1. Learning objective: WALT: understand entropy WILF: an explanation of spontaneous endothermic reactions 07/07/2016 Watch the ice melt… …exciting isn’t it! Why is it happening? H 2 O (s)  H 2 O (l) –Is this process exothermic or endothermic? –How do you know? –Why should this be surprising now you think about it?

2. icewater What’s different?

3. Entropy (S) A degree of disorder or randomness in a system During any process the entropy of the Universe increases  S total =  S system +  S surroundings H 2 O (s)  H 2 O (l)  H = + 6 kJmol -1 –The entropy of the water increases –The surroundings cool down – becoming less disordered but… –…the increase in entropy in the beaker is greater than the decrease in entropy of the surroundings Disorder: gas>liquid>solid

4. Crystals Crystals are very organised! Crystalline substances have low entropy At 0 K (absolute 0) a perfect crystal has zero entropy Dissolving increases entropy: Link to crystal dissolving Link to crystal dissolving Crystal: one arrangement of particles Solution: lots of arrangements of particles

5. Experiments to show positive  S 1.Mix a few spatulas of ammonium nitrate with distilled water 2.Add a few spatulas of ammonium carbonate to ethanoic acid 3.Set fire to a 1 cm strip of magnesium ribbon – make sure fans are on and window is open! 4.Watch the reaction between solid hydrated barium hydroxide and ammonium chloride In each case describe  H and explain how  S total has increased

6. Equations 1.NH 4 NO 3 (s) + (aq)  NH 4 + (aq) + NO 3 - (aq) 2.(NH 4 ) 2 CO 3 (s) + 2 CH 3 COOH (aq)  2 CH 3 COO - (aq) + 2 NH 4 + (aq) + CO 2 (g) + H 2 O (l) 3.2 Mg (s) + O 2 (g)  2 MgO (s) 4.Ba(OH) 2.8H 2 O (s) + 2 NH 4 Cl (s)  BaCl 2 (s) + 10 H 2 O (l) + 2 NH 3 (g)

7.  H and  S system Compound  H f /kJmol -1 S/Jmol -1 K -1 Ba(OH) 2.8H 2 O (s)-3345427 NH 4 Cl (s)-31495 NH 3 (g)-46192 H 2 O (l)-28670 BaCl 2 (s)-859124 1.Calculate  H using a Hess cycle 2.Calculate entropy of reactants and products and hence  S system

8. HH Ba(OH) 2.8H 2 O (s) + 2 NH 4 Cl (s)  BaCl 2 (s) + 10 H 2 O (l) + 2 NH 3 (g)  H 1  H 2 Ba (s) + 13 H 2 (g) + 5 O 2 (g) + N 2 (g) + Cl 2 (g) 1.  H reaction = -  H 1 +  H 2 2.  H 1 = -3345 + 2 x -314 = -3973 kJmol -1 3.  H 2 = -859 + 10 x -286 + 2 x -46 = -3811 kJmol -1 1.= -(-3973) + (-3811) = + 162 kJmol -1

9. SS S reactants = 427 + 2 x 95 = 617 Jmol -1 K -1 S products = 124 + 10 x 70 + 2 x 192 = 1208 Jmol -1 K -1  S = 1208 – 617 = 591 Jmol -1 K -1

10.  S surroundings  S surroundings = -ΔH system /T surroundings  S surroundings = -162 x 10 3 /298  S surroundings = -56 Jmol -1 K -1  S total =  S system +  S surroundings  S total = 591+ (-56)  S total = 535 Jmol -1 K -1 Notice the reaction occurred despite the fact that  S surroundings is negative What will the effect of changing the temperature be? Use your data books to calculate  S total for the other three reactions – comment on the values for  S total,  S system,  S surroundings &  H

11.  S, temperature and K Can you see a link between entropy change, temperature and position of equilibrium? Think about: –the solubility of ionic compounds as temperature increases –what happens to  S total for the oxidation of magnesium as the temperature rises