Spontaneity, Entropy, and Free Energy Thermodynamics Spontaneity, Entropy, and Free Energy

First Law of Thermodynamics Law of Conservation of Energy Energy can change forms Not “lost”, but changed Discuss things like How much energy is exchanged? Where does the energy go? (calorimeter) What form is the energy?

Spontaneous Processes A process is spontaneous if it occurs without outside intervention. We discuss the direction of the reaction Says nothing of the kinetics or rate For example: A ball rolls down hill, but never spontaneously rolls uphill. Iron exposed to water rusts. Rust does not spontaneously turn into iron A container will fill uniformly with a gas; the gas does not spontaneously pool at one end.

Spontaneous Processes Spontaneous processes are those that can proceed without any outside intervention. The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously return to vessel B.

Kinetics The reaction pathway Thermodynamics the initial and final states

2nd Law of Thermo Entropy in the universe is increasing The driving force for spontaneous processes is an increase in Entropy Natural tendency is to go from ordered to disordered Take a deck of cards. Throw them into air. When you put them back, what are the chances they are all in order? But there is a chance, however unlikely.

2nd Law Entropy is a function that describes the number of possible arrangements Available to a particular system Nature proceeds toward the states that have the highest probability of existing The driving force is “probability”

Let’s Look at a Simple System Four atoms of an ideal gas Three possible arrangements How many ways can each state be achieved?

Examine All Possibilities (Pg 795)

Possibilities The arrangement with two on each side is most likely to occur By the ratio of 6:4:1

Probability of finding all the Molecules in the Left Bulb as a function of the total number of molecules

Unlikely to Occur 1 in 10 or not likely to occur But it is possible! 2 x 1023 But it is possible!

Positional Entropy A gas expands into a vacuum Because the expanded state has the highest positional probability or entropy of all the states available to the gas Illustrated by changes of state The larger the intermolecular distances, the more states available The more states, the more entropy

Coffee Cup Explain on a molecular level how a hot cup of coffee cools to room temperature What is the possibility of this whole process going in reverse? But it is possible! Next time your coffee is cold, just wait for it to get hot.

Entropy on the Molecular Scale Ludwig Boltzmann described the concept of entropy on the molecular level. Temperature is a measure of the average kinetic energy of the molecules in a sample.

Entropy on the Molecular Scale Molecules exhibit several types of motion: Translational: Movement of the entire molecule from one place to another. Vibrational: Periodic motion of atoms within a molecule. Rotational: Rotation of the molecule on about an axis or rotation about  bonds. All of these are considered microstates of a system. 

Entropy on the Molecular Scale Each molecule has a specific number of microstates, W, associated with it. Entropy is S = k lnW where k is the Boltzmann constant, 1.38  1023 J/K.

Entropy on the Molecular Scale The change in entropy for a process, then, is S = k lnWfinal  k lnWinitial Wfinal Winitial S = k ln Entropy increases with the number of microstates in the system.

Standard Entropies Larger and more complex molecules have greater entropies.

Entropy Kinetic-molecular view For an ideal gas at one atmosphere of pressure, as the temperature is lowered, the volume will be reduced. At 0 K, the molecules will have no energy of motion. There is only one possible arrangement for the molecules. Ideal gas at one atm and 0 K.

Entropy and temperature The entropy of an ideal gas at constant pressure increases with increasing temperature. This is because the volume increases. 0 K T1 T2 T3

Entropy and temperature There are other reasons for entropy to increase with increasing temperature. Increased temperature will result in a greater distribution of molecular speeds. speed number T3 T2 T1 T1 < T2 < T3

Entropy and temperature Increased temperature also results in more energy levels in atoms and molecules being occupied. For molecules, this means that they will be able rotate and their bonds can vibrate. This further increases entropy.

Examples of Entropy What has more entropy Gas or liquid? Solid or liquid? Homogeneous solution or separate mixture sugar dissolved in water or sugar and water The more random or lack of order The more entropy Do you have it? Iodine vapor condensing on cold glass? Gas at 1 atm or 1 x 10-2 atm? S = Sfinal – Sintial

2nd Law Restated In any spontaneous process, there is always an increase in the entropy of the universe Suniverse = Ssystem + Ssurroundings If S univ > 0, process is spontaneous. If S univ < 0, process is non-spontaneous. The process is spontaneous in the other direction. If S univ = 0, process has no tendency to occur or is at equilibrium.

How can complex molecules assemble in a bacteria? The created order is in the bacteria. The energy needed for this activity is supplied from an external source. The Universe gains entropy while the cell is organized. Most of our energy comes from the sun. The constant influx of energy supplies the energy to overcome entropy…. for the time being!

The Sun is Entropic! Stars produce light in all directions This energy is spread through the universe Sounds entropic Think about a star that is 1 million light years away.

Star

Star Further away

Chaos Theory Chaotic events tend to organize themselves Best example is a whirlpool. (toilet) The particles organize themselves in order to become disorganized more efficiently

How can we determine if a process is spontaneous? Suniverse = Ssystem + Ssurroundings The sign of Ssurr depends on direction of heat flow exothermic process adds energy to the universe The universe now has more random motion So the universe experiences an increase in entropy Suniverse > 0 or positive.

Ssurrounding Magnitude of Ssurr depends on the temperature If the surroundings have a low temp, additional energy makes a big difference If the surroundings have a high temperature, additional heat does not add much more energy (entropy) it has little effect. (little change, small Ssurr)

Entropy Continued The tendency for a system to lower its energy becomes more important at lower temperatures. Driving Force Provided by energy flow Magnitude of the Entropy change of The surroundings Quantity of heat (J) temperature (K)

Entropy depends on Enthalpy The change in Enthalpy, H, which is the direction and magnitude of heat exchanged Energy of system is proportional to its temp in kelvin in an isothermal system. Change in enthalpy exotherm = neg endotherm = pos J = - H = Ssurr K T

Spontaneity S system Surrounding Universe Spontaneous? + Yes - No (process in opposite direction ? Yes if Ssys > Ssurr Ssys < Ssurr

Gibbs Free Energy There is a “war” between order and disorder Enthalpy and Entropy The sun is the source of our energy It drives our enthalpic world If the sun were to stop, how long would live still exist. In a million years would things still look the same?

G = H - T S This war can be described mathematically G is Gibbs Free Energy Gibbs Free Energy is the energy “free” to do work We will use this to determine the “force” behind reactions Remember the second law!

Free Energy G = Gibbs Free Energy G = H – TS In processes where temp is constant G = H - T S We are referring to the system No subscripts needed

Free Energy G = H - T Ssys If we divide by –T -G = - H + Ssys - H = Ssurr T T T -G = Ssurr + Ssys = Suniv at constant T, P T At what value of G , is Suniv > 0 or “spontaneous”

Spontaneity Again Processes are spontaneous H2O(s)  H2O (l) H = 6.03 x 103 J/mol S = 22.1 J/K • mol If they have a positive Suniv If they have a negative G , at constant P,T G = H - T Ssys Spontaneous processes have negative G

Is Water Melting Spontaneous? Will this be spontaneous at -10, 0, or 10oC? H2O(s)  H2O (l) H = 6.03 x 103 J/mol S = 22.1 J/K • mol G = H - T Ssys

Calculate Sunv and G G -10 263 -22.9 -0.8 5.81 + 2.2 x102 273 -22.1 H = 6.03 x 103 J/mol S = 22.1 J/K • mol G = H - T Ssys T (°C) T K -H = Ssurr S + Ssurr=Sunv TS X 103 G -10 263 -22.9 -0.8 5.81 + 2.2 x102 273 -22.1 6.03 10 283 -21.3 +0.8 6.25 - 2.2 x102

The spontaneity of the process depends on the temp Result Positive Negative Spontaneous at All temps Spontaneous at High Temps (exotherm not important) Spontaneous at Low Temps (Exotherm is important) Not Spontaneous Process Reverse spontaneous at all temps

Gibbs Free Energy If DG is negative, the forward reaction is spontaneous. If DG is 0, the system is at equilibrium. If G is positive, the reaction is spontaneous in the reverse direction.

G < 0 for spontaneous process G = 0 for equilibrium process Br2(l)  Br2(g) At what temp is the following process spontaneous at 1 atm? What is the normal boiling point of liquid Br2? H = 31.0 kJ/mol S = 93.0 J / K • mol G < 0 for spontaneous process G = 0 for equilibrium process G = H - T Ssys 0 = H - T Ssys = 31.0 x 103 – T (93.0) T = 333K T > 333 K Ssys is dominant. Liquid vaporizes T = 333 K G = 0, liquid and vapor coexist (normal BP) (exothermic processes dominant) T < 333 K H is dominant. Liquid forms.

What About Reactions? Chemistry is all about the changes that occur. How can we use thermo and entropy to evaluate the changes around us?

Which has greater positional entropy?

@ Constant Temperature and Pressure Why would we use this as a constraint on a thermodynamic system? 2nd law Suniv = Ssys + Ssurr No temp change means no Ssurr 4NH3(g) + 5O2 (g)  4NO(g) + 6H2O(g) Is this process thermodynamically favored?

How about this? Al2O3(s) + 3H2(g) → 2Al(s) + 3H2O(g) Same amount of gas on both sides. Entropy would appear equal. Its actually +179J/K. Why? Water is more complex a molecule than hydrogen. More ways it can move = more entropy.

And this? Cdiamond → Cgraphite ∆Go = -3kJ So how come we still have diamonds?

Third Law of Thermodynamics When can perfect order be achieved? What conditions would have to be necessary to first achieve it, and the keep it that way? The only time the entropy is zero is when you have a perfect crystal at 0K Any rise in temperature will create movement and therefore raise entropy.

Other information As with enthalpy which is a state function, Ho = np Hf products - nr Hf reactants So too with entropy and free energy. They are both state functions So = np S products - nr S reactants Go = np Gf products - nr Gf reactants free energy of formations for an element in its standard state is zero. Also free energy and entropy for reactions can be added like Hess’s Law.

Free Energy & the Equilibrium Constant Recall that G and K (equilibrium constant) apply to standard conditions. However, G and Q (reaction quotient) apply to any conditions. It is useful to determine whether substances under any conditions will react: Where R is the ideal gas constant, 8.314 J/mol•K

Free Energy & the Equilibrium Constant At equilibrium, Q = K and G = 0, so From the above we can conclude: If G < 0, then K > 1. If G = 0, then K = 1. If G > 0, then K < 1.

G = - RT lnK Free Energy & the Equilibrium Constant Solving for the equilibrium constant, K , G = - RT lnK

G and work G is the value of all free energy from a reaction. Therefore its value is equal to the maximum work possible from a reaction. (if -) If G is positive, what does it tell us? Used for efficiency. Will never be 100%, why?

Summary of Thermo 1st law says you can’t win, only break even. 2nd law says you can’t break even. Explains energy crisis!