Energetics

• Enthalpy change

  the heat energy change at constant pressure

  units: kJ mol-1

  enthalpy (H) is the heat energy that is stored in a chemical system.
• exothermic reactions

  heat energy is given out to the surroundings, so temperature of the surroundings increases. The chemicals lose heat energy so enthalpy change is negative= ^H=-ve
• important exothermic reactions

  • combustion of fuels
  • respiration
• endothermic reactions


  heat energy is taken in from the surroundings, so temperature of the surroundings decreases. the chemicals gain heat energy so ^H=+ve

  endothermic reactions often require an input of heat energy
• endothermic reaction examples

  • thermal decomposition of calcium carbonate
  • CaCO3 -> CaO + CO2
  • Photosynthesis
  • 6CO2 + 6H2O -> C6H12O6 + 6O2
• Enthalpy profile diagrams can be used to illustrate the enthalpy change for a reaction. As well as the products, reactions and ^H, these diagrams also show the activation energy, Ea
• Activation energy, Ea

  minimum energy required to start a reaction by breaking of bonds
• exothermic enthalpy profile diagrams

  eventhough the products have a lower energy than the reactants, energy is required to break the bonds and start the reaction. Once the energy barrier has been overcome, the net energy released will provide the energy needed to overcome the activation energy

  
• endothermic enthalpy profile diagram

  Again, energy is required to break the bonds and start the reaction. Most endothermic reactions must be heated continuously in order to provide the necessary energy

  
• bond enthalpy

  heat energy required to break one mole of a given covalent bond in the molecules in the gaseous state. For example, Cl-Cl -> 2Cl* (g)= bond enthalpy of the Cl-Cl bond
• why are bond enthalpies always positive?

  energy is required to break bonds
• Bond enthalpies give indications of the relative strength of a covalent bond so if it's higher bond enthalpies = must be stronger bond.
• Mean bond enthalpies

  the heat energy required to break one mole of a covalent bond, averaged for that type of bond in a range of different compounds
• Why do chemists use mean bond enthalpy values?

  the same covalent bond may appear in different compounds e.g C-H (alkanes, alkenes, benzene) and the value of the bond enthalpy will be slightly different
• How to calculate mean bond enthalpies

  • energy is required to break bonds
  • energy is released when new bonds form
  • the enthalpy change of a reaction is the difference between these two values
  • equation:

  
• calculate the enthalpy change for the complete combustion of methane

  Mean bond enthalpy values: C-H:+413, O=O:+497, C=O:+805, O-H:+463

  • energy required to break bonds
  • (4x413) + (2x497) = 2646
  • energy released when bonds are formed
  • (2x805) + (4x463) = 3462
  • 2646-3462=-816 kJ mol-1
• what are limitations of bond energy calculations?

  Using mean bond enthalpies to calculate the enthalpy change of a reaction often leads to value less accurate than a value obtained from Hess's law

  • bond enthalpies are mean values which are taken from a range of compounds
  • bond enthalpies apply to gaseous reactions only
  • can only be used on covalent substances
• standard enthalpy changes


  experimentally determined values for a particular reaction under a standard set of conditions
• standard conditions

  temp- 298K, pressure- 100KPa
• standard state

  the physical state of a substance under standard conditions. For example, O2 (g)
• standard enthalpy of formation

  the enthalpy change that occurs when one mole of a compound is formed from its constituent elements with all reactants and products in their standard states
• write the equation to show the standard enthalpy of formation of ethanol

  2C (s) + 3H2 (g) + 1/2 O2 (g) -> C2H5OH (l)
• • By definition, ΔH_f⦵ of an element in its standard state is zero
  • the more negative ΔH_f⦵, the more stable is the compound
• standard enthalpy of combustion

  the enthalpy change that occurs when one mole of a compound reacts completely in oxygen with all reactants and products in their standard states
• equation example for standard enthalpy of combustion

  C3H8 (g) + 5O2 (g) -> 3CO2 (g) + 4H2O (l)
• Calorimetry

  Method used to determine enthalpy changes by experiment. Involves measuring the temperature change of a given amount of water as the reaction occurs and converting this to a quantity of heat energy
• When water is heated, 4.18J are required to heat 1g of water by 1 kelvin- this is the specific heat capacity of water
• Heat energy change

  Mass of water (g) x specific heat capacity (4.18Jk-1g-1) x delta T in k ( temperature change)

  q= mc^T

  note: mass of water is equal to the volume in cm3
• Method to calculate heat energy change


  • q= m x c x delta T
  • divide q by 1000 to convert J to kJ
  • heat given out or taken in by one mole= g(in kJ)/no. Of moles (of limiting reactant)
  • If temp increases, ^T is negative
  • if temp decreases, ^T is positive
• Why is the experimental value less exothermic than the date book value for enthalpy change of combustion of methanol?

  • not all of the heat energy released from the combustion would’ve been transferred to the H2O. Some heat energy would’ve been lost to the other surroundings i.e air around apparatus
  • not all the combustion would’ve been complete I.e there would be incomplete combustion
  • non standard conditions may have been used
• Cooling curves


  Can be used as a method of accounting for heat loss with reactions in solutions
• 1. plot a graph of temperature against using the results and determine the maximum temperature change accompanying the reaction. Extrapolate back to when the reactant is added to establish the max temp rise
  2. write a balance equation
  3. calculate enthalpy change
• Percentage error


  Uncertainty/measured value x100

  for a thermometer x by 2
• First law of thermodynamics

  Energy can’t be created or destroyed but it can be changed from one form to another
• Its not always possible to measure the enthalpy change of a reaction directly, why?

  • activation energy may be too high
  • reaction is too slow
• From Hess's law: enthalpy change for route 1= enthalpy change for route 2. so deltaH1= deltaH2 + deltaH3.

  therefore if we know two of the enthalpy changes we can calculate the third
• enthalpy cycle


  Hess's law is used to calculate enthalpy changes which cannot be measured directly using an enthalpy cycle

  
• how to use Hess's law in calculations

  1. write a balanced equation for the reaction in the question- this is top row of the energy cycle
  2. complete the energy cycle by identifying how the species involved in the equation relate to the data provided
  3. identify the two routes around the cycle and apply Hess's law
  4. re-arrange to find the enthalpy change required
• Hess’ law


  the enthalpy change for a chemical reaction is independent of the route taken