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Energetics

Energetics. Energetics – ‘Chemical changes caused by energy’. Chemical Reactions. Chemical reactions can be classified as either exothermic or endothermic. Exothermic – chemical reactions that result in the release of heat to the surroundings

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Energetics

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  1. Energetics Energetics – ‘Chemical changes caused by energy’

  2. Chemical Reactions • Chemical reactions can be classified as either exothermic or endothermic. • Exothermic – chemical reactions that result in the release of heat to the surroundings • Endothermic – chemical reactions where heat is absorbed from the surroundings Potassium chloride + Water Sodium Hydroxide + Water Classify as exo or endothermic:

  3. Enthalpy ΔH • The heat absorbed/released by a reaction is described using the term enthalpy. • The symbol for a change in enthalpy is ΔH • The enthalpy of a system (H) cannot be measured, but the enthalpy change (ΔH ) from an initial (reactants) to final (products) state can be measured Delta = Change H = Heat

  4. Enthalpy Diagrams CH4 + 2O2 CO2 + 2H2O ΔH is negative for exothermic reactions (products – reactants) CH4 + 2O2 Enthalpy No scale as we cannot measure H, we can only measure the change Reactants ΔH = -890kJmol-1 CO2 + 2H2O Products Reaction Progress

  5. Enthalpy Diagrams ΔH is positive for endothermic reactions (products – reactants) N2 + O2 2NO 2NO Enthalpy Products ΔH = +181 kJmol-1 N2 + O2 Reactants Reaction Progress Lower enthalpy = more stable Are the products or reactants more stable?

  6. Activation Energy • CH4 + 2O2 CO2 + 2H2O • The minimum energy colliding particles must have before collision results in a chemical reaction is called the activation energy Methane gas is unstable, a small amount of energy (spark) can cause combustion to occur

  7. Activation Energy CH4 + 2O2 CO2 + 2H2O CH4 + 2O2 Enthalpy Reactants In general, the higher the activation energy the slower the reaction ΔH CO2 + 2H2O Products Reaction Progress

  8. Activation Energy N2 + O2 2NO 2NO Products Enthalpy ΔH N2 + O2 Reactants Reaction Progress

  9. Measuring Enthalpy Changes • When experimentally measuring the ΔH of an exothermic reaction it is important to consider specific heat capacity. This substance has a specific heat capacity of 0.86 J g-1 °c-1 Substance (1g) The amount of energy required to raise the temperature of a substance by 1°C What will the temperature increase be? 2.16 J

  10. T • q = heat energy (J) • m = mass (g) • c = specific heat capacity (J g-1 °c-1) • ΔT = change in temperature (°C) 125J of heat energy is used to increase the temperature of 111g of iron by 2.5°C. What is the specific heat capacity of iron? C = q m x ΔT 125 111 x 2.5 C = 0.45 J g-1 °c-1

  11. Measuring ΔH of combustion reactions Calculate ΔH when 1 mole of ethanol is burnt Mass of water = 150g Initial temperature of water = 19.5°C Maximum temperature of water = 45.7°C Initial mass of ethanol = 121.67g Final mass of ethanol = 120.62g Specific heat capacity of water = 4.18 J g-1 °c-1 Molar mass of ethanol = 46.08 g mol-1 T q = 150 x 4.18 x (45.7-19.5) = 16400J Mass of ethanol burnt = 121.67 – 120.62 = 1.05g Number of moles = mass / molar mass 1.05 / 46.08 = 0.0228 mol ΔH for 1 mole = 16400 / 0.0228 = 721 000 J mol-1 = 721kJ mol-1 This is an exothermic reaction: ΔH= -721kJ mol-1

  12. Enthalpy Changes in Solution • Method: known amounts of reagents with known temperatures are mixed together in a polystyrene cup. The maximum/minimum temperature is recorded. • Water is the most common solvent, therefore the specific heat capacity of the solution is assumed to be the same as water. Solute + solvent = solution

  13. ΔHsol = enthalpy change of solution when 1 mole of substance dissolves. Can be endothermic or exothermic. • ΔHn = enthalpy change of neutralisation when 1 mole of water is produced (acid + base) is always exothermic. Experiments 1)100ml HCl (2M) + 150ml NaOH (1.5M) 2) 11.4g KCl in 200ml H2O (record volume)

  14. Calculate ΔH for the experiment: • Remember to factor in 1 mole • Look at the balanced equation for the ratios. Example: KOH + HCl KCl + H2O You can use the examples on page 188-190 to help you. Complete the questions on page 191. 0.100mol 0.120mol Coefficient = 1 for reactants and products. Therefore, there is more HCl than can react with KOH. 0.100mol of KCl and H2O will be produced as it’s a 1:1 ratio Limiting reactant 0.02mol excess

  15. ΔHsol • 11.4g KCl in 200ml H2O • q = mcΔT • Calculate ΔT based on your experimental data • q = 200 x 4.18 x ΔT • ΔHsolcalculations require an answer for 1 mole of a solute being dissolved. • Number of moles of KCl = 11.4 / 74.55 = 0.153 moles • ΔHsol= q / 0.153 • Convert answer to kJ mol-1 (endothermic or exothermic)

  16. ΔHn • 100ml HCl (2M) + 150ml NaOH (1.5M) • q = mcΔT • Calculate ΔT based on your experimental data • When calculating ΔHnwe assume that the solution has the properties of water. • Specific heat capacity = 4.18 J g-1 °c-1 • 1ml = 1g • ΔHncalculations are for when 1 mole of H2O is produced Therefore, ΔHn= (1/0.2) x q kJ mol-1 (negative) 0.1 x 2.0 = 0.200 moles 0.15 x 1.5 = 0.225 moles 0.200 moles 0.200 moles

  17. Hess’s Law • Hess’s law is used when it is not possible to experimentally measure ΔH • Hess’s law states that ΔH is independent of the pathway between the initial and final states. ΔH2 + ΔH3 = ΔH1

  18. Enthalpy changes are always stated under standard conditions (1 atmospheric pressure/25°C) • ΔHꝋ = standard enthalpy change • ΔHꝋr = standard enthalpy change of a reaction, using the molar amounts as shown in the stoiciometric equation. • Textbooks commonly use ΔH as standard conditions are implied and it is always calculated based on a reaction

  19. Manipulating Equations Calculate ΔHfor the following reaction 2CO(g) + O2(g) 2CO2(g) 2C(s) + O2(g) 2CO(g) ΔH= -222kJ mol-1 C(s) + O2(g) CO2(g)ΔH= -394kJ mol-1 2CO(g) 2C(s) + O2(g)ΔH= +222kJ mol-1 2C(s) + 2O2(g) 2CO2(g) ΔH= -788kJ mol-1 2CO(g) + O2(g) 2CO2(g) ΔH= -566kJ mol-1

  20. Calculate ΔHfor the following reaction: H2O(l) H2O(g) 2H2(g) + O2(g) 2H2O(l) ΔH= -572kJ mol-1 2H2(g) + O2(g) 2H2O(g)ΔH= -484kJ mol-1 H2O(l) H2(g) + H2(g) + H2O(g) H2O(l) H2O(g)

  21. Calculate ΔH for the following reaction: C2H4(g) + H2O(g) C2H5OH(l) 1)H2O(l) H2O(g) 2) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) 3)C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l) ΔH1= +44kJ mol-1 ΔH2= -1371kJ mol-1 ΔH3= -1409kJ mol-1

  22. Given the following equations and Ho values, determine the heat of reaction (kJ) at 298 K for the reaction: 2 SO2(g)+ 2 P(s)+ 5 Cl2(g) 2 SOCl2(l)+ 2 POCl3(l) SOCl2(l)+ H2O(l) SO2(g)+ 2 HCl(g) = 10.3 kJ mol-1 PCl3(l)+ 1/2 O2(g) POCl3(l) = -325.7 kJ mol-1 P(s)+ 3/2 Cl2(g) PCl3(l) = -306.7 kJ mol-1 4 HCl(g)+ O2(g) 2 Cl2(g)+ 2 H2O(l) = -202.6 kJ mol-1

  23. Bond Enthalpies • Bond enthalpy is the enthalpy change when 1 mole of covalent bonds, in a gaseous molecule, are broken under standard conditions. • Allows us to calculate ∆H using the energy stored between bonds. • Breaking bonds requires energy : endothermic +∆H • Making bonds releases energy : exothermic -∆H

  24. Average bond enthalpy Average amount of energy needed to break a particular bond in a range of molecules

  25. Calculating ∆H from bond enthalpies C2H4(g) + Br2(g) C2H4Br2(g) • Drawing the structural formula helps to clearly see all the bonds. • For calculation purposes we imagine that all the bonds in the reactants are broken; giving us the total energy from broken bonds. • We can then calculate the total energy for bonds made. ∆H = Σ(bonds broken) – Σ(bonds made) All values are put in a positive for this method

  26. Calculating a bond enthalpy from ΔHr • Calculate the Br-F bond enthalpy. Br2(g) + 3F2(g) 2BrF3(g) ΔHꝋ = -545 kJ mol-1 ∆H = Σ(bonds broken) – Σ(bonds made) -545 = [193 + (3 x 158)] – [6 x Br-F] 6 x Br-F = 193 + (3 x 158) + 545 6 x Br-F = 1212 Br-F = 1212/6 = 202 kJ mol-1

  27. Enthalpies of atomisation • In order to use bond enthalpies substances must be in a gaseous state. • If a substance is not in the gaseous state we must use its enthalpy change of atomisation value. • ΔHꝋat– the enthalpy change when 1 mole of gaseous atoms is formed from the element under standard conditions (always endothermic as energy is needed to break bonds) Also known as sublimation 3C(s) + 4H2(g) C3H8(g) C(s) C(g) ΔHꝋat= 715kJmol-1 ((3x715) + (4x436)) – ((2x348) + (8x412)) = -103 kJ mol-1

  28. Enthalpy Change of Formation ΔHꝋf • The enthalpy change when 1 mole of the substance is formed from its elements in their standard states under standard conditions. C(s) + 2H2(g) CH4(g) • Can be endothermic or exothermic

  29. Review of ΔH ΔHc 1 mole CH4(g) + 2O2(g) CO2(g) + 2H2O(g) KCl(s) + H2O(l) HCl(aq) + KOH(aq) HCl(aq) + NaOH(aq)NaCl(aq) + H2O(l) C2H4(g) + H2O(l) C2H5OH(l) Na(s) Na(g) ΔHsol 1 mole ΔHn 1 mole ΔHf 1 mole ΔHat 1 mole

  30. Enthalpy changes for ionic compounds The enthalpy change when one electron is removed from each atom in 1 mole of gaseous atoms under standard conditions • First ionisation energy M(g) M+(g) + e- • Second ionisation energy M+(g) M2+(g) + e- • First electron affinity X(g) + e- X-(g) • Second electron affinity X(g)-+ e- X2-(g) • Lattice enthalpy (ΔHꝋlatt)) The enthalpy change when one electron is added to each atom in 1 mole of gaseous atoms under standard conditions The enthalpy change when 1 mole of an ionic compound is broken apart into its constituent gaseous ions under standard conditions NaCl(s) Na+(g) + Cl-(g)

  31. All ionisation energies are endothermic • The first electron affinity is exothermic for virtually all elements • The second electron affinity is always endothermic due to repulsion between the negative charges • Lattice enthalpy is an endothermic process; energy must be supplied to separate the ions against the attractive forces holding them in the lattice

  32. Born-Haber Cycles • The Born-Haber cycle is an enthalpy level diagram breaking down the formation of an ionic compound using simpler steps. • The Born-Haber cycle can be used to find an unknown enthalpy value by applying Hess’s law Example: Calculate ΔHꝋlattfor Na+(g) + Cl-(g)

  33. Na+(g) + e- + Cl(g) Bond Dissociation(Cl2) 1stElectron Affinity (Cl) enthalpy Indirect route Na+(g) + Cl-(g) Na+(g) + e- + ½Cl2(g) 1st Ionisation(Na) Na(g) + ½Cl2(g) ∆Hlatt ∆Hat (Na) Direct Route Na(s) + ½Cl2(g) ∆Hf NaCl(s)

  34. Draw a Born-Haber cycle for magnesium oxide and use it to calculate the 2nd electron affinity of O

  35. Mg2+(g) + 2e- + O(g) Mg2+(g) + O2-(g) enthalpy Bond Dissociation(O2) 1stElectron Affinity (O) 2nd Electron Affinity (O) Mg2+(g) + 2e- + ½O2(g) Mg2+(g) + O-(g) 2nd ionisation (Mg) Mg+(g) + e-+ ½O2(g) 1st ionisation (Mg) ∆Hlatt Mg(g) + ½O2(g) ∆Hat (Mg) Mg(s) + ½O2(g) ∆Hf MgO(s)

  36. Using the values given in the table below, construct a Born-Haber cycle and calculate the ΔHꝋlatt for CaF2

  37. Ca2+(g) + 2e- + 2F(g) enthalpy Bond Dissociation 1stElectron Affinity Ca2+(g) + 2e- + F2(g) Ca2+(g) + 2F-(g) 2nd ionisation Ca+(g) + e- + F2(g) 1st ionisation Ca(g) + F2(g) ∆Hlatt ∆Hat Ca(s) + F2(g) ∆Hf CaF2(s)

  38. Comparison of Lattice Enthalpies • Lattice enthalpy is a result of the electrostatic attraction between oppositely charged ions. • The most important factor in determining lattice enthalpy is the charge on the ions. • The greater the attraction, the greater the required energy to break apart the lattice. • Force of attraction increases along the series: 1+/1- < 1+/2 - < 2+/2 - Arrange the following compounds in order of increasing lattice enthalpy: BaCl2 LiFCaO

  39. The size of the ions also affects the lattice enthalpy. • Smaller ions have a greater attraction between them due to the smaller ionic radii (less distance between charges). • The greatest lattice enthalpy is obtained for small, highly charge ions.

  40. Comparing theoretical and experimental lattice enthalpies • If the theoretical value is exactly the same as the experimental value there is complete ionic bonding. • A difference between the values indicates covalency. • Covalency is caused by polarisation of the anion (-) by a small, highly charged cation (+) • Example MgI2 – the Mg2+ ion pulls electron density away from the I– • This effect is especially large if the anion is large

  41. Entropy • Entropy is a measure of the randomness or disorder of a system • The greater the number of possible micro states that greater the entropy • Entropy is given the symbol S. The units are J K-1 mol-1 • Entropy change = ∆S • Positive ∆S indicates an increase in entropy • Negative ∆S indicates a decrease in entropy More molecules/atoms in the system = more ways that it can be arranged (microstates) More possible micro states = more entropy

  42. As temperature increases the kinetic energy of the particles increases. The faster the particles move the greater the number of possible micro states. Increasing the volume of a system or the number moles that makes up a system increases entropy

  43. Predicting Entropy Change • Predict whether ∆S will be positive or negative: • N2(g) + 3H2(g) 2NH3(g) • CaCO3(s)CaO(s) + CO2(g) • CH4(g) + 2O2(g) CO2(g) + 2H2O(l) • C2H4(g) + H2(g) C2H6(g) • F2(g) + Cl2(g) 2ClF(g)

  44. Calculating Entropy Change • ∆S = total entropy of products – total entropy of reactants a) CH4(g) + 2O2(g) CO2(g) + 2H2O(l) b) 2Cu(NO3)2(s) 2CuO(s) + 4NO2(g) + O2(g) c) 4BCl3(l) + 3SF4(g) 4BF3(g) + 3SCl2(g) + 3Cl2(g)

  45. Spontaneity • A spontaneous reaction is one that occurs without any outside influence • We can predict whether a reaction will occur spontaneously using Gibbs Free Energy and the second law of thermodynamics. • Second Law of Thermodynamics – for a process to occur spontaneously it must result in an increase in the entropy of the universe. ∆Suniverse= ∆Ssurroundings + ∆Ssystem

  46. Gibbs free energy (G) - The energy associated with a chemical reaction that can be used to do work. • The change in Gibbs free energy ∆G under standard conditions can be calculated: • ∆G = ∆H - T∆S Units = kJ/mol T must be in Kelvin • For a reaction to be spontaneous ∆G must be less than 0 Spontaneous at high temperatures Spontaneous Spontaneous at low temperatures Not Spontaneous

  47. Calculating ∆G • Method 1 ∆G = ∆H - T ∆S C2H2(g) + 2H2(g) C2H6(g) ∆H = -313kJ/mol ∆S = -233J K-1 mol-1 298K As the value of ∆G is negative the reaction is spontaneous

  48. Method 2 – Using standard free energy of formation ∆G = ∆Gf(products) - ∆Gf(reactants) ∆Gf for an element in its standard state = 0 All other ∆Gf values can be found in tables 2SO2(g) + O2(g) 2SO3(g) ∆Gf SO2(g) = -300kJ/mol ∆GfSO3(g) = -370kJ/mol As the value of ∆G is negative the reaction is spontaneous

  49. Which of the following does not represent an increase in entropy? Solid to liquid Increase in volume Solid to gas Gas to liquid Liquid to gas • Melting of gold • NaCl dissolving in water • Sublimation of CO2 • Condensation of water • Boiling water

  50. Which of the following represents a negative change in entropy? • Melting of ice • NaCl dissolving in water • Sublimation of CO2 • heating of water • None of the above

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