Section 3 Energetics

Modern College AL Chemistry Notes (2009 – 10)

Section 3

Name: ______________________________ Class: _______________ Class No.: ____________ Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)

Prepared by Mr. Chau Chi Keung, Richard

Section 3

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Modern College AL Chemistry Notes (2009 – 10)

Section 3

3.1 What is Energetics? 1. 2. 3. 4. 5. 6. 7.

Energetics(能學): The study of the energy changes. Energy: The ability to do work. Energy can be changed from one form to the other forms. (e.g. light, heat, kinetic, potential etc) Heat is a form of kinetic energy: it is the kinetic energy associated with the motion of atoms and molecules. Thermochemistry: The study of heat changes in chemical reactions. In thermochemistry, the total energy stored in a substance is called ‘Heat content’ or ‘Enthalpy’ (Symbol: H). The enthalpy of a substance is impossible to measure, but the enthalpy change (ΔH)(焓變) of a reaction can be measured experimentally. In the study of energetics, the region or particular quantity of matter of interest is defined as the system (體系). The surroundings (外界) are defined as everything other than the system.

A. Internal energy and Enthalpy  The measurement of the heat change of a system may be carried out at constant volume, called the change in internal energy of the system.  If the system is under constant pressure, this is called enthalpy change of the system.  The enthalpy change (ΔH) of a reaction is the heat exchange with the surroundings at constant pressure before and after the reaction. Consider the following reaction: Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g) Enthalpy change at constant pressure = Change in internal energy + Work done on the surroundings

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Modern College AL Chemistry Notes (2009 – 10)

Section 3

For the enthalpy change of the above reaction: ΔH = Enthalpy of products – Enthalpy of reactants So ΔH of exothermic reaction is ________________ (Hp < Hr). ΔH of endothermic reaction is _______________ (Hp > Hr).  Law of conservation of energy(能量守恆定律): Energy cannot be destroyed or created but can be changed from one form into another. The total energy in the universe is constant. (能量不能被創造或者被消滅，但可以由某一形式轉換為另一形式。而宇宙的總能量 為一恆定值。) This law is also called the First Law of Thermodynamics (熱力學第一定律).

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B. Exothermic and endothermic reaction An exothermic reaction (放熱反應) is a reaction that releases heat to the surroundings. (Stored chemical energy is converted to heat energy) An endothermic reaction ( 吸 熱 反 應 ) is a reaction that absorbs heat from surroundings. (Heat energy is converted to chemical energy) All chemical reactions involve a rearrangement of atoms. Chemical bonds in the reactants are broken and new bonds are formed in the products.  The energy of chemical bonds is a form of potential energy, arising from the positions of atoms and molecules with respect to one another.  Therefore, the energy change during a chemical reaction involves interconversion between heat (kinetic) energy and potential energy. Bond breaking: Energy absorbing process. (Absorbs energy from surroundings) (拆散化學鍵 

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時需要吸收能量)  Bond forming: Energy releasing process. (Gives out energy to surroundings) 成化學鍵時會釋放能量) In exothermic reactions: Energy needed for bond breaking < Energy released in bond forming (in the reactants) ( in the products) ∴Heat is released to the surroundings, so the temperature ___________. In endothermic reactions: Prepared by Mr. Chau Chi Keung, Richard

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(形

Modern College AL Chemistry Notes (2009 – 10)

Section 3

Energy needed for bond breaking > Energy released in bond forming (in the reactants) ( in the products) ∴Heat is absorbed from the surroundings, so the temperature ___________.

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Chemical reactions occur spontaneously (自發進行) are usually exothermic. Exothermic processes Neutralization (中和作用) e.g. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) Combustion e.g. 2H2(g) + O2(g) → 2H2O(l) The Contact Process (接觸法) for sulphur trioxide production 2SO2(g) + O2(g) → 2SO3 (g) Dehydration by concentrated H2SO4 e.g. C6H12O6 → 6C + 6H2O Thermite reaction (鋁熱反應) e.g. 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)

Endothermic processes Ionization of a weak electrolyte (弱電解質的電離) e.g. CH3COOH(aq) → H+(aq) + CH3COO–(aq) Thermal decomposition (受熱分解) of metal/ammonium carbonate e.g. CaCO3(s) → CaO(s) + CO2(g) Melting of ice

Producing sugar by photosynthesis Cooking an egg

3.2 Standard Enthalpy Changes, ΔHθ Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)

Section 3

A. Definitions and criteria  The enthalpy of a substance depends on the physical states of the reactants and products. So we need to standardize the data under a standard state and conditions (標準態及條件): Standard Conditions (標準條件):  Temperature: 298K  For gases: pressure of exactly 1 atm  For solutions: concentration of exactly 1 moldm–3  For liquid or solid, it must be pure  Elements or compounds must be in physical states which they exist at 1 atm and 298K (元素或化合物必須處於其一般物理狀態) Standard enthalpy change of reaction (標 準 反 應 焓 變 )(ΔHθ) refers to enthalpy change specifically when the molar quantities of reactants shown in a balanced chemical equation completely react to form products under standard conditions.  Unit of ΔH/ΔHθ: usually kJmol–1  Study the following thermochemical equations (熱化學方程式): 2H2(g) + O2(g) → 2H2O(l), ΔHθ = –572 kJmol–1 The above equation tells us: 1. 2 moles of hydrogen reacts with 1 mole of oxygen to produces 2 moles of water. 2. The reaction is under standard condition (at 298 K and 1 atm), and it releases 572 kJ of heat energy.  If half amount of reactant used, the enthalpy change under standard condition will also be halved (當反應物和生成物份量減半，焓變值亦會相應減半。). Compare: 2H2(g) + O2(g) → 2H2O(l), ΔHθ = –572 kJmol–1 

H2(g) +

1 O2(g) → H2O(l), ΔHθ = –286 kJmol–1 2

Enthalpy change also depend on state of the reactants and products: 2H2(g) + O2(g) → 2H2O(g), ΔHθ = –484 kJmol–1

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It should be noted that:  For all thermochemical equations, all physical states of the species involved must be 

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clearly stated. (書寫熱化學方程式時，必須標明物態符號。) Many reactions do not occur under standard conditions (e.g. petrol will not burn unless being ignited by an electrical spark or a flame). Determination of standard enthalpy change of such reactions may require an indirect method. If the element has different allotropic forms (同素異形體)(e.g. carbon can exist as either graphite or diamond), the form involved must be clearly stated since changes in allotropic form may involve energy change e.g. C(graphite) → C(diamond) ΔH = +1.9 kJmol–1

Example 1 Complete the following table by selecting the specific terms which describe the standard Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10) enthalpy changes in the following reactions: Reaction C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) ΔHθ = –1366.7 kJmol-1 MgCl2(s) + 500H2O(l) → MgCl2(aq, 500H2O) ΔHθ = –152.9 kJmol-1 Mg(s) + C(s) +

Section 3

Type(s) of standard enthalpy change

3 O2(g) → MgCO3(s) 2

ΔHθ = –1069.7 kJmol-1 Mg(s) +

1 O2(g) → MgO(s) 2

ΔHθ = –601.7 kJmol-1 2HNO3(aq) + Ba(OH)2(aq) → Ba(NO3)2(aq) + 2H2O(l) θ ΔH = –116 kJmol-1

Various types of Standard Enthalpy changes  In order to find the value of standard enthalpy changes of reactions, enthalpy changes of certain types of reactions are labelled: (1) Standard Enthalpy change of neutralization (ΔHθneu)(標準中和焓變)  The enthalpy change when ONE MOLE of water is formed from the neutralization reaction under standard conditions. (在標準條件下，酸和鹼互相中和生成一摩爾水的焓變。) H+(aq) + OH–(aq) → H2O(l) ΔHθneu = –57.3 kJmol–1

The standard enthalpy change of neutralization is always negative because neutralization is exothermic. (由於中和反應是放熱的，所以標準中和焓變必定是負值)  The value between strong acid and strong alkalis is always about –57.3 kJmol–1 regardless of what the acids and alkalis are.  NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ΔHθneu = –57.3 kJmol–1  2HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2H2O(l) ΔHθneu = –114.6 kJmol–1  The value will be less negative if weak acids and/or weak alkalis are used instead (當 使 用 

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Modern College AL Chemistry Notes (2009 – 10)

Section 3

弱酸或者弱鹼時，標準中和焓變會負得比較小), for example:  NaOH(aq) + CH3COOH(aq) → CH3COONa(aq) + H2O(l) ΔHθneu = –55.2 kJmol–1  HCN(aq) + NaOH(aq) → NaCN(aq) + H2O(l) ΔHθneu = –11.7 kJmol–1  This is because some energy is used for complete dissociation of weak acids and/or weak alkalis. (部份能量用於把弱酸或者弱鹼完全電離) HCN(aq) → H+(aq) + CN–(aq) ΔH = +45.6 kJmol–1 H+(aq) + OH-(aq) → H2O(l) ΔH = –57.3 kJmol–1 ∴HCN(aq) + NaOH(aq) → NaCN(aq) + H2O(l) ΔHθneu = –11.7 kJmol–1 (2) Standard Enthalpy change of a solution (ΔHθsoln)(標準溶解焓變)  The enthalpy change when one mole of substance is completely dissolved in an infinite amount of solvent (usually water) under standard conditions. (在標準條件下，把一摩爾 

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溶質完全溶於水，形成一無限稀釋溶液時的焓變。) For example, LiCl(s) → Li+(aq) + Cl–(aq) ΔHθsoln = –37 kJmol–1

As the solute is dissolved in a large amount of excess of solvent, further addition of solvent produces no further enthalpy changes.

(3) Standard enthalpy change of formation(ΔHθf)(標準生成焓變)  The enthalpy change when one mole of substance is formed from its elements in their standard states at standard conditions. (在標準條件下，由組成某物質的成分元素 （在其標準態下），生成一摩爾該物質時的焓變。) Example 2 Given that the standard enthalpy change of formation of ethanol is –278 kJmol–1, we can write the thermochemical equation as follows: 2C(graphite) + 3H2(g) + 

1 O2(g) → C2H5OH(l) 2

ΔHθf = –278 kJmol–1

It should be noted that the enthalpy change of formation of an element is zero (對於一元素 而言，其生成焓變必然是 0。) Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)

Section 3

(e.g. ΔHfθ[O2(g)] = 0 kJmol–1) Example 3 Write thermochemical equation for the following compounds: (i) CaCO3(s) ΔHfθ = –1207 kJmol–1 (ii) H2SO4(l) ΔHfθ = –811 kJmol–1 (iii) H2O(l) ΔHfθ = –285.8 kJmol–1

(4) Standard enthalpy change of combustion (ΔHθc)(標準燃燒焓變)  The enthalpy change when one mole of a substance undergoes complete combustion (i.e. being oxidized by oxygen) in the standard state. (在標準條件下，把一摩爾物質在氧 中完全燃燒時的焓變。)  For example, complete combustion of methane CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔHcθ = -890.5 kJmol–1 Example 4 Write thermochemical equation for the standard enthalpy change of combustion of the following substance. (i) Benzene, C6H6(l) ΔHcθ = –3273 kJmol–1 (ii) C12H22O11(s) ΔHcθ = –5644 kJmol–1 (iii) Propane, C3H8(g) ΔHcθ = –2220 kJmol–1

 

Note that all enthalpies of combustion must have a negative sign since combustion is exothermic. (由於燃燒反應是放熱的，所以任何燃燒焓變必定是負值。) Different standard enthalpy changes may refer to the same equation, for example: C(graphite) + O2(g) → CO2(g) ΔHθ = –393.5 kJmol–1  The above standard enthalpy change can be a standard enthalpy change of _____________ of ____________ or standard enthalpy change of ______________ of ______________.

B. Experimental determination of enthalpy changes by calorimetry Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)

Section 3

1. Determination of enthalpy change of neutralization (P.148 – 150)  A calorimeter (量熱器) is used for measuring the heat transferred during a reaction

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Polystyrene cup is used since its heat capacity is low (∴absorb less heat energy) and is a good insulator (∴↓heat loss). (發泡膠的比熱容量低，吸收熱能較少，故此是合適的絕緣體。) Vacuum flask (真空瓶): an even more accurate set-up which can further minimize heat loss General Procedures: 1. The initial temperature of an acid and an alkali are recorded respectively. 2. Known and same volume of acid and alkali is added to the calorimeter. 3. The reaction mixture is stirred 4. Temperature is measured at regular intervals (e.g. every 5 cm3 of acid/alkali added) 5. A graph of temperature against volume of acid/alkali added is plotted to find the maximum temperature. 5. Using the graph, the temperature change is calculated 6. Using the following equation, calculate the heat change of the above reaction:

Heat evolved = (m1c1 + m2c2) ΔT Where m1 = mass of the solution, m2 = mass of the calorimeter, c1 = specific heat capacity of the solution, c2 = specific heat capacity of the calorimeter ΔT = temperature change of the reaction 7. Calculate the enthalpy change of the neutralization. ∆H =

Heat evolved No. of mole of water formed

****Textbook example 6.4A (P.149)

Possible errors (可能的誤差):

Improvement (改善方法):

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Modern College AL Chemistry Notes (2009 – 10)

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Heat loss from the system to surroundings Replace the polystyrene cup with a vacuum due to conduction, convection, radiation flask calorimeter. and evaporation. (熱能因為傳導、對流、輻 射或蒸發散失到外界。) The experiment is not carried out under Perform the experiment at 298K and 1 atm. standard conditions. Specific heat capacity of the solution may not be equal to that of water (4.18 Jg-1K-1) (溶液的比熱容未必等於水的比熱容。) Heat capacity of the polystyrene cup is ignored. Thermometer is not precise enough (溫度計不夠準確)

Determine the exact specific heat capacity of the solutions. Determine the exact heat capacity of the polystyrene cup as well. Use a Beckmann thermometer or digital devices for higher accuracy in measuring temperatures (利用貝克曼溫度計或者電 子儀器，使量度出來的溫度更準確。)

Common experimental set-up and data treatment:

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Modern College AL Chemistry Notes (2009 – 10)

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Example 5 (a) When 100 g of water at 98.0℃ were added to a calorimeter at 17.5℃, the temperature rose to 82.5℃. Calculate the heat capacity of the calorimeter. (b) 250 cm3 of 0.4M sodium hydroxide and 250 cm3 of 0.4M hydrochloric acid, both at the same temperature, were mixed in the calorimeter. A temperature rise of 2.6℃ was recorded. In another similar experiment, 250 cm3 of 0.4M ethanoic acid was used instead of hydrochloric acid. The temperature rise was found to be 2.0℃. (i) Calculate the enthalpy changes of neutralization in both experiments. (ii) Explain the difference of the two values obtained in (a). Given: Specific heat capacity of final solution = specific heat capacity of water = 4.18 Jg-1K-1 Density of final solution = Density of water = 1.0 gcm-3 (a) Heat received by the calorimeter = Heat released by water C (82.5 – 17.5) = 100(4.18)(98.0 – 82.5) C = 99.7 JK–1 (b) No. of mole of H+ = No.of mole of OH- =

250 ×0.4 = 0.1 1000

∴ No. of mole of water formed = 0.1 Mass of final solution = (250 + 250) cm-3 × 1.0 gcm-3 = 500 g (i) For NaOH vs. HCl, Heat from neutralization = Heat received by the water and calorimeter = 500(4.18)(2.6) + 99.7(2.6) = 5693 J ∴ ΔHneu = −

5693 = –56930 Jmol–1 = –56.93 kJmol–1 0.1

(ii) For NaOH vs. CH3COOH, Heat from neutralization = 500(4.18)(2.0) + 99.7(2.0) = 4379 J ∴ ΔHneu = −

4379 = –43790 Jmol–1 = –43.79 kJmol–1 0.1

Ethanoic acid was only partially ionized in water. Some of the heat from the neutralization process was absorbed for complete ionization of ethanoic acid molecules. (乙酸是弱酸，在水中只能部份電離。故此部份從中和作用釋出的熱能會 被吸收，用於把乙酸完全電離。)

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Modern College AL Chemistry Notes (2009 – 10)

Section 3

Example 6 A student tried to determine the enthalpy change of neutralization by adding 25.0 cm3 of 1.0 M HNO3 in a polystyrene cup and adding 25.0 cm3 of 1.0 M NH3 into it. The temperature rise recorded was 3.11 oC. Given that the mass of the polystyrene cup is 250 g, the specific heat capacities of water and the polystyrene cup are 4200 Jkg–1K–1 and 800 Jkg–1K–1 respectively. Determine the enthalpy change of neutralization of nitric acid and aqueous ammonia. (Density of water = 1 g cm–3)

2. Determination of enthalpy change of combustion (P.150 – 152)  The following diagram shows a flame calorimeter which is used to determine the enthalpy change of combustion of a liquid fuel (e.g. ethanol).

Procedures: 1. Record the mass of ethanol and the mass of spirit lamp. 2. Record the initial temperature of water in the calorimeter. 3. The spirit lamp is ignited for about 10 minutes. 4. Record the final temperature and the increase in water temperature is calculated. 5. The mass of spirit lamp after the experiment is recorded to find the mass of ethanol used. 6. Using the following equation, calculate the heat change of the above reaction: Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)

Section 3

Heat evolved = (m1c1 + m2c2) ΔT Where m1 = mass of the water in the calorimeter, m2 = mass of the calorimeter, c1 = specific heat capacity of water, c2 = specific heat capacity of the calorimeter ΔT = temperature change of the reaction 7. Calculate the enthalpy change of combustion: ∆H =

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Possible errors: Heat is carried away by draughts (氣流把熱能帶走) and never enters the calorimeter. Heat lost from the top and sides of the calorimeter. (熱能從量熱器的頂部及邊緣散失。) Heat energy is used to heat up the beaker rather than water. Incomplete combustion. (不完全燃燒)

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Improvements: A lid and polystyrene boards as shields are used to reduce heat loss to the surroundings. (把

1. 2. 3. 4.

1.

Heat evolved No. of mole of ethanol used

量熱器用蓋蓋好，並用發泡膠版圍繞量熱器的四周，減少熱能流失。) 2. Assuming the calorimeter/beaker takes the same increase in temperature, calculate the heat energy absorbed by the calorimeter/beaker as well. The improved experimental set-up is shown below:

Example 7 Determine the enthalpy change of combustion of ethanol using the following data: Mass of spirit lamp before experiment: 53.59 g Mass of spirit lamp after experiment: 52.80 g Mass of water in copper calorimeter = 50 g Mass of copper calorimeter without water = 380 g Initial temperature of water = 20.0℃ (293K) Final temperature of water = 42.5℃ (315.5K) The specific heat capacities of water and copper are 4200 Jkg–1K–1 and 2100 Jkg–1K–1 Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)

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respectively. Relative molecular mass of ethanol = 46.0 Heat evolved by combustion of ethanol = Heat absorbed by the water and calorimeter = [(0.05)(4200) + (0.38)(2100)](315.5 – 293) = 22680 J From the equation, C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(l) Mass of ethanol burnt = (53.59 – 52.80) = 0.79 g ∴ No. of mole of ethanol burnt = ∴ΔHc = −

0.79 = 0.0172 46

22680 = –1318605 Jmol–1 = –1318.6 kJmol–1 0.0172

Example 8 In an experiment, 1.60 g of methanol was burnt and 50.0 g of water was heated up, raising the temperature by 33.2 oC. Given that the specific heat capacities of water and copper calorimeter are 4200 Jkg–1K–1 and 2100 Jkg–1K–1 respectively and the mass of the calorimeter is 400 g, calculate the enthalpy change of combustion of methanol.

3. Determination of enthalpy change of solution  Experimental determination of enthalpy change of solution is very similar to that of enthalpy change of neutralization. Example 6 When 8.0 g of anhydrous copper(II) sulphate are added to 100 cm3 water in a polystyrene cup, there is a temperature rise of 8.0℃ in the solution. (a) Calculate the molar enthalpy change of solution of anhydrous copper(II) sulphate. Given: Molar heat capacity of water = 76.5 J K-1 mol-1; Molar mass of water = 18.0 g mol-1; Molar mass of anhydrous copper(II) sulphate = 159.5 g mol-1 Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10) Heat evolved =

100 × 76 .5 ×8 = 3400 J 18

No. of mole of CuSO4(s) dissolved = ∴ ΔHsoln = −

Section 3

8.0 = 0.0502 159 .5

3400 = – 67729J = – 67.73 kJmol-1 0.0502

Hint: In this reaction, temperature increases ⇒ Exothermic ⇒ Negative sign for ΔH (b) State the assumptions (假設) made in calculating the enthalpy change of solution in (a). 1. The heat capacities of polystyrene cup and thermometer are negligible. (發泡膠杯 溫 度計的熱容量可忽略不計。) 2. The specific heat capacity of copper(II) sulphate solution is the same as that of water. (硫酸銅(II)溶液的比熱容量和水的比熱容量值相等。) 3. The density of copper(II) sulphate solution is the same as that of water (1.0 g cm -3). (硫酸銅(II)溶液的密度和水的密度相等。) 4. Anhydrous copper(II) sulphate dissolves in water quickly such that heat loss to the surroundings can be negligible (無 水 硫 酸 銅 (II)於 水 中 快 速 溶 解 ， 故 此 溶 解 時 的 熱能流失小至可被忽略。). 5. Further dilution of the solution would cause no further heat change. (c) Indicate the three sources of error in this experiment and suggest how these errors could be minimized. Any three of the following: Possible errors: Improvement: Heat loss from the system to the Replace the polystyrene cup with a vacuum surroundings due to conduction, flask calorimeter. convection and evaporation. Specific heat capacity of copper(II) Determine the exact specific heat capacity of sulphate may not be equal to that of water. the copper(II) sulphate solution. Heat capacity of the polystyrene cup is Determine the exact heat capacity of the ignored. polystyrene cup as well. Thermometer is not precise enough

Use a Beckmann thermometer or digital devices for higher accuracy in measuring temperatures Anhydrous copper(II) sulphate may have Gently heat the sample in an oven for several absorbed moisture from air (實 驗 中 無 水 hours and keep it in a desiccator (乾燥皿). 硫酸銅(II)可能會吸收空氣中的濕氣。).

Example 9 Prepared by Mr. Chau Chi Keung, Richard

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Modern College AL Chemistry Notes (2009 – 10)

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When 0.05 mol of silver nitrate was added to 50 g of water in a polystyrene cup, a temperature drop of 5.2 oC was recorded. Assuming that there was no heat absorption by the polystyrene cup, calculate the enthalpy change of solution of silver nitrate. (Specific heat capacity of water = 4200 Jkg–1K–1)

4. Determination of enthalpy change of reaction (P.152) Read the following example:

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Modern College AL Chemistry Notes (2009 – 10)

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Modern College AL Chemistry Notes (2009 – 10)

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3.3 Hess’s Law – Finding ΔH Values Indirectly A. What is Hess’s Law? 

It is not possible to measure the enthalpy change of some reactions directly because  The extent of reaction cannot be controlled ( 反 應 進 度 無 法 控 制 ) (e.g. C(s) + 1/2O2(g) → CO(g), which represents an incomplete combustion of carbon).  The rate of reaction is very slow under standard conditions (於標準條件下速率太慢)  The reaction may involve the formation of side products (有副產品生成)

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To find the enthalpy change of these reactions, we can apply Hess’s Law ( 赫 斯 定 律 ), which was formulated by Germain Hess (1802 – 1850), a Russian Chemist. Hess’s law of constant heat summation: The total enthalpy change accompanying a chemical reaction is independent of the route by which the chemical reaction takes place. (化學反應的總焓變，與其所循的途徑是無關的。) Which means, The standard enthalpy change of a reaction depends on only the difference in standard enthalpy between the reactants and the products but not on the route of the reaction.

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Hess’s law is actually the natural consequence of the law of conservation of energy.  As shown in the following figure, the conversion from X to Y can proceed in two different routes:  Route 1: Direct conversion, enthalpy change = ΔH1.  Route 2: Indirect route (X → A → B → Y), enthalpy change =ΔH2+ΔH3+ΔH4. A(s) ΔH2

ΔH3 B(s)

ΔH4

X(s)

Y(s) ΔH1 By Hess’s Law: ΔH1= ΔH2+ΔH3+ΔH4

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Modern College AL Chemistry Notes (2009 – 10) 

Section 3

If ΔH1 ≠ ΔH2 + ΔH3 + ΔH4, it would be possible that energy is created or destroyed by making X from Y by route 1 and then converting X to Y by route 2. This would contradict to the Law of Conservation of Energy.

B. Applications of Hess’s Law? 



Hess’s law can be used to determine enthalpy changes which are not easily obtainable by experiment. (利用赫斯定律，可以計算出一些難以在實驗室進行的反應的焓 變。) An enthalpy level diagram (焓 階 圖 ) or a Born-Haber cycle ( 波 恩 – 哈 柏 循 環 ) (also known as “enthalpy cycle”) can be used to link the enthalpy changes of different routes.  Enthalpy level diagram: Relate substances together in terms of enthalpy changes of reactions ( 藉反應的焓變，把不同物質連繫起來。 ). For example, consider the oxidation of C(graphite) to CO2(g):



Enthalpy cycle / Born-Haber cycle: Relate the various equations involved in a reaction ( 把 反 應 中 所 涉 及 的 不 同 方 程 式 連 繫 起 來 。 ). Again, take the oxidation of C(graphite) to CO2(g) as an example:

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Example 10 (a) Explain why direct determination of the following reaction is impossible. C(s) +

1 O2(g) → CO(g) 2

(b) Calculate the standard enthalpy change for the reaction in (a). Given the following information: Standard enthalpy change of formation of carbon dioxide is –394 kJmol–1 Standard enthalpy change of combustion of carbon monoxide is –283 kJmol–1 (a) This is because the reaction represents an incomplete combustion of carbon. Carbon may react with oxygen to form carbon dioxide and hence the extent of this reaction cannot be controlled. (b)From the given information, we have, C(graphite) + O2(g) → CO2(g) ΔHθf = –394 kJmol–1 CO(g) + 1/2O2(g) → CO2(g) ΔHθc = –283 kJmol–1 Approach 1: Enthalpy cycle (Born – Haber cycle)

Approach 2: Enthalpy level diagram

Hints on using Hess’s Law I 1. For all thermochemical equations, all physical states of the species involved must be clearly stated (必須標明物態符號). 2. When using an enthalpy level diagram, you may organize your

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information as follows:  Put all reactants on the top  Put the given reaction on one side, while the target reaction on the other side  Exothermic: ↓; Endothermic: ↑ 3. Never insert any numerical value on the vertical axis of an enthalpy level diagram since the absolute value of enthalpy is impossible to measure. Example 11 The chemical equation for the formation of MgCO3 from its constituent elements is as follows: Mg(s) + C(graphite) +

3 O2(g) → MgCO3(s) 2

(a) Suggest two reasons on why the enthalpy change of formation of MgCO3(s) cannot be determined directly. (b) Calculate the enthalpy change of formation of MgCO3(s) by using the following data: Mg(s) +2H+(aq) → Mg2+(aq) + H2(g) ΔH 1 = –434 kJmol–1 MgCO3(s) + 2H+(aq) → Mg2+(aq) +H2O(l) + CO2(g) ΔH 2 = –44 kJmol–1 C(graphite) + O2(g) → CO2(g) ΔH 3 = –394 kJmol–1 H2 +

1 2

O2(g) → H2O(l)

ΔH 4 = –286 kJmol–1

(a) Any two of the following:  The extent of the reaction cannot be controlled.  Side reactions are possible.  Direct combustion of magnesium may be too violent to handle. (b) By Hess’s Law

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Hints on using Hess’s Law II 1. It should be noted that if a reaction is reversed, the sign of ΔH is also reversed, the numerical value is the same (當 反應 方 向 掉 轉 時， 焓變 數 值相等，而正負相反。). For example: Ag+(g) + Cl–(g) → AgCl(s) ΔH = – 769 kJmol–1 AgCl(s) → Ag+(g) + Cl–(g) ΔH = + 769 kJmol–1 2. Magnitude of ΔH is proportional to the quantities of reactants and products in the reaction (焓變值大小與反應物和生成物的份量成正比。 ). For example: Ag+(g) + Cl–(g) → AgCl(s) ΔH = – 769 kJmol–1 2Ag+(g) + 2Cl–(g) → 2AgCl(s) ΔH =2 × –769 kJmol–1 = –1538 kJmol–1 3. When drawing a Born-Haber cycle, you may have to:  Reverse some reaction equations to have the required reactants and products.  Multiply some equations to get the correct number of reactants and products. 4. Always show all your working steps clearly. During examinations, as long as your working is clear and correct, you will still have some marks even though you make a early mistake in your calculation. Example 12 It is impossible to measure the enthalpy change of hydration of magnesium sulphate because this process is very slow and cannot be controlled. Therefore, this enthalpy change is usually determined indirectly by measuring the enthalpy change of solution of two related solids. Experiment A 3.01 g of anhydrous magnesium sulphate were added to 50.0 cm3 of water in a polystyrene cup and the solution stirred to dissolve the solid as quickly as possible, there was a maximum rise in temperature of the solution by 11.0℃. Experiment B 6.16 g of magnesium sulphate-7-water were added to 50.0 cm3 of water in a polystyrene cup and the solution stirred to dissolve the solid as quickly as possible, there was a maximum fall in temperature of the solution by 1.5℃. (a) Calculate the enthalpy of solution of anhydrous magnesium sulphate. (b) Calculate the enthalpy of solution of magnesium sulphate-7-water. (c) Calculate the enthalpy change of hydration (水合焓變) of magnesium sulphate. Prepared by Mr. Chau Chi Keung, Richard

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(d) Plot your results on an enthalpy level diagram. Assume that the heat capacity of the polystyrene cup is negligible. Specific heat capacity of water = 4.18 J g-1K-1 Density of water = 1.00 gcm-3 ; Molar mass of MgSO4 = 120.4 g mol-1 Molar mass of MgSO4.7H2O = 246.4 gmol-1

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(a).Heat evolved = mcΔT = 50 × 4.18 × 11.0 = 2299 J No. of mole of MgSO4(s) = ∴ ΔHsoln = −

3.01 = 0.025 120 .4

2299 0.025

= – 91960 Jmol-1 = – 91.96 kJmol-1 (b). Heat absorbed = mcΔT = 50 × 4.18 × 1.5 = 313.5 J No. of mole of MgSO4‧ 7H2O(s) = ∴ ΔHsoln =

6.16 = 0.025 246 .4

313 .5 = 12540 Jmol–1 = 12.54 kJmol–1 0.025

(c). MgSO4(s) + 7H2O(s)

MgSO4‧ 7H2O(s)

(d)

Example 13 From calorimetric measurements, the standard molar enthalpies of combustion at 298K for Graphite = –393.5 kJmol–1 Hydrogen = –285.8kJmol–1 Ethanol = –1371 kJmol–1 Calculate the standard molar enthalpy of formation for ethanol at 298K.

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Modern College AL Chemistry Notes (2009 – 10) 2C(graphite) + 3H2(g) + 1/2O2(g)

Section 3

C2H5OH(l)

Note: Actually there exists a short cut in calculating ΔH. Read the next example: Example 14 Calculate the standard enthalpy change for the reaction 2H2S(g) + SO2(g) → 3S(s) + 2H2O(l) Using only the standard heat of formation data: Compound ΔHθf /kJmol–1 H2S(g) –20.6 SO2(g) –296.9 H2O(l) –285.9 From the above data, we have: (1) H2(g) +S(s) → H2S(g) ΔH1 = –20.6 kJmol–1 (2) S(s) + O2(g) → SO2(g) ΔH2 = –296.9 kJmol–1 (3) H2 + 1/2O2(g) → H2O(l) ΔH3 = –285.9 kJmol–1 (1) × 2, and then reverse the reaction, we have, (4) 2H2S(g) → 2H2(g) + 2S(s) ΔH4 = +41.2 kJmol–1 Reverse (2), we have, (5) SO2(g) → S(s) + O2(g) ΔH5 = +296.9 kJmol–1 (3) × 2, we have, (6) 2H2 + O2(g) → 2H2O(l) ΔH6 = –571.8 kJmol–1 For the reaction, ΔHθ =ΔH4 + ΔH5 +ΔH6 = +41.2 + 296.9 + (–571.8) = –233.7 kJmol–1 To sum up, enthalpy change of reactions can be found by three methods: Enthalpy cycle(Born–Haber cycle) (the most common method) Energy level diagram By algebraic method (a short cut)

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Example 15 Find the standard enthalpy change of formation of solid sodium hydroxide by Na(s) + 1/2H2(g) + 1/2O2(g) → NaOH(s) ΔHfθ = ? (a) Using algebraic method from the following equations; ΔHθ kJmol–1 (1) Na(s) + H2O(l) → NaOH(aq) + 1/2H2(g) –79.1 (2) H2(g) + 1/2O2(g) → H2O(l) –285.8 (3) NaOH(s) → NaOH(aq) –62.8 ______________________________________________________________________________ Reverse (3), we have, (4) NaOH(aq) → NaOH(s) ΔH4 = 62.8 kJmol–1 (1) + (2) + (4), we have, Na(s) + 1/2H2(g) + 1/2O2(g) → NaOH(s) ΔHfθ = ΔH1 + ΔH2 + ΔH4 = –79.1 + (–285.8) + 62.8 = –302.1 kJmol–1 (b)Drawing an enthalpy level diagram to represent the above processes;

Enthalpy

(c) Using an enthalpy cycle;

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C. Calculation of enthalpy change of reaction from ΔHf  From an enthalpy cycle, it can be shown that: ΔHθ = ∑ [ΔHθf (products)] – ∑ [ΔHθf (reactants)]

Example 16 By using the following thermochemical data, calculate the standard enthalpy change of thermal decomposition of calcium carbonate. ‘ CaCO3(s) → CO2(g) + CaO(g) Standard enthalpy change of formation of CaO: –635.1 kJ mol–1 Standard enthalpy change of formation of CO2: –393.5 kJ mol–1 Standard enthalpy change of formation of CaCO3: –1207.6 kJ mol–1

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Example 17 Given that the standard enthalpy changes of formation (in kJ mol–1): CO2(g) = –393.5; H2O(l) = –285.9; CH4(g) = –74.8 Calculate the standard enthalpy change of combustion of methane For combustion of methane, we have,

Example 18 Ethanol reacts with phosphorus pentachloride (PCl5) according to the reaction: C2H5OH(l) + PCl5(s) → C2H5Cl(l) + POCl3(l) + HCl(g) Calculate the standard enthalpy change of reaction, using the following data: ΔHθf [C2H5OH (l)] = –277 kJ mol–1 ΔHθf [PCl5 (s)] = –443 kJ mol–1 ΔHθf [C2H5Cl (l)] = –136 kJ mol–1 ΔHθf [POCl3 (l)] = –597 kJ mol–1 ΔHθf [HCl (g)] = –92 kJ mol–1

‘You may read Chapter 6 in your textbook for more examples and exercises’. Prepared by Mr. Chau Chi Keung, Richard

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D. Calculation of enthalpy change of formation from ΔHc  From an enthalpy cycle, it can be shown that: ΔHθf = ∑ [ΔHθc (elements)] – ΔHθc (compound)

Example 19 Given the following information, find the standard enthalpy change of formation of methane. C(graphite) + O2(g) →CO2(g) ∆ Hc[C(graphite)] = –393.5 kJmol–1 H2(g) +

1 O2(g) →H2O(l) ∆ Hc[H2(g)] = –285.8 kJmol–1 2

CH4(g) + 2O2(g) →CO2(g) + 2H2O(l) ∆ Hc[CH4(g)] = –20 kJmol–1 Direct measurement of ΔHf [CH4(g)] is impossible because carbon(graphite) and hydrogen do not combine directly, and methane does not decompose directly to form carbon(graphite) and hydrogen. Since methane contains carbon and hydrogen only, they can be related to carbon dioxide and water by the combustion of methane and its constituent elements as shown in the diagram below.

ΔHθf = ∑ [ΔHθc (elements)] – ∑ ΔHθc (compound) ∴ΔHf [CH4(g)] =ΔH1 + 2ΔHf – ΔH3 = [–393.5 + 2 × (–285.8) – (–890.4)] = –74.7 kJmol–1 Example 20 Find the standard enthalpy change of formation of butane gas (C4H10(g)). Given: ΔHθc [C(graphite)] = –393.5 kJmol–1 ΔHθc [H2(g)] = –285.8 kJmol–1 ΔHθc [C4H10(g)] = –2877 kJmol–1

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E. ΔH as a measure of energetic stability (能量穩定性 )  A universal rule: All substances in the universe tend to keep their energy content as low as possible (一般而言，宇宙間所有物質都傾向於降低它們的能量。).  Chemical reactions occur spontaneously are usually exothermic.  Many compounds are formed exothermically from their elements (有不少化合物在 



從其組成元素生成時都會放出熱能。). ΔHθf value of a compound provides us information about its energetic stability relative to its constituent elements.  If ΔHθf is negative, the compound formed is more stable than its elements (如 果 某 化合物的標準生成焓變是負值， 那就代表該化合物比 它 的組成元素更 穩 定。). However, this energetic stability is only a comparison between a particular substance and its elements. For example: H2(g) + O2(g) → H2O2(l), ΔHθ = –188 kJmol–1  From the equation, H2O2 is more stable relative to its elements, but it does not imply that H2O2 is stable with respect to water and oxygen. H2O2(l) → H2O(l) +

1 O2(g) ΔHθ = –98 kJmol–1 2

Therefore, H2O(l) will be even more stable relative to H2(g) and O2(g) when compared with H2O2(l). Hence, it is important to specify the stability of a compound with respect to what substances it refers. 





Moreover, it must be noted that ΔH value is not the only factor which determines the feasibility (可行性) of a chemical reaction.  ΔH is only a state function (狀態函數). It just tells us the current state of the system (e.g. amount of substance, temperature and pressure), but not the way in which the system got to that state.  For example, it tells us nothing about the rate of reaction (沒 有 提 供 任 何關 於 反 

應速率的資訊。). Negative ΔH value only tells us that the reaction has the tendency to take place. the

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more negative the ΔH, the more likely the reaction is feasible. However, it does not imply that the reaction will occur at an observable rate. (反應的 焓變是負值只代表該反應傾向能夠發生，但不代表它能夠以一合理速率 進行。)





The reactants have to gain a certain amount of energy so that bond breaking can take place in the reactants. This energy is called the activation energy (Eact)(活 化 能 ), which represents the energy barrier which the reactants have to overcome before a chemical reaction can occur (see the figures below). In other words, a reaction will not occur in the absence of Ea, regardless to the magnitude or sign of ΔH of the reaction.



This explain why diamond and graphite do not spontaneously decompose into carbon dioxide although their ΔHc values are highly negative (i.e. “energetically unstable”). C(diamond) + O2(g) → CO2(g) ΔHθc = –395.4 kJmol–1 C(graphite) + O2(g) → CO2(g) ΔHθc = –393.5 kJmol–1  However, Eact required for the breakdown of the giant covalent network in diamond or graphite is very high (要拆散金剛石或石墨的巨型共價結構需要極大的活化能。).  The rate will be extremely slow. (∴They are said to be “kinetically stable”.)  The concept of activation energy will be discussed in great detail in Section 5 (Chemical Kinetics).



Also, the kinetic factor also explains why there are some compounds can be formed endothermically without breaking up into their elements. For example: Ethyne (C2H2): 2C(graphite) + H2(g) → C2H2(g) ΔHθf = +227 kJmol–1 Nitrogen monoxide (NO):  

1 1 N2(g) + O2(g) → NO(g) ΔHθf = +90 kJmol–1 2 2

Decomposition of these compounds requires high activation energies Rate of breakdown is practically unobservable

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∴They are kinetically stable (於動力學而言是穩定的)

3.4 Spontaneity of changes – Concept of entropy (熵) A. Spontaneous changes 

 

A process is said to be spontaneous (自發的) if it happens naturally.  No external forces are needed to keep the process going, although some external aid is needed to get the process started (e.g. a spark or a flame is needed to ignite gasoline)  Can be either a physical change or a chemical change  Note: spontaneous ≠ instantaneous (即時的) Non-spontaneous process: does not happen unless we make it happen  It requires a continuous supply of energy in order to happen Spontaneous exothermic changes are very common. However, some spontaneously changes are not necessarily to be exothermic. They can be endothermic (自發進行的反應 不一定是放熱反應，也可以是吸熱反應。). For example:  Melting of ice at 0°C  Dissolution of some ammonium salts (溶解某些銨鹽，例如氯化銨和硝酸銨) NH4Cl(s) + 500H2O(l) → NH4Cl(aq, 500H2O) ΔHθsoln = + 14.9 kJmol–1 NH4NO3(s) + 500H2O(l) → NH4NO3(aq, 500H2O) ΔHθsoln = + 25.7 kJmol–1  Decomposition of nitrogen pentoxide (N2O5) into nitrogen dioxide and oxygen N2O5(s) → 2NO2(g) + O2(g) ΔHθ = + 109.5 kJmol–1  Reaction between solid barium hydroxide octahydrate and ammonium nitrate(V) Ba(OH)2．8H2O(s) + 2NH4NO3(s) → Ba(NO3)2(aq) + 2NH3(aq) + 10H2O ΔHθ = + 62.3 kJmol–1

B. Concept of entropy(熵 )  All substances in nature tend to keep their energy content as low as possible. However, this rule is not enough to explain whether a process can occur naturally or not.  In fact, there exists another universal rule at the same time: All substances and particles in the universe tend to be more disordered (or random). (i.e. having more ways of being arranged)  The state with a high degree of randomness is always favourable in nature  Spontaneous changes involves an increase of randomness  Take melting of ice as an example:  When ice melts, the regularly packed water molecules become free to move in the liquid state. The system becomes less ordered.  This process occurs spontaneously and endothermically.  Another example is the decomposition of ammonium carbonate: (NH4)2CO3(s) → 2 NH3(g) + CO2(g) + H2O(g) ΔHθ = + 68 kJ mol–1 Prepared by Mr. Chau Chi Keung, Richard

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One mole of a solid with an ordered crystalline structure is replaced by four moles of gas molecules. The system becomes less ordered.  This process occurs spontaneously and endothermically.  The randomness in arrangement of a system is measured by a quantity called entropy (S) (熵是一個量度某體系的亂度或無序程度的熱力學函數。)  The change in the randomness is called entropy change (亂度的變化則稱為熵變)(ΔS).  Entropy describes the extent to which atoms, molecules or ions are distributed in a disorderly manner. 

Solid

Liquid

Gas

Entropy increases Entropy of solid < Entropy of liquid << Entropy of gas



Second Law of Thermodynamics (熱力學第 二定律 ): There is always an increase in the total entropy of the universe.

ΔSUniverse = ΔSSystem + ΔSSurroundings > 0 

  

A perfect crystal at temperature zero K (i.e. absolute zero) has perfect order and zero entropy (which is actually the Third Law of Thermodynamics). As the substance is heated, the entropy increases. Heating increases the motion of particles and therefore the degree of disorder. Entropy can also be increased by providing more space (i.e. increasing volume) into which particles can spread. The entropy added can be described by the following equation: ΔS =

Q (*) T

Where Q is the heat absorbed and T is the absolute temperature (in K).

C. Entropy Change (ΔS)  

Like enthalpy (H), entropy is a state function, which means it depends only on the present state of the system, not on the path it took to arrive at that state. Hence, the entropy change (ΔS) of a system depends only on the difference between its final and initial states. That is: Prepared by Mr. Chau Chi Keung, Richard

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ΔS = S final – S initial (∆ S = S 最終 – S 起始 )

 1.

  

ΔS can be used to indicate the relative degree of disorder of the system. For example: Positive entropy: ΔS = S final – S initial = +ve e.g. 1 Ice → Water (Lower entropy) (Higher entropy)  Water molecules in liquid water have more freedom to move than in ice  ∴ ΔS = Swater – Sice > 0

e.g. 2 CaCO3(s) → CaO(s) + CO2(g) (Lower entropy) (Higher entropy) 1 mole of calcium carbonate decompose to give 1 mole of calcium oxide and 1 mole of carbon dioxide Entropy of gas >> Entropy of solid ∴ ΔS = (SCaO(s) + SCO2(g)) – SCaCO3(s) > 0 2. Negative entropy: ΔS = S final – S initial = –ve e.g. 1 Water → Ice (Higher entropy) (Lower entropy) e.g. 2 Haber process (哈柏法) N2(g) + 3H2(g) → 2 NH3(g) (Higher entropy) (Lower entropy)  The entropy of a system is largely determined by the numbers of moles of gaseous molecules (More gaseous molecules, higher entropy).  Entropy decreases since 4 moles of reactant molecules are converted into 2 moles of product molecules

(a) (b) (c) (d) (e) (f)

Example 21 Predict whether the following changes or chemical reactions involve an increase or a decrease in entropy. Explain your prediction as well. Dissolving salt in water to form salt solution Complete combustion of carbon Condensation of steam on a cold mirror. Complete combustion of propane Oxidation of sulphur dioxide to sulphur trioxide. Polymerization of ethene to form poly(ethene). (g) Precipitation of chloride ions by adding silver nitrate(V) solution. (a) Increase. The solid with ordered crystalline structure dissolves to gives ions which Prepared by Mr. Chau Chi Keung, Richard

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are randomly distributed in solution. (b) Increase. The solid carbon with orderly arranged carbon atoms is converted into gaseous carbon dioxide molecules which have much more freedom to move. (c) Decrease. Water molecules in liquid water have less freedom to move than in steam. (d) Decrease. 6 moles of gaseous reactant molecules are converted into 3 moles of gaseous product molecules.

D. Free Energy Change (ΔG)   



Entropy is temperature dependent. At a higher temperature, the entropy of the system is higher. At a lower temperature, the entropy of the system is lower. Hence, the enthalpy, temperature and entropy are all related to the spontaneity of a process. By combining all these factors, the driving force for a process is called free energy (自 由 能 )(G), or more specifically Gibbs free energy ( 吉 布 斯 自 由 能 ), as a tribute to Josiah Willard Gibbs (1839 – 1903), an American mathematical physicist. G links enthalpy, temperature and entropy together in one equation:

G = H – TS 

For the change in the free energy of the system that occurs during a process, Gibbs suggested that:

ΔG = ΔH – Δ(TS) ⇒ ΔG = ΔH – TΔS (provided that T is constant)

Gibbs’ Derivation: Given a chemical reaction taking place at constant temperature and pressure.

ΔHSystem = –ΔHSurroundings ∆H T

The equation (*) on P.26 can be rewritten as ∆S = ∴∆S Surrouding

s

=

∆H Surrouding T

s

=−

∆H system T

(**)

From the Second Law of Thermodynamics, we have, Prepared by Mr. Chau Chi Keung, Richard

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ΔSUniverse = ΔSSystem + ΔSSurroundings Then, from (**), we have, ΔS Universe = ΔS System −

ΔH System T

Gibbs’ Derivation (cont’d): After multiplying both sides by –T, we have,

–TΔSUniverse = –TΔSSystem + ΔHSystem = ΔHSystem – TΔSSystem The term –TΔSUniverse is called the free energy change (自 由 能 變 ) at constant temperature and pressure and is denoted as ΔG (unit: Jmol-1).

ΔG = –TΔSUniverse Hence, based on the reaction/process itself (i.e. the system),

ΔG = ΔH – TΔS



By the Second Law of Thermodynamics, the total entropy of the universe is always increasing (i.e. ΔSUniverse > 0).  ΔG = –TΔSUniverse  ∴ At constant temperature and pressure, ΔG < 0 For all spontaneous processes (對 





於任何自發進行的反應，其自由能變必為負值。) Conversely, if ΔG > 0, the process must be non-spontaneous, energy from an external source is required in order to make it happen. (Note: The reverse of that process will be spontaneous). (相反，如果反應的自由能變為正值，則該反應 無法自發進行，必須有外界的干預和能量的輸入才能令它發生。但注意其 逆反應則可以自發進行。) If ΔG = 0, the system is said to be at equilibrium (ΔG = 0 時 ， 反 應 達 致 平 衡 ), neither the forward nor the reverse process is favored (i.e. no net change).

Again, take melting of ice as an example:  The packing of water molecules in ice is more ordered than that in water. Thus, melting is a process associated with an increase in randomness (i.e. ΔS > 0). (在冰中 水分子的排列比在水中的水分子較有規律，所以冰溶化會令體系的亂度增加， 故熵變是正值。) Prepared by Mr. Chau Chi Keung, Richard

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Enthalpy change (ΔH) of melting = 6010 kJmol–1 Entropy change (ΔS) of melting = 22.0 kJmol–1

∴At any temperature above 0°C, ice begins to melt (∵ΔG < 0)





The following table summarizes the relationship among the exothermicity (ΔH), entropy change (ΔS) and spontaneity (ΔG) of a process. ΔH ΔS –ve +ve

ΔG = ΔH – TΔS –ve

Result Spontaneous at all temperatures.

Example Exothermic reactions with production of gases (e.g. burning of fuels)

+ve

–ve

Spontaneous at all temperatures.

+ve –ve

+ve

Non-spontaneous at all temperatures (its reverse process is spontaneous at all temperatures)

+ve +ve

ΔH < TΔS at high temperatures

+ve +ve

ΔH > TΔS at low temperatures

Only spontaneous at high temperatures (entropy is dominant) Non-spontaneous

–ve –ve

ΔH < TΔS at low temperatures

–ve –ve

ΔH > TΔS at high temperatures

CHCl3 dissolves in CCl4 (Here ΔH = 0 since the intermolecular forces of the two species are similar. The reduction of Gibbs free energy arises solely from the increase in entropy.) Endothermic reactions in which the number of moles of gas molecules decreases (e.g. photosynthesis) Endothermic reactions in which the number of moles of gas molecules increases (e.g. evaporation of a volatile liquid, thermal decomposition of metal carbonate) Exothermic reactions in which the number of moles of gas molecules decreases (e.g. Haber process*)

0

Only spontaneous at low temperatures (exothermicity is dominant) Non-spontaneous

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temperatures (about 300 – 550°C) based on the kinetic concerns. 

It should be noted that when both ΔH and ΔS have the same sign, the temperature becomes critical in determining whether or not a reaction is spontaneous.  Same as the case in ΔH, ΔG only provides information about the energetic stability of a system. It tells us nothing about the reaction rate. (和 ΔH 一樣，ΔG 只能讓我 們知道體系的能量穩定性，它並沒有提供任何關於反應速率的資訊。) Example 22 It is often said that a reaction is favourable if there is an increase in entropy in the system. Consider the following reaction: H+(aq) + OH–(aq) → H2O(l) (a) Does the entropy of this system increase or decrease? Explain briefly. (b) Is this a spontaneous reaction? Why? (a) Decrease. 2 moles of freely moving ions are converted into 1 mole of liquid water. This reduces the degree of randomness of the system. (2 摩 爾 的 自 由 移 動 的 離 子 被 轉 化 為 1 摩爾的水分子，使體系的亂度下降。) (b) It is still spontaneous. By the relation ΔG = ΔH – TΔS, although ΔS is negative, the highly negative value of ΔH (–57.3 kJmol–1) can compensate the decrease in entropy of the system so that the overall ΔG value can still be negative. (根 據 公 式 ΔG = ΔH – TΔS， 當 ΔG < 0， 反 應就能自發進行。在這個例子中，雖然 體系中熵值下降，但由於焓變是一 相 當大的負值 (–57.3 kJmol–1)，大得足以抵銷熵值下降的影響，令 ΔG 最終能維 持負值。) Example 23 Consider the following reaction: NH4Cl(s) + aq. → NH4+(aq) + Cl–(aq)

ΔGθ = –6.7 kJ mol–1 ; ΔHθ = + 16 kJ mol–1

Explain why does ammonium chloride dissolve spontaneously under standard conditions even under standard conditions even though ΔH for the process is positive. It is because when one mole of NH4Cl(s) is dissolved in water, the solid with ordered crystalline structure dissolves to gives ions which are randomly distributed in solution. Thus the entropy of the system increase. (當 氯 化 銨 溶 於 水 時 ， 原 本 1 摩 爾 具 有 有 秩 序的晶體結構的氯化銨固體被轉化為 2 摩爾分佈在溶液中、無秩序地運動的離子 所以體系中的熵值增加。) Under standard condition, the entropy change ΔS is positive so that ΔG can still be negative according to the relation ΔG = ΔH – TΔS. Hence, even though ΔH for the process is positive, ammonium chloride can dissolve spontaneously. (雖 然在 標準 條件 下 ΔH > 0，但因為 ΔS > 0，根據公式 ΔG = ΔH – TΔS，ΔG 最終仍然能維持負值 。也 就是說體系中熵值的增加足以抵銷焓(能量)的增加。所以氯化銨能夠自發地溶於 水。) Example 24 Prepared by Mr. Chau Chi Keung, Richard

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Consider the following reaction: H2O(l) → H2O(s)

ΔGθ = +0.6 kJ mol–1 ; ΔHθ = –44.1 kJ mol–1

Why does liquid water not freeze under standard conditions even though ΔH is negative? Although ΔH of the process is negative, the entropy of the system decrease as water molecules in ice have less freedom to move than those in liquid water. Under standard conditions, according to the relation ΔG = ΔH – TΔS, ΔS is negative so that ΔG is still positive. Hence, the process is non-spontaneous.

Reading for Interest – Why does that happen, Daddy?

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Supplementary Notes: Essay on Thermochemistry (AL 1990 Paper 1C Q.1) Write an essay on thermochemistry. Your essay should include experimental and theoretical considerations, together with the application of thermochemical data. (20 marks) You may organize the points your essay as shown below [Total marks of C.K.: 10] 1. What is thermochemistry? [1] 

Thermochemistry: The study concerning the energy changes of chemical reactions.



In thermochemistry, the total energy stored in a substance is called enthalpy (H), and the energy change of chemical reaction at constant pressure is called the enthalpy change (ΔH).



When heat is absorbed, the reaction is said to be endothermic. For an exothermic reaction, heat is released to the surroundings from the reaction.

2. Definition of standard enthalpy changes [2] 

Enthalpy changes under standard conditions (must include definition of “standard conditions”), with appropriate notations.



[1]

Any two examples (neutralization, formation, solution or combustion)

[0.5@]

3. Description of practical methods to determine ΔH [Max: 3] 

Describe any two method you learnt in your lesson notes (with experimental details) [Max: 2, 1@]



Sources of error and improvement

[1]

4. Description of indirect method of measuring ΔH based on Hess‘s law

[Max: 3]



State Hess’s Law and its scope of application

[1]



Example (e.g. ΔHhyd of a hydrated salt such as MgSO4．7H2O(s)), with calculation details (e.g. use of Born-Haber cycle) [1]



An additional example

[1]

5. Uses of thermochemical data [2, 1@] 

Relation between ΔH of equilibrium constant (Kc) of a reaction – The Van’t Hoff Equation (You will learn this in Section 6) ln

K2 ∆H  1 1   =− −  (C is a constant)  K1 R  T2 T1  

♦

∆ H is the enthalpy change of reaction; R is the universal gas constant 8.314 JK–1mol–1); T is the temperature in Kelvin.

♦

The effect of temperature change on equilibrium

(R =



Energetic stability of a compound



Use of bond enthalpy terms to estimate the enthalpy change of a reaction (You will learn this in Section 4)

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Marks for “Chemical Knowledge”

Maximum mark for “Organization”

8 – 10

6

7

5

6

4

4–5

3

2–3

2

0–1

1

Section 3

Marks for “Chemical Maximum mark for Knowledge” “Presentation” 7 – 10 4 5–6 3 3–4 2 2 or below 1 In general: Your Final mark ≈ Marks on “Chemical Knowledge” × 2 ( ± 1)

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