Energy = force x distance (Joules) In chemical reactions, we need energy usually in the form of heat. Energy is absorbed to break the bonds of the reactants and energy is given out when new bonds are formed in the products. Exothermic reactionsEndothermic reactions

 Heat is the energy transferred between objects that are at different temperatures.  The amount of heat transferred depends on the amount of the substance. ◦ Energy is measured in units called joules (J).

 Temperature is a measure of “hotness” of a substance and represent the average kinetic energy of the particles in a substance.  It does not depend on the amount of the substance. Do both beakers contain the same amount of heat?

 All chemical reactions are accompanied by some form of energy change  ExothermicEnergy is given out  Endothermic Energy is absorbed  Activity : observing exothermic and endothermic reactions

 Enthalpy (H) is the heat content that is stored in a chemical system.  We measure the change in enthalpy ∆ H i.e. the amount of heat released or absorbed when a chemical reaction occurs at constant pressure, measured in kilojoules per mole (kJmol -1 ). ∆ H = H (products) – H (reactants)

Enthalpy  For exothermic reactions, the reactants have more energy than the products, and the enthalpy change, ∆ H = H (products) - H (reactants)  ∆ H is negative since H (products) < H (reactants)  There is an enthalpy decrease and heat is released to the surroundings

 Self-heating cans ◦ CaO (s) + H₂O (l)  Ca(OH)₂ (aq)  Combustion reactions ◦ CH₄ (g) + 2O₂ (g)  CO₂ (g) + 2H₂O (l)  neutralization (acid + base) ◦ NaOH(aq) + HCl(aq)  NaCl(aq) + H₂O(l)  Respiration ◦ C₆H₁₂O₆ (aq) + 6O₂ (g)  6CO₂ (g) + 6H₂O (l)

 For endothermic reactions, the reactants have less energy than the products, and the enthalpy change, ∆ H = H (products) - H (reactants)  ∆ H is positive since H (products) < H (reactants)  There is an enthalpy increase and heat is absorbed from the surroundings Enthalpy

 Self-cooling beer can ◦ H ₂O (l)  H₂O (g)  Thermal decomposition  CaCO₃ (s)  CaO (s) + CO ₂ (g)  Photosynthesis  6CO₂ (g) + 6H₂O (l)  C₆H₁₂O₆ (aq) + 6O₂ (g)

 Amount of heat required to raise the temperature of a unit mass of a substance by 1 degree or 1 kelvin.  Uint : Jg -1 0 C -1 The specific heat capacity of alminium is 0.90 Jg -1 0 C -1. If 0.90J of energy is put into 1g of aluminium, the temperature will be raised by 1 0 C. Calculating heat absorbed and released q = c × m × ΔT q = heat absorbed or released c = specific heat capacity of substance m = mass of substance in grams ΔT = change in temperature in Celsius

 Heat given off by a process is measured through the temperture change in another substance (usually water).  Due to the law of conservation of energy, any energy given off in a process must be absorbed by something else, we assume that the energy given out will be absorbed by the water and cause a temperature change.  calculate the heat through the equation Q = mc ΔT

How much heat is required to increase the temperature of 20 grams of nickel (specific heat capacity 440Jkg -1 0 C -1 ) from 50 0 C to 70 0 C?

 The standard enthalpy change of combustion for a substance is the heat released when 1 mole of a pure substance is completely burnt in excess oxygen under standard conditions.  Example, CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2H 2 O(g) ΔH Ɵ c =-698 kJmol -1  The heat given out is used to heat another substance,e.g. water with a known specific heat capacity.  The experiment set-up can be used to determine the enthalpy change when 1 mole of a liquid is burnt. Example : refer to page 185

 Loss of heat to the surroundings (exothermic reaction); absorption of heat from the surroundings (endothermic reaction). This can be reduced by insulating the calorimeter.  Using incorrect specific heat capacity in the calculation of heat change. If copper can is used, the s.h.c. of copper must be accounted for.  Others include – e.g incomplete combustion. Some of the ethanol could be used to produce CO & soot & water (less heat is given out) Use bomb calorimeter – heavily insulated & substance is ignited electronically with good supply of oxygen

 If 1g of methanol is burned to heat 100g of water, raising its temperature by 42K, calculate the enthalpy change when 1 mole of methanol is burnt. Note: Specific heat capacity of water is 4.18 Jg -1 0 C -1 Practice questions page 187 #1-4

Enthalpy change of neutralisation (ΔH n )  The standard enthalpy change of neutraisation is the enthalpy change that takes place when 1 mole of H + is completely neutralised by an alkali under standard conditions. Example, NaOH(g) + HCl(g)  NaCl(g) + H 2 O(l) ΔH Ɵ =-57 kJmol -1  The enthalpy change of neutralisation of a strong acid and a strong alkali is almost the same as they undergo complete ionisationof ions in water.  Reaction between strong acid and strong base involves H + (aq) + OH - (aq)  H 2 O(l) ΔH Ɵ =-57 kJmol -1 For sulfuric acid, the enthalpy of neutralisation equation is ½ H 2 SO 4 (aq) + KOH(aq)  ½K 2 SO 4 (aq) + H 2 O(l) ΔH Ɵ =-57 kJmol -1 Example : refer to page 188

Enthalpy change of neutralisation ( ΔH n )  The standard enthalpy change of neutraisation is the enthalpy change that takes place when 1 mole of H + is completely neutralised by an alkali under standard conditions.  Example, NaOH(g) + HCl(g)  NaCl(g) + H 2 O(l) ΔH Ɵ =-57 kJmol -1 The enthalpy change of neutralisation of a strong acid and a strong alkali is almost the same as they undergo complete ionisationof ions in water. Enthalpy change of solution (ΔH sol ) - The enthalpy change when 1 mol of solute is dissolved in excess solvent to form a solution of ‘infnite dilution’ under standard conditions. NH 4 NO 3 (s) in excess water  NH 4 + (aq)+ NO 3 - (aq) Example : refer to page 188

For neutralisation between a weak acid, a weak base or both, the enthalpy of neutraisation will be smaller than -57 kJmol -1 (less exothermic) CH 3 COOH(aq) + NaOH(aq)  CH 3 COONa(aq) + H 2 O(l) ΔH Ɵ =-55.2 kJmol -1  Some of the energy released is used to ionise the acid.

 200.0cm 3 of M HCl is mixed with 100.0cm 3 of M NaOH. The temperature rose by C. If both solutions were originally at the same temp, calculate the enthalpy change of neutralisation. Assume that the density of the solution is 1 gcm -3 and the specific heat capacity is 4.18J Jg -1 0 C kJmol-1

The experimental change of neutralisation is kJmol -1 The accepted literature value is kJmol -1 (1) Heat loss to the environment. (2)Assumptions that (a)the denisty of NaOH and HCl solutions are the same as water. (b)the specific heat capacity of the mixture are the same as that of water

When 3 g of sodium carbonate are added to 50 cm 3 of 1.0 M HCl, the temperature rises from 22.0 °C to 28.5°C. Calculate the enthalpy change for the reaction. Assume that the density of the solution is 1 gcm -3 and the specific heat capacity is 4.18J Jg -1 0 C -1. Example : refer to page 189 dissolving ammonium chloride

The experimental change of solution is kJmol -1 The accepted literature value is 15.2 kJmol -1 (1) Absorption of h eat from the environment. (2)Assumptions that the specific heat capacity of the solution is the same as that of water (3)The mass of ammonium chloride is not taken into consideration when working out the heat energy released.

100.0 cm 3 of mol dm -3 copper II sulphate solution is placed in a styrofoam cup g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction. First step Make sure you understand the graph. Extrapolate to determine the change in temperature. The extrapolation is necessary to compensate for heat loss while the reaction is occurring. Why would powdered zinc be used? Example

Determine the limiting reactant Calculate Q Calculate the enthalpy for the reaction cm 3 of mol dm -3 copper II sulphate solution is placed in a styrofoam cup g of powdered zinc is added and a single replacement reaction occurs. The temperature of the solution over time is shown in the graph below. Determine the enthalpy value for this reaction.

The following measurements are taken:  Mass of cold water (g)  Temperature rise of the water ( 0 C)  The loss of mass of the fuel (g) We know that it takes 4.18J of energy to raise the temperature of 1g of water by 1 0 C. This is called the specific heat capacity of water, c, and has a value of 4.18Jg -1 K -1. Hence, energy transferred can be calculated using: Energy transfer = mcΔT (joules)  If one mole of the fuel has a mass of M grams, then:  Enthalpy transfer = m x 4.18 x T x M/y  where y is mass loss of fuel.

Given that: Vol of water = 100 cm 3 Temp rise = C Mass of methanol burned = 0.75g Specific heat capacity of water = 4.18 Jg -10 C -1 Calculate the molar enthalpy change of the combustion of methanol. What is the big assumption made with this type of experiment?

States that  If a reaction consists of a number of steps, the overall enthalpy change is equal to the sum of enthalpy of individual steps.  the overall enthalpy change in a reaction is constant, not dependent on the pathway take.

 measured under standard conditions: pressure of 1 atmosphere (1.013 x 10 5 Pa), temperature of 25 0 C (298K) and concentration of 1 moldm -1. e.g. N 2(g) + 3H 2(g)  2NH 3(g) Δ H Ɵ = -92 kJmol -1 The enthalpy change of reaction is -92 kJmol kJ of heat energy are given out when 1 mol of nitrogen reacs with 3 mols of hydrogen to form 2 mols of ammonia.

Calculate the enthalpy change for the formation of sodium chloride solution from solid sodium hydroxide. NaOH(aq) NaOH(s) NaCl(s) + H 2 O(l) + HCl(aq) 1. Indirect path: NaOH(s) + (aq)  NaOH(aq) ΔH Ɵ 1 =-43kJmol NaOH(aq) + HCl (aq)  NaCl(aq) + H 2 O(l) ΔH Ɵ 1 =-57kJmol NaOH(s) + HCl (aq)  NaCl(aq) + H 2 O(l). ΔH2ΔH2 ΔH1ΔH1 Indirect path Direct path + HCl(aq) + H 2 O(l)

Calculate the enthalpy change for the combustion of carbon monoxide to form carbon dioxide. C(s) + O 2 (g)  CO 2 (g) ΔH Ɵ =-394 kJmol -1 2C(s) + O 2 (g)  2CO(g) ΔH Ɵ = -222kJmol -1 2CO(s) + O 2 (g)  2CO 2 (g) 2CO(g) + O 2 (g) 2CO 2 (g) ΔHƟΔHƟ ΔH Ɵ = -(-222)+2(-394) = -566kJmol -1

Example : refer to page 196 evaporation of water & 197 formation of ethanol from ethene

Calculate the enthalpy change for the thermal decomposition of calcium carbonate. CaCO 3 (s)  CaO(s) + CO 2 (g) CaCO 3 (s) +2HCl(aq)  CaCl 2 (aq) + H 2 O(l) ΔH Ɵ 1 =-17 kJmol -1 CaO(s) +2HCl(aq)  CaCl 2 (aq) + H 2 O(l) ΔH Ɵ 1 =-195kJmol -1 CaCO 3 (s) CaO(s) +CO 2 (g) CaCl 2 (aq) + H 2 O(l) +CO 2 (g) ΔHΔH -195kJmol -1 Direct path Indirect path + 2HCl(aq) -17 kJmol HCl(aq)

Calculate the enthalpy of hydration of anhydrous copper(II)sulfate change. CuSO 4 (s) +5H 2 O(l)  CuSO 4. 5H 2 O (s) CuSO 4 (s) +5H 2 O(l) CuSO 4. 5H 2 O (s) Cu 2+ (aq) + SO 4 2- (aq) ΔHΔH ΔH2ΔH2 ΔH1ΔH1 Direct pathway Indirect pathway

From the following data at 25 0 C and 1 atmosphere pressure: Eqn 1: 2CO 2 (g)  2CO(g) + O 2 (g) ΔH Ɵ =566 kJmol -1 Eqn 2: 3CO(g) + O 3 (g)  3CO 2 (g) ΔH Ɵ =-992 kJmol -1 Calculate the enthalpy change calculated for the conversion of oxygen to 1 mole of ozone,i.e. for the reaction O 2 (g)  O 3 (g)

Calculate the enthalpy change for the conversion of graphite to diamond under standard thermodynamic conditions. C (s,graphite) + O 2 (g)  CO 2 (g) ΔH Ɵ =-393 kJmol -1 C (s, diamond) + O 2 (g)  CO 2 (g) ΔH Ɵ =-395 kJmol -1

Practice questions page 199 #7-9

 Enthalpy changes can also be calculated directly from bond enthalpies.  The bond enthalpy is the amount of energy required to break one mole of a specified covalent bond in the gaseous state.  For diatomic molecule the bond enthalpy is defined as the enthalpy change for the process X-Y(g) X(g) + Y(g) [gaseous state]

 Bond enthalpy can only be calculated for substances in the gaseous state. Br 2 (l)  2Br(g) ΔH Ɵ = 224 kJmol -1 Br 2 (l) 2Br(g) Br 2 (g) Br-Br bond enthalpy ΔH Ɵ vap enthalpy change of vaporisation 2 x ΔH Ɵ at atomisation Energy must be supplied to break the van der Waals’ forces between the Bromine molecules and to break the Br-Br bonds. Endothermic process

 Ave bond enthalpies are enthalpies calculated from a range of compounds,eg C-H bond enthalpy is based on the ave bond energies in CH 4, alkanes and other hydrocarbons.

Bond Ave bond enthalpy, ΔH Ɵ (Kjmol -1 ) Bond length (nm) H-H C-C C-H O-H N-H N-N C=C O=O C Ξ C NΞNNΞN 944 Refer to page 201

When a hydrocarbon e.g. methane (CH 4 ) burns, CH 4 + O 2  CO 2 + H 2 O What happens?

C H H H H + O O O O C H H H H O O O O C H H H H O O OO ENERGY Enthalpy Level (KJ) Progress of Reaction 2 O=O Bond Breaking Bond Forming 4 C-H 4 H-O2 C=O CH 4 + 2O 2  CO 2 + 2H 2 O

C H H H H + O O O O C H H H H O O OO Why is this an exothermic reaction (produces heat)?

Bond Ave Bond Enthalpy (kJ/mol) C-H412 H-O463 O=O496 C=O743 CH 4 + 2O 2  CO 2 + 2H 2 O Energy absorbed when bonds are broken = (4 x C-H + 2 x O=O) Energy given out when bonds are formed = ( 2 x C=O + 4 x H-O) = 4 x x 496 = 2640 kJ/mol C H H H H + O O O O C H H H H O O OO Break  Form = 2 x x 464 = 3338 kJ/mol

Energy absorbed when bonds are broken (a) = 2640 kJ/mol Energy released when bonds are formed (b) = 3338 kJ/mol Enthalpy change, ΔH = ∑(bonds broken) - ∑(bonds made) = a + (-b) = 2640 – 3338 = -698 kJ/mol Why is this an exothermic reaction (produces heat)?

What can be said about the hydrogenation reaction of ethene? H H C=C (g) + H-H (g)  H-C-C-H (g) H H HH HH

What can be said about the combustion of hydrazine in oxygen? H H N-N (g) + O=O (g)  N ΞN (g) + 2 O (g) H H HH

Example Calculate the mean Cl-F bond enthalpy given that Cl 2(g) + 3F 2(g)  2ClF 3(g) ΔH Ɵ = -164 kJmol -1 Bond enthalpy for Cl-Cl = 242 kJmol -1 and F-F = 158 kJmol -1

Standard enthalpy change of atomisation is the enthalpy change when 1 mole of gaseous atoms is formed from the element under standard conditions. Example C(s)  C(g) Calculate the enthalpy change for the process 3 C (s) + 4H 2(g)  C 3 H 8(g) ΔH Ɵ = -164 kJmol -1 Bond enthalpy for C-H = 412 kJmol -1, H-H = 436 kJmol -1 and C-C = 348 kJmol -1 (ΔH Ɵ at ) ΔH Ɵ at = 715 kJmol -1 Practice questions page 206 #10,12,13

The combustion of both C and CO to form CO 2 can be measured easily but the combustion of C to CO cannot. This can be represented by the energy cycle. ΔH x = -393 – (-283) = kJmol -1 ΔHxΔHx ½O 2 (g) -393kJmol -1 C(s)+ ½O 2 (g) CO(g) CO 2 (g) -283kJmol -1 ½O 2 (g)

Calculate the standard enthalpy change when one mole of methane is formed from its elements in their standard states. The standard enthalpies of combustion of carbon, hydrogen and methane are -393, -286 and -890 kJmol -1 respectively.

Dissolving sugar. Sugar molecules are dispersed throughout the solution and are moving around. More disordered or random. Other examples: melting ice H 2 O(s)  H 2 O(l) evaporating water H 2 O(l)  H 2 O(g)

Increasing entropy Entropy (S) : amount of disorder Unit : JK -1 mol -1 S Ɵ : standard entropy Δ S Ɵ : entropy change If Δ S Ɵ > 0 => increase in entropy => increase in disorder E.g. H 2 O(l )  H 2 O(g) Δ S Ɵ =+119 JK -1 mol -1 If Δ S Ɵ decrease in entropy => decrease in disorder E.g. N H 3 (g ) + HCl(g)  NH 4 Cl(s) Δ S Ɵ = JK -1 mol -1

(1) State of matter ◦ Gas (most) particle motion is more random in a gas ◦ Liquid (middle) particle motion is less random than a gas but more than a solid ◦ Solid (least) particle motion is restricted. Possible positions for molecules are restricted ◦ Examples  (changing state) H 2 O(l)  H 2 O(g)  (changing state) H 2 O(s)  H 2 O(l) ΔS(gas) > ΔS(liquid) > ΔS(solid)

(2) Temperature ◦ Comparing two gasses, one at 20  C and one at 80  C ◦ Molecules in the 80  C gas have more kinetic energy, they are moving more and colliding more

(3) The number of molecules ◦ More molecules means more possible positions relative to the other molecules  (more moles and change of state) Li 2 CO 3 (s)  Li 2 O(s) + CO 2 (g)  (more moles) MgSO 4  8H 2 O  Mg 2+ (aq) + SO 4 2- (aq) + 8H 2 O(l) (4) More complex molecules have higher entropy values

ReactionEntropy (increase/ decrease) ΔS Ɵ ( + / - ) Explanation N 2 (g) + 3H 2 (g)  2NH 3 (g)4 moles of gas to 2 moles of gas CaCO 3 (s)  CaO(s) + CO 2 (g)1 mole of solid to 1 mole of solid + 1 mole of gas CH 4 (g) + 2O 2 (g)  CO 2 (g) +2H 2 O(l)3 moles of gas to 1 mole of gas C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)2 moles of gas to 1 mole of gas Is there an increase or decrease in disorder of the system? Is there an increase or decrease in the no. of moles of gas? For reaction where the no. of moles of gas is the same on both sides, the ΔS = 0. E.g. F 2 (g) + Cl 2 (g)  2ClF(g) Practice Qn 24from pg 230 (textbook

Which of the following reactions has the largest ΔS value?  CO 2 (g) + 3H 2 (g)  CH 3 OH(g) + H 2 O(g)  2Al(s) + 3S(s)  Al 2 S 3 (s)  CH 4 (g) + H 2 O(l)  3H 2 (g) + CO(g)  2S(s) + 3O 2 (g)  2SO 3 (g)

Entropy change = total entropy of products – total entropy of reactants ΔS Ɵ = ∑ ΔS Ɵ products - ∑ ΔS Ɵ reactants E.g. Calculate the standard entropy change for the reaction CH 4 (g) + 2O 2(g)  CO 2(g) + 2H 2 O (l) Use standard entropy from pg 230 (textbook)

 Spontaneous reaction – one that occurs without any outside influence (no input of energy)  A spontaneous reaction does not have to happen quickly.  E.g. 4Na(s) + O 2(g)  2Na 2 O (s) can happen by itself.

 When heat is released in a chemical reaction, the surrounding is hotter and particles move around more => entropy increases. The entropy change,ΔS is related to the enthalpy change, ΔH of the system. ΔG= ΔH – TΔS ΔG : free energy change ΔG Ɵ : standard free energy change For a reaction to be spontaneous, ΔG < 0

 Temperature, T should be in Kelvin, K 0°C = 273K (273.15K)  Check the units of ΔS, entropy is often given in JK –1 mol –1 but must be converted to kJK –1 mol –1 ΔG ө = ΔH ө – TΔS ө  ө = standard conditions, 25°C (298K) and 1 atm (101.3 kPa)

Given that the changes in enthalpy and entropy are kJ and 277 J/K respectively for the reaction given below, calculate the change in Gibbs energy. Then state whether the reaction is spontaneous at 25  C C 6 H 12 O 6 (aq)  2C 2 H 5 OH(aq) + 2CO 2 (g)

ΔG Ɵ = ∑ ΔG f Ɵ products - ∑ ΔG f Ɵ reactants standard free energy of formation : free energy change for the formation of 1 mole of substance from its elements in their standard states & under standard conditions. Calculate ΔG Ɵ for the reaction CaCO 3 (s)  CaO(s) + CO 2 (g) given that Substance ΔG f Ɵ /kJmol -1 CaCO CaO- 604 CO Practice Qn from pg 235 (textbook

For a spontaneous reaction ΔG is negative (–) For a non–spontaneous reaction ΔG is positive (+) ΔH : Enthalpy Change ΔS : Entropy Change