Entropy - The Drive Toward Maximum Randomness Chemical reactions are driven by 2 main forces: the drive to minimize stored energy (enthalpy) the drive.

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Presentation transcript:

Entropy - The Drive Toward Maximum Randomness Chemical reactions are driven by 2 main forces: the drive to minimize stored energy (enthalpy) the drive to maximize randomness (entropy) Substances, over time, will become as random and disorganized as possible

Observe what happens to randomness when a substance is heated and/or changes phases.

Entropy is the word used to describe this randomness because of the drive toward maximum entropy a highly organized state has a low probability the probability that a given substance will exist in a specific state is known as its molar entropy and has the symbol S with units of J/molK the larger the number the greater the probability of existence

For each of the following is entropy increasing or decreasing. H 2 O (l) + heat -----> H 2 O (g) Increasing C 2 H 6 (g) + 7/2O 2 (g) ----> 2CO 2 (g) + 3H 2 O(g) Increasing, 4.5 molecules to 5 molecules CaCO 3 (s) ---> CaO(s) + CO 2 (g) Increasing, solid to gas, 1 particle to 2 particles 2Mg(s) + O 2 (g) ---> 2MgO(s) Decreasing, 3 particles to 2 particles, g to s AgNO 3 (aq) + KCl(aq) --> AgCl(s) + KNO 3 (aq) Decreasing, 4 mobile ions to 2 mobile ions Movie1 Movie2

Entropy Changes  S =  S(products) -  S(reactants) S values for elements are NOT assigned an arbitrary value of zero When  S is negative - entropy is decreasing, this change is not favoured (less randomness)  S is positive- entropy is increasing, this change is favoured (more randomness)

Chemical Reactions have a tendency to be spontaneous when the  H is negative and the  S is positive. In other words the randomness of the system is increasing and the amount of stored energy is decreasing.

Using Entropy values from pg. 798 consider the combustion of liquid benzene (C 6 H 6 ). Water vapour is the product. C 6 H 6 (l) +7.5O 2 (g) =>6CO 2 (g) + 3H 2 O (g)  S =  S(products) -  S(reactants) = (6 mol x J/molK + 3 mol x J/molK) - (1 mol x 173 J/molK mol x J/molK) = 136 J/ K since  S is positive the entropy drive is favourable

Free Energy In nature enthalpy and entropy operate simultaneously and to determine if a given chemical reaction will or will not occur we must look at the two driving forces together If  H is -ve and  S is +ve = spontaneous If  H is +ve and  S is -ve = not spontaneous If one drive is favourable and the other is not it is not clear whether it is or isn’t spontaneous

When enthalpy and entropy oppose one another, their relative importance in determining spontaneity is far from obvious. In addition, temperature is a 3rd factor that influences the direction of a spontaneous change.

Chemical Reactions are spontaneous when the answer to this equation is negative  H - TSTS where T is temperature in K so when T  S >  H the reaction is spontaneous. The difference is called Gibb’s (  G) Free Energy GG H - TSTS  G =

At what temperature is this reaction spontaneous? H 2 O(l) -----> H 2 O(g) Recall when T  S >  H the reaction is spontaneous. Go to the tables to solve for T. T >  H/  S T>(-242 kJ/mol x 1 mol - (-286 kJ/mol x 1 mol ) J/ molK x 1 mol J/ molK x 1 mol T> (44 kJ)/( kJ/K) T> 3.7 x 10 2 K

Homework Read pg Questions Pg. 498 # 1 Pg. 508 #2 Pg. 512 # 1-11