5.1 Exothermic and endothermic reactions

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5.1 Exothermic and endothermic reactions

• If a reaction produces heat (increases the
  temperature of the surroundings) it is
  exothermic.
• If the temperature of the reaction mixture
  decreases (ie heat is absorbed) then the reaction
  is endothermic.
• Exothermic -gt a reaction which produces heat (?H
  has a negative value by convention, -ve)
• Endothermic -gt a reaction which absorbs heat (?H
  has a positive value by convention, ve)

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Enthalpy of reaction, ?H The change in internal
(chemical) energy (H) in a reaction

• The most stable state is where all energy has
  been released. When going to a more stable state,
  energy will be released, and when going to a less
  stable state, energy will be gained (from the
  surroundings).
• On an enthalpy level diagram, higher positions
  will be less stable (with more internal energy)
  therefore, if the product is lower, heat is
  released (more stable, ?H is -ve) but if it is
  higher, heat is gained (less stable, ?H is ve).

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• Formation of bonds causes an energy release
  (exothermic).
• Breaking of bonds requires energy (endothermic).

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Calculation of enthalpy changes

• Change in energy mass x specific heat capacity
  x change in temperature
• E m x c x ?T
• where m is the mass of water present (kilograms),
  and c 4.18 kJ Kg-1 K-1.

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Hess' Law

• Hess' Law states that the total enthalpy
  change between given reactants and products is
  that same regardless of any intermediate steps
  (or the reaction pathway).

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• Example Calculate the enthalpy of formation
  of methane.
• Example 1
• Data
• ?H combustion (carbon) -394kj
• ?H combustion (hydrogen) -242kj
• ?Hcombustion (methane) -891kj

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• If ?Hf methane is represented by the equation
• C(s) H2(g) ? CH4 (g)
• This equation can be constructed using the
  equations for combustion of the reactants (carbon
  and hydrogen) and the product (methane)
• C(s) O2(g) ? CO2(g)
  -394k..enthalpy 1  
• H2(g) ½ O2(g) ? H2O(l)
  -242kj..enthalpy 2
• multiply this equation by 2 to get   
• 2H2(g) O2(g) ? 2H2O(l) -484kj
  .enthalpy 3
• Add the first and third equation together to
  get 
• C(s) 2O2(g) 2H2(g) ?CO2(g) 2H2O(l) -878kj
  
• 
  enthalpy (1 3) 4
• Now take away the eq for the combustion of
  methane 
• CH4 (g) 2O2(g) ? CO2(g) 2H2O(l)
  -891kj enthalpy 5  
• And after rearrangement ( take the CH4 to the
  right hand side) the result is the equation for
  the formation of methane 
• C(s) H2(g) ? CH4 (g)
  13kj enthalpy ( 4 - 5)
• 
  

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• Example 1
• Monoclinic sulphur is formed in volcanic regions
  by reaction between sulphur dioxide and hydrogen
  sulphide according to the equation
• SO2 2H2S --gt
  2H2O 3S
• Draw an enthalpy diagram or cycle and calculate
  the standard enthalpy change for this reaction.
• Here are some values.
• Standard
  enthalpy of formation (all in kJ)
• 
  H2O(l) -286
• 
  H2S(g) - 20.2
• Standard
  enthalpy of combustion of S (monoclinic)
  -297.2kJ
• equation 1 H2 ½ O2 --gt H2O    
  -286 kJ
• equation 2 H2 S --gt H2S     
  -20.2kJ
• equation 3 S O2 --gt SO2    
   -297.2 kJ
• multiply E2 by 2 (equation 4) H2 2S --gt
  2H2S     -40.4 kJ
• add equation 3 4... 2H2 3S O2 --gt
  SO2 2H2S      -337.6 kJ
• multiply E1 by 3(equation 5) 2H2 O2 --gt
  2H2O    -572 kJ
• subtract 4 from 5-3S ---gt 2H2O - SO2 -
  2H2S      -234.4 kJ
• rearrangeSO2 2H2S --gt 2H2O 3S     -234.4 kJ

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Born Haber Cycles

• Born Haber cycles are the application of Hess'
  law to ionic systems. An ionic solid consists of
  a giant structure of ions held together in a
  giant lattice.
• Application of Hess law tells us that the
  enthalpy of formation of an ionic crystal is
  equal to the sum of the energies of formation of
  the ions plus the enthalpy of the lattice.
• it is a several step process that is best
  represented by a diagram showing the individual
  steps as endothermic upwards and exothermic
  downwards.