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Put these substances in order of increasing entropy ( DO NOT LOOK UP VALUES) MgCO 3 (s) He (g) Pb (s) C (diamond) NH3 (g) C (graphite) Mg(s) H 2 0 (l)

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Presentation on theme: "Put these substances in order of increasing entropy ( DO NOT LOOK UP VALUES) MgCO 3 (s) He (g) Pb (s) C (diamond) NH3 (g) C (graphite) Mg(s) H 2 0 (l)"— Presentation transcript:

1 Put these substances in order of increasing entropy ( DO NOT LOOK UP VALUES) MgCO 3 (s) He (g) Pb (s) C (diamond) NH3 (g) C (graphite) Mg(s) H 2 0 (l) C 2 H 5 OH (l) O 2 (g) MgO (s)

2 Entropy – look up the standard entropy values Standard entropies, S o /Jmol -1 K -1 C (diamond) C (graphite) Mg(s) Pb (s) MgO (s) MgCO 3 (s) H 2 0 (l) C 2 H 5 OH (l) He (g) NH 3 (g) O 2 (g)

3 Entropy Standard entropies, S o /Jmol -1 K -1 C (diamond)2.4 C (graphite)5.7 Mg(s)32.7 Pb (s)64.8 MgO (s)26.9 MgCO 3 (s)65.7 H 2 0 (l)69.9 C 2 H 5 OH (l)160.7 He (g)126.0 NH 3 (g)192.3 O 2 (g)205.0

4 Calculating ∆Ssystem ∆S o system = S o products - S o reactants Calculate the ∆S o system for the following reaction: Ammonium chlorideAmmonia + Hydrogen chloride chloride

5 ∆S o and spontaneous reactions If: ∆S > 0 the reaction can go This is an illustration of the 2 nd Law of thermodynamics The total entropy of the universe always increases in a natural change.

6 Hold on a minute!!! If this is true then how is it that reactions like combustion of magnesium in air are spontaneous even if entropy seems to decrease!

7 Using standard entropies to calculate the entropy change of the system 2Mg (s) + O 2 (g)  2MgO (s) S o [Mg(s)] = + 32.7 Jmol -1 K -1 S o [1/2 O2(g)] = + 102.5 Jmol -1 K -1 S o [MgO(s)] = + 26.9 Jmol -1 K -1 N.B. These values are ALWAYS PER MOLE OF ATOMS! ∆ S o sytem = S o products - S o reactants ∆ S o system = +(2*26.9) – (2*32.7) – (2*102.5) = -216.6Jmol -1 K -1 This entropy change is negative but the reaction is spontaneous. Why?

8 Calculating the entropy change of the surroundings ∆ S surroundings = -∆ H reaction T ∆ H formation [MgO] = -601.7 kJmol -1 Therefore, ∆ H reaction [MgO] = -601.7*2 kJmol -1 ∆S o surroundings = -(-1203.4)*1000 298 = +4038 Jmol -1 K -1

9 Calculating the total entropy change ∆S o total = ∆S o system + ∆S o surroundings For our magnesium example, ∆S o total = (-216.6 + 4083) Jmol -1 K -1 = +3821.4 Jmol -1 K -1 The reaction should go spontaneously at 298K Remember this when dealing with entropy and equilibrium

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