| Sections | |||
|---|---|---|---|
| Spontaneous Reaction | 10 mins | 0 completed | Learn |
| First Law of Thermodynamics | 6 mins | 0 completed | Learn Summary |
| Entropy | 58 mins | 0 completed | Learn Summary |
| Gibbs Free Energy | 30 mins | 0 completed | Learn |
| Additional Practice |
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| Boltzmann Equation |
| Additional Guides |
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| Second and Third Laws of Thermodynamics |
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 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 Conversion
STEP 2: Plug in your values into the Gibbs-Free Energy formula to obtain your answer.
Calculating 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.
Gibbs 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.
Temperature 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 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
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.
Energetically 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.
Additional Problems
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.
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?
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.
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.
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
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
Fill in the blanks:
If ΔG°< 0, then K is _____. If ΔG° > 0, then K is _____. If ΔG° = 0, then K is ______.
(a) > 1, < 1, = 1
(b) < 1, > 1, = 1
(c) < 0, > 0, = 0
(d) > 0, < 0, = 0
(e) < 1, > 1, = 0
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
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
If Q > K, what must be true regarding ΔG?
A. it is positive
B. it equals zero
C. it is negative
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
Every sample of a pure element, regardless of its physical state, is assigned zero Gibbs free energy of formation.
a) True
b) False
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
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
Calculate the equilibrium constant, K, for the reaction at 298 K:
N2O4 (g) ⇌ 2 NO2 (g)
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.
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?
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
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)
The process by which plants convert CO2 and H2O into C6H12O6 and O2 is called photosynthesis. This process is nonspontaneous, so energy must be provided to make the reaction occur, and that energy is provided by sunlight. “Chemosynthesis” refers to the biological conversion of simple carbon-containing molecules (e.g. CO2 or methane, CH4) into more complex energy-rich molecules using non-biological (inorganic) molecules as the energy source, rather than sunlight, as in photosynthesis. For example, giant tube worms found near deep-ocean sulfur vents contain bacteria that convert carbon dioxide to simple sugars using hydrogen sulfide as the energy source according to the chemical equation
12H2S + 6CO2 → C6H12O6 + 6H2O + 12S
A. Using the data provided below, calculate the following thermodynamic parameters for the above reaction.
1. ΔH°
2. ΔS°
3. ΔG°
B. Note the value you calculated for ΔG°. The normal expectation is that the reaction is spontaneous, but the calculations indicate that the reaction is nonspontaneous at 25°C.
At what temperature will the reaction become spontaneous?
When the reaction becomes spontaneous, will it be driven by enthalpy or entropy?
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?
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?
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
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
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
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
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 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
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
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.
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)
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 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
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.
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
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
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.
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 has ΔG of = 0 at 25 oC?
A) CO2 (l) B) H2O (l) C) Hg (s) D) O2 (g) E) NH3 (g)
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
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).
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
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
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
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
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
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
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
Assuming the following ΔH°rxn and ΔS°rxn values are independent of temperature, which of reactions I, II, and III is/are spontaneous at 400°C?
a) I
b) II
c) III
d) I and II
e) II and III
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
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
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.
The compound benzoic acid, C6H5COOH, melts and 122.4°C. If the enthalpy of fusion of this compound is 17.3 kJ/mol, what is the entropy of the fusion.a) 31.3 J/(mol•K)b) 34.1 J/(mol•K)c) 38.0 J/(mol•K)d) 40.2 J/(mol•K)e) 43.8 J/(mol•K)
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.
Which is the best estimate for the boiling point of benzene (°C) given that ΔH° of vaporization is 31 kJ/mol and ΔS° of vaporization is 90 J/mol K?a. 25b. 45c. 65d. 15e. 5
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/molC2H6 (g) ΔG° f = -32.9 kJ/mol A. 9.07 x 10 -1B. 97.2C. 1.24 x 10 31D. 2.74 x 10 42
When can Q > 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
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
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
The reaction A(g) ⇌ B(g) has an equilibrium constant of Kp = 2.3x105. What can you conclude about the sign of ΔG°rxn for the reaction?a) ΔG°rxn = 0b) ΔG°rxn < 0c) ΔG°rxn > 0d) No conclusion can be made
The equilibrium constant for the following reaction is 5.0 × 108 at 25 °C.N2(g) + 3 H2(g) ⇌ 2 NH3(g)The value of ΔG° is ____ kJ/mol.
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)?
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 ∆G° of reaction is zero at point C.
Which combination of ∆G◦ and K is possible at standard conditions?1. ∆G◦ = 99.3 kJ, K = 1.022. ∆G◦ = 73.4 J, K = 1.38 × 10 73. ∆G◦ = −41.1 kJ, K = 0.9714. ∆G◦ = −33.3 J, K = 1.01 5. ∆G◦ = −44.6 J, K = 5.62 × 10 −18
Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine. The data refer to 298 K.SO2(g) + Cl2(g) → SO2Cl2(g)Substance: SO2(g) Cl2(g) SO2Cl2(g)ΔH°f (kJ/mol): –296.8 0 –364.0ΔG°f (kJ/mol): –300.1 0 –320.0S° (J/K•mol): 248.2 223.0 311.9What is the value of ΔG° for this reaction at 600 K?A) -162.8 kJB) -40.1 kJC) -28.4 kJD) 28.4 kJE) 162.8 kJ
The temperature at which the following process reaches equilibrium at 1.0 atm is the normal boiling point of hydrogen peroxide. (R = 8.314 J•K-1•mol-1)H2O2(l) ⇌ H2O2(g)Use the following thermodynamic information at 298 K to determine this temperature.Substance: H 2O2(l) H2O2(g)ΔH°f (kJ/mol): -187.7 -136.3ΔG°f (kJ/mol): -120.4 -105.6S° (J/K•mol): 109.6 232.7A) 120°CB) 144°CC) 196°CD) 418°CE) 585°C
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
For bromine, ΔH ovap = 30.91 kJ/mol and ΔS ovap = 93.23 J K -1 mol-1 at 25 oC. What is the normal boiling point for bromine? a) 332 oCb) 25 oCc) 124 oCd) 58 oC
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 reaction2 NO2(g) ⇌ N2O4(g) ?1. 6.882. 0.1453. 7.01 × 10−94. 1.005. None of these6. 1.02 × 10−107. 9.72 × 1098. 0.657
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.
A weak base has a ΔG° of 24 kJ/mole. What is the approximate value for the K b?A. 10-4B. 104C. 10-24D. 1024
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 kJB. +266.3 kJC. -16.9 kJD. +207.8 kJE. +17.3 kJ
Calculate ΔG° for the following reaction3 NO2(g) + H2O(l) → 2 HNO3(l) + NO(g)Given the following free energies of formation ΔG°f(kJ/mol)H2O(l) –237.2HNO3(l) –79.9NO(g) 86.7NO2(g) 51.8A. -414 kJB. 8.7 kJC. -192 kJD. 192 kJE. -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, nonspontaneousB. ΔG = -97.5 kJ, spontaneousC. ΔG = 40.6 kJ, nonspontaneousD. ΔG = -69.1 kJ, spontaneousE. 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°CΔH° = -98.2kJΔS° = 70.1 J/KA. 3.46 x 1017B. 1.3 x 10-21C. 7.7 x 1020D. 8.6 x 104E. 20.9
If Q > K, what must be true regarding ΔG?A. it is positiveB. it equals zeroC. it is negative
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 isA. 152B. 1.01C. 1.08 x 1016D. 0.00657E. 9.21 x 10-17
What is the value of K for this aqueous reaction at 298 K? ΔG = 13.75 KJ/molA + B ⇋ C + D
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 be) Both a and b and c
Complete the table below by filling in the missing information. Note that T = 25.0 °C.
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
What is the value of K for this aqueous reaction at 298 K?ΔG = 13.75 kJ/molA+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
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.
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= -6.70M(s)=0O2(g)= 0a) 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)
The heat of vaporization for 1.0 mole of water at 100°C and 1.0 atm is 40.6 kJ/mol. What is ΔS for the processH2O(g) → H2O(l) at 100°C.a) -109 J/K molb) 109 J/K molc) 406 J/K mold) -406 J/K mol
Calculate the standard change in Gibbs free energy for the following reaction at 25°C.3H2(g) + Fe2O3(s) → 2Fe(s) + 3H2O(g)
What is the difference between ΔG° and ΔG° prime?
The chemical reaction that causes chromium to corrode in air is given by:4Cr + 3O2 → 2Cr2O3In whichΔH°rxn = -2256 kJΔS°rxn = -549.1 J/KAt what temperature (Teq) do the forward and reverse corrosion reactions occur in equilibrium?Express your answer as an integer and include the appropriate units.
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 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.
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
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
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
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?
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
Methanol burns in oxygen to form carbon dioxide and water. 2CH3OH+3O2 → 2CO2+4H2O Calculate the rxn of ΔH, G & S 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) ⇌ CO (g) + H2O (g)Δ Hº = 41.15 kJΔ Sº = 42.35 J/K
A reaction has a standard free-energy change of -14.90 kJ mol -1 (-3.561 kcal mol-1). Calculate the equilibrium constant for the reaction at 25°C.
Consider the reaction HCl (g) + NH3 (g) → NH4Cl (s) Using the Standard thermodynamic data, calculate the equilibrium constant for this reaction at 298.15 K. ANSWER: ____________
At 1500°C the equilibrium constant for the reaction CO(g) + 2H2(g) ⇌ CH3OH(g) has the value Kp = 1.4 x 10-7. Calculate ΔGº for this reaction at 1500°C.a) -233 kJ/molb) 233 kJ/molc) -105 kJ/mold) 105/mole) 1.07 kJ/mol
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 = kJ
Consider the oxidation of SO2 to SO3: SO2 (g) + 1/2 O2 (g) → SO3 (g) Calculate ΔGºrxn at 25 C. Express your answer to one decimal place with the appropriate units. Determine whether the reaction is spontaneous at standard conditions.
Given the thermodynamic data in the table below, calculate the equilibrium constant (at 298 K) for the reaction:2SO2 (g) + O2(g) ⇌ 2SO3(g)a) 2.37 x 1024b) 1.06c) 1.95d) 3.92 x 1023e) More information is needed.
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 ___________ kJ.This reaction is (reactant, product) ____________ favored under standard conditions at 311 K.Assume that ΔH° and ΔS° are independent of temperature.
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.
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. [malate] = 1.17 mM[oxaloacetate] = 0.210 mM[NAD+] = 480 mM[NADH] = 190mM
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) 150b) 9.3 x 1015c) 8.4 x 104d) 1.1 x 10–16e) 1.4 x 108
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
What is the ΔSo if the ΔHº is +88.00 kJ and ΔGº is +11.21 kJ? a) +0.258 J/Kb) -258 J/K c) +258 J/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?a) The system is at equilibrium b) The system is spontaneous in the reverse direction. c) The system is spontaneous as written.
Calculate the standard change in Gibbs free energy (kJ) for the following reaction at 25 °C. 3NO2 (g) + H2O (l) → 2HNO3 (l) + NO (g)
Complete this statement ΔG = ΔGo when K = 0Q = 1Q = 0Q = KK = 1
Use the data from this table of thermodynamic properties to calculate the maximum amount of work that can be obtained from the combustion of 1.00 mole of ethane, CH3CH3(g) at 25 °C and standard conditions.
Calculate the Kp for the following reaction at 25º C: H2(g) + I2(g) → 2HI(g) ΔGº = 2.60 kJ/mol
Part A: Use the data given here to calculate the values of ΔG°rxn (kJ/mol) at 25 °C for the reaction described by the equation: A + B ⇌ C A ΔG°f = +387.7 kJ/molB ΔG°f = +529.5 kJ/molC ΔG°f = +402.0 kJ/molPart B: If ΔH°rxn and ΔS°rxn are both positive values, what drives the spontaneous reaction and in what direction at standard conditions?a) enthalpy-driven to the leftb) entropy-driven to the leftc) entropy-driven to the rightd) enthalpy-driven to the right
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 (kj•mol-1 ) of each reaction below using values from this table.(1) H2 (g) + O2 (g) ⇌ H2O2 (l)(2) H2O (l) + 1/2 O2 (g) ⇌ H2O2 (l)Which method requires less energy under standard conditions?reaction (1)reaction (2)
For the reaction2 HBr(g) → H2(g) + Br2(l) ΔH°= 72.6 kJ and ΔS° = -114.5 J/KThe equilibrium constant for this reaction at 251.0 K is _____.Assume that ΔH° and ΔS° are independent of temperature.
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.ΔG°rxn = _______kJ•mol-1Finally, determine which direction the reaction is spontaneous as written at 25.0 °C and standard pressure a) forward b) reverse c) both d) neither
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.
Calculate the standard entropy, ΔSorxn of the following reaction at 25.0 0C. The standard enthalpy of the reaction, ΔHorxn, is -44.2 kJ/mol.
Yeast can produce ethanol by the fermentation of glucose (C 6H12O6), which is the basis for the production of most alcoholic beverages. C6H12O6 (aq) → 2 C2H5OH (l) + 2 CO2 (g). Calculate ∆rHº in kJ, ∆rSº in J/K, and ∆rGº for the reaction at 25 ºC.
For the above reaction at standard conditions (298º K, 1.00 atm pressure, 1.00 M), ΔH is -14.6 kJ/mol, and ΔS is -5.06 J/(mol*K). What is the value of ΔG (in kJ/mol) at these conditions?
For the following reaction, the change in free energy, ∆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
Determine ΔG° (kJ) 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.
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.
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.
Calculate the standard change in Gibbs free energy (kJ/mol) for the following reaction at 25 °C. Use the ΔG°f values found in the Chempendix. Fe2O3 (s) + 2Al (s) → Al2O3 (s) + 2Fe (s)
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 (kJ/mol) 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?
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
Calculate the equilibrium constant at 25º C for the following reaction:2NH3(g) + CO2(g) → NH2CONH2(aq) + H2O(l) ΔGº = -13.6 kJExpress your answer using two significant figures.
Calculate ΔG° for the following reaction at 298 K. A + B → 2C
Given the following informationA + B → 2D ΔH° = 724.9 kJ ΔS° = 307.0 J/KC → D ΔH° = 402.0 kJ ΔS° = -139.0 J/KCalculate ΔG° for the following reaction at 298 K.A + B → 2C
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.