 2 H 2 O ( l ) N a O H ( a q ) + H C l ( a q ) → N a C l ( a q ) + H 2 O ( l ) \mathrm{NaOH}(aq) + \mathrm{HCl}(aq) \to \mathrm{NaCl}(aq) + \mathrm{H}_2\mathrm{O}(l) NaOH ( a q ) + HCl ( a q ) → NaCl ( a q ) + H 2 O ( l )
Answer: B. A solid produces two gases, increasing the number of particles and the disorder.
Question 9 (Paper 2 style) Using the following data, calculate the enthalpy of reaction for:
C H 4 ( g ) + 2 O 2 ( g ) → C O 2 ( g ) + 2 H 2 O ( l ) \mathrm{CH}_4(g) + 2\mathrm{O}_2(g) \to \mathrm{CO}_2(g) + 2\mathrm{H}_2\mathrm{O}(l) CH 4 ( g ) + 2 O 2 ( g ) → CO 2 ( g ) + 2 H 2 O ( l )
Given bond enthalpies (kJ/mol): C—H = 413 = 413 = 413 = 495 = 495 = 495 = 743 = 743 = 743 = 463 = 463 = 463
Bonds broken : 4 ( C − − H ) + 2 ( O = O ) = 4 ( 413 ) + 2 ( 495 ) = 1652 + 990 = 2642 k J / m o l 4(\mathrm{C--H}) + 2(\mathrm{O=O}) = 4(413) + 2(495) = 1652 + 990 = 2642\mathrm{ kJ/mol} 4 ( C − − H ) + 2 ( O = O ) = 4 ( 413 ) + 2 ( 495 ) = 1652 + 990 = 2642 kJ/mol
Bonds formed : 2 ( C = O ) + 4 ( O − − H ) = 2 ( 743 ) + 4 ( 463 ) = 1486 + 1852 = 3338 k J / m o l 2(\mathrm{C=O}) + 4(\mathrm{O--H}) = 2(743) + 4(463) = 1486 + 1852 = 3338\mathrm{ kJ/mol} 2 ( C = O ) + 4 ( O − − H ) = 2 ( 743 ) + 4 ( 463 ) = 1486 + 1852 = 3338 kJ/mol
Δ H = 2642 − 3338 = − 696 k J / m o l \Delta H = 2642 - 3338 = -696\mathrm{ kJ/mol} Δ H = 2642 − 3338 = − 696 kJ/mol Question 10 (Paper 2 style) For the reaction: N 2 O ( g ) → N 2 ( g ) + O 2 ( g ) \mathrm{N}_2\mathrm{O}(g) \to \mathrm{N}_2(g) + \mathrm{O}_2(g) N 2 O ( g ) → N 2 ( g ) + O 2 ( g ) Δ H = − 163 k J / m o l \Delta H = -163\mathrm{ kJ/mol} Δ H = − 163 kJ/mol Δ S = + 149 J / ( m o l ⋅ K ) \Delta S = +149\mathrm{ J/(mol}\cdot\mathrm{K)} Δ S = + 149 J/ ( mol ⋅ K )
(a) Calculate Δ G ∘ \Delta G^\circ Δ G ∘ 298 K 298\mathrm{ K} 298 K
Δ G ∘ = − 163000 − 298 × 149 = − 163000 − 44402 = − 207402 J / m o l = − 207.4 k J / m o l \Delta G^\circ = -163000 - 298 \times 149 = -163000 - 44402 = -207402\mathrm{ J/mol} = -207.4\mathrm{ kJ/mol} Δ G ∘ = − 163000 − 298 × 149 = − 163000 − 44402 = − 207402 J/mol = − 207.4 kJ/mol Since Δ G ∘ < 0 \Delta G^\circ \lt 0 Δ G ∘ < 0 298 K 298\mathrm{ K} 298 K
(b) At what temperature does Δ G ∘ \Delta G^\circ Δ G ∘
Δ G ∘ = 0 w h e n T = Δ H Δ S = − 163000 149 = − 1094 K \Delta G^\circ = 0 \mathrm{ when } T = \frac{\Delta H}{\Delta S} = \frac{-163000}{149} = -1094\mathrm{ K} Δ G ∘ = 0 when T = Δ S Δ H = 149 − 163000 = − 1094 K Since both Δ H \Delta H Δ H Δ S \Delta S Δ S 1094 K 1094\mathrm{ K} 1094 K Δ G ∘ \Delta G^\circ Δ G ∘
Practice Problems Question 1: Calorimetry Calculation 50.0 c m 3 50.0\mathrm{ cm}^3 50.0 cm 3 1.0 M 1.0\mathrm{ M} 1.0 M H C l \mathrm{HCl} HCl 50.0 c m 3 50.0\mathrm{ cm}^3 50.0 cm 3 1.0 M 1.0\mathrm{ M} 1.0 M N a O H \mathrm{NaOH} NaOH 22.0 ° C 22.0\degree\mathrm{C} 22.0° C 28.8 ° C 28.8\degree\mathrm{C} 28.8° C
Answer q = m c Δ T = 100.0 × 4.18 × 6.8 = 2842 J = 2.842 k J q = mc\Delta T = 100.0 \times 4.18 \times 6.8 = 2842\mathrm{ J} = 2.842\mathrm{ kJ} q = m c Δ T = 100.0 × 4.18 × 6.8 = 2842 J = 2.842 kJ
n ( H 2 O ) = 0.0500 × 1.0 = 0.0500 m o l n(\mathrm{H}_2\mathrm{O}) = 0.0500 \times 1.0 = 0.0500\mathrm{ mol} n ( H 2 O ) = 0.0500 × 1.0 = 0.0500 mol
Δ H = − 2.842 0.0500 = − 56.8 k J / m o l \Delta H = -\frac{2.842}{0.0500} = -56.8\mathrm{ kJ/mol} Δ H = − 0.0500 2.842 = − 56.8 kJ/mol
Question 2: Hess's Law with Formation Enthalpies Using standard enthalpies of formation, calculate Δ H r ∘ \Delta H_r^\circ Δ H r ∘
C 3 H 8 ( g ) + 5 O 2 ( g ) → 3 C O 2 ( g ) + 4 H 2 O ( l ) \mathrm{C}_3\mathrm{H}_8(g) + 5\mathrm{O}_2(g) \to 3\mathrm{CO}_2(g) + 4\mathrm{H}_2\mathrm{O}(l) C 3 H 8 ( g ) + 5 O 2 ( g ) → 3 CO 2 ( g ) + 4 H 2 O ( l )
Given: Δ H f ∘ ( C 3 H 8 ) = − 104 k J / m o l \Delta H_f^\circ(\mathrm{C}_3\mathrm{H}_8) = -104\mathrm{ kJ/mol} Δ H f ∘ ( C 3 H 8 ) = − 104 kJ/mol Δ H f ∘ ( C O 2 ) = − 394 k J / m o l \Delta H_f^\circ(\mathrm{CO}_2) = -394\mathrm{ kJ/mol} Δ H f ∘ ( CO 2 ) = − 394 kJ/mol Δ H f ∘ ( H 2 O ) = − 286 k J / m o l \Delta H_f^\circ(\mathrm{H}_2\mathrm{O}) = -286\mathrm{ kJ/mol} Δ H f ∘ ( H 2 O ) = − 286 kJ/mol
Answer Δ H r ∘ = ∑ Δ H f ∘ ( p r o d u c t s ) − ∑ Δ H f ∘ ( r e a c t a n t s ) \Delta H_r^\circ = \sum \Delta H_f^\circ(\mathrm{products}) - \sum \Delta H_f^\circ(\mathrm{reactants}) Δ H r ∘ = ∑ Δ H f ∘ ( products ) − ∑ Δ H f ∘ ( reactants )
= [ 3 ( − 394 ) + 4 ( − 286 ) ] − [ ( − 104 ) + 5 ( 0 ) ] = [3(-394) + 4(-286)] - [(-104) + 5(0)] = [ 3 ( − 394 ) + 4 ( − 286 )] − [( − 104 ) + 5 ( 0 )]
= ( − 1182 − 1144 ) − ( − 104 ) = − 2326 + 104 = − 2222 k J / m o l = (-1182 - 1144) - (-104) = -2326 + 104 = -2222\mathrm{ kJ/mol} = ( − 1182 − 1144 ) − ( − 104 ) = − 2326 + 104 = − 2222 kJ/mol
Question 3: Bond Enthalpy Calculation Using average bond enthalpies, calculate Δ H \Delta H Δ H
N 2 ( g ) + 3 H 2 ( g ) → 2 N H 3 ( g ) \mathrm{N}_2(g) + 3\mathrm{H}_2(g) \to 2\mathrm{NH}_3(g) N 2 ( g ) + 3 H 2 ( g ) → 2 NH 3 ( g )
Given: N ≡ N = 945 k J / m o l \mathrm{N} \equiv \mathrm{N} = 945\mathrm{ kJ/mol} N ≡ N = 945 kJ/mol H − H = 436 k J / m o l \mathrm{H}-\mathrm{H} = 436\mathrm{ kJ/mol} H − H = 436 kJ/mol N − H = 391 k J / m o l \mathrm{N}-\mathrm{H} = 391\mathrm{ kJ/mol} N − H = 391 kJ/mol
Answer Bonds broken: 1 ( N ≡ N ) + 3 ( H − H ) = 945 + 3 ( 436 ) = 945 + 1308 = 2253 k J / m o l 1(\mathrm{N} \equiv \mathrm{N}) + 3(\mathrm{H}-\mathrm{H}) = 945 + 3(436) = 945 + 1308 = 2253\mathrm{ kJ/mol} 1 ( N ≡ N ) + 3 ( H − H ) = 945 + 3 ( 436 ) = 945 + 1308 = 2253 kJ/mol
Bonds formed: 6 ( N − H ) = 6 × 391 = 2346 k J / m o l 6(\mathrm{N}-\mathrm{H}) = 6 \times 391 = 2346\mathrm{ kJ/mol} 6 ( N − H ) = 6 × 391 = 2346 kJ/mol
Δ H = 2253 − 2346 = − 93 k J / m o l \Delta H = 2253 - 2346 = -93\mathrm{ kJ/mol} Δ H = 2253 − 2346 = − 93 kJ/mol
The actual value is − 92 k J / m o l -92\mathrm{ kJ/mol} − 92 kJ/mol
Question 4: Gibbs Free Energy and Spontaneity For the decomposition of calcium carbonate:
C a C O 3 ( s ) → C a O ( s ) + C O 2 ( g ) \mathrm{CaCO}_3(s) \to \mathrm{CaO}(s) + \mathrm{CO}_2(g) CaCO 3 ( s ) → CaO ( s ) + CO 2 ( g )
Δ H = + 178 k J / m o l \Delta H = +178\mathrm{ kJ/mol} Δ H = + 178 kJ/mol Δ S = + 161 J / ( m o l ⋅ K ) \Delta S = +161\mathrm{ J/(mol \cdot K)} Δ S = + 161 J/ ( mol ⋅ K )
(a) Calculate Δ G \Delta G Δ G 298 K 298\mathrm{ K} 298 K
(b) Calculate the minimum temperature at which the reaction becomes spontaneous.
Answer (a) Δ G = Δ H − T Δ S = 178000 − 298 × 161 = 178000 − 47978 = + 130 022 J / m o l = + 130 k J / m o l \Delta G = \Delta H - T\Delta S = 178000 - 298 \times 161 = 178000 - 47978 = +130\,022\mathrm{ J/mol} = +130\mathrm{ kJ/mol} Δ G = Δ H − T Δ S = 178000 − 298 × 161 = 178000 − 47978 = + 130 022 J/mol = + 130 kJ/mol
Since Δ G > 0 \Delta G \gt 0 Δ G > 0 not spontaneous at 298 K 298\mathrm{ K} 298 K
(b) At Δ G = 0 \Delta G = 0 Δ G = 0
T = Δ H Δ S = 178000 161 = 1106 K T = \frac{\Delta H}{\Delta S} = \frac{178000}{161} = 1106\mathrm{ K} T = Δ S Δ H = 161 178000 = 1106 K
The reaction becomes spontaneous above 1106 K 1106\mathrm{ K} 1106 K 833 ° C 833\degree\mathrm{C} 833° C
Question 5: Entropy Change Prediction Predict the sign of Δ S \Delta S Δ S
(a) N H 4 C l ( s ) → N H 3 ( g ) + H C l ( g ) \mathrm{NH}_4\mathrm{Cl}(s) \to \mathrm{NH}_3(g) + \mathrm{HCl}(g) NH 4 Cl ( s ) → NH 3 ( g ) + HCl ( g )
(b) 2 N O ( g ) + O 2 ( g ) → 2 N O 2 ( g ) 2\mathrm{NO}(g) + \mathrm{O}_2(g) \to 2\mathrm{NO}_2(g) 2 NO ( g ) + O 2 ( g ) → 2 NO 2 ( g )
(c) N a C l ( s ) → N a + ( a q ) + C l − ( a q ) \mathrm{NaCl}(s) \to \mathrm{Na}^+(aq) + \mathrm{Cl}^-(aq) NaCl ( s ) → Na + ( a q ) + Cl − ( a q )
Answer (a) Positive Δ S \Delta S Δ S
(b) Negative Δ S \Delta S Δ S
(c) Positive Δ S \Delta S Δ S
For the A-Level treatment of this topic, see Thermodynamics & Energetics .
Common Pitfalls Assuming that a strong acid always has a lower pH than a weak acid without considering concentration.
Drawing structural formulae incorrectly — check the number of bonds each atom can form and the overall charge.
Writing half-equations without balancing charges or atoms — always check electrons, hydrogen ions, and water molecules.
Confusing enthalpy of formation with enthalpy of combustion, or using the wrong sign convention.
Cross-References Topic Site Link [Thermodynamics (Chemistry)] A-Level View [Thermodynamics (Chemistry)] IB View [Thermodynamics (Chemistry)] DSE View
Worked Examples Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.