Chapter 14:  Thermodynamics

 

 

Section 14-1: Predicting the Sign of the Entropy Change, S

Section 14-2: Calculating S Using Standard Entropies, S°

Section 14-3: The Second Law of Thermodynamics and Gibbs Free Energy Change, G

Section 14-4: The Effect of Temperature on G and the Spontaneity of Reactions

Section 14-5: Calculating G from Standard Gibbs Free Energies of Formation, Gf°

Chapter 14 Practice Exercises and Review Quizzes

 

 

 

Section 14-1:  Predicting the Sign of the Entropy Change, S

 

Whereas thermochemistry focuses primarily on the heat involved in chemical reactions, thermodynamics explores the role of energy more broadly.  Entropy (S) is essentially an indication of the level of disorder or chaos in a given system.  The following processes generally occur with an increase in entropy and, thus, a positive entropy change (S):

 

1. The temperature of a substance increases.

2. A solid is converted to a liquid, solution, or gas (sublimation).

3. A liquid is converted to a gas.

4. A chemical reaction in which the number of moles of gas increases (the sum of coefficients of gases among the products on the right side is larger than the sum of coefficients of gases among the reactants on the left side).  If the number of moles of gas does not change during a chemical reaction, more information is generally needed to determine the sign of S.

 

The following processes generally occur with a decrease in entropy and, thus, a negative S:

 

1. The temperature of a substance decreases.

2. A solid is formed from a liquid, solution, or gas.

3. A gas is converted to a liquid.

4. A chemical reaction in which the number of moles of gas decreases.

 

Sample Exercise 14A:

 

Predict the sign of S for each process:

 

(a) Water cools from 35C to 25C.

(b) Liquid ethanol freezes.

(c) Solid iodine dissolves in carbon disulfide.

(d) 3 S (s) + 2 H2O (g) SO2 (g) + 2 H2S (g)    

(e) Bromine vapor condenses.

(f) CO (g) + H2O (g) CO2 (g) + H2 (g)

 

Solution:

 

(a) Temperature decreases, so S < 0.

(b) Liquid to solid, so S < 0.

(c) Solid to solution, so S > 0.

(d) Although the total moles decrease (3 + 2 1 + 2), the moles of gas increase (2 1 + 2).  Therefore, S > 0.

(e) Gas to liquid, so S < 0.

(f) The moles of gas are unchanged (1 + 1 1 + 1), so more information is needed to determine the sign of S.   

 

 

Section 14-2:  Calculating S Using Standard Entropies, S

 

We can calculate the standard entropy change, S, for a reaction in the unit J/molK using the standard entropies of the reactants and products as follows:

 

 

 

Sample Exercise 14B:

 

Predict the sign of S and then calculate the value of S for the reaction N2 (g) + 3 F2 (g) 2 NF3 (g) using the following information:

 

Substance

S (J/molK)

F2 (g)

203

N2 (g)

192

NF3 (g)

261

 

Solution:

 

S is expected to be negative because the moles of gas decrease during the reaction (1 + 3 2).

 

S = [2(261) - 1(192) - 3(203)] J/molK = -279 J/molK

 

 

 

Section 14-3:  The Second Law of Thermodynamics and Gibbs Free Energy Change, G

 

A spontaneous process is one that will continue to progress without outside intervention.  According to the Second Law of Thermodynamics, the entropy of the universe must increase as a result of a spontaneous process and, therefore, the entropy change of the universe must be positive as a result of a spontaneous process.  For a chemical reaction, we can consider the entropy change of the universe to be the sum of the entropy change of the reaction plus the entropy change of the surroundings, which includes everything but the reaction particles.  Therefore, the Second Law of Thermodynamics can be expressed for chemical reactions as follows:

 

spontaneous reaction:  Suniverse = Srxn + Ssurroundings > 0

 

Ssurroundings is related to Hrxn and the Kelvin temperature as follows:  Ssurroundings = -Hrxn/T.  If we substitute for Ssurroundings in the equation above, we obtain:

 

spontaneous reaction:  Srxn - Hrxn/T > 0

 

If we multiply both sides of the equation by (-T), we obtain:

 

spontaneous reaction:  HrxnTSrxn < 0

 

The quantity HrxnTSrxn is known as the Gibbs free energy change (or Gibbs energy change), Grxn.  Therefore, Grxn will be negative for a spontaneous reaction:

 

spontaneous reaction:  Grxn = HrxnTSrxn < 0

 

Note that if Grxn and Hrxn are expressed in kJ/mol, Srxn must first be converted from J/molK to kJ/molK before using the equation above.

 

Sample Exercise 14C:

 

For a certain reaction at 75C, H = 118 kJ/mol and S = 175 J/molK.  Calculate G for the reaction at 75C and determine if the reaction is spontaneous at this temperature.

 

Solution:

 

 

Since G > 0, the reaction is nonspontaneous at 75C.

 

 

Section 14-4:  The Effect of Temperature on G and the Spontaneity of Reactions

 

Given that the sign of the Kelvin temperature cannot be negative, we see mathematically that:

 

1. A reaction with H < 0 and S > 0 will have G = H – TS < 0 at all temperatures.  Therefore, the reaction will be spontaneous at all temperatures.  Since both the negative sign of H and the positive sign of S lead to the reaction being favorable, the reaction is both enthalpy and entropy driven.

 

2. A reaction with H > 0 and S < 0 will have G = H – TS > 0 at all temperatures.  Therefore, the reaction will be nonspontaneous at all temperatures.

 

3. A reaction with H > 0 and S > 0 will have G = H – TS < 0 only at relatively high temperatures.  Therefore, the reaction will only be spontaneous at relatively high temperatures.  Since only the positive sign of S (but not the positive sign of H) leads to the reaction being favorable, the reaction is entropy driven.

 

4. A reaction with H < 0 and S < 0 will have G = H – TS < 0 only at relatively low temperatures.  Therefore, the reaction will only be spontaneous at relatively low temperatures.  Since only the negative sign of H (but not the negative sign of S) leads to the reaction being favorable, the reaction is enthalpy driven.

 

The discussion above is summarized in the following table:

 

H

S

G = H – TS

Reaction is

(-)

(+)

(-) at all T

spontaneous at all T,

both enthalpy and entropy driven

(+)

(-)

(+) at all T

nonspontaneous at all T

 

(+)

(+)

(-) at high T only

spontaneous at high T only,

entropy driven

(-)

(-)

(-) at low T only

spontaneous at low T only,

enthalpy driven

 

 

Sample Exercise 14D:

 

Determine whether reactions with the following H and S values will be spontaneous at all temperatures, nonspontaneous at all temperatures, spontaneous at high temperatures only, or spontaneous at low temperatures only.  Also indicate the driving force for each spontaneous reaction:

 

(a) H = 42 kJ/mol, S = 86 J/molK

(b) H = -242 kJ/mol, S = 357 J/molK

(c) H = -124 kJ/mol, S = -289 J/molK

(d) H = 26 kJ/mol, S = -33 J/molK

 

Solution:

 

(a) spontaneous at high T only, entropy driven

(b) spontaneous at all T, both enthalpy and entropy driven

(c) spontaneous at low T only, enthalpy driven

(d) nonspontaneous at all T

 

 

For reactions that are neither spontaneous at all temperatures nor nonspontaneous at all temperatures, we can estimate the cutoff temperature at which the reaction changes from spontaneous to nonspontaneous by setting G equal to zero in the equation G = H – TS and then solving for the temperature.  Note that the calculated cutoff temperature is only an estimate because the H and S values used in the calculation may change appreciably as the reaction temperature increases or decreases.  

 

Sample Exercise 14E:

 

For a reaction with H = -77.1 kJ/mol and S = -121 J/molK, estimate the cutoff temperature in C at which the reaction changes from spontaneous to nonspontaneous and also specify if the reaction is spontaneous above or below this cutoff temperature.

 

Solution:

 

 

Therefore, the cutoff temperature in C is 637 K – 273 = 364C.  Since both H and S are negative, the reaction will only be spontaneous below 364C.

 

 

 

Section 14-5:  Calculating G from Standard Gibbs Free Energies of Formation, Gf

 

We can calculate the standard Gibbs free energy change, G, for a reaction directly using the standard Gibbs free energies of formation of the reactants and products as follows:

 

 

Note that Gf = 0 kJ/mol for all elements in their standard states.

 

Sample Exercise 14F:

 

Calculate G for the reaction 2 H2S (g) + 3 O2 (g) 2 H2O (l) + 2 SO2 (g) at 25C using the following information:

 

Compound

Gf (kJ/mol) at 25C

H2S (g)

-33.6

H2O (l)

-237.1

SO2 (g)

-300.2

 

Solution:

 

Gf = 0 kJ/mol for the standard state element O2 (g), so G = [2(-237.2) + 2(-300.2) - 2(-33.6) - 3(0)] kJ/mol = -1007.4 kJ/mol

 

Note that G values calculated using this method are only valid at the temperature specified with the Gf data.  At other temperatures, we can estimate G using the equation G = H - TS.

 

 

Chapter 14 Practice Exercises and Review Quizzes:

 

14-1) Predict the sign of S for each process:

 

(a) Ice melts.

(b) 2 C8H18 (l) + 25 O2 (g) 16 CO2 (g) + 18 H2O (l)

(c) Helium gas is heated from room temperature to 200C.

(d) Dry ice (solid carbon dioxide) sublimes.

(e) Sodium chloride crystallizes from a salt water solution.

(f) Liquid methanol evaporates.

 

Click for Solution

 

14-1)

(a) Solid to liquid, so S > 0.

(b) Although the total moles increase (2 + 25 16 + 18), the moles of gas decrease (25 16).  Therefore, S < 0.

(c) Temperature increases, so S > 0.

(d) Solid to gas, so S > 0.

(e) Solution to solid, so S < 0.

(f) Liquid to gas, so S > 0.

 

 

14-2) Predict the sign of S and then calculate the value of S for the reaction CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g) using the following information:

 

Substance

S (J/molK)

CH4 (g)

186.2

CO2 (g)

213.6

H2O (g)

188.7

O2 (g)

205.0

 

Click for Solution

 

14-2) The sign of S is difficult to predict and likely close to zero because the moles of gas are unchanged (1 + 2 1 + 2). 

 

S = [1(213.6) + 2(188.7) - 1(186.2) – 2(205.0) J/molK = -5.2 J/molK

 

 

14-3) For a certain reaction at 115C, H = -88 kJ/mol and S = -170. J/molK.  Calculate G for the reaction at 115C and determine if the reaction is spontaneous at this temperature.

 

Click for Solution

 

14-3)

 

Since G <0, the reaction is spontaneous at 115C.

 

 

14-4) Determine whether reactions with the following H and S values will be spontaneous at all temperatures, nonspontaneous at all temperatures, spontaneous at high temperatures only, or spontaneous at low temperatures only.  Also indicate the driving force for each spontaneous reaction:

 

(a) H = 609 kJ/mol, S = -322 J/molK

(b) H = -92 kJ/mol, S = -46 J/molK

(c) H = 65 kJ/mol, S = 73 J/molK

(d) H = -402 kJ/mol, S = 149 J/molK

 

Click for Solution

 

14-4)

(a) nonspontaneous at all T

(b) spontaneous at low T only, enthalpy driven

(c) spontaneous at high T only, entropy driven

(d) spontaneous at all T, both enthalpy and entropy driven

 

 

14-5) For a reaction with H = 47.3 kJ/mol and S = 120. J/molK, estimate the cutoff temperature in C at which the reaction changes from spontaneous to nonspontaneous and also specify if the reaction is spontaneous above or below this cutoff temperature.

 

Click for Solution

 

14-5)

 

Therefore, the cutoff temperature in C is 394 K – 273 = 121C.  Since both H and S are positive, the reaction will only be spontaneous above 121C.

 

 

14-6)

Calculate G for the reaction Fe3O4 (s) + 4H2 (g) 3 Fe (s) + 4 H2O (g) at 25C using the following information:

 

Compound

Gf (kJ/mol) at 25C

Fe3O4 (s)

-1015

H2O (g)

-229

 

Click for Solution

 

14-6) Gf = 0 kJ/mol for the standard state elements H2 (g) and Fe (s), so G = [3(0) + 4(-229) – 1(-1015) – 4(0)] kJ/mol = 99 kJ/mol

 

 

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