Energetics

Exothermic and Endothermic reactions 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 reactions Endothermic reactions

Heat and Temperature 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).

It does not depend on the amount of the substance. 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?

Energy Changes in Chemical Reactions All chemical reactions are accompanied by some form of energy change Exothermic Energy is given out Endothermic Energy is absorbed Activity : observing exothermic and endothermic reactions

Enthalpy (H)and Enthapy change(ΔH) 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 Level Diagram -Exothermic Change 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 Enthalpy

Examples of Exothermic Reactions 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)

Enthalpy Level Diagram -Endothermic Change 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

Examples of Endothermic Reactions 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)

Specific heat capacity Amount of heat required to raise the temperature of a unit mass of a substance by 1 degree or 1 kelvin. Uint : Jg-1 0C-1 The specific heat capacity of alminium is 0.90 Jg-1 0C-1 . If 0.90J of energy is put into 1g of aluminium, the temperature will be raised by 10C. 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

Calorimetry 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 = mcΔT

Example How much heat is required to increase the temperature of 20 grams of nickel (specific heat capacity 440Jkg-1 0C-1) from 500C to 700C?

Enthalpy change of combustion reactions 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, CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ΔHƟc=-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

Problems with calorimetry 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

Example 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 0C-1 Cal no of moles of methanol = 1/32 = 0.0312 Q = 100 x 4.18 x 42 Enthalpy change, delta H = (100 x 4.18 x 42)/ 0.0312 -563kJmol-1 Practice questions page 187 #1-4

Enthalpy change in solutions Enthalpy change of neutralisation (ΔHn) 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) + H2O(l) ΔHƟ=-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)  H2O(l) ΔHƟ=-57 kJmol-1 For sulfuric acid, the enthalpy of neutralisation equation is ½ H2SO4(aq) + KOH(aq)  ½K2SO4(aq) + H2O(l) ΔHƟ=-57 kJmol-1 Example : refer to page 188

Enthalpy change in solutions Enthalpy change of neutralisation (ΔHn) 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) + H2O(l) ΔHƟ=-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 (ΔHsol) The enthalpy change when 1 mol of solute is dissolved in excess solvent to form a solution of ‘infnite dilution’ under standard conditions. NH4 NO3(s) in excess water  NH4 + (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) CH3COOH(aq) + NaOH(aq)  CH3COONa(aq) + H2O(l) ΔHƟ=-55.2 kJmol-1 Some of the energy released is used to ionise the acid.

Example 200.0cm3 of 0.150 M HCl is mixed with 100.0cm3 of 0.350 M NaOH. The temperature rose by 1.360C. 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 0C-1. Total vol = 300 Density of water = 1g/cm3 Enthalpy = 300 x 4.18 x 1.36 = 1705.44J No. of moles of HCl = 0.2 x 0.15M = 0.03 No. of moles of NaOH = .100 x 0.35M = 0.035 Therefore, no. of moles of water formed= 0.03 Heat released = 1705.44/0.03 = 56.848 kJmol-1 Enthalpy change of neutralisation = - 56.8 kJmol-1 (exothermic ) -56.8kJmol-1

Possible errors The experimental change of neutralisation is -56.8 kJmol-1 The accepted literature value is -57.2 kJmol-1 (1) Heat loss to the environment. Assumptions that the denisty of NaOH and HCl solutions are the same as water. the specific heat capacity of the mixture are the same as that of water

Example When 3 g of sodium carbonate are added to 50 cm3 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 0C-1. Vol of water = 50 cm3 Q = 50 x 4.18 x (28.5 – 22) No. of moles of HCl = 0.05 No. of moles of Na2CO3 = 3/106 = 0.0283 No. of moles of water = 0.0283 Heat of change = 1358.5/0.0283= 48kJmol-1 Example : refer to page 189 dissolving ammonium chloride

Possible errors (page 189) The experimental change of solution is +13.8 kJmol-1 The accepted literature value is 15.2 kJmol-1 (1) Absorption of heat from the environment. Assumptions that the specific heat capacity of the solution is the same as that of water The mass of ammonium chloride is not taken into consideration when working out the heat energy released.

Example 100.0 cm3 of 0.100 mol dm-3 copper II sulphate solution is placed in a styrofoam cup. 1.30 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. Actual Highest temp reached could be 26 deg Extrapolate to get the temp reached if no heat is lost to surroundings Delta T = 28 – 17 The extrapolation is necessary to compensate for heat loss while the reaction is occurring. Why would powdered zinc be used?

100.0 cm3 of 0.100 mol dm-3 copper II sulphate solution is placed in a styrofoam cup. 1.30 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. Determine the limiting reactant Calculate Q Calculate the enthalpy for the reaction.

Enthalpy changes of combustion of fuels The following measurements are taken: Mass of cold water (g) Temperature rise of the water (0C) 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 10C. This is called the specific heat capacity of water, c, and has a value of 4.18Jg-1K-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.

Example Given that: Vol of water = 100 cm3 Temp rise = 34.50C Mass of methanol burned = 0.75g Specific heat capacity of water = 4.18 Jg-10C-1 Calculate the molar enthalpy change of the combustion of methanol. What is the big assumption made with this type of experiment?

Hess’s Law 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.

Standard enthalpy changes, ΔHƟ measured under standard conditions: pressure of 1 atmosphere (1.013 x 105 Pa), temperature of 250C (298K) and concentration of 1 moldm-1. e.g. N2(g) + 3H2(g)  2NH3(g) ΔHƟ = -92 kJmol-1 The enthalpy change of reaction is -92 kJmol-1 92 kJ of heat energy are given out when 1 mol of nitrogen reacs with 3 mols of hydrogen to form 2 mols of ammonia.

Reaction in aqueous soln Calculate the enthalpy change for the formation of sodium chloride solution from solid sodium hydroxide. NaOH(aq) NaOH(s) NaCl(s) + H2O(l) + HCl(aq) 1. Indirect path: NaOH(s) + (aq)  NaOH(aq) ΔHƟ1=-43kJmol-1 2. NaOH(aq) + HCl (aq)  NaCl(aq) + H2O(l) ΔHƟ1=-57kJmol-1 3. NaOH(s) + HCl (aq)  NaCl(aq) + H2O(l). Indirect path + HCl(aq) + H2O(l) ΔH2 ΔH1 Direct path -100kJmol-1

Combustion reaction (using cycles) Calculate the enthalpy change for the combustion of carbon monoxide to form carbon dioxide. C(s) + O2(g)  CO2(g) ΔHƟ =-394 kJmol-1 2C(s) + O2(g)  2CO(g) ΔHƟ = -222kJmol-1 2CO(s) + O2(g)  2CO2(g) 2CO(g) + O2(g) 2CO2(g) ΔHƟ -111kJmol-1 The enthalpy for the reaction C(s) + 1/2O2(g) -> CO(g) cannot be found directly by experiment because CO2 is always formed when carbon reacts with only a limited amount of oxygen – unavoidable. However, the enthalpy changes of combustion of C and CO can be found experimentally. The reactions and their enthalpy changes can be linked using Hess law ΔHƟ = -(-222)+2(-394) = -566kJmol-1

Combustion reaction (manipulating equations) Example : refer to page 196 evaporation of water & 197 formation of ethanol from ethene

*Example : Decomposition reaction Calculate the enthalpy change for the thermal decomposition of calcium carbonate. CaCO3(s)  CaO(s) + CO2(g) CaCO3(s) +2HCl(aq)  CaCl2(aq) + H2O(l) ΔHƟ1=-17 kJmol-1 CaO(s) +2HCl(aq)  CaCl2(aq) + H2O(l) ΔHƟ1=-195kJmol-1 CaCO3(s) CaO(s) +CO2(g) CaCl2(aq) + H2O(l) +CO2(g) ΔH Direct path + 2HCl(aq) + 2HCl(aq) +178kJmol-1 The reaction is slow and a temp is required to bring it to completion. Direct measurement is not practical. 2 reactions that take place at rm temp are carried out nd their enthalpy change used to find the enthalpy of decomposition of calcium carbonate. -17 kJmol-1 -195kJmol-1 Indirect path

*Example : Enthalpy of hydration of an anhydrous salt. Calculate the enthalpy of hydration of anhydrous copper(II)sulfate change. CuSO4(s) +5H2O(l)  CuSO4.5H2O (s) CuSO4(s) +5H2O(l) CuSO4.5H2O (s) Cu2+(aq) + SO42- (aq) ΔH Direct pathway ΔH1 ΔH2 It cannot be found directly because when 5 moles of water is added, hydrated copper (II) sulphate is not produced in a controlled way. Can only be produced by crystallisation from a solution. The enthalpy change can be determined indirectly by finding the enthalpy of solution of both the anhydrous and the hydrated copper(II) sulfates. Indirect pathway

*Example From the following data at 250C and 1 atmosphere pressure: Eqn 1: 2CO2(g) 2CO(g) + O2(g) ΔHƟ=566 kJmol-1 Eqn 2: 3CO(g) + O3(g) 3CO2(g) ΔHƟ=-992 kJmol-1 Calculate the enthalpy change calculated for the conversion of oxygen to 1 mole of ozone,i.e. for the reaction O2(g) O3 (g) +143kJmol-1

*Example Calculate the enthalpy change for the conversion of graphite to diamond under standard thermodynamic conditions. C (s,graphite) + O2(g) CO2 (g) ΔHƟ=-393 kJmol-1 C (s, diamond) + O2(g) CO2(g) ΔHƟ=-395 kJmol-1 +2kJmol-1

Practice questions page 199 #7-9

Bond enthalpies (Bond energies) 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 Enthalpies Bond enthalpy can only be calculated for substances in the gaseous state. Br2(l)  2Br(g) ΔHƟ= 224 kJmol-1 atomisation 2 x ΔH Ɵat Br2(l) 2Br(g) Br2(g) ΔH Ɵvap enthalpy change of vaporisation Br-Br bond enthalpy Energy must be supplied to break the van der Waals’ forces between the Bromine molecules and to break the Br-Br bonds. Endothermic process

Average bond enthalpies Ave bond enthalpies are enthalpies calculated from a range of compounds,eg C-H bond enthalpy is based on the ave bond energies in CH4 , alkanes and other hydrocarbons.

Some average bond enthalpies Ave bond enthalpy, ΔHƟ (Kjmol-1) Bond length (nm) H-H 436 0.07 C-C 348 0.15 C-H 412 0.11 O-H 463 0.10 N-H 388 N-N 163 C=C 612 0.13 O=O 496 0.12 C Ξ C 837 NΞN 944 Refer to page 201

Bond breaking and Forming When a hydrocarbon e.g. methane (CH4) burns, CH4 + O2  CO2 + H2O What happens?

Energy Level Diagram CH4 + 2O2  CO2 + 2H2O Enthalpy Level (KJ) Bond Breaking O O C H ENERGY O H ENERGY O H C H + O H Bond Forming C H O 4 C-H 2 O=O 4 H-O 2 C=O CH4 + 2O2  CO2 + 2H2O Progress of Reaction Energy Level Diagram

Bond breaking and Forming CH4 + 2O2  CO2 + 2H2O C H + O C H O Why is this an exothermic reaction (produces heat)?

CH4 + 2O2  CO2 + 2H2O Break  Form C-H 412 H-O 463 O=O 496 C=O 743 Bond Ave Bond Enthalpy (kJ/mol) C-H 412 H-O 463 O=O 496 C=O 743 C H + O C H 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 412 + 2 x 496 = 2640 kJ/mol = 2 x 803 + 4 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 The forming of the bonds (2 x C=O and 4 x H-O) between the atoms in water gives out more heat than is required to break the bonds (4 x C-H and 2 x O=O) . The bonds in the products (2 x C=O and 4 x H-O) are stronger than those in the reactants (4 x C-H and 2 x O=O) since they require more energy to break them. Why is this an exothermic reaction (produces heat)?

Example 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 H H H H

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

The bond enthalpy from an enthalpy change of reaction Example Calculate the mean Cl-F bond enthalpy given that Cl2(g) + 3F2(g)  2ClF3(g) ΔHƟ= -164 kJmol-1 Bond enthalpy for Cl-Cl = 242 kJmol-1 and F-F = 158 kJmol-1

Using bond enthalpies & enthalpies of atomisation 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) + 4H2(g)  C3H8(g) ΔHƟ= -164 kJmol-1 Bond enthalpy for C-H = 412 kJmol-1 , H-H = 436 kJmol-1 and C-C = 348 kJmol-1 (ΔH Ɵat ) ΔH Ɵat = 715 kJmol-1 -103kJmol-1 Practice questions page 206 #10,12,13

Hess’s Law - example The combustion of both C and CO to form CO2 can be measured easily but the combustion of C to CO cannot. This can be represented by the energy cycle. ΔHx = -393 – (-283) = - 110 kJmol-1 ΔHx C(s)+ ½O2(g) CO(g) CO2(g) ½O2(g) -393kJmol-1 ½O2(g) -283kJmol-1

Hess’s Law - example 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. -75kJmol-1

Entropy : degree of disorder Dissolving sugar . Sugar molecules are dispersed throughout the solution and are moving around. More disordered or random. Other examples: melting ice H2O(s)  H2O(l) evaporating water H2O(l)  H2O(g)

The sign of entropy, S Increasing entropy Entropy (S) : amount of disorder Unit : JK-1mol-1 SƟ : standard entropy Δ SƟ : entropy change If Δ SƟ > 0 => increase in entropy => increase in disorder E.g. H2O(l )  H2O(g) Δ SƟ =+119JK-1mol-1 If Δ SƟ < 0 => decrease in entropy => decrease in disorder E.g. NH3(g ) + HCl(g)  NH4Cl(s) Δ SƟ = - 285JK-1mol-1

state of matter temperature number of molecules What are the factors that affect ENTROPY? state of matter temperature number of molecules

Factors affecting entropy (1) State of matter Gas particle motion is more random in a gas Liquid particle motion is less random than in a liquid than a gas but more than a solid Solid particle motion is restricted. Examples (changing state) H2O(l)  H2O(g) (changing state) H2O(s)  H2O(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

(4) More complex molecules have higher entropy values (3) The number of molecules More molecules means more possible positions relative to the other molecules (more moles and change of state) Li2CO3(s)  Li2O(s) + CO2(g) (more moles) MgSO48H2O  Mg2+(aq) + SO42-(aq) + 8H2O(l) (4) More complex molecules have higher entropy values

Predict the sign of ΔSƟ Is there an increase or decrease in disorder of the system? Is there an increase or decrease in the no. of moles of gas? Reaction Entropy (increase/ decrease) ΔSƟ ( + / - ) Explanation N2(g) + 3H2(g)  2NH3(g) 4 moles of gas to 2 moles of gas CaCO3(s)  CaO(s) + CO2(g) 1 mole of solid to 1 mole of solid + 1 mole of gas CH4(g) + 2O2(g)  CO2(g) +2H2O(l) 3 moles of gas to 1 mole of gas C2H4(g) + H2(g)  C2H6(g) 2 moles of gas to 1 mole of gas decrease - increase + decrease - decrease - For reaction where the no. of moles of gas is the same on both sides, the ΔS = 0. E.g. F2(g) + Cl2(g)  2ClF(g) Practice Qn 24from pg 230 (textbook

Example Which of the following reactions has the largest ΔS value?   CO2(g) + 3H2(g)  CH3OH(g) + H2O(g) 2Al(s) + 3S(s)  Al2S3(s) CH4(g) + H2O(l)  3H2(g) + CO(g) 2S(s) + 3O2(g)  2SO3(g) 3rd

Calculate ΔSƟ Entropy change = total entropy of products – total entropy of reactants ΔSƟ =∑ ΔSƟproducts - ∑ ΔSƟreactants E.g. Calculate the standard entropy change for the reaction CH4 (g) + 2O2(g)  CO2(g) + 2H2O(l) Use standard entropy from pg 230 (textbook)

Spontaneity 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) + O2(g)  2Na2O(s) can happen by itself.

Gibbs free energy, G 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 ΔGƟ : 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–1mol–1 but must be converted to kJK–1mol–1 ΔGө = ΔHө – TΔSө ө = standard conditions, 25°C (298K) and 1 atm (101.3 kPa)

Calculate ΔGƟ using ΔGƟ = ΔHƟ – TΔSƟ Given that the changes in enthalpy and entropy are -139 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 C6H12O6(aq)  2C2H5OH(aq) + 2CO2(g)

Calculate ΔGƟ using standard free energy of formation ΔGƟ =∑ ΔGfƟproducts - ∑ ΔGfƟreactants 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 ΔGƟ for the reaction CaCO3(s)  CaO(s) + CO2(g) given that Substance ΔGfƟ /kJmol -1 CaCO3 - 1129 CaO - 604 CO2 - 395 Practice Qn from pg 235 (textbook

Spontaneity of reactions – Gibbs Free Energy, ΔG ΔH : Enthalpy Change ΔS : Entropy Change For a spontaneous reaction ΔG is negative (–) For a non–spontaneous reaction ΔG is positive (+)