Ch. 17 - Chemical ThermodynamicsSee all chapters
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Ch. 17 - Chemical Thermodynamics
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Gibbs Free Energy

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Sections
Spontaneous Reaction
First Law of Thermodynamics
Entropy
Gibbs Free Energy
Additional Practice
Second and Third Laws of Thermodynamics
Boltzman 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      
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    
At 25 oC, ΔHo = 1.895 kJ and ΔSo = – 3.363 J/K for the transition  C (graphite) →  C (diamond) Based on their data a) Graphite is more stable than diamond below 290 oC and diamond is more stable than graphite above 290 oC.  b) Graphite cannot be converted to diamond at 1 atm pressure.  c) Diamond is more stable than graphite at all temperatures at 1 atm.  d) Diamond is more stable than graphite below 290 oC and graphite is more stable than diamond above 290 oC. 
For each of the following chemical processes, state whether ΔS and ΔG are positive, negative,essentially zero, or undeterminable (based on the information provided).  
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
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
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.
Based on your knowledge of carbon allotropes, You can say that ∆G◦f of graphite is ________ and ∆G◦rxn of diamond → graphite is _______. 1. zero, negative  2. positive, positive 3. large, zero 4. zero, large 5. large, small 6. small, negative
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
Estimate ΔG°rxn for the following reaction at 775 K. 2 Hg(g) + O2(g) → 2 HgO(s) ΔH°= -304.2 kJ; ΔS°= -414.2 J/K A) -625 kJ B) -181 kJ C) +17 kJ D) +321 kJ E) -110 kJ
Use the data below to calculate the standard free-energy change for the reaction. P4 (g) + 6 Cl2 (g) → 4 PCl3 (g) run at 298 K. ΔGf° (P4) = 24.4 kJ/mol ΔGf° (PCl3) = -269.6 kJ/mol
Use Hess's law to calculate ΔG°rxn using the following information. CO(g) → C(s) + 1/2 O2(g)         ΔG°rxn = ?   CO2(g) → C(s) + O2(g)              ΔG°rxn = +394.4kJ CO(g) + 1/2 O2(g) → CO2(g)      ΔG°rxn = -257.2 kJ   A) -60.0 kJ B) +651.6 kJ C) -265.8 kJ D) +137.2 kJ E) +523.0 kJ  
In the gas phase, methyl isocyanate (CH 3NC) isomerizes to acetonitrile (CH 3CN), H3C−N≡C (g) ⇄ H3C−C≡N (g) with ΔHo = –89.5 kJ/mol and ΔGo = –73.8 kJ/mol at 25oC. What is the equilibrium constant, Kp, for this reaction at 100oC, assuming that the enthalpy and entropy of reaction are independent of temperature. a) 4.62 x 10 ‐11 b) 1.68 x 10 ‐10 c) 5.96 x 109 d) 5.36 x 109 e) 8.64 x 1012
Calculate ΔG at 298 K for a mixture of 1.0 atm N 2, 3.0 atm H2, and 0.50 atm NH3 being used in the Haber process: The Gibbs energy of formation of ammonia is -16.4 kJ/mol.  N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)
Which expression is true at standard concentrations:             A.  ΔG = 0             B.  ΔG° = 0             C.  ΔG = ΔG°             D.  Q = K             E.  K = 0
Which of the following reactions will have the largest 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 standard enthalpy and entropy changes of reaction are –30.20 kJ and +9.410 J/K, respectively, and are independent of temperature. Calculate equilibrium constant of the reaction at 250.0  ̊C. a) 1.01 b) 3.15 c) 3.2x103 d) 1.4x106 e) 3.9x108
Determine the equilibrium constant for the following reaction at 298 K. SO3(g) + H2O(g) → H2SO4(l) ΔG° = -90.5 kJ A) 1.37 x 10 -16 B) 4.78 x 1011 C) 7.31 x 1015 D) 9.11 x 10 -8 E) 0.964
What is the ΔG (kJ/mol) for a reaction at 25 Celsius that is: Mg3(PO4)2 (s) → 3 Mg2+(aq) + 2 PO43− (aq)       ΔG° = 137.0 kJ/mol If there is initially 0.65 M Mg2+ (aq) and 0.43 M PO43−(aq) in solution? A. 129.6 B. 134.0 C. 137.0 D. 140.0 E. 144.4
Using the listed information calculate ΔG° for the reaction: a) –454.8 kJ b) –386.5 kJ c) –297.1 kJ d) –148.3 kJ e) –394.4 kJ
What is the ∆G (in kJ/mol) for a reaction at 255 Celsius that has a ∆H = 110 kJ/mol and a ∆S of 655 J/mol-K? A. -456 B. -236 C. -57 D. 236 E. 456
To examine the extent that water spontaneously decomposes to its elements at room temperature, consider that the standard Gibbs energy of formation (∆G ̊f) of H2O (l) is –237.20 kJ/mol. Use this information to calculate the equilibrium constant for the decomposition of water at 298 K: 2 H2O (l) ⇄ 2 H2 (g) + O2 (g) a) 7.0 x 10 –84 b) 2.6 x 10 –42 c) 0.91 d) 3.8 x 1041 e) 1.4 x 1083
The definition of the  standard (molar) Gibbs energy of formation (ΔG f°) parallels that of ΔHf°; since ‘formation’ signifies a particular kind of reaction, what is the reaction corresponding to ΔGf° of CO2(g)? Given that ΔGf°(CO2(g)) = -394.4 kJ/mol, what can we say about the stability of carbon dioxide (relative to its elements)?
A reaction is proceeding toward equilibrium. At a certain stage, the concentrations of reactants and products are such that ∆G = ∆G°. What conclusion can reasonably be drawn about the reaction at this time? a) K &gt; Q b) K &lt; Q c) K = Q d) K = 1 e) Q = 1
Every sample of a pure element, regardless of its physical state, is assigned zero Gibbs free energy of formation. a) True b) False
Consider the following compounds and their standard free energies of formation: Which of these liquids is/are thermodynamically unstable? 1. 2 and 4 only 2. 1, 3, and 5 only  3. 2 and 3 only 4. 1 and 4 only 5. 2 only
Which of the following is a description of a phase change for which ΔG is positive.  A. Liquid water forming steam at standard pressure and 94°C B. Liquid water forming ice at standard pressure and -7°C C. Steam forming liquid water standard pressure and 30°C D. Ice forming liquid water at standard pressure and 5°C  
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, H 2, and 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
The standard free energy of formation of AgCl(s) is -110 kJ/mol. ΔG° for the reaction 2 AgCl(s) ⇌ 2 Ag(s) + Cl2(g) is  a) -110 kJ b) -220 kJ c) 110 kJ d) 220 kJ e) none of these
A weak base has a ΔG° of 24 kJ/mole. What is the approximate value for the K b? A. 10-4 B. 104 C. 10-24 D. 1024
If Q &gt; K, what must be true regarding ΔG? A.  it is positive           B.  it equals zero         C.  it is negative
Calculate the standard free energy of formation of mercury (II) oxide at 298 K given the data below. a) +58.5 kJ⋅mol −1  b) +117.1 kJ⋅mol −1  c) –58.5 kJ⋅mol −1  d) –123.1 kJ⋅mol −1  e) –117.1 kJ⋅mol −1 
The standard free energy of formation of CS 2 (l) is 65.27 kJ⋅mol −1 at 298 K. This means that at 298 K a) CS2 (l) will not spontaneously form C (s) + 2 S (s). b) CS2 (l) is thermodynamically unstable. c) CS2 (l) is thermodynamically stable. d) No catalyst can be found to decompose CS 2 (l) into its elements. e) CS2 (l) has a negative entropy.
The standard Gibbs Free Energy of formation for N2 (g) at 25°C is: A. Positive  B. Negative C. 0 kJ/mol D. Not enough information is given. 
Hot and cold packs are commercially available for both medical treatment and food storage. Ammonium chloride, NH4Cl, is commonly used in cold packs and is activated when it is dissolved in solution. What are the signs for ∆H, ∆S, ∆G (at high T) and ∆G (low T)? 
The standard free energy change for a chemical reaction is –71.3 kJ/mol. What is the equilibrium constant for the reaction at 30°C?
For the following reaction: 2 HBr (g)      →      H2(g) + Br2(g)  If HBr, H2 and Br2 are 0.80 atm, 1.1 atm and 1.4 atm respectively at 30°C , calculate ∆G for the reaction when ∆G° is equal to -103.6 J.
Using data tabulated below, calculate ΔG r° at 25°C for C2H4 (g) + 3 O2 (g) → 2 CO2 (g) + 2 H2O (l). Species      ΔHf° (kJ mol –1)       S° (J K –1 mol –1)      ΔGf° (kJ mol –1) C2H4(g)            52.26                         219.45                  68.12 CO2(g)           –393.51                       213.63               –394.36 H2O(l)            –285.83                         69.91                –237.18 O2(g)                                                  205.03 a) 1314.0 kJ  b) –699.7 kJ  c) –1195.0 kJ  d) –1314.0 kJ  e) –1331.2 kJ 
What is true if ln K is negative?   A) ΔG°rxn is positive and the reaction is spontaneous in the forward direction. B) ΔG°rxn is negative and the reaction is spontaneous in the forward direction. C) ΔG°rxn is negative and the reaction is spontaneous in the reverse direction. D) ΔG°rxn is positive and the reaction is spontaneous in the reverse direction. E) ΔG°rxn is zero and the reaction is at equilibrium.
Use Hess’s law to calculate ∆G°rxn using the following information. NO(g) + O(g) → NO2(g), ∆G°rxn = ? 2O3(g) → 3O2(g), ∆G°rxn = +489.6kJ O2(g) → 2O(g), ∆G°rxn = +463.4kJ NO(g) + O3 → NO2(g) + O2(g), ∆G°rxn = -199.5kJ 277.0 kJ -225.7 kJ 753.5 kJ -1152.5 kJ -676.0 kJ
A reaction at equilibrium has a value of ΔG° = -13.2 kJ/mol at 43.0°C. The value of the equilibrium constant K at this temperature is A. 152 B. 1.01 C. 1.08 x 1016 D. 0.00657 E. 9.21 x 10-17
What is the value of K for this aqueous reaction at 298 K? ΔG = 13.75 KJ/mol A + B ⇋ C + D    
Calculate the free energy change for the reaction at 30°C. Is it spontaneous?2Ca(s) + O2(g) → 2CaO(s)Give thatΔH = -1269.8 kJΔS = -346.6 J/K
For which of the following pure chemical substances (at T = 25.0 °C) will ΔG° f, the free energy of formation, be equal to 0.00 kJ/mol? a) O2(g) b) O3(g) c) SO2(g) d) Both a and b e) Both a and b and c
Thermodynamic data are given below (at T = 25. °C) and may be of use in doing this problem.Phosphoric acid (H3PO4) may be prepared by adding tetraphosphorus decaoxide (P4O10) to water (H2O), by the process                P 4O10(s) + 6 H2O(I) → 4 H3PO4(s)a) What are ΔS°rxn and ΔG°rxn for the below reaction, at T = 25.0 °C?b) Give the correctly balanced formation reaction for H 3PO4(s).
Which statement below is true about the standard Gibbs energy -33.3kJ at 25°C for the reaction.N2 (g) + 3 H2 (g) → 2 NH3 (g)                      ΔG° = -33.3kJa. ΔG° corresponds to 1 atm of N 2 reaction with 3 atm of H2 to form 2 atm of NH 3.b. ΔG° corresponds to 1 atm of N 2 reaction with 1 atm of H2 to form 2 atm of NH  3.c. ΔG° corresponds to 1 atm of each N 2, H2, and NH3 reacting to 100% completion.d. ΔG° corresponds to 1 atm of each N 2, H2, and NH3 reacting until equilibrium is established.e. None of the above are true.
What is the value of K for this aqueous reaction at 298 K? ΔG = 13.75 kJ/mol A+B ⇌ C+D
What is the value of K for this aqueous reaction at 298 K?A+B ⇌ C+D          ΔG° = 10.33 kJ/mol
Consider the oxidation of NO to NO 2 :NO(g) + 1/2O2(g)→NO2(g).Calculate ΔG°rxn at 25°C.Express your answer with the appropriate units. Determine whether the reaction is spontaneous at standard conditions. 
The following equation represents the decomposition of a generic diatomic element in its standard state.1/2X2 (g) ⇌ X (g)Assume that the standard molar Gibbs energy of formation of X(g) is 4.82 kJ·mol–1 at 2000 K and –58.08 kJ·mol–1 at 3000 K.a. Determine the value of K (the thermodynamic equilibrium constant) at each temperature.b. Assuming that ΔH°rxn is independent of temperature, determine the value of ΔH°rxn from these data. 
Given the following information A + B → 2D   ΔH° = 724.9 kJ     ΔS° = 307.0 J/K C → D          ΔH° = 402.0 kJ      ΔS° = -139.0 J/K Calculate ΔG° for the following reaction at 298 K. A + B → 2C
When can Q &gt; K but ΔG = 0? a) When the reaction is going forwards after being equilibriumb) When the reaction is going backward towards equilibrium c) For an every endothermic reaction when Q = K d) When the reaction is going backward after being at equilibrium e) never
In the Haber process, N2 (g) + 3H2 (g) → 2NH3 (g)ΔG° at 298 K for this reaction is -33,300 J/mol.The value of ΔG at 298 K for a non-equilibrium reaction mixture that consists of 1.9 atm N  2, 1.6 atm H2, and 0.65 atm NH 3 is __________.a. -40,500 Jb. -1,800 Jc. -7.25 × 106 Jd. -3.86 × 106 Je. -104,500 J 
Calcium carbonate can be converted to quick lime by the following reaction:CaCO3(s) → CaO(s) + CO2(g) Given that:       for CaCO3(s)  at 25°C     ΔHf° = -1206.9 kJ/mol  ΔS° = 92.9 J/mol K,                        for CaO(s)                     ΔHf° = -635.5 kJ/mol      ΔS° = 39.8 J/mol  K                        for CO2(g))                     ΔHf° = -393.5 kJ/mol     ΔS° = 213.6 J/mol  K Use these data to calculate ΔGo  for the reaction.A)  225.7 kJ/mole         B)   130.1 kJ/mole   C) -130.1 kJ/mole    D)    -225.7 kJ/mole      E)  none of these 
At constant temperature and pressure, how is ΔSuniv related to ΔGsys? a) ΔGsys = TΔSuniv b) ΔGsys = -TΔSuniv c) ΔGsys = -ΔSuniv d) ΔGsys = -(ΔSuniv/T) e) ΔGsys = ΔSuniv f) ΔGsys = ΔSuniv/T
The solubility product constant at 25°C for AgI(s) in water has the value 8.3 x 10 –17. Calculate ΔGrxn at 25°C for the process AgI(s) → Ag+(aq) + I– (aq) where [Ag+] = 9.1 x 10–9 and [I–] = 9.1 x 10–9.
What is the difference between ΔG° and ΔG° prime? 
The diagram for the free energy of the reactionA(g) + B(g) ⇌ AB(g)The reaction progress starts on the left with pure reactants, A and B, each at 1 atm and moves to pure product, AB, also at 1 atm on the right. Select the true statements.a. The difference between the top left of the curve and the minimum of the curve corresponds to ΔG standard.b. The "x" on the graph corresponds to ΔG standard.c. The minimum on the graph corresponds to the equilibrium position of the reaction.d. The entropy change for the reaction is positive.e. At equilibrium, all of A and B have reacted to form pure AB.  
Calculate ΔG° for the following reaction at 298 K. A + B → 2C
Use standard free energies of formation to calculate ΔGºrxn for the balanced chemical equation:Ca(s) + N2O(g) → CaO(s) + N2(g)Substance                 ΔG ∘f (kJ/mol)N2O(g)                          103.7CaO(s)                          -603.3
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M2O3(s) → 2M(s) + 3/2 O2(g) Info given for Gf(kJ/mol): M2O3= -6.70 M(s)=0 O2(g)= 0 a) What is the standard change in Gibbs energy for rxn as written in forward direction? (kJ/mol)     b) What is the equilibrium constant (K) of this rxn, as written in forward direction at 298K?     c) What is the equilibrium pressure of O2(g) over M(s) at 298K? (atm)
Which of the following statements best describes an endothermic reaction?a) The reaction must be spontaneous at all temperaturesb) The reaction may become spontaneous at high T if ∆S rxn &gt; 0c) The reaction may become spontaneous at low T if ∆S rxn &lt; 0d) The reaction may become spontaneous at high T if ∆S rxn &lt; 0
Methanol burns in oxygen to form carbon dioxide and water. 2CH3OH+3O2 → 2CO2+4H2O Calculate the rxn of ΔH, G &amp; S at 25°C.
Consider the reaction and table of thermodynamic data at 298 K: What is the value of equilibrium constant for the reaction at 25  ̊C. a) 150 b) 9.3 x 1015 c) 8.4 x 104 d) 1.1 x 10–16 e) 1.4 x 108
If Q &gt; K, what must be true regarding ΔG?A. it is positiveB. it equals zeroC. it is negative
Consider the reaction: FeO(s) + Fe(s) + O2(g)→ Fe2O3(s) Given the following table of thermodynamic data, determine the temperature (in C°) which the reaction is nonspontaneous. a. This reaction is spontaneous at all temperatures. b. 6180.1 c. 756.3 d. 2438 e. 1235
Consider the oxidation of NO to NO2:NO(g)+1/2O2(g)→NO2(g)Standard thermodynamic quantities for selected substances at 25°CCalculate ΔG°rxn at 70°C.Express the free energy change to three significant figures and include the appropriate units. 
What is the equilibrium constant of the reaction at 25°C if ΔG° = -13.0 kJ?2NO(g) + Br2(g) ⇌ 2NOBr(g)a. 190b. 5.3c. 44d. 10-13.0 
The diagram below represents a spontaneous reaction (ΔG°&lt;0). Fill in tthe blanks below
The Kb for methylamine, CH3NH2, at 25°C is 4.4 x 10-4a. Write the chemical equation for the equilibrium that corresponds to K b.b. By using the value of Kb, calculate ΔG° for the equilibrium in part a.c. What is the value of ΔG at equilibrium?d. What is the value of ΔG when [H +] = 1.6 ×10-8 M, [CH3NH3+] = 5.5 ×10-4 M, and [CH3NH2] = 0.130 M?
Calculate the standard change in Gibbs free energy for the following reaction at 25°C. 3H2(g) + Fe2O3(s) → 2Fe(s) + 3H2O(g)
The diagram shows the free energy change of the reaction. A(g) + B(g) ⇌ AB(g) The reaction progress starts on the left with pure reactants, A and B, at 1 atm and moves to pure product, AB, also at 1 atm on the right. Select the true statements.a. The "x" on the graph corresponds to ΔG of this reaction.b. The minimum on the graph corresponds to the equilibrium position of the reaction.c. At equilibrium, all of A and B have reacted to form pure AB.d. The difference between the top left of the curve and the bottom of the curve corresponds to ΔG of this reaction.e. The entropy change for the reaction is positive.
When the oxide of generic metal M is heated at 25°C, only a negligible amount of M is produced. MO2(s) ⇌ M(s) + O2(g) ΔG = 288.5 kJ/mol When the reaction is coupled to the conversion of graphite to carbon dioxide, it becomes spontaneous. i) What is the chemical equation of this coupled process? Show that the reaction is in equilibrium, include physical states, and represent graphite as C(s)?     ii) What is the thermodynamic equilibrium constant for the coupled reaction?
Calculate ΔH∘rxn, ΔS°rxn, ΔG°rxn at 25 °C for the reaction:2NH3(g) → N2H4(g) + H2(g)
A particular reaction has ΔG = –350kJ and ΔS = –350J/K. Which of the following can be said about this reaction at 25 ºC?a. work produced is less than the heat produced, because some of the heat cannot be converted to workb. work produced is less than the heat produced, because some of the work cannot be converted to heatc. heat produced is less than the work produced because some of the heat cannot be converted to workd. heat produced is less than the work produced, because some of the work cannot be converted to heat
Calculate ∆G° rxn using the following information, assume room temperature. 2 HNO3 (aq) + NO (g) → 3 NO2 (g) + H2O (l) H°f(kJ/mol)  -207.0      91.3      33.2     -285.8 S°(J/mol•K) 146.0      210.8      240.1      70.0 -186 kJ -85.5 kJ 50.8 kJ 222 kJ -151 kJ  
Calculate ΔG° for the reaction NH4NO3(s) → NH4+(aq) + NO3−(aq) to determine whether the reaction is spontaneous or not. Temperature: 298.15 K and ΔS°: 108.7 J/K•molCompound          ΔHf° (kJ/mol)NH4NO3(s)          −365.56NH4+(aq)             −132.51NO3−(aq)             −205.0A.  4.4 J , nonspontaneousB.  – 4.4 kJ, spontaneousC.  28.5 kJ spontaneousD.  −28.5 J nonspontaneousE.  5.6 kJ nonspontaneous 
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.M2O3(s) → 2M(s) + 3/2 O2(g)Informaton given for G f(kJ/mol):M2O3 = -6.70M(s) = 0O2(g) = 0a. What is the standard change in Gibbs energy for rxn as written in forward direction?b. What is the equilibrium constant (K) of this rxn, as written in forward direction at 298K?c. What is the equilibrium pressure of O 2(g) over M(s) at 298K?
Consider the following isomerization reactions of some simple sugars and values for their standard Gibbs free energy ΔG∘:reaction A:glucose-1-phosphate⟶ glucose-6-phosphate, ΔG∘=−7.28 kJ/molreaction B: fructose-6-phosphate⟶⟶glucose-6-phosphate,ΔG∘=−1.67 kJ/molCalculate the equilibrium constant K for the isomerization of glucose-1-phosphate to fructose-6-phosphate at 298 K. Express your answer numerically using two significant figures.
What is the value of the equilibrium constant, K, for a reaction for which ΔG is equal to -5.20 KJ at 50°C?  
The reaction SO2(g) + 2H2S(g) ⇌ 3S(s) + 2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant gases. The standard free energy of each substance are ΔGf°S(s) = 0 kJ/mol ΔGf°H2O(g) = -228.57 kJ/mol ΔGf°SO2(g) = -300.4 kJ/molΔGf° H2S (g) = -33.01kJ/molWhat is the equilibrium constant for the reaction at 298K?In principle, is this reaction a feasible method of removing SO2?If Pressure of SO2 = Pressure of H2S and the vapor pressure of water is 26 torr, calculate the equilibrium SO 2 pressure in the system at 298 K.Would you expect the process to be more or less effective at higher temperatures?
Aniline, C6H5NH2 is a weak base related to ammonia. It reacts with water as shown in the following equation. C6H5NH2 (aq) + H2O (l) → C6H5NH4+(aq) + OH- (aq)      Kb (aq) = 4.27 x 10 -10 (a) Calculate ΔG° for the base ionization of aniline. (kJ/mol) (b) Calculate ΔG° for the base ionization of ammonia.(kJ/mol)
For the reaction:2KClO3(s) → 2KCl(s) + 3O2(g)Calculate ΔGºrxn at 25ºC.Express your answer using one decimal place.
Which of the following conditions is always true at equilibrium?
Calculate the Kp for the following reaction at 25º C: H2(g) + I2(g) ---&gt;  2HI(g) Delta Gº = 2.60 kJ/mol
For the following reaction, the change in free energy, delta G°, is 4.73 kJ/mol at 25 C. Determine the equilibrium constant.N2O4 (g) ⇌ 2 NO2(g)a) -1.91b) 0.148c) 6.74d) 1.91
Consider the oxidation of SO2 to SO3: SO2 (g) + 1/2 O2 (g) --&gt; SO3 (g) Calculate Delta Gºrxn at 25 C. Express your answer to one decimal place with the appropriate units. Determine whether the reaction is spontaneous at standard conditions.
Use the data given here to calculate the values of ΔG°rxn at 25 °C for the reaction described by the equation: A + B ⇌ C If ΔH°rxn degree and ΔS°rxn degree are both positive values, what drives the spontaneous reaction and in what direction at standard conditions?
Hydrogen peroxide can be prepared in several ways. One method is the reaction between hydrogen and oxygen, another method is the reaction between water and oxygen. Calculate the ΔG°rxn of each reaction below using values from this table.H2 (g) + O2 (g) ⇌ H2O2 (l)H2O (l) + 1/2 O2 ⇌ (g) H2O2 (l)Which method requires less energy under standard conditions?
At 1500 C the equilibrium constant for the reaction CO(g) + 2H2(g) &lt;--&gt;CH3OH(g) has the value Kp = 1.4 times 10^-7. Calculate Delta Gº  for this reaction at 1500 C.-233 kJ/mol233 kJ/mol-105 kJ/mol105/mol1.07 kJ/mol
For the reaction2 HBr(g) → H2(g) + Br2(l)ΔH°= 72.6 kJ and ΔS° = -114.5 J/K The equilibrium constant for this reaction at 251.0 K is _____.Assume that ΔH° and ΔS° are independent of temperature.
Yeast can produce ethanol by the fermentation of glucose (C6H12O6), which is the basis for the production of most alcoholic beverages. C6H12O6 (aq) → 2 C2H5OH (l) + 2 CO2 (g) Calculate ∆Hº, ∆Sº, and ∆Gº for the reaction at 25 ºC. Thermodynamic Data: ∆Hº = kJ ∆Sº = J/K ∆Gº = kJ
Calculate the standard entropy, Δ S°rxn, of the following reaction at 25.0 °C using the data in this table. The standard enthalpy of the reaction, ΔH°rxn, is -44.2 kJ mol-1 C2H4 (g) + H2O (l) →C2H5OH(l)ΔS °rxn = ________ J K-1 mol-1 Then, calculate the standard Gibbs free energy of the reaction, Δ G °rxn. deltaΔG°rxn = _______kJ mol^-1Finally, determine which direction the reaction is spontaneous as written at 25.0 °C and standard pressure forward reverse both neither
At 25 °C, the equilibrium partial pressures for the following reaction were found to be PA = 5.54 atm, PB = 4.35 atm, Pc = 4.28 atm, and PD = 5.91 atm.3A(g) + 3B(g) → C(g) + 3D (g)What is the standard change in Gibbs free energy of this reaction at 25 °C?
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.M2O3(s) ⇌ 2M(s) + 3/2 O2(g)What is the standard change in Gibbs energy for the reaction, as written, in the forward direction?What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?What is the equilibrium pressure of O2(g) over M(s) at 298 K?
Calculate the equilibrium constant at 25º C for the following reaction:2NH3(g) + CO2(g) ---&gt; NH2CONH2(aq) + H2O(l)  Delta Gº = -13.6 kJExpress your answer using two significant figures
Calculate the ΔG° (in kJ) for the following reaction from the equlibrium constant, KC, 9.5995 at 3, 260°C. H2 (g) + I2 (g) → 2HI (g) (Write answer in decimal form to 2 decimal places)
Calculate ΔrG° for the reaction below at 25.0°C CH4 (g) + H2O (g) → 3H2 (g) + CO (g)ΔfG° [CH4 (g)] = -50.8 kJ/mol, ΔfG° [H2O (g)] = -228.6 kJ/mol, ΔfG° [H2 (g)] = 0.0 kJ/mol, and ΔfG° [CO (g)] = -137.2 kJ/mol. (a) -416.3 kJ/mol-rxn (b) -142.2 kJ/mol-rxn (c) +142.2 kJ/mol-rxn (d) +315.0 kJ/mol-rxn (e) +416.3 kJ/mol-rxn
A reaction has a standard free-energy change of -14.90 kJ mol-1 (-3.561 kcal mo-1). Calculate the equilibrium constant for the reaction at 25°C.
For a particular reaction, Δ H° is -28.4 kJ and ΔS is -87.9 J/K. Assuming these values change very little with temperature, over what temperature range is the reaction spontaneous in the forward direction? The reaction is spontaneous for temperatures
What is the value of K for this aqueous reaction at 298 K? A + B ⇌             G° = 29.30 kJ/mol 
For a particular reaction at 112.0°C, ΔG = -157.33 kJ/mol, and ΔS = 770.16 J/(mol • K). Calculate ΔG for this reaction at -100.5°C.
Given the thermodynamic data in the table below, calculate the equilibrium constant (at 298 K) for the reaction:2SO2 (g) + O2(g) ⇌2SO3(g)
For the reaction Ca(OH)2(aq) + 2HCI(aq) → CaCI2(s) + 2H2O(I) delta Hº = -30.2 kJ and delta Sº = 205.9 J/K.The equilibrium constant for this reaction at 329.0º K is ______Assume that delta Hº and delta Sº are independent of temperature.
Yeast can produce ethanol by the fermentation of glucose (C6H12O6), which is the basis for the production of most alcoholic beverages. C6H12O6 (aq) --&gt; 2 C2H5OH (l) + 2 CO2 (g) Calculate delta H , delta S , and delta G degree for the reaction at 25 C
Given the thermodynamic data below, calculate the value of the equilibrium constant for the reaction shown at 25.0º C. H2 (g) + CO2 (g) &lt;--&gt;CO (g) H2O (g)Δ Hº = 41.15 kJΔ Sº = 42.35 J/K .
For the above reaction at standard conditions (298º K, 1.00 atm pressure, 1.00 M), Delta H is -14.6 kJ/mol, and Delta S is -5.06 J/(mol*K). What is the value of Delta G (in kJ/mol) at these conditions?
A reactionA(aq) + B(aq) ⇌ C(aq)has a standard free-energy change of -5.07 kJ/mol at 25°C.What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively?[A] = _____ M[B] = _____ M[C] = _____ MHow would your answers above change if the reaction had a standard free-energy change of +5.07 kJ/mol?•All concentrations would be higher•There would be no change to the answers.•All concentrations would be lower.•There would be more A and B but less C.•There would be less A and B but more C&gt; 
An aqueous reaction at 350 K has an equilibrium constant (Kc) of 2.65 times 10^-6. What is the delta Gº of the reaction at 350 K?
Calculate the value of K, at 298 K, for each value of Delta G degree Delta G degree = 7.5 kJ/mol Delta G degree = -9.0 kJ/mol
A chemist fills reaction vessel with 0.828g aluminum hydroxide (Al OH)3) solid, 0.801 M aluminum (Al3+) aqueous solution, and 0.563M hydroxide (OH-) aqueous solution at a temperature of 25.0° C. Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:Al(OH)3(s) ⇌ AI3+(aq) + 3OH-(aq)Round your answer to the nearest kilojoule.
The following gas phase system is at equilibrium:CCl4(g) + CH4(g)⇌ 2 CH2Cl2(g)The pressure of each gas is measured at 539 K:P(CCl4) = 177 mmHgP(CH4) = 5.22 x 10-2 mmHgP(CH2Cl2) = 194 mmHgWhat is the Gibbs free energy change at this temperature?
Determine ΔG° for the following reaction:Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)Use the following reactions with known ΔG°rxn values:2Fe(s) + 3/2 O2(g) → Fe2O3(s), ΔG°rxn = -742.2 kJCO(g) + 1/2 O2(g) → CO2(g), ΔG°rxn = -257.2 kJ Express your answer using one decimal place.
What is the ΔG° (kJ/mol) for the reaction shown below of silver chloride dissolving at 25°C? Ksp = 1.8 x 10-10 M2. AgCI(s) → Ag+(aq) + Cl-(aq)  ΔG° = ____ kJ/mol (Answer to two significant figures.)
Calculate the standard free energy of the following reaction at 25 °C. a. For the reaction H2 (g) + I2 (s) → 2HI (g) b. For the reaction 2NO (g) + 3H2O (g) → 2NH3 (g) + 5/2O2 (g) 
Consider the reaction. CaCO3 (s) → CaO (s) + CO2 (g) Estimate ΔG° for this reaction at each temperature. (Assume that ΔH° and ΔS° do not change too much within the given temperature range.) Part A290 K Express your answer using one decimal place. Part B 1065 K Express your answer using one decimal place. Part C1445 K Express your answer using one decimal place.
For an endothermic reaction with ΔH°rxn = 92.7 kJ and ΔS°rxn = 7.1 J/K, above what temperature will the reaction occur spontaneously?
2Ca (s) + O2 (g) → 2CaO (s) ΔH°rxn = -1269.8 kJ; ΔS°rxn = - 364.6 J/K Part ACalculate the free energy change for the reaction at 33°C. Express your answer using four significant figures.Part B Is the reaction spontaneous?                                  (i) spontaneous                                  (ii) nonspontaneous
The partial pressure of carbon dioxide is 0.402 atm. Calculate the ΔG (in kJ) for the conversion limestone (CaCO3) to quicklime (CaO) at 797°C.                                        CaCO3 (s) → CaO (s) + CO2 (g) ΔH° 298 (in kJ/mol)       -1206.9         -635.5      -393.51 ΔS° 298 (in J/mol*K)       92.9              40.0         213.6 (Write answer in decimal form to two decimal places)
For the reaction 2HBr (g) → H2 (g) + Br2 (l) ΔG° = 104.8 kJ and ΔH° = 72.6 kJ at 281 K and 1 atm. This reaction is (reactant, product) _________  favored under standard conditions at 281 K.The entropy change for the reaction of 1.80 moles of HBr(g) at this temperature would be________ J/K. 
The chemical reaction that causes aluminum to corrode in air is given by 4Al + 3O2 → 2Al2O3Part AWhat is the standard Gibbs free energy for this reaction? Express your answer as an integer and include the appropriate units. Part BWhat is the Gibbs free energy for this reaction at 5400 K? Assume that ΔH and ΔS do not change with temperature. Express your answer to two decimal places and include the appropriate units. Part CAt what temperature Teq do the forward and reverse corrosion reaction occur in equilibrium?Express your answer as an integer and include the appropriate units.
For a particular reaction at 15 1.7°C, ΔG = - 80761 kJ/mol, and ΔS = 188.82 J/(mol • K). Calculate ΔG for this reaction at -9.5°C.
For the reaction Fe3O4 (s) + 4H2 (g) → 3Fe (s) + 4H2O (g) ΔH° = 151.2 kJ and ΔS° = 169.4 J/K The standard free energy change for the reaction of 2.33 moles of Fe3O4(g) at 311 K, 1 atm would be ___________.This reaction is (reactant, product) ____________ favored under standard conditions at 311 K.Assume that ΔH° and ΔS° are independent of temperature. 
The value of Kp for a gas-phase reaction is halved when the temperature is lowered from 400 K to 300 K at a fixed pressure. What is the value of ΔH°rxn for this reaction? If Kp at 300 K is 150, what is ΔS°rxn?
Consider a process at 298 K with ΔH = +16.6 kJ/mol and ΔS = +93.1 J/K mol Which of the following is true? • ΔG = -11.1kJ/mol and the process will not be spontaneous • The spontaneity of the reaction cannot be determined from the information given • ΔG = -51.2 kJ/mol and the process will be spontaneous • ΔG = -51.2 kJ/mol and the process will not be spontaneous • ΔG = -11.1 kJ/mol and the process will be spontaneous     
At 25°C, the equilibrium partial pressures for the following reaction were found to be PA = 4.56 atm, PB = 4.76 atm, PC = 4.07 atm, and PD = 4.73 atm. 3A(g) + 3B(g) ⇌ C(g) + 3D(g) What is the standard change in Gibbs free energy of this reaction at 25°C? 
If ΔrG° &gt; 0 for a reaction at all temperatures, then ΔrH° is _____ and ΔrS° is _____. (a) negative, positive (b) positive, negative (c) negative, negative (d) positive, positive (e) positive, either positive or negative
For the reaction 2H2S (g) + 3O2 (g) → 2H20 (l) + 2SO2 (g)  ΔH° = -1124.0 kJ and ΔS° = -390.7 J/K The standard free energy change for the reaction of 1.91 moles of H2S(g) at 339 K, 1 atm would be ________ kJ.This reaction is (reactant, product) _________ favored under standard conditions at 339 K.Assume that ΔH° and ΔS° are independent of temperature.  
For the reaction 2H2 (g) + O2 (g) → 2H2O (g) ΔG° = -455.6 kJ and ΔS° = -88.9 J/K at 315 K and 1 atm. This reaction is (reactant, product) _________  favored under standard conditions at 315 K.The standard enthalpy change for the reaction of 2.47 moles of H2(g) at this temperature would be _________ kJ.
Consider the following reaction: CH3OH(g) ⇌ CO(g) + 2H2 (g) Calculate ΔG for this reaction at 25°C under the following conditions: PCH3OH = 0.890 atm PCO = 0.120 atm PH2 = 0.200 atm Δ G =
A (l) → A (g)                       ΔH°vap = -42.55 kJ/mol Calculate the entropy of vaporization, ΔSvap, for A(/) at 250°C given that the boiling point of A is 76.53°C, and the molar heat capacity of A() is 116.59 J/(mol K). Assume that the molar heat capacity of A(g) is 6.7% of that of A(l). Calculate the standard Gibbs free energy of vaporization, ΔG°vap, at 25.0°C. Determine the equilibrium constant, K, for the vaporization at 179.0°C. 
The equilibrium constant, Kp, for the reaction shown below at 25.0°C is 9.18 x 10-22. Calculate the change in standard Gibbs energy for this reaction. C(s) + CO2(g) ⇌ 2CO(g) a. +1.2 x 105 kJ b. +1.2 x 102 kJ c. -1.2 x 105 kJ d. +1.0 x 104 J e. -2.4 x 103 J
 Part AAcetylene C2H2 can be convened to ethane. C2H6, by a process known as hydrogenation. The reaction is C2H2 (g) + 2H2 (g) ⇌ C2H5 (g) Given the following data at standard conditions (all pressures equal to 1 atm and the common reference temperature 298 K), what is the value of Kp for this reaction? Express your answer using two significant figures.
Calculate the standard change in Gibbs free energy for the following reaction at 25 °C. ΔG°f values can be found here. 3NO2 (g) + H2O (l) → 2HNO3 (l) + NO (g) 
The following equation represents the decomposition of a generic diatomic element in its standard state. 1/2 X2 (g) → X (g) Assume that the standard molar Gibbs energy of formation of X (g) is 5.14 kJ • mol-1 at 2000 K and -51.10 kJ • mol-1 at 3000 K. Determine the value of K (the thermodynamic equilibrium constant) at each temperature. Assuming that ΔH°rxn is independent of temperature, determine the value of ΔH°rxn from these data. 
Calculate the standard entropy, ΔS°rxn, of the following reaction at 25.0 °C using the data in this table. The standard enthalpy of the reaction, ΔH°rxn is -44.2 kJ • mol-1 C2H4 (g) + H2O (l) → C2H5OH (l) Then, calculate the standard Gibbs free energy of the reaction, ΔG°rxn. Finally, determine which direction the reaction is spontaneous as written at 25.0 °C and standard pressure. 
Calculate the standard change in Gibbs free energy for the following reaction at 25 °C. Use the  ΔG°f values found in the Chempendix. Fe2O3 (s) + 2Al (s) → Al2O3 (s) + 2Fe (s) 
The values of ΔH°rxn and ΔS°rxn for the reaction 2NO (g) + O2 (g) → 2NO2 (g) are -12.0 kJ and -147 J/K. Calculate ΔG° at 298 K for this reaction. 
Given that S (g) + O2 (g) → SO2 (g)           ΔrG° = -300.1 kJ/mol-rxn 2S (g) + 3O2 (g) → 2SO3 (g)    ΔrG° = -742.1 kJ/mol-rxn calculate ΔJG° of the following reaction: SO2 (g) + 1/2O2 (g) → SO3 (g) (a) -1042.2 kJ/mol-rxn (b) -71.0 kJ/mol-rxn (c) +2.47 kJ/mol-rxn (d) +71.0 kJ/mol-rxn (e) +1042.2 kJ/mol-rxn
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M2O3 ⇌ 2M(s) + 3/2O2(g) What is the standard change in Gibbs energy for the reaction, as written, in the forward direction? What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K? What is the equilibrium pressure of O2(g) over M(s) at 298 K?
For the reactionH2 (g) + C2H4 (g) → C2H6 (g)  ΔG° = -100.7 kJ and ΔS° = -120.7 J/K at 301 K and 1 atm. This reaction is (reactant, product) ____________ favored under standard conditions at 301 K.The standard enthalpy change for the reaction of 2.40 moles of H2(e) at this temperature would be ___________ kJ.
A reaction has a standard free-energy change of -14.10 kJ mol-1 (-3.370 kcal mol-1). Calculate the equilibrium constant for the reaction at 25°C.
The pH of a saturated solution of nickel(II) hydroxide is found to be 10.55. Use the solubility of this sparingly soluble salt to find the value of Ksp and ΔG°rxn.
Given the following information A + B → 2D               ΔH° = -743.7 kJ               ΔS° = 339.0 J/K C → D                       ΔH ° = 509. 0 kJ              ΔS° = -233.0 J/K calculate ΔG° for the following reaction at 298 K.A + B → 2C
A critical reaction in the production of energy to do work or drive chemical reactions in biological systems is the hydrolysis of adenosine triphosphate, ATP, to adenosine diphosphate, ADP, as described by ATP (aq) + H2O (l) → ADP(aq) + HPO 42- (aq)for which ΔG°rxn = -30.5 kJ/mol at 37.0 °C and pH 7.0. Calculate the value of ΔG  rxn in a biological cell in which [ATP] = 5.0 mM, [ADP] = 0.80 mM, and [HPO42-] = 5.0 mM. Is the hydrolysis of ATP spontaneous under these conditions?
What are E°cell and ΔG° of a redox reaction at 25°C for which n = 1 and K = 7.0 x 104?
Consider the reaction CaCO3 (g) → CaO (g) + CO2 (g) Using the Standard thermodynamic data m the tables linked above, calculate the equilibrium constant for this reaction at 298.15 K. ANSWER: ____________
What is the value of K for this aqueous reaction at 298 K? A + B ⇌ C + D          ΔG° = 21.88 kJ/mol
Use the standard free energy of formation data in your textbook to determine the free energy change (in kJ) for the following reaction, which was run under standard state conditions and 25 °C. Fe2O3 (s) + 3CO (g) → 2Fe (s) + 3CO2 (g)
Calculate the standard change in Gibbs free energy for the following reaction at 25°C. ΔG°f values can be found here. 2C2H6 (g) + 7O2 (g) → 4CO2 (g) + 6H2O (g)
For a gaseous reaction, standard conditions are 298 K and a partial pressure of 1 atm for all species. For the reaction N2 (g) + 3H2 (g) ⇌ 2NH3 (g) the standard change in Gibbs free energy is ΔG° = -69.0 kJ/mol. What is ΔG for this reaction at 298 K when the partial pressures are PN2 = 0.250 atm, PH2 = 0.400 atm, and PNH3 = 0.850 atm
The value of Ksp at 25°C for Agl(s) in water is 8.3 x 10-17 M. Calculate ΔG at 319 K for the process:Agl(s) ⇌ Ag+(aq, 2.5 x 10-16 M) + I -(aq, 5.6 x 10-9 M)
For a particular reaction at 127.1°C, ΔG = -500.31 kJ/mol, and ΔS = 868.25 J/(mol • K). Calculate ΔG for this reaction at 12.0°C.
For the reaction 2CO (g) + 2NO (g) → 2CO2 (g) + N2 (g) ΔH° = -746.6 kJ and Δ S° = -198.0 J/K The standard free energy change for the reaction of 1.80 moles of CO(g) at 258 K, 1 atm would be _______ kJ.This reaction is (reactant, product) ________ favored under standard conditions at 258 K.Assume that ΔH° and ΔS° are independent of temperature. 
At 25°C, the equilibrium partial pressures for the following reaction were found to be PA = 4.15 atm, PB = 5.27 atm, PC = 5.14 atm, and PD = 4.61 atm. 3A (g) + 2B (g) ⇌ C (g) + 2D (g) What is the standard change in Gibbs free energy of this reaction at 25°C?
Be sure to answer all parts.What are E°cell and ΔG° of a redox reaction at 25°C for which n = 2 and K = 45: Enter your answer in scientific notation.
From the values given below, calculate the values of ΔG° and ΔH° for the following reaction at 25°C. Which of the two oxides is more stable at 25°C and PO2 = 1 atm? 2Co3O4 (s) → 6CoO (s) + O2 (g) 
Calculate ΔG°rxn and E°cell for a redox reaction with n = 3 that has an equilibrium constant of K = 5.3 x 10-2. You may what to reference (Pages 929 - 933) section 20.5 while completing this problem. Part AExpress your answer using two significant figures. Part BExpress your answer using two significant figures.
Consider the reaction HCl (g) + NH3 (g) → NH4Cl (s) Using the Standard thermodynamic data m the tables linked above, calculate the equilibrium constant for this reaction at 298.15K. ANSWER: ____________
A reaction has a standard free-energy change or -18.20 kJ mol-1 (-4.350 kcal mol-1). Calculate the equilibrium constant for the reaction at 25°C.
Given the following information A + B → 2D             ΔH° = 787.4 kJ              ΔS° = 289.0 J/K C → D                     ΔH° = 417.0 kJ               ΔS° = -140.0 J/K Calculate ΔG° for the following reaction at 298 K. A + B → 2C 
Enter your answer in the provided box. Bromine monochloride is formed from the elements: Cl2 (g) + Br2 (g) → 2BrCl (g)           ΔH°rxn  = -1.35 kJ/mol   ΔG°rxn = -0.88 kJ/mol Calculate S° of BrCl(g). 
The equilibrium constant for a reaction is 0.48 at 25 ° C. What is the value of delta G degree (kJ/mol) at this temperature?•1.8•-4.2•1.5x 102 •4.2•More information is needed.
Calculate ΔG° (in kJ) for the following reaction from the equilibrium constant at the temperature given. CH3NH2 (aq) + H2O (l) &lt;-&gt; CH3NH3+ (aq) + OH - (aq)      K = 4.40 x 10 -4 at T = 25°C
At 25°C, the equilibrium partial pressures for the following reaction were found to be PA = 4.19 atm, PB = 5.69 atm, PC = 4.31 atrm, and PD = 4.01 atm. 3A (g) + 2B (g) ⇌ C (g) + 2D (g) What is the standard change in Gibbs free energy of this reaction at 25°C?
Use the data below to calculate ΔG°rxn for the reaction: A + 3B → C + 2D Data:Reaction 1: 3C → A + 3B ΔG°rxn1 = -365.0kJ/molReaction 2: C → D ΔG°rxn2 = 390 0 kJ/mol
The standard free-energy change in the reaction of water with ethyl acetate (CH3CO2C2H5) to give ethanol (C2H5OH) and acetic acid (CH3COOH) is -19.7 kJ. From this value and the values of the standard free-energy of formation of liquid ethanol. acetic acid, and water, calculate ΔG°f for ethyl acetate.
In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation: Glu + ATP → G6P + ADP        ΔG°298 = -17 kJIn this process, ATP becomes ADP summarized by the following equation: ATP → ADP      ΔG°298 = -30 kJDetermine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process:
Carbon dioxide dissolves in water to form carbonic acid. Estimate the thermodynamic equilibrium constant for this reaction using the ΔG°f values in the table. 
The molar enthalpy of fusion of solid aluminum is 10.8 kJ mol-1, and the molar entropy of fusion is 11.6 J K-1mol-1. (a) Calculate the Gibbs free energy change for the melting of 1.00 mol of aluminum at 959 K. (b) Calculate the Gibbs free energy change for the conversion of 2.38 mol of solid aluminum to liquid aluminum at 959 K. (c) Will aluminum melt spontaneously at 959 K? 
For the reaction 2Fe (s) + 3Cl (g) → 2FeCl3 (s) ΔG° = -688.5 kJ and ΔS° = -440.3 J/K at 251 K and i atm. This reaction is (reactant, product) ____________ favored under standard conditions at 251 K.The standard enthalpychanige for the reaction of 2.07 moles of Fe (s) at this temperature would be __________ kJ.
For a spontaneous reaction, it is always true that a. the change in enthalpy &lt; 0b. the change in free energy &lt; 0c. the change in free energy &gt; 0d. the equilibrium constant &lt; 1 
Given the following information A + B → 2D            ΔH° = 676.2 kJ                 ΔS° = 355.0 J/K C → D                    ΔH° = 44 1.0 kJ                 ΔS° = 241.0 J/K calculate ΔG° for the following reaction at 298 K. A + B → 2C 
Rxn 1: 2CH3OH (g) + 3O2 (g) → 2CO2 (g) + 4H2O(g) ΔS°rxn =                                            ΔG°rxn = Rxn 2: CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l) ΔS°rxn =                                            ΔG°rxn = Rxn 3: 2H2 (g) + O2 (g) → 2H2O(g) ΔS°rxn =                                            ΔG°rxn =
A reaction A (aq) + B (aq) ⇌ C (aq) has a standard free-energy change of -3.21 kJ/mol at 25°C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively? How would your answers above change if the reaction had a standard free-energy change of +3.21 kJ/mol?                (a) There would be no change to the answers.                (b) All concentrations would be lower.                (c) All concentrations would be higher.                (d) There would be less A and B but more C.                (e) There would be more A and B but less C.
Consider the malate dehydrogenase reaction from the citric acid cycle. Given the following concentrations, calculate the free energy change for this reaction at 37.0 °C ( 310 K).  △G° for the reaction is +29.7 kJ/mol. Assume that the reaction occurs at pH 7. 
For the reaction HCHO (g) + 2/3O3 (g) → H2O (g) + CO2 (g) the value of ΔG° is -620.6 kJ at 25°C. Other data are as follows (at 25°C): Calculate the absolute entropy, S°, per mole of O3(g). 
For a particular reaction at 119.6°C, ΔG = -998.32 kJ/mol, and ΔS = 253.60 J/(mol- K). Calculate ΔG for this reaction at 9.9°C.
The transition metals form a class of compounds called metal carbonyls, an example of which is the tetrahedral complex Ni(CO)4. Given the following thermodynamic data (at 298 K): Calculate the equilibrium constant for the formation of Ni(CO) 4 (g) from nickel metal and CO gas. 
Which of the following conditions is always true at equilibrium?So there are two boxes where I have to sort the choices in. Box number 1 is called True at equilibrium. Box number 2 is called Not necessarily true at equilibrium.Choices area)ΔG = 0b)ΔGo = 0c)ΔG = ΔGod)Q = 0  e)ΔGo = 1f)Q = 1g)K = 1
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal.                                     M    3O4 (s) ⇌ 3 M (s) + 2 O2 (g) i) What is the standard change in Gibbs energy for the reaction, as written, in the forward direction?ii) What is the equilibrium constant of this reaction, as written, in the forward direction at 298 K?iii) What is the equilibrium pressure of O2 (g) over M (s) at 298 K? 
What is the standard Gibbs free energy for the transformation of diamond to graphite at 298 K?
For a particular reaction, ΔH 111.4 kJ and ΔS- 25.0 J/K. Calculate ΔG for this reaction at 298º K. What can be said about the spontaneity of the reaction at 298º K?The system is at equilibrium The system is spontaneous in the reverse direction. The system is spontaneous as written.
Complete this statement ΔG = ΔGo when K = 0Q = 1Q = 0Q = KK = 1