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Energetics

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.

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Energetics

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  1. Energetics

  2. 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

  3. 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).

  4. 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?

  5. 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

  6. 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)

  7. 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

  8. 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)

  9. 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

  10. 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)

  11. 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

  12. 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

  13. 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?

  14. 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

  15. 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

  16. 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 Practice questions page 187 #1-4

  17. 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

  18. 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)+ NO3 -(aq) Example : refer to page 188

  19. 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.

  20. 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-10C-1. -56.8kJmol-1

  21. 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

  22. 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-10C-1. Example : refer to page 189 dissolving ammonium chloride

  23. 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.

  24. 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. The extrapolation is necessary to compensate for heat loss while the reaction is occurring. Why would powdered zinc be used?

  25. 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.

  26. 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.

  27. 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?

  28. 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.

  29. 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.

  30. 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

  31. 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 ΔHƟ = -(-222)+2(-394) = -566kJmol-1

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

  33. *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) -17 kJmol-1 -195kJmol-1 Indirect path

  34. *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 Indirect pathway

  35. *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)

  36. *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

  37. Practice questions page 199 #7-9

  38. 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]

  39. 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

  40. 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.

  41. Some average bond enthalpies Refer to page 201

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

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

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

  45. H O O + C O O H H H O O C O H H O H H Break  Form CH4 + 2O2 CO2 + 2H2O 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

  46. 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)?

  47. 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

  48. 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

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