Unit 1 - Enthalpy

Enthalpy Changes & Activation Energy
1. every substance contains an energy known as enthalpy (H)
  1. this enthalpy changes when reactants turn into products
    1. the enthalpy change (/\H) is the difference in energy between products & reactants
2. when reactant molecules collide, they form a highly energetic arrangement called an activated complex
  1. the activation energy is the minimum energy required for reactant molecules to form an activated complex
    1. the activated complex can form products or turn back to reactants
      1. Energy is released when new bonds are formed in the products
Potential Energy Diagrams
1. exothermic reactions
  1. heat is given out
    1. /\H = Hproducts - Hreactants
      1. Hproducts < Hreactants
        therefore, the enthalpy change is negative (-ve)
2. endothermic reactions
  1. heat is taken in
    1. /\H = Hproducts - Hreactants
      1. Hproducts > Hreactants
        therefore, the enthalpy change is positive (+ve)
Effect of a Catalyst
1. catalysts lower the Ea
  1. more product can be formed as more particles have the energy needed to overcome the Ea barrier
    1. catalysts have no effect on the enthalpy change
Enthalpy of Combustion
1. the enthalpy of combustion is when 1 mole of a substance burns completely in oxygen
2. energy produced by a known mass of fuel burning can be calculated
  1. Eh = cm/\T
    1. c = specific heat capacity of water
    2. m = the mass of water being heated (kg)
    3. /\T = the temperature difference in degrees in centigrade
PPA 3
1. determine the combustion of ethanol
  1. burner contains ethanol
2. enthalpy given for ethanol in data book is higher than result due to heat loss in the surroundings or incomplete combustion
  1. ethanol is flammable - shouldn't be placed near a Bunsen burner
Enthalpy of Solution
1. the enthalpy of solution is the enthalpy change when 1 mole of a substance dissolves in water
  1. Example: KOH(s) = K+(aq) + OH-(aq)
    1. Eh = cm/\T = 4.18 x 0.1 x 12`C = 5.01kJ
      1. M = n x GFM = 1 x 56.1 = 56.1kJ
        5g KOH = 5.01kJ 56.1g KOH = 56.1 x .01/5 = 56.3kJ mol ^-1
Enthalpy of Neutralisation
1. the enthalpy of neutralisation is the energy change when an acid is neutralised to form 1 mole of water
  1. Example: HCl(aq) + NaOHkJ(aq) = NaCl(aq) = H2O(l)
    1. vol. H2O = vol. acid + vol. alkali = 100 + 100 = 200cm3 (0.2kg - 1cm3 H20 = 1g)
      1. Eh = cm/\T = 4.18 x 0.2 x 6`C = 5.02kJ
        mole ratio = 1:1
        moles of acid = c x V = 1 x 0.1 = 0.1moles => moles of H2O = 0.1moles
        0.1moles H2O = 5.02kJ 1 mole of H2O = 1 x 5.02kJ/0.1 = -50.2kJ mol^-1
Writing Standard Enthalpy Equation
1. combustion
  1. there must ONLY be 1 mole in the balanced equation
    1. C2H6 + 3 1/2O2 = 2CO2 + 3H2O
2. solution
  1. only 1 mole
    1. (NH4)2SO4(s) = 2NH4+(aq) + SO4 2-(aq)
3. neutralisation
  1. 1 mole
    1. HCL(aq) + NaOH(aq) = NaCl(aq) + H2O(l)

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Unit 1 - Enthalpy

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	Criado por Rosie:) mais de 10 anos atrás