Entropy, Spontaneity, and Free Energy 15.3 and 15.4.

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Presentation transcript:

Entropy, Spontaneity, and Free Energy 15.3 and 15.4

Spontaneity ● In chemical processes, to know whether something will happen on its own or not. ● Often more complicated to predict than you’d think. ● Spontaneous processes: Processes that happen by themselves without any intervention. ● These processes may not happen very quickly, but they will eventually happen by themselves. Examples: ● The combustion of gasoline: This process is spontaneous but VERY, VERY slow without the addition of heat. ● The rusting of a nail: It happens slowly under normal conditions but the process can be speeded to be almost instantaneous under the right conditions.

Spontaneous Physical and Chemical Processes A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0 C and ice melts above 0 0 C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust spontaneous nonspontaneous 18.2

Why Things Happen Spontaneously? ● Tendency for energy to move from areas of high to low ● So, any process in which the amount of energy for the system decreases (as energy flows to the surroundings) would be favorable? Thought experiments - Lighting a match ● Simple activity: Strike a match against the rough surface on the back of the matchbook. ● What did you observe? ● Was this process spontaneous?

Another Playing cards ● Pretend in front of you are thirteen playing cards. Arrange them numerically from the lowest number to the highest (aces are high for this activity). ● Was this process spontaneous? Why? ● Now, shuffle the cards to put them in random order. ● Was this process spontaneous? Why?

Yet Another Pennies ● In front of you is a container that holds 20 pennies. Arrange them so that they are all heads up on the table. ● Was this process spontaneous? Why? ● Now arrange them so they’re randomly either heads or tails on the table. ● Was this process spontaneous? Why?

Last One Dissolving sodium hydroxide in water ● In a small container holding small pieces of sodium hydroxide. Fill a small beaker halfway with water and dissolve several small pieces of sodium hydroxide. ● What would you observe? ● Was this process spontaneous? Why? ● Are spontaneous processes more or less likely to result in the generation of heat? Explain? ● Are spontaneous processes more or less likely to result in an increase in randomness of the thing being studied? Explain?

Entropy (S) - The first energetic factor ● Entropy is a thermodynamic measure of the randomness in the universe. ● As a general rule of the universe processes that produce randomness are favored over processes that decrease the amount of randomness in the universe. ● The Second Law of Thermodynamics: All processes result in an overall increase in randomness for the universe. ● How much the randomness increases or decreases for a system during a process can be measured energetically and is a thermodynamic quantity. ● As with enthalpy, we cannot know the exact entropy for a system. However, we can know how much the entropy of a system changes during the process. As a result, the most useful term is the change in entropy, called ∆S. ∆S = S final – S initial ● and this value is positive for any spontaneous process

Entropy (S) is a measure of the randomness or disorder of a system. order S disorder S If the change from initial to final results in an increase in randomness ΔS > 0 For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state S solid < S liquid << S gas H 2 O (s) H 2 O (l) ΔS >

First Law of Thermodynamics Energy can be converted from one form to another but energy cannot be created or destroyed. Second Law of Thermodynamics The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process. ΔS univ = ΔS sys + ΔS surr > 0 Spontaneous process: ΔS univ = ΔS sys + ΔS surr = 0 Equilibrium process: 18.4

Entropy Changes in the System (ΔS sys ) aA + bB cC + dD ΔS0ΔS0 rxn dS 0 (D) cS 0 (C) = [+] - bS 0 (B) aS 0 (A) [+] ΔS0ΔS0 rxn S 0 (products) = Σ S 0 (reactants) Σ - The standard entropy of reaction (ΔS 0 ) is the entropy change for a reaction carried out at 1 atm and 25 0 C. rxn 18.4 What is the standard entropy change for the following reaction at 25 0 C? 2CO (g) + O 2 (g) 2CO 2 (g) S 0 (CO) = J/K mol S 0 (O 2 ) = J/K mol S 0 (CO 2 ) = J/K mol ΔS0ΔS0 rxn = 2 x S 0 (CO 2 ) – [2 x S 0 (CO) + S 0 (O 2 )] ΔS0ΔS0 rxn = – [ ] = J/K mol

Entropy Changes in the System (ΔS sys ) 18.4 When gases are produced (or consumed) If a reaction produces more gas molecules than it consumes, ΔS 0 > 0. If the total number of gas molecules diminishes, ΔS 0 < 0. If there is no net change in the total number of gas molecules, then ΔS 0 may be positive or negative BUT ΔS 0 will be a small number. What is the sign of the entropy change for the following reaction? 2Zn (s) + O 2 (g) 2ZnO (s) The total number of gas molecules goes down, ΔS is negative.

Predicting whether something will be spontaneous or not: ● Phase changes result in significant changes in entropy: ● Phase changes that result in greater molecular freedom have a positive ∆S, and those that result in less molecular freedom have a negative ∆S. Example: For melting, ∆S = +, for freezing ∆S = - ● Dissolving solids and liquids results in a positive ∆S. ● When you dissolve a solid particles are no longer locked in place and can now move freely, leading to greater randomness. ● When liquids are miscible, mixed liquids are more random than the pure liquids. As a result, ∆S is positive. ● Dissolving a gas results in a negative ∆S. ● When you dissolve a gas a lot of its freedom to move around is taken away. As a result, this makes it have less random.

Predicting whether something will be spontaneous or not: ● By examining states of products and reactants in a process, you can usually tell if the entropy for a reaction is positive or negative. ● If a gas is made from non-gases, it’ll probably be positive. ● Example: 2 H 2 O 2(l) → 2 H 2 O (l) + O 2(g) ∆S = + ● If it’s the other way around, ∆S will be negative ● If solids or liquids are made from gases, it’ll probably be negative. ● Example: HCl (g) + NH 3(g) → NH 4 OH ­(s) ∆S = - ● If it’s the other way around, ∆S will be positive. ● If the number of gas molecules increase, it’ll be positive – if it’s the other way around it will be negative: ● Example: 2 H 2(g) + O 2(g) → 2 H 2 O (g) ∆S = -

Enthalpy - The second energetic factor ● If a process takes place and the system is at lower energy than it was before, this also fulfills the tendency of the universe to spread around its energy. ● As a result, exothermic reactions are more likely to be spontaneous than endothermic reactions.

Put everything we’ve learned together and you get the following equation: ● ∆G = ∆H – T∆S ● ∆G is called the “Gibbs free energy”: the amount of energy that is available for a system to do work. ● When ∆G is negative, a process is spontaneous. ● When ∆G is positive, a process is not spontaneous. ● When ∆G = 0, the forward rate of reaction and reverse rate of reaction are the same. The system is said to be at equilibrium. ● Because ∆G determines whether a process is spontaneous or not, we can use ∆H and ∆S to determine whether a process is spontaneous or not.

Determining ∆G ● As with ∆H values, there are several ways of determining ∆G for a process. We will focus on one, which involves solving the equation: ∆G = ∆H – T∆S

You’ve got Problems ∆G = ∆H – T∆S Example: Is the reaction N 2(g) + 3 H 2(g) → 2 NH 3(g) spontaneous at 450 K? ∆H 0 rxn = kJ and ∆S 0 rxn = -197 J/K. ● ∆H rxn = kJ = -91,800 J (need to convert to J to get same units!) ● T∆S rxn = (450 K)(-197 J/K) = -88,650 J ● ∆G = ∆H – T∆S = -91,800 – (-88,650) = -3,150 J ● The reaction is spontaneous at 450 K!

Does this make Sense? ● ∆H is negative, which means that heat is released by this reaction. This tends to drive reactions toward spontaneity. ● ∆S is negative, which means that the amount of randomness in the system decreases. ● Problem: Does this violate the second law of thermodynamics (which says that all processes must undergo an increase of entropy)? ● No! Though the amount of entropy in the system decreases, the energy given off by the system causes the surroundings to randomize more than enough to make up for it. That’s why the ∆H part of this equation is important. ● This does make sense!

Relationship between ∆H, ∆S, and ∆G ● Overall, the relationship between ∆H, ∆S, and ∆G can be expressed by the following chart: ∆H is positive∆H is negative ∆S is positive ∆G only negative (reaction only spontaneous at high temperatures) ∆G always negative (reaction always spontaneous) ∆S is negative∆G always positive (reaction never spontaneous) ∆G only negative (reaction only spontaneous) at low temperatures