Ch. 17 - Chemical ThermodynamicsWorksheetSee all chapters
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Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
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Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
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Ch. 17 - Chemical Thermodynamics
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Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds
Sections
Spontaneous Reaction
First Law of Thermodynamics
Entropy
Gibbs Free Energy
Additional Practice
Second and Third Laws of Thermodynamics
Boltzmann Equation
GuideLearnAdditional Practice
Jules Bruno

Gibbs Free Energy is used to determine the amount of reusable energy in a thermodynamic system (chemical reaction). Its value can help to determine the direction a reaction takes in order to reach spontaneity. 

Gibbs Free Energy, Spontaneity and Entropy

According to the 2nd Law of Thermodynamics, the entropy of the universe always increases for a spontaneous process. This is illustrated by the equation: 

Second-Law-ThermodynamicsSecond Law of Thermodynamics

Gibbs Free Energy is a thermodynamic property that can be used in determining the direction of a spontaneous process at either constant temperature (isothermal) or constant pressure (isobaric). By rearranging the 2nd Law of Thermodynamics we obtain the following equation to define Gibbs-Free Energy: 

ΔGº = ΔHº-TΔSº


PRACTICE: For a particular reaction, ΔH = + 111.4 kJ and ΔS = – 25.0 J/K. Calculate ΔG for this reaction at 298 K. What can be said about the spontaneity of the reaction at 298 K?

a) ΔG = – 103.95 kJ; The system is at equilibrium 

b) ΔG = + 118.85 kJ; The system is spontaneous in the reverse direction. 

c) ΔG = + 118.85 kJ; The system is spontaneous as written.


STEP 1: Change the units of entropy from joules (J) to kilojoules (kJ) because both enthalpy (ΔH) and Gibbs-Free energy (ΔG) are in kJ. 

Entropy-Value-ConversionEntropy Value Conversion


STEP 2: Plug in your values into the Gibbs-Free Energy formula to obtain your answer. 

Gibbs-Function-Free-Enthalpy-Potential-Gibbs-Energy-non-spontaneuos-formationCalculating Gibbs Free Energy


STEP 3: Whenever Gibbs-Free energy (ΔG) is a positive value this means that the chemical reaction will be non-spontaneous in the forward direction. However chemical reactions always wish to move in the direction that makes them spontaneous. This means the reaction will move in the reverse direction to become spontaneous. Option B is the correct answer. 


Spontaneity & Temperature

The researchers Josiah Willard Gibbs and Hermann von Helmholtz theorized that at constant pressure there is a strong correlation between temperature and Gibbs-Free Energy. If we know the signs of enthalpy (ΔH) and entropy (ΔS) then varying temperatures can predict spontaneity. 

Standard-Free-Energy-Gibbs-TemperatureGibbs Free Energy Table & Spontaneity

+ΔH  +ΔS

When both enthalpy and entropy are positive values then the reaction becomes more spontaneous (ΔG < 0) as the temperature increases.


+ΔH –ΔS

When enthalpy is positive and entropy is negative then the reaction will always be non-spontaneous (ΔG > 0) at all temperatures.  


–ΔH +ΔS

When enthalpy is negative and entropy is positive then the reaction will always be spontaneous (ΔG < 0) at all temperatures. 


–ΔH –ΔS

When both enthalpy and entropy are negative values then the reaction becomes more spontaneous (ΔG < 0) as the temperature decreases.


PRACTICE: Which statement best describes the following endothermic reaction?

C2H6(g) → C2H4(g) + H2(g)

This reaction is ________.

a) spontaneous at all temperatures

b) spontaneous only at a high temperature

c) spontaneous only at a low temperature

d) nonspontaneous at all temperatures

STEP 1: Identify the sign of enthalpy (ΔH). 

We are told that the chemical reaction is exothermic. This means that enthalpy is a negative value. 


STEP 2: Identify the sign of entropy (ΔS). 

Entropy represents chaos or disorder. As a result of one reactant molecule producing multiple product molecules there is an increase in entropy. This means that entropy is a positive value. 


STEP 3: Match the signs of enthalpy (ΔH) and entropy (ΔS) to our spontaneity grid. 

Standard-Free-Energy-Gibbs-Temperature-Generator-free-energy-definitionTemperature and Gibbs Free Energy

ANSWER: –ΔH +ΔS

Since enthalpy is negative and entropy is positive the reaction will be spontaneous (ΔG < 0) at all temperatures. Option A is the correct answer. 


Endergonic and Exergonic Reactions

Exergonic reactions have a release of energy that results in the formation of less-energetic products with greater stability. In addition the Gibbs Free Energy value will be negative, the process will be spontaneous and the forward direction will be favored in the formation of products. 

Exergonic-Reaction-Equation-system-irreversibleExergonic Energy Diagram

Endergonic reactions deal with the absorption of energy that results in the formation of more-energetic and less stable products. In addition the Gibbs Free Energy value will be positive, the process will be non-spontaneous, and the reverse direction will be favored in the formation of reactants. 

Endergonic-Energy-Diagram-thermodynamic-system-completely-reversible-workEndergonic Energy Diagram

At equilibrium there is no net change in the energy between reactants and products. This results in a Gibbs Free Energy value equal to zero and no direction is predominantly favored. 

Neutral-Gibbs-DiagramEnergetically Neutral Diagram

Gibbs Free energy is at the center of our understanding of spontaneity, but isn’t the only useful variable. Other variables such as the equilibrium constant (K), cell potential (ΔE), entropy (ΔS), and the internal energy (ΔU) of a process can also be examined. 

Jules Bruno

Jules felt a void in his life after his English degree from Duke, so he started tutoring in 2007 and got a B.S. in Chemistry from FIU. He’s exceptionally skilled at making concepts dead simple and helping students in covalent bonds of knowledge.

Previous SectionEntropy
Additional Problems
Consider the reaction below, what is ΔG when the pressures of the gases are as follows at 25 °C? H2 = 0.20 atm; Cl 2 = 0.30 atm; HCl = 0.90 atm. The value for ΔG° is −190 kJ/mol                         H2(g) + Cl2(g) → 2 HCl(g) A.  −190.0 kJ/mol B.  6.4 kJ/mol C.  45.5 J/mol D.  183.5 kJ/mol E.  −183.5 kJ/mol  
Consider the reaction 2 Fe2O3(s) + 3 C(s) → 4 Fe(s) + 3 CO2(g), ΔH° = 462 kJ, ΔS° = 558 J • K -1. Calculate the equilibrium constant for this reaction at 525°C. 1. 1.9 x 106 2. 2.8 x 10-2  3. 8.07 x 10-2 4. 3.04 x 10-3 5. 5.20 x 10-7      
As O2 (l) is cooled at 1 atm, it freezes at 54.5 K to form Solid I. At a lower temperature, Solid I rearranges to Solid II, which has a different crystal structure. Thermal measurements show that ΔH for the I→II phase transition = -743.1 J/mol, and ΔS for the same transition = -17.0 J/K mol. At what temperature are Solids I and II in equilibrium? A. 2.06 K B. 43.7 K C. 31.5 K D. 53.4 K E. They can never be in equilibrium because they are both solids.
Determine the equilibrium constant for the following reaction at 298 K. Cl(g) + O3(g) → ClO(g) + O2(g)                                      ΔG° = −34.5 kJ/mol A) 0.986          B) 4.98 × 10−4           C) 5.66 × 105                      D) 1.12 × 106                     E) 8.96 × 10−7
For the reaction: A (l)  + 2 D (g) → 3 X (g) + Z (s) Having ΔG° = -2400 kJ at 25°C, the equilibrium mixture _____________.   a. will consist almost exclusively of A and D b. will consist almost exclusively of A and Z c. will consist almost exclusively of X and Z d. will consist of significant amounts of A, D, X, and Z e. Has a composition predictable only if one knows T and ΔH° and ΔS°
Water gas, a commercial fuel, is made by the reaction of hot coke with steam. C (s) + H2O (g) → CO (g) + H2 (g) When equilibrium is established at 800°C the concentrations of CO, H2, H2O are 4.00 x 10 -2, 4.00 x 10 -2, and 1.00 x 10 -2 mole/liter, respectively. What is the value of ΔG° for this reaction at 800°C? A. 109 kJ B. -43.5 kJ C. 193 kJ D. 16.3 kJ E. none of these
Calculate ΔG°rxn for the following reaction at 1000°C. 2CO(g) + 2NO(g) → N2(g) + 2CO2(g) ΔH° = -748.6 kJ; ΔS° = -197.8 J/K A) -496 kJ B) +1000 kJ C) -1000 kJ D) +496 kJ E) -551 kJ
Given the following free energies of formation calculate Kp at 298 K for: C2H2 (g) + 2 H2 (g) → C2H6 (g) C2H2 (g)          ΔG° f  = 209.2 kJ/mol C2H2 (g)          ΔG° f  = -32.9 kJ/mol   A. 9.07 x 10 -1 B. 97.2 C. 1.24 x 10 31 D. 2.74 x 10 42   
Use Hess's law to calculate ΔG°rxn using the following information. H2O(g) + C(s) → CO(g) + H2(g)          ΔG°rxn = ? H2(g) + ½ O2(g)→ H2O(g)                 ΔG°rxn = -228.6 kJ C(s)+ ½ O2(g) → CO(g)                     ΔG°rxn = -137.2 kJ A) -365.8 kJ B) +365.8 kJ C) -91.4 kJ D) +91.4 kJ E) more information is required
Which of the following reactions will have the smallest equilibrium constant (K) at 298 K? A) CaCO3(s) → CaO(s) + CO2(g)                           ΔG° =+131.1 kJ B) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)          ΔG° = -28.0 kJ C) 3 O2(g) → 2 O3(g)                                             ΔG° = +326 kJ D) 2 Hg(g) + O2(g) → 2 HgO(s)                              ΔG° = -180.8 kJ E) It is not possible to determine without more information.
The figure represents a reaction at 298 K. Which statement is true? 1. At point B, the reaction will shift to the right to reach equilibrium.  2. The reactants possess less free energy than the products. 3. K is less than 1. 4. The ∆Go  of reaction is zero at point C.
Which combination of ∆G◦ and K is possible at standard conditions? 1. ∆G◦ = 99.3 kJ, K = 1.02 2. ∆G◦ = 73.4 J, K = 1.38 × 10 7 3. ∆G◦ = −41.1 kJ, K = 0.971 4. ∆G◦ = −33.3 J, K = 1.01  5. ∆G◦ = −44.6 J, K = 5.62 × 10 −18  
Consider the reaction below. A2 (g) + 3 B2 (g) → 2 AB3 (g) A flask is allowed to come to equilibrium at 584 °C, and is found to contain 0.420 atm of A  2; 0.780 atm of B2 and 1.362 atm of AB3. What is the correct value ΔG° for this reaction? a. -156.9 kJ b. -10.8 kJ c. -10.2 kJ d. -15.9 kJ e. -6.9 kJ      
For the reaction: 2C(graphite) + H2(g) → C2H2(g) ΔG° = +209.2kJ at 25°C. If P(H2) = 100 atm, and P(C2H2) = 0.10 atm, calculate ΔG for reaction. A. +192.1 kJ B. +266.3 kJ C. -16.9 kJ D. +207.8 kJ E. +17.3 kJ    
HI has a normal boiling point of -35.4 oC and its ΔHvap is 21.16 kJ/mol. Calculate the molar entropy of vaporization (ΔSvap). 
Fill in the blanks: If ΔG°&lt; 0, then K is _____. If ΔG° &gt; 0, then K is _____. If ΔG° = 0, then K is ______.    (a) &gt; 1, &lt; 1, = 1     (b) &lt; 1, &gt; 1, = 1     (c) &lt; 0, &gt; 0, = 0 (d) &gt; 0, &lt; 0, = 0     (e) &lt; 1, &gt; 1, = 0
Which of the following has ΔG of  = 0 at 25 oC? A)  CO2 (l)    B)  H2O (l)    C)  Hg (s)    D)  O2 (g)    E)  NH3 (g)
The standard molar Gibbs free energy of formation of NO 2 (g) at 298 K is 51.30 kJ·mol−1 and that of N2O4 (g) is 97.82 kJ·mol−1. What is the equilibrium constant at 25°C for the reaction 2 NO2(g) ⇌ N2O4(g) ? 1. 6.88 2. 0.145 3. 7.01 × 10−9 4. 1.00 5. None of these 6. 1.02 × 10−10 7. 9.72 × 109 8. 0.657
Use Hess's law to calculate ΔG°rxn using the following information. ClO(g) + O3(g) → Cl(g) + 2 O2(g)    ΔG°rxn = ?   2 O3(g)→ 3 O2(g)                            ΔG°rxn = +489.6 kJ Cl(g) + O3(g) → ClO(g) + O2(g)       ΔG°rxn = -34.5 kJ a) -472.4 kJ b) -210.3 kJ c) +455.1 kJ d) +262.1 kJ e) +524.1 kJ
A certain reaction is spontaneous at 72 °C. If the enthalpy change for the reaction is 19 kJ/mol, what must be the minimum value of ΔS (in J/K-mol) for this reaction?
You calculate the value of ΔG° for a chemical reaction and get a positive value. Which would be the most accurate way to interpret this result? 1. If a mixture of reactants and products is created and left to equilibrate, the equilibrium mixture will contain more reactant than product.  2. If a mixture of reactants and products is created, we cannot say anything about its composition at equilibrium but we can say it will reach equilibrium very rapidly.  3. The reaction will not occur under any circumstances. 4. If a mixture of reactants and products is created and left to equilibrate, the equilibrium mixture will contain more product than reactant. 
Calculate ΔG° for the following reaction 3NO2(g) + H2O(l) →  2HNO3(l) + NO(g) Given the following free energies of formation                        ΔG°f(kJ/mol) H2O(l)                -237.2 HNO3(l)             -79.9 NO(g)                86.7 NO2(g)              51.8 A. -414 kJ B. 8.7  kJ C. -192 kJ D. 192  kJ E. -155 kJ  
The equilibrium constant for the reaction:  AgBr(s) ⇌  Ag+(aq) + Br-(aq) is the solubility product constant, Ksp= 7.7 x 10 -13 at 25°C. Calculate ΔG for the reaction when [Ag+] = 1.0 x 10-2 M and [Br-] = 1.0 x 10-3 M. Is the reaction spontaneous or nonspontaneous at these concentrations? A. ΔG = 69.1 kJ, nonspontaneous B. ΔG = -97.5 kJ, spontaneous C. ΔG = 40.6 kJ, nonspontaneous D. ΔG = -69.1 kJ, spontaneous E. ΔG = 97.5 kJ, nonspontaneous
The reaction rates of many spontaneous reactions are actually very slow. Which of these statements is the best explanation for this observation? A) Kp for the reaction is less than one. B) The activation energy of the reaction is large. C) ΔG° for the reaction is positive. D) Such reactions are endothermic. E) The entropy change is negative. 
Hydrogen peroxide (H2O2) decomposes according to the equation: H2O2(l) → H2O(l) + 1/2 O2(g) From the following data calculate Kp for this reaction at 25°C ΔH° = -98.2kJ ΔS° = 70.1 J/K A. 3.46 x 1017 B. 1.3 x 10-21 C. 7.7 x 1020 D. 8.6 x 104 E. 20.9
Which of following has ΔG°f = 0 at 25°C? A) H2O(l) B) H2O(g) C) Na(s) D) O3(g) E) O(g)
Sulfuryl  dichloride is formed when sulfur dioxide reacts with chlorine. The data refer to 298 K. SO2(g) + Cl2(g) → SO2Cl2(g) Substance:              SO 2(g)           Cl 2(g)          SO2Cl2(g) ΔH°f (kJ/mol):         -296.8                0                -364.0 ΔG°f (kJ/mol):         -300.1                0                -320.0 S°   (J/K•mol):           248.2             223.0           311.9 What is the value of ΔG° for this reaction at 600 K? A) -162.8 kJ B) -40.1 kJ C) -28.4 kJ D) 28.4 kJ E) 162.8 kJ  
The U.S. government now requires that automobile fuels consist of a renewable component. One such biofuel, ethanol, may be produced by fermentation of the glucose, C6H12O6(s), from feedstocks such as corn and sugar cane: C6H12O6(s) → 2 C2H5OH(l) + 2 CO2(g) Given the following thermodynamic data, what is the standard Gibbs free energy change for the above reaction at 298 K?                          ΔHf° (kJ·mol  –1 )           S°(J·K –1 ·mol –1 ) C6H12O6(s)        -1274.5                        212.7 C2H5OH(l)          -277.7                         160.7 CO2(g)               -393.5                          213.7 (a) –92.0 kJ (b) –228 kJ (c) –19.5 kJ (d) –116 kJ (e) 92.0 kJ
What is the value of the equilibrium constant K, for a reaction for which ΔG ° is equal to -5.20 kJ at 50 °C? a) 0.144 b) 0.287 c) 6.93 d) 86.4
For the reaction NH4Cl (s) → NH3 (g) + HCl (g) ΔH ° = +176 kJ and ΔG ° = +91.2 kJ at 298 K. What is the value of ΔG at 1000 K? a) -109 kJ b) -64 kJ c) +64 kJ d) +109 kJ      
The standard molar enthalpy of formation of NO2(g) is 33.2 kJ/mol at 25◦C and that of N2O4(g) is 9.16 kJ/mol. At 25◦C their absolute entropies are 240.0 and 304.2 J/mol K, respectively. Use the data to calculate the standard Gibbs free energy change for the reaction N2O4(g) → 2 NO2(g) at 25 ◦C. 1. 41.5 kJ/mol 2. 21.3 kJ/mol 3. 4.1 kJ/mol 4. 11.4 kJ/mol 5. 4.85 kJ/mol
A. Using the thermodynamic data provided for dissolving Na 2SO4(s) in water, that is, for the chemical process Na2SO4(s) → 2Na+(aq) + SO42-(aq)         Calculate ΔH°.   B. When I calculated ΔS°, I got -11.84 J/K•mol. Using this value and your answer to part A, calculate ΔG° at 25°C.     C. Which term is responsible for the temperature dependence of ΔG: enthalpy (ΔH°), entropy (ΔS°), both, or neither (ΔG is not temperature dependent)? Will increasing the temperature make ΔG larger (more positive), smaller (more negative), or have no effect? D. So based on your answer to part C, will the solubility of Na 2SO4 increase or decrease with increasing temperature?              
Calculate the equilibrium constant, K, for the reaction at 298 K: N2O4 (g) ⇌  2 NO2 (g)  
The reaction A(g) ⇌ B(g) has an equilibrium constant of Kp = 2.3x10-5. What can you conclude about the sign of ΔG°rxn for the reaction? a) ΔG°rxn = 0 b) ΔG°rxn &lt; 0 c) ΔG°rxn &gt; 0 d) No conclusion can be made
Iron metal will react with oxygen gas to form a variety of iron oxides. This oxidation reaction is typically referred to as the iron “rusting”. The fact that this reaction is spontaneous at room temperature tells you that  1. iron oxides have a negative Gibbs energy of formation  2. iron oxides have a positive enthalpy of formation 3. iron oxides have a higher standard entropy compared to oxygen and iron 4. the 2nd law of thermodynamics has been violated
The equilibrium constant for the following reaction is 5.0x10 8 at 25 °C. N2 (g) + 3 H2 (g) ⇌ 2 NH3 (g) The value of ΔG° is ____ kJ/mol.
The solubility product constant at 25°C for AgI (s) in water has the value of 8.3x10  -17. Calculate ΔGrxn at 25 °C for the process: AgI (s) ⇌ Ag+ (aq) + I - (aq) where [Ag+] = 9.1 x10-9 and [I-] = 9.1x10-9.  
Given the following free energies of formation calculate K p at 298 K for: C2H2 (g) + 2 H2 (g) → C2H6 (g) C2H2(g)      ΔG°f = 209.2 kJ/mol C2H6(g)      ΔG°f = -32.9 kJ/mol a) 9.07 x 10 –1 b) 97.2 c) 1.24 x 10 31 d) 2.74 x 10 42 e) None of these is within a factor of 10 of the correct answer.  
Consider the following reactions for dissolving two different chloride salts in water:  a. Calculate ΔG° for both of these reactions using the thermodynamic parameters below (assume 298K).       b. Using the results of the above calculations, what is the Ksp of these salts?     c. Which salt is more soluble? AlCl3(s) or NaCl(s)   d. Assuming each solution reaches saturation, which salt should I add to increase the boiling point of water more? AlCl3(s) or NaCl(s)  
Calculate the standard free-energy change for the formation of NO (g) from N 2 (g) and O2 (g) at 298 K: N2 (g) + O2 (g) → 2 NO (g) Given that ΔH° = 180.7 kJ and ΔS° = 24.7 J/K. Is the reaction spontaneous under these conditions?
The equilibrium constant for the reaction AgBr(s) ⇄ Ag +(aq) + Br– (aq) is the solubility product constant, Ksp = 7.7 x 10–13 at 25oC. Calculate ΔG for the reaction when [Ag+] = 1.0 x 10–2 M and [Br–] = 1.0 x 10–3 M. Is the precipitation of AgBr spontaneous or nonspontaneous at these concentrations? a) ΔG = 69.1 kJ/mol, spontaneous b) ΔG = ‐69.1 kJ/mol, non‐spontaneous c) ΔG = 97.5 kJ/mol, non‐spontaneous d) ΔG = 40.6 kJ/mol, spontaneous e) ΔG = ‐97.5 kJ/mol, spontaneous