THERMODYNAMICS - mymission.lamission.edupaziras\Chem102\Chap_17.pdf · Chemistry 102 Chapter 17 5 HEAT (q) AND WORK (w) RELATIONSHIP IN A PHYSICAL SYSTEM A gas is enclosed in Heat

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Chemistry 102 Chapter 17

1

THERMODYNAMICS

Thermodynamics is concerned with the energy changes that accompany chemical and physical

processes.

Two conditions must be fulfilled in order to observe a chemical or physical change:

1. The change must be possible (determined by Thermodynamics)

2. The change must occur at a reasonable speed (determined by Kinetics)

Important Terms in Thermodynamics:

1. System and Surroundings:

System : The part of the universe we happen to be studying

Surroundings: Everything outside the system

Three types of systems can be present:

Open: exchange of energy and matter

Closed: exchange of energy but not matter

Isolated: No exchange of energy or matter

The State of the system can be evaluated in terms of the fixed properties of the:

System Components of the system

Temperature

Pressure

Number of moles

Physical State of each system


2

2. Adiabatic and Isothermal Changes:

Adiabatic Change Isothermal Change

Takes place in a system that is insulated from

the surroundings.

Heat cannot flow between system and

surroundings.

The temperature of the system changes:

Temp. increases if rxn is exothermic.

Temp. decreases if rxn is endothermic

Takes place in a system that is not

insulated from the surroundings.

Heat can flow between system

and surroundings and as such the

temperature can be maintained

constant.

A change that occurs at constant

temperature.

3. State Function

A quantity whose value depends only on the current state of the system, and does

not depend on the prior history of the system.

Example: Temperature, Internal Energy


3

4. Internal Energy

Internal Energy (U) is a state function.

It is a property of a system that depends only on its present sate, determined by variables such as temperature and pressure

Kinetic Energy Potential Energy Internal Energy, U = of the particles of the particles

making the system making the system

results from the results from

energy of motion

of

electrons atoms molecules chemical bonding attractions

between atoms molecules


4

FIRST LAW OF THERMODYNAMICS

The 1st Law of Thermodynamics is the “Law of Conservation of Energy” applied to

thermodynamic systems.

When a system changes from one state to another (ex: warming), its internal energy

changes from one definite energy to another.

Change in Internal Energy : U = Uf Ui where: Uf = final value of the internal energy

Ui = initial value of the internal energy

The Changes in Internal Energy, U :

are more important in thermodynamics than the absolute values of Internal Energy

are measured by noting the exchanges of energy between the system and the surroundings

Heat = q Work = w

The energy that moves into or out of the

system because of a temperature

difference between the system and its

surroundings

The energy exchange that results when a force F

moves an object through a distance, d

Work = Force x Distance

If heat is lost U drops

q is negative

If work is done by the system U drops

w is negative

If heat is absorbed U rises

q is positive

If work is done on the system U rises

w is positive

Examples:

Determine the sign of the work for each of the changes shown below (carried out at constant P):

a) 2 SO2 (g) + O2 (g) 2 SO3 (g) w =

b) CaCO3 (s) CaO (s) + CO2 (g) w =

c) CO2 (g) + H2O (l) + CaCO3 (s) Ca2+

(aq) + 2 HCO3– (aq) w =


5

HEAT (q) AND WORK (w) RELATIONSHIP IN A PHYSICAL SYSTEM

A gas is enclosed in Heat passes from the The gas pressure In lifting the

a vessel equipped the surroundings to the increases and causes weight, the system

with a movable vessel and the temperature the gas to expand. does work = the

piston, momentarily of the gas is increased This will cause to lift energy gained by

fixed in position Assume q = + 165 J the piston and the piston and weight

weight on top of it Assume w = - 92 J

The system gains Internal Energy, U from the heat absorbed : q = + 165 J

The system loses internal energy via the work done: w = - 92 J

The change in

Internal Energy = U = (+165 J) + (- 92 J) = + 73 J

of the System


6

FIRST LAW OF THERMODYNAMICS

The Change in Internal Energy of a system, U, equals q + w U = q + w

Consider a system where work is done on the system and heat is lost by the system:

A gas in enclosed in a vessel

equipped with a movable piston,

momentarily fixed in position.

Work is done on the system by

pushing down the piston and thus

compressing the gas.

The gas pressure increases and

causes the gas to warm up.

This will cause heat to flow out

of the system.

Assume w = + 92 J Assume q = - 165 J

The system gains Internal Energy, U from the work done on it : w = + 92 J

The system loses internal energy via the heat lost: q = - 165 J

The change in Internal Energy of the System = U = q + w

U = (-165 J) + (+ 92 J) = - 73 J

Heat (q) and Work (w) relationship in a chemical system

Consider the following exothermic chemical reaction carried out in a beaker open to the

atmosphere:

Zn (s) + 2 HCl (aq) → ZnCl2 (aq) + H2 (g) qp = - 152.4 kJ/mol Zn

Zn (s) + 2 H+ (aq) → ZnCl2 (aq) + H2 (g) qp = - 152.4 kJ/mol Zn

Note: qp = indicates that the process occurs at constant pressure (atmospheric pressure)

The H2 gas that is produced:

increases the volume of the system

must push against the atmosphere in order to evolve

It follows that work must be done by the system.

Can the work done by the system be measured and/or calculated?


7

PRESSURE-VOLUME WORK

height = h = A

V

The work done by the system in expanding = (Force of Gravity) x (Distance the piston moves)

V F

w = -F x h = - F x = - x VA A

Negative sign is given because work “w” is done by the system and represents energy lost by it.

It follows: w = - P V

Assume that the pressure exerted by the

atmosphere is replaced by a piston and

weights, whose downward force of gravity,

F, creates a pressure on the gas equivalent to

that of the atmosphere

F = force of gravity= atmospheric pressure

F

V = increase in volume of the system

due to the production of H2 gas

V = (cross sectional area) x( height)

Atmospheric

pressure


8

Examples:

1. In the reaction shown below, 1.00mol of Zn reacts with excess HCl to produce 1.00 mol of H2 gas.

At 25C and 1.00 atm, this amount of H2 occupies 24.5 L. Calculate the change in Internal Energy

(∆U) for this reaction. (1 Latm = 0.101 kJ)

Zn (s) + 2H3O+

(aq) Zn2+

(aq) + 2 H2O (l) + H2 (g) q = –152.4 kJ/mol

∆U = q + w

q = –152.4 kJ

w = ?

w done by system (to push back atmosphere) = –PV

= –(1.00 atm)(24.5 L)(0.101 kJ/1 Latm)

= –2.47 kJ

U = qp + w = ( –152.4 kJ ) + (–2.47 kJ) = –154.9 kJ

2. What is U when 1.00 mol of liquid water vaporizes at 100C? (H0 vap = 40.66 kJ/mol at 100 C)


9

Examples:

3. Consider the combustion (burning) of methane, CH4, in oxygen:

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)

The heat of reaction at 25 C and 1.00 atm is –890.2 kJ.

a) What is the change in volume, V when 1.00 mol CH4 reacts with 2.00 mol O2?

(Ignore the volume of liquid water formed in the reaction)

b) What is w for this change?

c) Calculate U for the change indicated by the chemical equation.

a) Find the change in volume (V)

Volume occupied by 1 mole of any gas at STP = 22.4 L

Volume occupied by 1 mole of any of the gases in the equation at 25 C:

298 KV = 22.4 L x = 24.45 L

273 K

CH4(g) + 2 O2(g) CO2(g) + 2 H2O (l) H = –890.2 kJ

3 moles of gas 1 mole of gas

Gaseous Volume decreases: 3 moles of gas 1 mole of gas

V = Volume occupied by 3 moles – Volume occupied by 1 mole

V = (3 x 24.45 L) – 24.45 L = 48.9 L

b) Calculate the work (w) done on the system

w = + P V = (1.00 atm)(48.9 L)(0.101 kJ/1 Latm) = 4.94 kJ

c) Calculate change in internal energy, U

= + 4.94 kJ

qp = –890.2 kJ

U = qp + w = (–890.2 kJ) + (+ 4.94 kJ) = –885.3 kJ

Conclusion: Energy leaves the system


10

4. What is U for the following reaction at 25 C?

N2 (g) + 3 H2 (g) 2 NH3 (g) H = – 91.8 kJ

4 moles of gas 2 moles of gas

Volume occupied by 1 mole of any of the gases in the equation at 25 C:

298 KV = 22.4 L x = 24.45 L

273 K

The decrease in Volume:

V == 2 x (24.45 L) = 48.9 L

Since the Volume decreases, work is done on the chemical system by the atmosphere:

w = + P V = (1.00 atm) x (48.9 L)(0.101 kJ/1 Latm) = + 4.94 kJ

U = qp + w = (–91.8 kJ) + (+ 4.94 kJ) = – 86.9 kJ

Conclusion: Energy leaves the system

5. Calculate the amount of work done in each of the following reactions. In each case, is the work

done by or on the system?

a) oxidation of HCl at 200 C

4 HCl (g) + O2 (g) 2 Cl2 (g) + 2 H2O (g)

b) The decomposition of NO2 at 300 C

2 NO (g) N2(g) + O2 (g)


11

ENTHALPY AND ENTHALPY CHANGE

Recall:

Enthalpy = The Heat of Reaction, H, at constant pressure (qp)

Actually: Enthalpy = H = U + PV

U (internal Energy), P (pressure), V (volume) are state functions.

If follows: H (Enthalpy) is also a state function

Consequences:

1. For a given temperature and pressure, a given amount of a substance has a definite enthalpy.

2. If the enthalpies of substances are known, the change of enthalpy for a reaction, H, can be

easily calculated.

H = Final Enthalpy – Initial Enthalpy = Hf – Hi

At constant pressure, P : Hf = Uf + PVf and Hi = Ui + PVi

H = (Uf + PVf) – (Ui + PVI) = (Uf – Ui) + (PVf – PVi ) = U + PV

H = U + PV and as found earlier: U = qp – P V

It follows: H = (qp – P V) + P V = qp – P V + P V = qp

Conclusion: H = qp (Chapter 6)

The Enthalpy or Reaction equals the Heat of Reaction at constant pressureApplications:

Heat absorbed (endothermic reaction) or

given off (exothermic reaction) by a H0 = Standard Enthalpy Change

chemical reaction

H

0 = Standard Enthalpy Change for a Reaction can be calculated if the

Standard Enthalpies of formation, H0

f for both reactants and products are known.

H0

f = Standard Enthalpy of Formation of a Substance

= Enthalpy change for the formation of one mole of substance in its standard

state from its elements in their reference form ( most stable form at 250C and

1.00 atm) and in their standard states. (Given in Table 6.2 and Appendix C)


12

Examples:

1. Calculate the heat absorbed or evolved in the reaction between NH3 gas and CO2 gas that yields

aqueous urea (NH2CONH2) and liquid water.

2 NH3 (g) + CO2 (g) CO(NH2)2 (aq) + H2O (l) H0 = ?

From Appendix C:

H0

f (NH3) gas = – 45.9 kJ/mol

H0

f (CO2) gas = – 393.5 kJ/mol

H0

f [CO(NH2)2] aqueous = – 319.2 kJ/mol

H0

f (H2O) liquid = – 285.8 kJ/mol

2 NH3 (g) + CO2 (g) CO(NH2)2 (aq) + H2O (l)

2 x (– 45.9 kJ/mol) (– 393.5 kJ/mol) (– 319.2 kJ/mol) (– 285.8 kJ/mol)

H

0 = n H

0f (products) – n H

0f (reactants)

H

0 = [(– 319.2) + (– 285.8 kJ)] – [2 x (– 45.9 ) + (– 393.5)] = – 119.7 kJ

Since H0 is negative, we conclude that the reaction is exothermic.

2. Use data in appendix C to find H0 for the reactions shown below:

a) NH4NO3 (s) NO2 (g) + 2 H2O (l)

b) SiO2 (s) + 3 C (graphite) SiC (s) + 2 CO (g)


13

THE SECOND LAW OF THERMODYNAMICS

Spontaneous Processes: Physical or chemical changes that occurs by themselves, without outside

assistance

Spontaneous Physical Change: A ball rolling from the top of a hill

High Potential Energy

Low Potential Energy

Spontaneous Chemical Change: The reaction of solid K(potassium) with liquid water

spontaneous

2 K(s) + 2 H2O (l) 2 KOH(aq) + H2(g)

In order for spontaneous changes to go in the opposite direction work would have to be expended, and

the changes would be called non-spontaneous.

Non-spontaneous Physical Change: Rolling a ball uphill

Non-spontaneous Chemical Change Change: Obtaining K(s) from KOH(aq)

non-spontaneous

2 KOH (aq) + H2 (g) 2 K(s) + 2 H2O(l)

This process requires several chemical reactions.

NOTE:

The Second Law of Thermodynamics determines if a reaction is spontaneous or not.

Exothermic Reactions (H 0) are not necessarily spontaneous and Endothermic Reactions (H 0)

are not necessarily non-spontaneous


14

ENTROPY (S)

Entropy is the thermodynamic quantity that is a measure of the randomness or disorder in a system.

Entropy is a state function: - the quantity of entropy in a substance depends only on variables that

determine the state of the substance (temperature and pressure)

Entropy is measured in J/K (SI unit)

Entropy increases when the disorder in a sample of the substance increases:

Example: Consider the melting of ice at O0C and 1 atm.

MELTING

ICE LIQUID WATER

H2O molecules occupy regular fixed positions

(crystal lattice)

Ordered crystalline structure

Lower Entropy

Si = Initial Entropy

= Entropy of Ice (41 J/K)

H2O molecules move about freely, giving a

disordered structure

Less-ordered structure

Higher Entropy

Sif= Final Entropy

= Entropy of liquid water (63 J/K)

melting at 00C and 1 atm

H2O(s) H2O(l) S = 22 J/K

Si = 41 J/K Sf = 63 J/K

In any natural process the degree of disorder (Entropy) increases.

Natural processes are changes that occur by themselves (spontaneous changes).

Second Law of Thermodynamics in reference to the system and its surroundings

The Total Entropy of a System and its Surroundings always increases for a Spontaneous Process.

ENTROPY ENERGY

is created during a spontaneous change can be neither created or destroyed

during a spontaneous change


15

SECOND LAW OF THERMODYNAMICS IN REFERENCE TO THE SYSTEM ONLY

Consider a spontaneous change and the changes that occur in the system at a given temperature:

1. Entropy is created 2. Heat (q) flows into or out of the system

Entropy accompanies the heat flow

Entropy Change q

associated with

heat flow (q) T

Entropy created Change in entropy

S = during + associated with

the spontaneous process heat flow

S = Entropy Created + Entropy Flow

q

S = Entropy Created +

T


16

q

S = Entropy Created +

T

For a Spontaneous change:

q

S 0 Entropy Created 0 may be positive or negative

T

q

It follows: S

T

For a Spontaneous Process at a Given Temperature, the Change in Entropy of the System is greater than the

Heat divided by the Absolute Temperature


17

ENTROPY CHANGE FOR A PHASE TRANSITION

The entropy of a sample of matter increases as it changes state from a solid to a liquid or from a

liquid to a gas. This is due to the increase in disorder of molecules in liquid compared to solid and

gas compared to liquid.

We can therefore predict the sign of ∆S for processes involving changes of state (or phase). In

general, entropy increases (∆S > 0) for each of the following:

the phase transition from a solid to a liquid

the phase transition from a solid to a gas

the phase transition from a liquid to a gas

an increase in the number of moles of a gas during a chemical reaction

Examples:

1. Predict the sign of ∆S for each of the following processes:

a) the boiling of water ∆S =

b) I2 (g) → I2 (s) ∆S =

c) CaCO3 (s) → CaO (s) + CO2 (g) ∆S =


18

ENTROPY CHANGE FOR A PHASE TRANSITION

The criterion for spontaneity is an increase in the entropy of the universe. However, there are

several spontaneous processes that include a decrease in entropy. For example, when water freezes

below 0C, the entropy of the water decreases, yet the process is spontaneous.

The 2nd

Law of Thermodynamics states that for a spontaneous process, the entropy of universe

increases (∆Suniv > 0). In the example above, even though the entropy of water decreases, the

entropy of the universe must increase in order for the process to be spontaneous.

Similar to the distinction of a system and surroundings in thermodynamics, we can distinguish

between the entropy of the system and the surroundings. In the example of the freezing water, ∆Ssys

is the entropy change of water, and the ∆Ssurr is the entropy change of the surroundings.

The entropy change of the universe is then the sum of the entropy changes of the system and the

surrounding.

∆S univ = ∆Ssys + ∆Ssurr

If a spontaneous process includes a decrease in ∆Ssys , then it would follow that there must be a

greater increase in ∆Ssurr in order for the ∆Suniv to increase.

Since freezing of water is an exothermic process, it would follow that the heat given off by the

process increases the entropy of the surroundings by a greater amount than the decrease of entropy

of the system, making it a spontaneous process.

To summarize:

An exothermic process increase the entropy of the surroundings.

An endothermic process decreases the entropy of the surroundings.


19

QUANTIFYING ENTROPY CHANGE OF SURROUNDINGS & SYSTEM

We know that when a system exchanges heat with surroundings, it changes the entropy of the

surroundings. At constant pressure, qsys can be used to quantify the entropy change of the

surroundings.

In general,

A process that emits heat into the surroundings (qsys = negative) increases the entropy of

the surroundings (positive ∆Ssurr)

A process that absorbs heat from the surroundings (qsys = positive) decreases the entropy

of the surroundings (negative ∆Ssurr)

The magnitude of the change in entropy of the surroundings is proportional to the magnitude of

the qsys

∆Ssurr α – qsys

It can also be shown that for a given amount of heat exchanged with the surroundings, the

magnitude of ∆Ssurr is inversely proportional to the temperature

∆Ssurr α 1/T

Combining the above relationships: sys sys

surr

- q - HS = =

T T (constant P, T)

Since the sum of ∆Ssys and ∆Ssurr must be positive for a spontaneous process, it would then

follow that for a spontaneous process

sys

sys

qS >

T

In an equilibrium process, such as a phase change, no significant amount of entropy is created by

molecular disorder, and almost all the change in entropy is due to the absorption of heat.

Therefore,

sys

sys

qS =

T (equilibrium process)


20

Examples: 1. Calculate S for the melting of ice (Heat absorbed = Heat of Fusion = Hfus = 6.0 kJ/mol)

3

fussurr

H 6.0x10 JS = = = 22 J/K

T 273 K

2. Liquid ethanol, C2H6O (l), at 250C has an entropy of 161 J/(mol x K). If the heat of vaporization,

Hvap at 25 C is 42.3 kJ/mol, what is the entropy of the vapor in equilibrium with the liquid at 25 C?

When the liquid evaporates, it absorbs heat: Hvap = 42.3 kJ/mol at 250C

q Hvap 42.3 kJ/mol

Entropy Change = S = = = = 141.9 J/K x mol T (25 + 273) K 298 K

The entropy of 1 mole of vapor is calculated using the entropy of 1 mole of liquid (161 J/K) to which the

entropy change resulting from the heat absorption (141.9 J/K) is added:

Entropy of Vapor = 161 J/K + 141.9 J/K = 303 J/K

3. Consider the combustion of propane gas:

C3H8 (g) + 5 O2 (g) → 3 CO2 (g) + 4 H2O (g) ∆Hsys = – 2044 kJ

a) Calculate the ∆Ssurr associated with this reaction occurring at 25C.

b) Determine the sign of ∆Ssys

c) Determine the sign of ∆Suniv. Will this reaction be spontaneous?


21

2ND

LAW OF THERMODYNAMICS AND SPONTANEITY OF REACTIONS

Consider the preparation of urea at constant temperature and pressure:

2 NH3 (g) + CO2(g) CO(NH2)2 (aq) + H2O (l)

Is this a spontaneous reaction? (Does it go from left to right as written?)

qp H H

For a spontaneous reaction: S = S

(constant T and P) T T T

H

For a spontaneous reaction: – S 0

(constant T and P) T

smaller larger

Multiplying each term by the positive quantity T yields:

For a spontaneous reaction: H – T S 0

(constant T and P)

If S is available and H is calculated (Table 6.2), one can predict if the reaction is spontaneous

or not:

H – T S 0 The reaction is spontaneous ( )

H – T S 0 The reaction is non-spontaneous ( )

The reaction is spontaneous ( )

H – T S = 0 The reaction is at equilibrium ( )

For a spontaneous reaction: H – T S 0 (constant T and P)


22

INTERPRETING SPONTANEITY

Two factors determine the spontaneity of a chemical change:

1. Nature’s tendency for the potential energy to be at a minimum.

Any change that tends to lower the potential energy tends to occur spontaneously

The quantity that is related to the lowering of the potential energy of changes that take

place at constant pressure is H

For exothermic reactions H 0

Exothermic reactions “tend” to occur spontaneously

2. Nature’s tendency toward disorder

Any change that tends to increase the disorder of a system tends to occur spontaneously

The quantity that is related to the increase in the degree of disorder for changes that take

place at constant pressure is S

The larger the entropy of a system, the greater its statistical probability

Reactions that go in the direction of greater entropy “tend” to occur spontaneously.

S = Sfinal – Sinitial When S 0 Spontaneity is favored

The entropy (S) of a system depends on several factors:

(a) Physical state of the components of the system

S (solid) S (liquid) S (gas)

(b) Temperature of the system

Entropy (S) increases with temperature

(c) Mixing of components

If the volume of the mixture is larger than the volumes of the components the Entropy (S)

increases

If two liquids are mixed to form a solution, S increases

If two gases are mixed, S increases only if the total volume increases


23

THIRD LAW OF THERMODYNAMICS

A substance that is perfectly crystalline at 0 K has an entropy of zero

This law helps determine the Entropy of different substances at different temperatures, if one

considers the following facts:

When the temperature of a substance is raised from 0 K,

the substance absorbs heat, and

the substance becomes more disordered

The entropy change that occurs when heat is absorbed, q

S = T

Examples:

1. What is S for a substance if it is heated from near 0.0 K to 2.0 K and the heat absorbed is 0.19 J?

0.19 J 0.19 J

S = = = 0.19 J/K (at 1.0 K)

T(average) 1.0 K

2. What is S for the same substance if it is heated from near 2.0 K to 4.0 K and the heat absorbed

is 0.88 J?

0.88 J 0.88 J

S = = = 0.29 J/K (at 3.0 K)

T(average) 3.0 K

Proceeding this way, the entropy at any temperature can be determined.

Standard entropy of methyl chloride, CH3Cl, at various temperatures

Notes:

1. The entropy increases gradually as

the temperature increases.

2. The entropy increases sharply

when a phase change occurs.


24

Standard Entropy of a Substance or Ion (Absolute Entropy), S0

S0 is the entropy value for the standard state of the species.

For a substance: Standard State is the pure substance at 1 atm

For a species in solution: Standard state is the 1 M solution

S0 is temperature dependent; it is usually given for 25

C

Entropy Changes (S) for a Reaction

A. S can sometimes be predicted

The entropy usually increases in the following situations:

(a) A reaction in which a molecule is broken into two or more smaller molecules.

Example: fermentation of glucose (grape sugar) to alcohol:

C6H12O6 (s) 2 C2H5OH (l) + 2 CO2 (g)

1 large molecule 4 smaller molecules

(b) A reaction in which there is an increase in moles of gas.

This may result from a molecules breaking up, in which case Rule (a) and Rule (b) are related.

Example: the burning of acetylene in oxygen:

2 C2H2 (g) + 3 O2(g) 4 CO2 (g) + 2 H2O (g)

5 moles of gas 6 moles of gas

n gas = 6 – 5 = +1

(c) A process in which a solid changes to a liquid or gas or a liquid changes to a gas

melting

Ice at 0 C Liquid Water at 0

C

Liquid vaporization

Ethyl alcohol Ethyl alcohol vapor at 25 C

at 25 C

sublimation

Solid CO2 at 25 C CO2 gas at 25 C


25

B. S can be calculated if the values of standard entropies are available

Examples:

1. Calculate the change in entropy, S0 for the fermentation of glucose.

The standard entropy of glucose is C6H12O6(s) is: 212 J/mol x K

The standard entropy for C2H5OH (l) is: 161 J/mol x K

(ethyl alcohol)

The standard entropy for CO2(g) is: 213.7 J/mol x K

C6H12O6 (s) 2 C2H5OH (l) + 2 CO2 (g)

S0

: 1 x (212) 2 x (161) 2 x (213.7)

S

0 = nS

0(products) - nS

0(reactants)

S

0 = [(2 x 161) + 2 x 213.7)] - [1 x 212] = 537 J/K

2. Predict the sign of S

0 for each reaction shown below. If you cannot predict the sign for

any reaction, state why.

a) C2H2 (g) + 2 H2 (g) C2H6 (g) S0 =

b) N2 (g) + O2 (g) 2 NO (g) S0 =

c) 2 C (s) + O2 (g) 2 CO (g) S0 =

3. Using S0 values in your textbook, calculate S

0 for the synthesis of ammonia.

N2 (g) + 3 H2 (g) 2 NH3 (g)


26

PREDICTING SPONTANEITY

Recall:

For a spontaneous reaction: H – T S 0

(constant T and P)

The value of H – T S can be used as a criterion for spontaneity:

If:

H – T S 0 The reaction is spontaneous from left to right ( )

The reaction is non-spontaneous from right to left ( )

H – T S 0 The reaction is non-spontaneous from left to right ( )

The reaction is spontaneous from right to left ( )

H – T S = 0 The reaction is at equilibrium ( )

Examples:

1. Consider the reaction by which urea is prepared from NH3 and CO2 , at 25 C

2 NH3 (g) + CO2 (g) CO(NH2)2 (aq) + H2O (l) H0 = – 119.7 kJ

Determine if the reaction is spontaneous from left to right as written.

The value of H – T S must be determined:

H0 = – 119.7 kJ

T = 25 + 273 = 298 K

S = ?

From Table 18.1:

S0(NH3) gas = 193 J/mol x K S

0 [CO(NH2)2] aqueous = 174 J/mol x K

S0 (CO2) gas = 214 J/mol x K S

0(H2O) liquid = 70. J/mol x K

2 NH3 (g) + CO2 (g) CO(NH2)2 (aq) + H2O (l)

S0

2 x 193 214 174 70

S0= n S

0(products) – n S

0(reactants) = [(174 + 70] – [2 x (193 ) + (214)] = – 0.356 kJ/K

H – T S = (119.7 kJ) (298 K) ( 0.356 kJ/K) = 13.6 kJ

The reaction is spontaneous under standard conditions from left to right, as written.


27

GIBBS FREE ENERGY

Willard Gibbs (American Physicist) brought together the H (Enthalpy Change) and S (Entropy

Change) into a single thermodynamic quantity G:

G = Gibbs Free Energy = H TS

As a reaction proceeds at a given temperature and pressure:

REACTANTS PRODUCTS

n Hf (reactants) m Hf (products) H= n Hf (products) - m Hf(reactants)

n S (reactants) mS (products) S = nS (products) - mS(reactants)

Result: For a change at constant T and P : G = H TS

For any spontaneous change that takes place in a system:

Free Energy G must be lowered (Gproducts Greactants)

G must be negative: G = n G (products) - m G(reactants) 0

– 0 +

G < 0 G > 0

Reaction is spontaneous

Product-favored reaction

Forward reaction is favored

Reaction is non-spontaneous

Reactant-favored reaction

Reverse reaction is favored

Conclusion:

G gives a composite of the two factors that contribute to spontaneity (H and S)


28

SPONTANEITY & FREE ENERGY

Consider the following situations: G = H TS

1. H is negative and S is positive G = (–) – T(+) = negative spontaneous (exothermic) (increase in entropy)

2. H is positive and S is negative G = (+) – T(–) = positive non-spontaneous (endothermic) (decrease in entropy)

3. H is positive and S is positive G = (+) – T(+) = ? (endothermic) (increase in entropy)

The temperature plays the determining role in controlling spontaneity:

(a) When T is large: TS H G = (+) – Tlarge(+) = negative spontaneous

(b) When T is small: TS H G = (+) – Tsmall(+) = positive non-spontaneous

4. H is negative and S is negative G = (–) – T(–) = ? (exothermic) (decrease in entropy)

The temperature plays the determining role in controlling spontaneity:

(a) When T is large: TS > H G = (–) – Tlarge(–) = positive non-spontaneous

(b) When T is small: TS < H G = (–) – Tsmall(–) = negative spontaneous

The effects of the algebraic signs of H and S and the effect of temperature on spontaneity can be

summarized as:

H S G = H – TS Outcome

(–) (+) negative Spontaneous at all temp.

(+) (–) positive Non-spontaneous regardless of temp.

(+) (+) negative only at high T Spontaneous only at high temp.

(–) (–) negative only at low T Spontaneous only at low temp.


29

FREE ENERGY AT STANDARD STATE

For interpreting thermodynamic data, standard states are chosen.

Standard States:

Standard states are indicated by a superscript degree sign on the symbol of the quantity:

G0 = standard free-energy change

H0f = standard enthalpy change = n H

0f (products) - m H

0f (reactants)

S0 = standard enthropy change = nS

0 (products) - mS

0(reactants)

Standard states are:

Pressure Concentration Temperature

1 atm pressure (for pure liquids and solids)

1 atm partial pressure (for gases) 1 M (for solutions 25 ºC (298 K)

Examples:

1. Calculate the standard free-energy change (G0) for the following reaction at 25

C, and determine

whether it is spontaneous or not.

CH4(g) + 2 O2 CO2(g) + 2 H2O(g)

H0

f : − 74.9 2 x 0 − 393.5 (2) x ( − 241.8)

S0 : 186.1 (2) x (205.0) 213.7 (2) x (188.7)

H0 = [- 395.5 + (2) x (-241.8)] kJ – [- 74.9 + 0] kJ = − 802.2 kJ

S0 = [213.7 + (2) x (188.7)] J/K – [186.1 + (2) x (205.0)] = − 5.0 J/K

G0 = H

0 - TS

0 = - 802.2 kJ – (298 K) (- 5.0 x 10

3 kJ/K) = – 800.7 kJ (spontaneous)

2. Calculate the standard free-energy change (G0) for the following reaction at 25

C, and determine

whether it is spontaneous or not.

N2 (g) + O2 (g) 2 NO (g)

H0f (kJ/mol) 0 0 90.25

S0 (J/mol.K) 191.5 205.0 210.7


30

STANDARD FREE ENERGY OF FORMATION (G0

f)

G0

f is the free-energy change that occurs when 1 mol of substance is formed from its elements in

their most stable states at 1 atm and 25 C

G0

f has units of kJ/mol of substance

G0

f is a constant, with values available in textbooks.

NOTE: Standard free energies of formation of elements in their most stable form = 0

G

0 can be calculated directly for any reaction by using:

G

0 = nG

0f (products) - mG

0f (products)

Examples:

1. Calculate the standard free energy of the following reaction at 25 C, using standard free energies of

formation given below:

Na2CO3 (s) + H+

(aq) 2 Na+

(aq) + HCO3(aq)

G

0f : – 1048.1 0 (2) x (– 261.9) – 587.1

G

0f = [(2) x (–261.9) + (–587.1)] – [(–1048.1) + 0] kJ = – 62.8 kJ

3. Calculate the standard free-energy change (G

0) for the reaction shown below, and determine whether it

is spontaneous or not. (Use G0

f values in your textbook).

2 NOCl (g) 2 NO (g) + Cl2 (g)


31

G0 AS A CRITERION FOR SPONTANEITY

G < –10 kJ –10 kJ <G < +10 kJ G > +10 kJ

Large negative number Small negative or positive

number Large positive number

Reaction is spontaneous Reaction gives an

equilibrium mixture

Forward reaction is non-spontaneous

Reverse reaction is spontaneous

Reactants Products Reactants Products Reactants Products

Reactants transform entirely

to products

Significant amount of both

reactants and products

present

Mostly reactants present

Examples:

1. Which of the following reactions are spontaneous in the direction written? (See table in your

textbook for G0

f values.

(A) C(graphite) + 2 H2 (g) CH4 (g)

Gf 0 : 0 0 – 50.8

G0 = [(–50.80 – (0)] kJ = –50.8 kJ (spontaneous reaction)

(B) Ag+

(aq) + I(aq) AgI (s)

Gf 0 : 77.1 –51.7 –66.3

G0 =

2. Consider the reaction of nitrogen, N2 and oxygen, O2 to form nitric oxide, NO:

N2 (g) + O2 (g) 2 NO (g)

The standard free energy of formation of NO is + 86.60 kJ/mol. Do you expect the reaction to be

spontaneous as written?

Recall: Standard free energies of formation of elements in their stablest form = 0

G0 = 2 (+ 86.60 kJ) = + 173.2 kJ (non-spontaneous reaction)


32

FREE ENERGY and USEFUL WORK

In a spontaneous reaction:

the “free” energy is lowered as reactants change to products,

the change in “free” energy is released as free-energy change( G)

this “G” can be harnessed to perform useful work

A spontaneous reaction can be used to obtain useful work

Examples:

1. The combustion of gasoline is used to move a car

2. The reaction in a battery generates electricity that can drive a motor

The free – energy change (G) is the maximum energy available (“free”) to do useful work

Maximum useful work = wmax = G

completely

In an ideal situation: G Useful work

converted into

converted

In real situations: G Some Useful Work + Some Entropy

into

Conclusions:

1. In theory, all of the free-energy decrease liberated during a spontaneous chemical change can be

used to do work (this would be wmax)

2. In practice, less work is obtained and the difference appears as an increase in entropy

Example:

Living systems are able to convert only about 40% of the free energy available in the oxidation

of glucose to other forms of stored chemical energy (for example , ATP).

The rest (60%) appears as heat, which ensures the body’s effective temperature control system.


33

FREE-ENERGY CHANGE DURING A SPONTANEOUS REACTION

(combustion of gasoline)

Gasoline + O2 CO2(g) + H2O(g)

Free energy decreases as the reaction proceeds (G 0)

At equilibrium, free energy is at a minimum

(equilibrium mixture is mostly products)

FREE-ENERGY CHANGE DURING A NON-SPONTANEOUS REACTION

(synthesis of NO from elements)

N2(g) + O2(g) 2 NO(g)

Free energy increases as the reaction proceeds (G 0)

There is a small decrease in free energy as the system goes to

equilibrium (some reaction occurs to give the equilibrium

mixture which consists mostly of reactants)

N2(g) + O2(g) NO(g)


34

G0 & THE EQUILIBRIUM CONSTANT

G determines the maximum amount of energy that is available to perform useful work as a system

passes from one state to another (G = Gproducts – Greactants)

As a reaction proceeds:

the system’s capacity to perform useful work decreases (G diminishes),

the system eventually reaches equilibrium and G 0 (Free Energy ceases to change)

The value of G determines where the system stands with respect to equilibrium:

When G < 0 When G = 0 When G > 0

The reaction is spontaneous

and proceeds in the forward

direction towards

equilibrium (G decreases)

The system is in a state of

dynamic equilibrium

The reaction is not

spontaneous in the forward

direction.

The reaction is spontaneous

in the reverse direction.

Qualitative Relationship between G0 and the Position of Equilibrium

The direction in which a reaction proceeds toward equilibrium is determined by where the system lies

with respect to the free-energy minimum.

G0

G

Direction of

reaction

Reactants Products

Position of

equilibrium

NOTE:

The reaction proceeds spontaneously only in a direction that gives rise to a decrease in free

energy (G is negative)


35

RELATIONSHIP BETWEEN G0 AND EQUILIBRIUM CONSTANT

G = G0 + RT lnQ

Q = Thermodynamic Reaction Quotient

for reactions involving gases Q is obtained from partial pressures.

for reactions in solutions, Q is obtained from molar concentrations

R = Gas Law Constant in units of energy = 8.314 J/mol K

G0 = Standard Free-Energy Change

G = Free-Energy Change when Reactants in nonstandard states are changed to products in

non-standard states

At equilibrium:

1. The free energy ceases to change: G = 0

2. The Thermodynamic Reaction Quotient, Q becomes equal

to the Thermodynamic Equilibrium Constant (K): Q = K

G = G0 + RT lnQ becomes: 0 = G

0 + RT lnK

By rearrangement: G

0 = – RT ln K or G

0 = – 2.303 RT log K

Basic Equations relating the standard

free-energy change to the equilibrium constant.

Natural logarithms (ln) and logarithms to the base 10 (log) are related by: ln x = 2.303 log x

G0 = - 2.303 RT log K Thermodynamic Equilibrium Constant

The equilibrium Constant in which:

the concentration of gases are expressed in partial pressures in atmospheres.

the concentration of solutes are expressed in molarities

It follows that:

1. For reactions involving only gases: K = Kp

2. For reactions involving only solutes in liquid solutions: K = Kc

3. For net ionic equations: K = Ksp


36

RELATIONSHIP OF G0, K, AND SPONTANEITY

G

0 = –2.303 RT log K

When K > 1 When K 1 When K <1

Reactants transform almost

entirely to products

Significant amounts of both

reactants and products are

present

Mostly reactants present

Reactants Products Reactants Products Reactants Products

log K > 0 log K 0 log K < 0

G0 < 0

G0 = 10 kJ

K = (0.018 – 57) G

0 > 0

Reaction is spontaneous Reaction gives an

equilibrium mixture

Forward reaction is non-spontaneous

Reverse reaction is spontaneous

Examples:

1. Use the standard free energies of formation given to determine the equilibrium constant (K)

for the following reaction at 25 C:

N2 (g) + 3 H2 (g) 2 NH3 (g)

G0

f (kJ/mol) 0 0 –16.66

G0 = [ 2 x G

0f (NH3)] – [ G

0f (N2) + 3 x G

0f (H2)] = 2 x (–16.66) – 0 = –33.32 kJ

G0 = –RT ln K ln K =

00

G /RTG K = e

RT

0ΔG=

RT

( 13.45 533, 320 J)= 13.45 K = e = 6.9 x 10

(8.314 J/K)(298 K)

2. Use the standard free energies in your text to calculate G0 and K for the following reaction

at 25 C:

2 H2 (g) + O2 (g) 2 H2O (g)


37

CHANGE OF FREE ENERGY WITH TEMPERATURE

How can G0 be found for temperatures other than standard temperatures (25

C)?

An approximate method used to calculate G0

T is based on the assumption that both H0 and

S0 are constant with respect to temperature (only approximately true)

Then: G0

T = H0 - TS

0 (a convenient approximation for G

0T )

NOTE: G0

T is strongly temperature dependent

G0

T = Change in free Energy for a substance:

at 1 atm of pressure (standard pressure) and

at the specified temperature, T (nonstandard temperature)

depends on the value of determined by

SPONTANEITY G0

T TEMPERATURE

It follows that SPONTANEITY IS TEMPERATURE DEPENDENT!

Meaning:

Some chemical changes may be non-spontaneous at one temperature but spontaneous at

another temperature.

Effect of Temperature on the Spontaneity of Reactions

G0

T = H0 - TS

0 For a Spontaneous Reaction: G

0T 0

H0 S

0 Temperature G

0 Spontaneity

(–) (+) no effect Negative

regardless of temp.

Spontaneous

at any temp.

(+) (–) no effect Positive

regardless of temp.

Non-spontaneous

at any temp.

(–) (–)

Low Negative Spontaneous

at low temp.

High Positive Non-spontaneous

at high temp.

(+) (+)

Low Positive Non-spontaneous

at low temp.

High Negative Spontaneous

at high temp.


38

Examples:

1. When glucose ferments, it produces ethyl alcohol and carbon dioxide. The reaction is exothermic.

Under what temperature conditions is this reaction spontaneous?

C6H12O6 (s) 2 C2H5OH (l) + 2 CO2(g) H0 0

glucose ethyl alcohol

1 molecule breaks 2 molecules of liquid

of and S0 0

solid into 2 molecules of gas

G

0T = H

0 - TS

0 = (Negative Number) – [(Temperature) x (Positive Number)]

G0

T = H0 - TS

0 = (–) – [ (+) ]

NOTE: T is absolute temperature in K cannot be negative

G0

T is negative (regardless of temperature)

Reaction is spontaneous at any temperature

2. Sodium Carbonate, Na2CO3, can be prepared by heating sodium hydrogen carbonate, NaHCO3.

Estimate the temperature at which NaHCO3 decomposes to products at 1 atm.

2 NaHCO3 (s) Na2CO3 + H2O (g) + CO2 (g)

First calculate the H0 and S

0 , using the given Hf

0 and S

0 values:

2 NaHCO3 (s) Na2CO3 + H2O (g) + CO2 (g)

Hf 0 : 2 x (–947.7) –1130.8 –241.826 –393.5

S0 : 2 x 102 139 188.72 213.7

H0

= [(–1130.8) + (–241.826) + (–393.5)] – [ (2) x (–947.7)] = 129.274 kJ

S0 = [(139) + (188.72) + (213.7)] – [ 92) x (102)]= 337.4 J/K = 0.3374 kJ/K

Substitute these values into G0 = H

0 – TS

0 ; Let G

0 = 0

H0 129.274 kJ

H0 – TS

0 = 0 T = = = 383 K (110

C) [T 383 K (110

C)]

S0 0.3374 kJ/K

The temperature at which NaHCO3 decomposes to products at 1 atm. should be at least = 383 K (110 C)


39

3. Oxygen was first prepared by heating mercury(II) oxide, HgO. Estimate the temperature at which

HgO decomposes to O2 at 1 atm.

2 HgO (s) 2 Hg (g) + O2(g)

First calculate the H0 and S

0 , using the given Hf

0 and S

0 values:

2 HgO(s) 2 Hg (g) + O2 (g)

Hf 0 : –90.8 61.3 0

S0 : 70.3 174.9 205.0

H

0 =

S0 =

Substitute these values into G0 = H

0 - TS

0 ; Let G

0 = 0

H0

H0 - TS

0 = 0 T = =

S0

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