Spontaneity, Entropy & Free Energy

Spontaneity, Entropy & Free Energy
First Law of Thermodynamics Basically the law of conservation of energy energy can be neither created nor destroyed i.e., the energy of the universe is constant the total energy is constant energy can be interchanged e.g. potential energy (stored in chemical bonds) can be converted to thermal energy in a chemical reaction CH4 + O2 --> CO2 + H2O + energy Doesn’t tell us why a reaction proceeds in a particular direction

Spontaneous Processes and Entropy Spontaneous processes occurs without outside intervention Spontaneous processes can be fast or slow

Kinetics depends on the pathway taken tells us the speed of the process depends on activation energy temperature concentration catalysts

Spontaneous Processes a ball rolls downhill, but the ball never spontaneously rolls uphill steel rusts, but the rust never spontaneously forms iron and oxygen a gas fills its container, but a gas will never spontaneously collect in one corner of the container. Water spontaneously freezes at temperatures below 0o C

What thermodynamic principle explains why these processes occur in one direction? ANSWER: ENTROPY The driving force for a spontaneous reaction is an increase in the entropy of the universe

Symbol: S A measure of randomness or disorder The natural progression is from order to disorder It is natural for disorder to increase Entropy is a thermodynamic function Describes the number of arrangements that are available to a system in a given state

The greater the number of possible arrangements, the greater the entropy of a system, i.e., there is a large positional probability. The positional probability or the entropy increases as a solid changes from a liquid or as a liquid changes to a gas

Ssolid < Sliquid < Sgas Choose the substance with the higher positional entropy: CO2(s) or CO2(g)? N2(g) at 1 atm and 25oC or N2(g) at .010 atm and 25oC?

Predict the sign of the entropy change solid sugar is added to water iodine vapor condenses onto a cold surface forming crystals

Second Law of Thermodynamics The entropy of the universe is increasing The universe is made up of the system and the surroundings DSuniverse = DSsystem + DSsurroundings

A process is spontaneous if the DSuniverse is positive If the DSuniverse is zero, there is no tendency for the reaction to occur

The effect of temperature on spontaneity H2O(l) --> H2O(g) water is the system, everything else is the surroundings DSsystem increases, i.e. DSsystem is positive, because there are more positions for the water molecules in the gas state than in the liquid state

What happens to the surrounding? Heat leaves the surroundings, entering the system to cause the liquid molecules to vaporize When heat leaves the surroundings, the motion of the molecules of the surroundings decrease, which results in a decrease in the entropy of the surroundings DSsurroundings is negative

Sign of DS depends on the heat flow Exothermic process: DSsurr >0 Endothermic process: DSsurr< 0 Magnitude of DS is determined by the temperature DSsurr = - DH T

Signs of Entropy Changes DSsys DSsurr DSuniv Spontaneous?

aka Gibbs Free Energy G another thermodynamic function related to spontaneity G = H - TS for a process that occurs at constant temperature (i.e. for the system): DG = DH - TDS

How does the free energy related to spontaneity? DG = DH - TDS - DG = - DH + DS (remember, - DH = DSsurr ) T T T -DG = DSsurr + DSsys (remember, DSsurr + DSsys = DSuniv) T -DG = DSuniv

DSuniv > 0 for a spontaneous reaction DG < 0 for a spontaneous reaction DG > 0 for a nonspontaneous reaction Useful to look at DG because many chemical reactions take place under constant pressure and temperature Now we have two ways to predict spontaneity!!

PRACTICE!!! H2O(s) --> H2O(l) Calculate DG at -10oC, 0oC, and 10oC
DHo = 6.03 x 103J/mole DSo = 22.1 J/K.mole Calculate DG at -10oC, 0oC, and 10oC At which temperatures is this process spontaneous? Does this make sense? Explain why or why not.

For the melting of ice DSsys and D Ssurr oppose each other spontaneity will depend on temperature DSo is positive because of the increase in positional entropy when the ice melts DSsurr is negative because the reaction is endothermic

Br2(l) --> Br2(g) DHo= 31.0 kJ/mol DSo = 93.0 J/K.mol At what temperatures is Br2(l) --> Br2(g) spontaneous?

Entropy Changes in Chemical Reactions Just like physical changes, entropy changes in the surroundings are determined by heat flow Entropy changes in the system are determined by positional entropy (the change in the number of possible arrangements)

N2 (g) + 3 H2(g) --> 2 NH3 (g) The entropy of the this system decreases because four reactant molecules form two product molecules there are less independent units in the system less positional disorder, i.e. fewer possible configurations

When a reaction involves gaseous molecules: the change in positional entropy is determined by the relative numbers of molecules of gaseous reactants and products I.e., if you have more product molecules than reactant molecules, DS will be positive

In thermodynamics, the change in a function is usually what is important usually we can’t assign an absolute value to a function like enthalpy or free energy we can usually determine the change in enthalpy and free energy

We can assign absolute entropy values, i.e., we can find S A perfect crystal at 0 K, while unattainable, represents a standard all molecular motion stops all particles are in their place the entropy of a perfect crystal at 0 K is zero = third law of thermodynamics

Increase the temperature of our perfect crystal molecular motion increases disorder increases entropy varies with temperature See thermodynamic tables for So values (at 298 K and 1 atm)

Entropy is a state function entropy does not depend on the pathway taken DSrxn = SnDSoproducts - SnDSoreactant

Calculate DSo at 25oC for 2NiS(s) + 3 O2(g) --> 2 SO2(g) + 2 NiO(s) Substance So(J/K.mol) SO2 248 NiO 38 O NiS 53

Calculate DSo for Al2O3(s) + 3 H2(g) --> 2 Al(s) + 3 H2O(g) Substance So (J/K.mol) Al2O3 51 H Al 28 H2O 189

What did you expect the DSo to be? Why is it large and positive? H2O is nonlinear and triatomic H2O has many rotational and vibrational motions H2 is linear and diatomic H2 has less rotational and vibrational motions The more complex the molecule, the higher the DSo

Free Energy and Chemical Reactions Standard Free Energy Change DGo the change in the free energy that occurs if the reactants in their standard states are changed to products in their standard states can’t be measured directly calculate from other values allows us to predict the tendency for a reaction to go

How do we calculate DGo? DGo = DHo - TDSo (for a reaction carried out at constant temperature) Use Hess’ Law Use DGof (standard free energy of formation) DGo = SnDGof (products) - SnDGof (reactants)

Calculate DGo for the reaction at 25oC 2SO2(g) + O2(g) --> 2 SO3(g) Substance DHof(kJ/mol) DSo (J/K.mol) SO2(g) SO O

Calculate DGo for the reaction Cdia --> Cgr using the following data: Cdia + O2 --> CO2(g) DGo = -397 kJ Cgr + O2 --> CO2(g) DGo = -394 kJ

Calculate DGo for the reaction 2CH3OH + 3 O2--> 2 CO2 + 4 H2O Substance DGof(kJ/mol) CH3OH -163 O2 0 CO H2O -229

The dependence of free energy on pressure How does pressure affect enthalpy and entropy? Pressure does not affect enthalpy Pressure does affect entropy because pressure depends on the volume 1 mole of a gas at 10.0 L has more positions available than 1 mole of a gas at 1.0 L Slarge volume > Ssmall volume Slow pressure > Shigh pressure

Given that G = DGo + RTln(P) where G is the free energy at some P (not necessarily 1 atm) where DGo is the free energy at 1 atm Ex: N2(g) + 3 H2(g) --> 2 NH3(g) (lots of equations…lots of equations…) DG = DGo + RT ln Q Q is the reaction quotient (from the law of mass action) T is the temperature in K R is the gas constant, J/mol.K

Calculate DG at 25o C for the reaction CO(g) + 2 H2(g) --> CH3OH where carbon monoxide is 5.0 atm and hydrogen gas at 3.0 atm are converted to liquid methanol. What does the answer tell us about this reaction under these conditions?

Free Energy and Equilibrium Equilibrium occurs at the lowest value of free energy available to the reaction system, i.e., when DG = 0 At equilibrium, DG = 0, Q = Keq so DG = 0 = DGo + RT ln Keq DGo = - RT ln Keq Use this equation to find Keq given DGo, or to find DGo given Keq

Relationship between DGo and Keq DGo Keq = 0 1 < 0 >1 > 0 < 1

For N2 + 3 H2 --> 2 NH3, DGo = kJ per mole of N2 consumed at 25oC. Predict the direction in which the reaction will shift to reach equilibrium a. PNH3 = 1.00 atm, PN2 = 1.47 atm, PH2 = 1.00 x 10-2 atm b. PNH3 = 1.00 atm, PN2 = 1.00 atm, PH2 = 1.00 atm

4Fe + 3 O2 <====> 2Fe2O3 Calculate the equilibrium constant using the following information: Substance DHof (kJ/mol) So(J/K.mol) Fe2O Fe O

Keq and temperature We used Le Chatelier’s Principle to determine how Keq would change when temperature changes Use DG to determine the new Keq at a new temperature DGo = -RT ln K = DHo - TDSo ln K = - DHo DSo R T R

Spontaneity, Entropy & Free Energy

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