# Spontaneity, Entropy, and Free Energy

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Spontaneity, Entropy, and Free Energy

Spontaneous Processes and Entropy

First Law of Thermodynamics
"Energy can neither be created nor destroyed", so while energy can be converted in a chemical process, the total energy remains constant The energy of the universe is constant

Spontaneous Processes
Processes that occur without outside intervention Spontaneous processes deal with the natural direction of a process and not with the speed at which it occurs-Spontaneous processes may be fast or slow Many forms of combustion are fast Conversion of diamond to graphite is slow

Entropy (S) A measure of the randomness or disorder
The driving force for a spontaneous process is an increase in the entropy of the universe Entropy is a thermodynamic function describing the number of arrangements that are available to a system Nature proceeds toward the states that have the highest probabilities of existing

Positional Entropy The probability of occurrence of a particular state depends on the number of ways (microstates) in which that arrangement can be achieved Ssolid < Sliquid << Sgas

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" If a process is spontaneous in one direction then it can’t be spontaneous in the reverse direction "The entropy of the universe is increasing" For a given change to be spontaneous, DSuniv must be positive We can predict whether a process is spontaneous by considering entropy changes that occur in the system and the surroundings nSuniv = nSsys + nSsurr

State functions-all depend only on the change between the initial and final states of a system, not on the process by which the change occurs For chemical reactions, this means that the thermodynamic state functions are independent of reaction pathways Enthalpy change (nH), entropy change (nS), and free-energy change (nG) are state functions.

Standard state conditions
Superscript circle represents standard state conditions All gases are at 1 atm pressure All liquids are pure All solids are pure All solutions are at 1M concentration Energy of formation of an element in its normal state is defined as zero T used for standard state values is almost invariably room temperature, 25oC (298K), though standard state values can be calculated for other temperatures.

Enthalpy Review When bonds are formed, energy is released
In order to break bonds, energy must be absorbed nH = Hproducts – Hreactants if products have stronger bonds than reactants, then products have lower enthalpy than reactants and are more stable-energy is released, so exothermic if products have weaker bonds than reactants, then products have higher enthalpy than reactants and are less stable-energy is absorbed, so endothermic since substances like lowest possible energy state which gives them the greatest stability, exothermic reactions are more likely to occur spontaneously than endothermic reactions

Will a reaction be spontaneous?
Enthalpy change Entropy change Spontaneous? Exothermic (nH < 0) Increase (nS > 0) Yes, always Exothermic (nH < 0) Decrease (nS < 0) Only at lower Ts Endothermic (nH > 0) Increase (nS > 0) Only at higher Ts Endothermic (nH > 0) Decrease (nS < 0) No, never

The Effect of Temperature on Spontaneity

Direction of Heat Flow Entropy changes in the surroundings are primarily determined by heat flow Exothermic reactions in a system at constant temperature increase the entropy of surroundings Endothermic reactions in a system at constant temperature decrease the entropy of surroundings The impact of the transfer of a given quantity of energy as heat to or from the surroundings will be greater at lower temperatures

Sign of nSsurr depends on the direction of the heat flow
+ for exothermic process – for endothermic process Magnitude of nSsurr depends on both the quantity of energy that flows as heat and the temperature at which the energy is transferred Entropy changes in the system are dominated by the change in the number of gaseous molecules Fewer gaseous molecules on the product side means a decrease in entropy, and vs. Molecular structure also plays a role-a complex molecule has a higher standard entropy than a simpler one

Free Energy (G), also called "Gibbs Free Energy"

Calculating Free Energy Change (constant temperature and pressure)
nG = nH - TnS H is enthalpy T is Kelvin temperature

Free Energy and Spontaneity
Reactions proceed in the direction that lowers their free energy (-nG)-process that occurs at constant temperature and pressure is spontaneous in the direction in which the free energy decreases At low temperature enthalpy is dominant At high temperature entropy is dominant

nH TnS nG Spontaneity Spontaneous at all temperatures Nonspontaneous at all temperatures - - ??? Spontaneous if the absolute value of nH is greater than the absolute value of TnS (low temperature) - Spontaneous at low temperatures + Nonspontanteous at high temperatures + + ??? Spontaneous if the absolute value of TnS is greater than the absolute value of nH (high temperature) + Nonspontaneous at low temperatures - Spontaneous at high temperatures

Entropy Changes in Chemical Reactions

Constant Temperature and Pressure
Reactions involving gaseous molecules The change in positional entropy is dominated by the relative numbers of molecules of gaseous reactants and products 2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g) 9 molecules 10 molecules nS increases

Third Law of Thermodynamics
"The entropy of a perfect crystal at O K is zero" (NO disorder, since everything is in perfect position) Entropy-randomness or disorder of the system Greater disorder, greater the entropy Since zero entropy is defined to be a solid crystal at 0K, all substances have positive value for entropy Liquids have higher entropy than solids Gases have higher entropy than liquids Particles in solution have higher entropy than solids Two moles of a substance has a higher entropy than one mole

Calculating Entropy Change in a Reaction
Entropy is an extensive property (a function of the number of moles) Generally, the more complex the molecule, the higher the standard entropy value nS0 = Σ nS0products - Σ nS0reactants

Free Energy and Chemical Reactions

Standard Free Energy Change
nG0 is the change in free energy that will occur if the reactants in their standard states are converted to the products in their standard states nG0 cannot be measured directly The more negative the value for nG0, the farther to the right the reaction will proceed in order to achieve equilibrium Equilibrium is the lowest possible free energy position for a reaction If nG is negative, the reaction is spontaneous If nG is positive, the reaction is not spontaneous If nG = 0, the reaction is at equilibrium

Calculating Free Energy Change
Method #1, for reactions at constant temperature: nG0 = nH0 - TnS0 Method #2, an adaptation of Hess's Law: Cdiamond(s) + O2(g)  CO2(g) nG0 = -397 kJ Cgraphite(s) + O2(g)  CO2(g) nG0 = -394 kJ Cdiamond(s)  Cgraphite(s) nG0 = -397 kJ - (-394 kJ) = -3kJ (Complete example is on page 798) Method #3, using standard free energy of formation (DGf0) Standard Free Energy of Formation is the change in free energy that accompanies the formation of 1 mole of that substance from its constituent elements with all reactants and products in their standard states nGf0 of an element in its standard state is zero nS0 = Σ nG0products - Σ nG0reactants

The Dependence of Free Energy on Pressure
Enthalpy, H is not pressure dependent

Entropy, S entropy depends on volume, so it also depends on pressure
S large volume > S small volume S low pressure > S high pressure Standard free energy change gives spontaneity of reaction when all concentrations of reactants and products are in standard state concentrations (1M) Free energy change, and thus spontaneity, of reaction will be different from standard free energy change if initial concentrations of reactants and products are not 1M

The standard free energy change can be related to energy change for other conditions by G = Go + RT ln(P) G0 is the free energy of the gas at a pressure of 1 atm G is the free energy of the gas at a pressure of P atm R is the universal gas constant, T is Kelvin temperature nG = nG0 + RT ln(Q) Q is the reaction quotient (from the law of mass action, section 13.5) R is the gas constant ( J/K ×mol) nG0 is the free energy change for the reaction with all reactants and products at a pressure of 1 atm nG is the free energy change for the reaction for the specified pressures of reactants and products

Free Energy and Equilibrium

Thermodynamic View of Equilibrium
Equilibrium point occurs at the lowest value of free energy available to the reaction system At equilibrium, nG = 0 and Q = K nG0 = -RT ln(K) nG0 K nG0 = 0 and K = 1, reactants and products equally favored at equilibrium nG0 < 0 and K > 1, products favored at equilibrium nG0 > 0 and K < 1, reactants are favored at equilibrium

Temperature Dependence of K
nG0 = -RT ln ( K) = nH0 -TnS0

Free Energy and Work Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy wmax = nG

Energy Released in many chemical and physical processes that is available to do work

Free energy-energy available to do work
Spontaneous change-skier going downhill, waterfall, fuel fire at gasoline refinery, book slips from your hand and will start to fall, ice cubes in a cold drink will gradually melt. Decrease in energy of system. Occur by themselves without outside assistance. Once conditions are right to begin, proceed on their own. Once begins, tendency to continue until it is finished. Some occur rapidly (pulling hand from hot stove), some slowly over many years (erosion of mountain), and some so slowly nothing appears to be happening (Gasoline-oxygen mixtures appear perfectly stable indefinitely at room T). If heated, gasoline-oxygen mixture’s rate of reaction increases-can react explosively until all of one or the other is totally consumed.

Give substantial amounts of products at equilibrium.
Spontaneous reactions naturally favor formation of products at specified conditions Give substantial amounts of products at equilibrium. All release free energy (exergonic) which lowers the energy of the system (exothermic) Exothermic changes have the tendency to proceed spontaneously The word tendency should be emphasized. Some exothermic reactions are not spontaneous, while not every nonspontaneous is endothermic. Salts dissolving in water-spontaneous change that is endothermic.

Nonspontaneous change-pile of bricks in morning later becomes brick wall.
Continues only as long as receives some outside assistance. Occurs only when some spontaneous event has occurred first. Nonspontaneous reactions do not favor formation of products at specified conditions (there are more reactions formed over products).

Universal phenomenon that something that brings about randomness is more likely to occur than something that brings about order. Your room does not spontaneously get more organized It gets messy after it is cleaned There is a tendency for things to become more disorganized and random. Throw new deck of cards that was in order into air-land disordered.

Enthalpy Enthalpy, H, (heat content of a substance): natural systems tend to go from a state of higher energy to a state of lower energy. A ball rolls down a hill spontaneously, but not up. The ball looses potential energy as it rolls downhill. At the bottom of the hill it has zero potential energy.

Enthalpy is the heat content of a substance.
The change in enthalpy in a reaction is the difference between the heat contained in the products and the heat contained in the reactants. In an endothermic reaction, heat is absorbed, which would create a positive enthalpy. In an exothermic reaction, heat is given off, which means the reaction would have a negative enthalpy.

Entropy Entropy, S, (disorder of a substance): natural processes tend to go from an orderly state to a disorderly state. Place 3 layers of different colored marbles in a container and shake it No longer ordered and no matter how much you shake box, chances getting it back to original pattern next to zero. Entropy describes the degree of randomness in a system. The larger the entropy value, the larger the degree of randomness of system. Enthalpy/entropy state functions-depends on state of system.

Any event w/increase in entropy tends to be spontaneous
∆S = Sfinal -Sinitial or, for chemical systems as ∆S=Sproducts - Sreactants If Sproducts > Sreactants is positive, there is an increase in entropy Tends to be spontaneous. Any event w/increase in entropy tends to be spontaneous

When nitroglycerine explodes, heat released and disorder increased
When nitroglycerine explodes, heat released and disorder increased. Both conditions are favorable.

Law of disorder-things move in direction of maximum disorder or randomness (chaos).
Entropy gas > entropy liquid > entropy solid. Increases when substance divided into parts (dissolve S in L). Tends to increase when # products molecules > # reactant molecules.

Physical State: Gas > Liquid > Solid
Entropy changes associated with changes in state can be predicted Increasing the disorder increases entropy Ssolids < Sliquids < Sgases H2O(l)  H2O(g) ∆Ssystem > 0

Temperature: An increase in the temperature of a substance is always accompanied by an increase in the random motion of its particles Increased temperature means increased kinetic energy which means faster movement, more possible arrangements and increased disorder. ∆Ssystem > 0

Number of Molecules: Entropy increases whenever reaction gives increase in number of gaseous particles. Assuming no change in physical state, the entropy of a system usually increases when the number of gaseous product particles is greater than the number of gaseous reactant particles Larger #number of gaseous particle, the more random arrangements are available 2SO3(g)  2SO2(g) + O2 (g) ∆Ssystem > 0

For same type of bonding, larger molecules possess higher entropy.
Molecular Size: For same type of bonding, larger molecules possess higher entropy. Relative Entropies: CH4 < C2H6 < C3H8 < C4H10

Gas particles have more entropy when they can move freely:
Mixing: Dissolving of gas in a solvent always results in a decrease in entropy Gas particles have more entropy when they can move freely: Sgas > Sdissolved gas CO2(g)  CO2 (aq) ∆Ssystem < 0

Mixing: With some exceptions, you can predict the change in entropy when a solid or a liquid dissolves to form a solution Solute particles become dispersed throughout the solvent, increasing randomness and disorder of the particles NaCl(s)  Na+(aq) + Cl-(aq) ∆Ssystem > 0

Expansion:

Think randomness:

Entropy, the universe and free energy
Spontaneous vs. Nonspontaneous Reactions

The law of disorder states that the entropy of the universe must increase as a result of a spontaneous reaction or process: ∆Suniverse > 0 for any spontaneous reaction. Since the universe equals the system plus the surroundings, any change in the entropy of the universe is the sum of changes occurring in the system and surroundings. ∆Suniverse = ∆Ssystem + ∆Ssurroundings

Changes in the system’s enthalpy and entropy affects ∆Suniverse
The reaction or process is exothermic which means ∆Hsystem is negative. Heat released raises T of surroundings so it increases the entropy of the surroundings, making ∆Ssurroundings positive. The entropy of the system increases, so ∆Ssystem is positive. Exothermic chemical reactions accompanied by an increase in entropy are all spontaneous .

Second Law of Thermodynamics:
Whenever a spontaneous event takes place in the universe, it is accompanied by an overall increase in entropy (in any isolated system, the degree of disorder can only increase). Movement towards order requires E-creation of disorganization requires no E-spontaneous Folding shirt takes more energy than throwing it on ground. Entropy of substance varies with T-lower T, lower entropy.

In a spontaneous process the entropy of the universe must increase:
∆Ssystem + ∆Ssurroundings > 0 Universe becomes more disordered.

Spontaneous reactions:
Reaction is spontaneous if reaction is exothermic or entropy of system may increase, or both may occur (free energy is released) Highly exergonic- Exothermic-heat released-heat content  (favorable) Entropy increases-more disorder (favorable) Slightly exergonic- exothermic (favorable) and offsets ii. Entropy decreases-more order (unfavorable) Exergonic- Endothermic-heat content increases (unfavorable) Entropy increases (favorable) and offsets i.

Nonspontaneous reactions:
Reaction is nonspontaneous if reaction is endothermic or entropy of system decreases, or both may occur Highly endergonic Endothermic-heat content increases (unfavorable) Entropy decreases (unfavorable) Slightly endergonic Endothermic (unfavorable) and offsets ii. Entropy increases (favorable) Endergonic Exothermic (favorable) Entropy decreases (unfavorable) and offsets i.

(to the left) (to the right)

Example: A mixture of 2.00 mol of H2, 1.00 mol of N2, and 2.00 mol of NH3 is placed in a 1.00-L container at 472oC. Would the reaction N2(g)  +  3 H2(g) { 2 NH3(g)  Kc = proceed toward the right or the left? Determine reaction quotient, Qc (expression having same form as Keq expression but whose concentration values are initial concentrations-given moles) and compare to Kc For the reaction above Qc = [NH3] [2.00]2 Qc = = = 0.500 [N2][H2] [1.00][2.00]3 Since Qc > Kc reaction proceeds left to reach equilibrium

Third Law of Thermodynamics:
At absolute zero, the entropy of a pure crystal is also zero. P=1 atm/T>100oC, water highly disordered gas w/very high entropy. If confined-molecules spread evenly throughout container and have constant motion. When system cooled, water vapor condenses to form a liquid and the entropy is lowered. Below 0oC, water molecules join together to form ice and are highly ordered, yet the order is not perfect Water still has some entropy because there is enough thermal energy left to cause them to vibrate and rotate within general area of lattice sites. Cool further, decrease thermal energy and entropy decrease Absolute zero, ice in state of perfect order-entropy will be zero.

How to calculate standard entropy change:
1 mol substance at 298K (25oC) and P = 1 atm-standard entropy, So. Units of J/Ko So = (sum of So of products)- (sum of So of reactants) So = Σ So products - Σ Soreactants How to calculate standard entropy change: Write the phase change as a balanced reaction. Find standard entropies, So, from table. Calculate the change in entropy by subtraction.

Example: Urea (from urine) hydrolyses slowly in the presence of water to produce ammonia and carbon dioxide. What is the standard entropy change, in J/K, for this reaction when 1 mole of urea reacts with water? CO(NH2)2(aq) + H2O(l)  CO2 (g) + 2NH3(g) So = [CO2 (g) + (2)NH3 (g)] - [CO(NH2) 2(aq) + H2O(l)]      = [213.6 J/K + (2)192.5 J/K] - [173.8 J/K J/K]      = (598.6 J/K)-(243.8 J/K) = J/K Since 2 molecules become 3 molecules, there is an increase in entropy, so the positive value.

Example: Evaluate the entropy change for the reaction CO(g) + 3 H2(g)  CH4(g) + H2O(g). So = (sum of So of products)-(sum of So of reactants) ∆S = [CH4(g) + H2O(g)] – [CO(g) + (3)H2(g)] ∆S = [ ] – [ (131)] J/K = -216 J (K mol)-1 Since 4 mol of reactants form 2 mol of products (all in gaseous state), there is a decrease in entropy, so the large negative change for the reaction.

3 mol gases on reactant side/1 mol of liquid on product side
And another: CO(g) + 2 H2(g)  CH3OH(l) 3 mol gases on reactant side/1 mol of liquid on product side Changing from more disordered to more ordered state. S (Sreactants - Sproducts) should be negative since Sreactants > Sproducts. So= [CH3OH(l) - CO(g)] – [(2)H2 (g)] = J/Kmol J/Kmol - (2)130.7 J/Kmol = J/K (negative as we had predicted).

Free Energy: Reactions can do work/serve as pathways for conversion of heat into work. Energy which is, or which can be, available to do useful work is called Gibbs free energy (G) or more often simply free energy. It is the comparison of changes of enthalpy & entropy during chemical reaction. It tells if there is energy available to do work.

G = H - T∆S where H is enthalpy, S is entropy, and T is absolute T
Free Energy: G = H - T∆S where H is enthalpy, S is entropy, and T is absolute T These factors can have either -/+ effect on whether or not a reaction will occur spontaneously or at all. All spontaneous processes move toward equilibrium. Reactions occur naturally to reach lower state of energy.

Change can only be spontaneous if it is accompanied by a decrease in free energy-G must be negative

H is negative/S is positive, G = H - T∆S = (-) - T(+)
Can predict at what temperature a nonspontaneous reaction will become spontaneous: H is negative/S is positive, G = H - T∆S = (-) - T(+) G will be negative regardless of value of the absolute temperature, T (can only be positive). Change will occur spontaneously at all T Increase in entropy-both factors favor spontaneity. H is positive/S is negative, G = H - T∆S = (+) - T(-) G will be positive at all temperatures and the change will always be nonspontaneous. Decrease in entropy-both factors work against spontaneity If H/S both positive, G = (+) - T(+) When H/S have the same sign, temperature becomes critical in determining spontaneity of an event. Only at relatively high T will value T∆S be larger than value of H so difference, G, is negative

Calculate free energy:
2 H2 + O2  2H2O Go = Gof (products) - Gof (reactants) Go = (2 x ) – (2 x ) = kJ/mol Spontaneous because it is negative.

Compute Go for the hydrolysis of urea, CO(NH2)2,
Example: Compute Go for the hydrolysis of urea, CO(NH2)2, CO(NH2)2(aq) + H2O(l)  CO2(g) + 2 NH3(g) Go = [CO2 + (2)NH3] - [CO(NH2)2 + H2O] = [ (2) ] - [ ] = ( kJ) - (-605.1) = kJ

Another example: What is Go for the combustion of ethyl alcohol (C2H5OH) to give CO2(g) and H2O(g)? Balance equation:C2H5OH(l) + 3O2(g) 2 CO2(g) + 3 H2O(g) Go = [(2)CO2(g) + (3)H2O(g)] - [C2H5OH(l) + (3)O2(g)] Go = [2 mol( kJ/mol) + 3 mol( kJ/mol)] = [1 mol( kJ/mol) + 3 mol(0 kJ/mol)] = ( kJ) - ( kJ) = kJ

http://college. hmco. com/cgi-bin/SaCGI. cgi/ace1app. cgi

2 NH4Cl(s) + CaO(s)  CaCl2(s) + H2O(l) + 2 NH3(g) -8.3 kJ/mole
Calculate Go in kJ for the following reactions, using the appropriate data tables.  Use Go = Gproducts - Greactants SO3(g) + H2O(l)  H2SO4(l)  -82.9 kJ/mole 2 NH4Cl(s) + CaO(s)  CaCl2(s) + H2O(l) + 2 NH3(g)  -8.3 kJ/mole CaSO4(s) + 2 HCl(g)  CaCl2(s) + H2SO4(l)  +74.1 kJ/mole C2H4(g) + H2O(l)  C2H5OH(l)  -5.9 kJ/mole Ca(s) + 2 H2SO4(l)  CaSO4(s) + SO2(g) + 2 H2O(l)  kJ/mole

Is the following reaction spontaneous at 250C?
2H2(g) + O2(g)  2H2O(g) Compute both ∆H and ∆ S ∆ H = (2* ∆ HH2O0) - (2* ∆ HH20 + 1* ∆ HO20) ∆ H = (2* ) - (2*0 + 1*0) ∆ H = kJ/mol ∆ S = (2* ∆ SH2O0) - (2* ∆ SH20 + 1* ∆ SO20) ∆ S = (2* 188.7) - (2* *205.0) ∆ S = J/mol*K = kJ/mol*K ∆ G = ∆ H - T* ∆ S ∆ G = kJ/mol - 298K*( kJ/mol*K) ∆ G = -457 kJ/mol The reaction is spontaneous, since ∆ G < 0

Example: Calculate the Go in kJ for the following reactions at 100oC using the appropriate data tables.  5 SO3(g) + 2 NH3(g)  2 NO(g) + 5 SO2(g) + 3 H2O(g)  delta H = kJ/mole delta S = J/mole K delta G = kJ/mole (spontaneous) N2O(g) + NO2(g)  2 NO(g delta H = kJ/mole delta S =  J/mole K delta G = kJ/mole (nonspontaneous)