12 Entropy

12 Entropy

• 1.
  Free Energy andEntropy
• 2.
  Some reactions whichhappen spontaneously are endothermic. The difference in enthalpy between the products and the reactants cannot be the only factor which decides if a chemical reaction takes place. It is often observed that reactions which occur spontaneously increase the randomness or disorder of a system For example when an ionic crystal dissolves it passes from the regular arrangement of a crystalline lattice to a random solution of ions
• 3.
  This is termedan increase in entropy (disorder) of the system. The 2 factors, enthalpy and entropy,combine to give the free energy of the system Free energy G = Enthalpy H – Temperature /K x Entropy S G = H - TS and ∆G = ∆H - T∆S
• 4.
  For a chemicalchange to occur ∆G must be negative . A change is therefore assisted by a decrease in enthalpy (∆H negative) and an increase in entropy (∆S positive) One method for finding the standard entropy change is to use Standard entropy change = sum of standard entropies of products - sum of standard entropies of reactants
• 5.
  Units: entropyis measured in J K -1 mol -1 If a reaction has a positive entropy overall there is an increase in disorder. A negative entropy indicates a decrease in disorder.
• 6.
  Which of theseprocesses lead to an increase in entropy? ice melting hydrogen peroxide dissociating into oxygen and water formation of a crystal lattice nitrogen and hydrogen reacting to form ammonia potassium chloride dissolving in water
• 7.
  Calculate the standardentropy change for the reaction of chlorine and ethene given the entropy values in J K -1 mol -1 Cl 2 223.5 C 2 H 4 219 CH 2 ClCH 2 Cl 208 C 2 H 4 + Cl 2 = CH 2 ClCH 2 Cl ΔS = 208 – (219 + 223) = - 234 J K -1 mol -1 Lets consider if this is a sensible answer! We started with 2 molecules of reactants and ended with only 1 molecule of product. There is a decrease in disorder (or an increase in order) so the overall entropy change is negative.
• 8.
  Calculations on GibbsFree Energy Using the equation ∆G = ∆H – T∆S calculate the change in free energy for the following reaction Fe 2 O 3(s) + 3H 2(g) -> 2Fe (s) + 3H 2 O (g) and determine if it will take place at a) 20 o C and b) 500 o C Fe 2 O 3 H 2 Fe H 2 O (g) standard enthalpy -822 0 0 -242 standard entropy 0.090 0.131 0.027 0.189 all values in kJ mol -1
• 9.
  ∆ H =(3 x – 242 - ( -822) ) = + 96 kJ mol -1 ∆S = 3 x 0.189 + (2 x 0.027) – (3 x 0.131 + 0.090) ∆S = +0.138 kJ mol-1 a) at 20 0 C ∆G = ∆H - T∆S  ∆G = +96 – (20 +273) x 0.138 = +43.26 kJ mol -1 b) at 500 o C ∆G = +96 – (500 +273) X 0.138 = -10.7 kJ mol -1 The reaction will occur spontaneously at 500 o C (when ∆G is negative) but not at 20 o C