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Understanding Free Energy and Entropy in Chemical Reactions

Learn about the connection between enthalpy, entropy, and free energy in chemical reactions. Explore how spontaneous reactions increase disorder and calculate standard entropy changes. Determine whether reactions occur at different temperatures.

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Understanding Free Energy and Entropy in Chemical Reactions

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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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