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Free Energy and Entropy

Free Energy and Entropy.

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Free Energy and Entropy

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  1. Free Energy and Entropy

  2. Some reactions which happen 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 termed an 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 SG = H - TSand ∆G = ∆H - T∆S

  4. For a chemical change 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: entropy is measured in J K-1mol-1If a reaction has a positive entropy overall there is an increase in disorder. A negative entropy indicates a decrease in disorder.

  6. Which of these processes 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 standard entropy change for the reaction of chlorine andethene given the entropy values inJ K-1mol-1Cl2223.5 C2H4 219 CH2ClCH2Cl 208 C2H4 + Cl2 = CH2ClCH2ClΔS = 208 – (219 + 223)= - 234 J K-1mol-1Lets 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 Gibbs Free Energy Using the equation ∆G = ∆H – T∆S calculate the change in free energy for the following reaction Fe2O3(s) + 3H2(g)→ 2Fe(s) + 3H2O(g) and determine if it will take place at a) 20oC and b) 500oC Fe2O3 H2 Fe H2O(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-1a) at 200C∆G = ∆H - T∆S∆G = +96 – (20 +273) x 0.138 = +43.26 kJ mol-1b) at 500oC∆G = +96 – (500 +273) X 0.138 = -10.7 kJ mol-1The reaction will occur spontaneously at 500oC (when ∆G is negative) but not at 20oC

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