Entropy SS. Spontaneity  Spontaneous reactions occur by themselves. Ice melting above 32 o F Water falling over a waterfall Iron rusting  Most spontaneous.

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Presentation transcript:

Entropy SS

Spontaneity  Spontaneous reactions occur by themselves. Ice melting above 32 o F Water falling over a waterfall Iron rusting  Most spontaneous reactions tend toward lower energy state and more disorder

Spon or Non-Spon? 1. Boiling water? 2. A match burning? 3. Decomposition of water? 4. Evaporation of water at room temperature? 5. Water freezing below 0 o C?

Disorder  Entropy is a measure of the amount of disorder in a system.  The greater the disorder the greater the entropy. H 2 O (l)  H 2 O (g)  H = 44 kJ/mol  The gas water molecules are more spread out, thus more disordered than the liquid water molecules.

Change in Entropy  S  S = S products – S reactants Positive  S = the system is more disordered following the reaction. Negative  S = the system becomes more ordered following the reaction. Disorder increases when the number of particles increases, or the state changes from solid to liquid to gas?

Positive or Negative  S? 1. N H 2  2NH 3 (-)  S: 4 moles of gas  2 moles of gas, entropy decreases 2. Sugar  Sugar + water (+)  S: change of state from solid to liquid, more disordered, entropy increases

2 nd Law of Thermodynamics  All spontaneous reactions increase the entropy of the universe.  S tot > 0  S tot >  S system +  S surrounding Which means that:  S universe > 0, the entropy of the universe is increasing

Gibbs Free Energy - DG  A measure of the spontaneity of a reaction  G o =  H o – T  S o  The change in free energy of a system depends on the enthalpy change, the temperature and entropy change. If  G o < 0: spontaneous If  G o > 0: non-spontaneous If  G o = 0: the system is at equilbrium

Sign Significance of  G =  H - T  S GG HH SS Spontaneity all temps. low temp. high temp. all temps. high temps. all temps.

Spon or Non-spon? C 6 H 12 O 6 (s) + 6 O 2 (s)  6CO 2 (s) + 6H 2 O(g)  H = kJ/mol  G =  H -T  S ? = (-) (+)  So  G will be negative at all values of T  How to oxidize sugar with potassium chlorate How to oxidize sugar with potassium chlorate

More Spontaneous Reactions  Bacon: the Other White Heat Bacon: the Other White Heat  Phosphine Phosphine  Braniac Thermite Braniac Thermite

 G and Equilibrium  G = equlibrium So,  H = T  equilibrium T =  H/  S is the temperature at which equilibrium occurs.

 G and Equilibrium  G =  G o + RT ln Q Q = reaction quotient (original concentrations) R = gas constant T = tempreature in kelvin

 G and Equilibrium equilb  G = 0 and Q = K so, 0 =  G o + RT ln K or  G o = - RT ln K

Steps (1) and (2) involve the formation of methane from its elements in their standard states. Step (3) can be estimated by adding the bond energies from for all the bond formed going from isolated gas phase atoms to the product. Bond Energies

Example: The standard enthalpy of formation of ethane is –84.5 kJ/mol. Estimate this value using average bond energies and atomization energies.

 The estimate differs from the tabulated value by 7% (-84.5 kJ/mol)  This error is remarkably small considering how easy the calculation was to perform