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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

  6. Enthalpy (H)and Enthapy change(ΔH) • Enthalpy (H) is the internal energy stored in the reactants. • The absolute value of enthalpy of reactants or products cannot be found but we can only measure the difference between them. • We measure the enthalpy change , ∆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. Standard enthalpy change(ΔHθ) The standard conditions for enthalpy changes are : • Temperature of 298K or 250C • Pressure of 100kPa (or 1 atm) • Concentration of 1 moldm-3 for all solutions • All substances in their standard states • in [kJmol-1]

  8. Enthalpy Level Diagram -Exothermic Reaction Energy of products & reactants • In reaction A + B  C + D, the enthalpy of the products (C&D) is < the enthalpy of the reactants (A&B). Energy change : ∆H = H(products) - H(reactants) ∆H is negative since H(products) < H(reactants) • Heat is lost to the surrounding • Temperature increases Enthalpy Map

  9. Examples of Exothermic Reactions • 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)

  10. Enthalpy Level Diagram -Endothermic Reaction Energy of products & reactants • In reaction A + B  C + D, the enthalpy of the products (C&D) is > the enthalpy of the reactants (A&B). Energy change : ∆H = H(products) - H(reactants) ∆H is positive since H(products) > H(reactants) • Heat is gained from the surrounding • Temperature decreases Enthalpy

  11. Examples of Endothermic Reactions • Thermal decomposition • CaCO₃ (s)  CaO (s) + CO ₂ (g) • Photosynthesis • 6CO₂ (g) + 6H₂O (l)  C₆H₁₂O₆ (aq) + 6O₂ (g)

  12. Specific heat capacity • Amount of heat required to raise the temperature of a unit mass of a substance by 1 degree or 1 kelvin. • Unit : 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 change (absorbed or released) [ J ] c = specific heat capacity of substance m = mass of substance in grams ΔT = change in temperature in Celsius

  13. 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 • Assumption : energy given out will be absorbed by the water and cause a temperature change. • calculate the heat through the equation Q = mcΔT

  14. Example 1. 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? 2. If 500J of heat is added to 100.0g of samples of each of the substances below, which will have the largest temperature increase?

  15. Standard Enthalpy Changes • Std enthalpy change of neutralisation • Overall enthalpy change (heat produced) when 1 mole of water is formed when an acid reacts with an alkali under std condition. E.g. HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) ΔHnθ=-394 kJmol-1 • Std enthalpy change of solution • Overall enthalpy change (heat produced or absorbed) when 1 mole of a solute is dissolved completely in excess solvent. E.g. NaOH(s) + (aq)  NaOH(aq) ΔHsθ=-394 kJmol-1

  16. (3) Std enthalpy change of combustion • Overall enthalpy change (heat produced) when 1 mole of the substance is completely burnt in oxygen under std condition. C(s) + O2(g)  CO2(g) ΔHcθ=-394 kJmol-1 (4) Std enthalpy heat of formation • Overall enthalpy change (heat produced or absorbed) when 1 mole of substance is formed from its elements in their standard states. 2C(g) + 3H2(g) + ½O2(g)  C2H5OH(l) ΔHfθ=-277 kJmol-1

  17. Enthalpy change of formation Example: Which of these are correct? (1) 2Fe(s) + O2(g)  Fe2O3(s) (2) 2C(s) + O2(g)  2CO(g) (3) C(s) + O2(g)  CO2(g) (4) C2H2(g)+ 2H2(g)  C2H6(g) (5) N2(g) + 3H2(g)  2NH3(g)

  18. Enthalpy change of combustion reactions • The standard enthalpy change of combustion for a substance is the enthalpy change (heat produced) 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

  19. Calculating heat of reaction from temp changes For exothermic reaction, heat produced by reaction is assumed to be absorbed by water, so temp of water increases. Heat change of reaction = - heat change of water = - mH2O x cH2O xΔTH2O For endothermic reaction, heat absorbed by reaction is assumed to be taken from the water, so temp of water decreases. Heat change of reaction = heat change of water = mH2O x cH2O xΔTH2O

  20. 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) • Enthalpy transfer = m x 4.18 x Δ T / No. of moles

  21. 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?

  22. 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

  23. 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

  24. Enthalpy changes of reaction in solution Enthalpy change of neutralisation (ΔHn) • The standard enthalpy change of neutraisation is the enthalpy change that takes place when 1 mole of water molecules is formed when an acid reacts with 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

  25. 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.

  26. 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

  27. 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

  28. 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

  29. 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.

  30. Enthalpy changes of reaction in solution 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?

  31. 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.

  32. Standard enthalpy changes, ΔHƟ N2(g) + 3H2(g) 2NH3(g) ΔHƟ = -92 kJmol-1 What does it mean? The enthalpy change of reaction is -92 kJmol-1 #

  33. Fat vs Metabolism Fats store plenty of energy compared to sugar. The body would store up to 67.5g of sugar complexes for the energy equivalent of 10g of fat. Calculate the energy in which a fat is converted to sugar: C18H36O2(s) + 8O2(g)  3C6H12O6(s) given C18H36O2(s) + 26O2(g)  18CO2(g) +H2O(l) ΔH1=-11407kJ C6H12O6(s) + 6O2(g)  6CO2(g) +6H2O(l) ΔH2=-2800kJ

  34. Hess’s Law • Hess's law can be applied to calculate enthalpies of reactions that are difficult to measure • Enthalpy is a state function • It depends only upon the initial and final state of the reactants/products and not on the specific pathway taken to get from the reactants to the products

  35. Hess’s Law Hess’s Law states that the enthalpy change for any chemical reaction is independent of the pathway provided the starting and final conditions, and reactants and products are the same.

  36. Example Consider these thermo-chemial equations: (1) NaOH(s) + (aq) NaOH(aq) ΔHƟ1=-43kJmol-1 (2) NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l) ΔHƟ2=-57kJmol-1 Calculate the enthaply change for (3) NaOH(s) + HCl(aq)  NaCl(aq) + H2O(l)

  37. methods • Using cycles • Manipulating equations • Equations • Enthalpy level diagrams

  38. Reaction in aqueous soln Calculate the enthalpy change for the formation of sodium chloride solution from solid sodium hydroxide. NaOH(aq) NaOH(s) NaCl(aq) + H2O(l) + HCl(aq) 1. Indirect path: NaOH(s) + (aq)  NaOH(aq) ΔHƟ1=-43kJmol-1 2. NaOH(aq) + HCl (aq)  NaCl(aq) + H2O(l) ΔHƟ2=-57kJmol-1 3. NaOH(s) + HCl (aq)  NaCl(aq) + H2O(l). Indirect path + HCl(aq) + H2O(l) ΔH2 ΔH1 Direct path

  39. *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) + CO2(g) ΔHƟ1=-17 kJmol-1 CaO(s) +2HCl(aq)  CaCl2(aq) + H2O(l) ΔHƟ1=-195kJmol-1

  40. 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(g) + O2(g)  2CO2(g)

  41. *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)

  42. *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

  43. Std enthalpy heat of formation • Overall enthalpy change (heat produced or absorbed) when 1 mole of substance is formed from its elements in their standard states. 2C(g) + 3H2(g) + ½O2(g)  C2H5OH(l) ΔHfθ=-277 kJmol-1 • It is zero for elements. • It can be positive or negative. • It is illustrated by equations which must balance to produce 1 mole of the substance.

  44. Example • The standard enthalpy of formation of strontium chloride is the enthalpy change for which of the following reactions? • Sr(g) + Cl2(g)  SrCl2(s) • Sr(s) + Cl2(g)  SrCl2(s) • Sr2+(g) + 2Cl-(g)  SrCl2(s) • Sr2+(aq) + 2Cl-(aq)  SrCl2(s)

  45. Example • The standard enthalpy of formation of magnesium bromide is the enthalpy change for which of the following reactions? • Mg2+(g) + 2Br-(g)  Mg2+(Br-)2(s) • Mg2+(g) + 2Br-(g)  Mg2+(Br-)2(g) • Mg(s) + Br2(g)  Mg2+(Br-)2(s) • Mg(s) + Br2(l)  Mg2+(Br-)2(s)

  46. Example • The standard enthalpy of combustion of hydrogen is -286kJmol-1 • What is the standard enthalpy of formation of water ?

  47. Hess’s Law • The enthalpy change of reaction = total enthalpy change of formation of product – total change of formation of reactant Hrxn =

  48. Example • Iron(III)oxide can be reduced to iron using hydrogen. Fe2O3(s) + 3H2(g)  2Fe(s) + 3H2O(g) Calculate the standard enthalpy change, ΔHθ

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