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Unit 5 Chapter 16: Spontaneity, Entropy, and Free Energy. Section 16.1: Spontaneous Processes and Entropy. Spontaneous Processes- processes that occur without outside intervention may be fast or slow thermodynamics tells us about the direction of a process but not the speed of a process.
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Unit 5 Chapter 16: Spontaneity, Entropy, and Free Energy
Section 16.1: Spontaneous Processes and Entropy Spontaneous Processes- processes that occur without outside intervention may be fast or slow thermodynamics tells us about the direction of a process but not the speed of a process.
Positional Probability-depends on the number of configurations in space that yield a particular state Higher positional probability = higher entropy How is change of state an example of positional probability? The smaller the volume, the less positions the particles can be in. Assignment: 19, 21, 23,
Section 16.2: Entropy and the Second Law of Thermodynamics Second law of thermodynamics- in any spontaneous process there is always an increase in the entropy of the universe. ∆Suniv = ∆Ssys + ∆Ssurr -∆Suniv = the process is NOT spontaneous (unless the rxn is reversed) +∆Suniv = process is spontaneous 0=∆Suniv = the process is at equilibrium
Section 16.3: The effect of temperature on spontaneity How are entropy changes in the surroundings determined? Determined by the flow of energy into and out of the system as heat (ENERGY IS CONSTANTLY CHANGING). What is the main reason the temperature influences the relative importance of the entropy of the system versus the entropy of the surroundings? Exothermicity. If the reaction is exothermic it will release heat, increasing the entropy of the surroundings.
Is the impact of energy change greater for high or low temperatures? Low temperatures Equations: ∆Ssurr = -(∆H/T) there must be a minus sign because H is from the system Assignment: 25, 26
Section 16.4: Free Energy Free energy- relates enthalpy, entropy and temperature G = H –TS ∆G = ∆H - T∆S What are the two functions that can be used to predict spontaneity? Entropy of the universe and free energy. ∆Suniv= -∆G/T (Derived from equations above)
How does temperature influence whether enthalpy or entropy determines spontaneity? H and S favor spontaneity in opposite directions, therefore temperature will be the deciding factor Assignment: 27, 29
Section 16.5: Entropy Changes in Chemical Reactions • Will a reaction tend to move toward fewer to greater numbers of molecules assuming all else is equal? • Yes. Positional entropy typically increases. • When a reaction involves gaseous molecules, how is positional entropy determined? • Total moles on either side of reaction
Third law of thermodynamics- The entropy of a perfect crystal at 0 Kelvin is zero. • Standard entropy values are measured under what conditions? • 298 K and 1 atm
What does this statement mean: entropy is a state function of the system. • It is not pathway dependent, therefore • ∆Sorxn= ∑npSoproducts - ∑nrSoreactants • Why is the entropy of 3 moles of water greater than the entropy of 3 moles of hydrogen gas? • Because it is a more complex molecule and it has many more possible arrangements • Assignment: 33, 35, 37
Section 16.6: Free Energy and Chemical Reactions Standard free energy change (∆Go) - The change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states. Why do we study standard free energy change? It will allow us to predict the tendency of these reactions to occur. Is free energy a state function? YES How to calculate ∆Go The most common way is to use ∆Go = ∆Ho - T∆So as long as the reaction is at constant Temperature
Another way to calculate ∆Go is to use the same process as Hess’s law G is a state function (independent of pathway), therefore we can manipulate reactions and ∆Go values to find the overall ∆Go
Another way to calculate ∆Go is to use standard free energy of formations (∆Gof). Standard free energy of formation (∆Gof)- the change in energy that accompanies the formation of 1 mole of a substance from its elements. ∆Go = ∑np∆Gof(products) - ∑nr∆Gof(reactants) Assignment: 51, 53, 55