Question 1

Determine Δ_rG° at 298 K using the following information: FeO(s)+ CO(g)→ Fe(s)+ CO₂(g)Δ_rH°= -11.0 kJ mol^-1; Δ_rS°= -17.4 J K^-1 mol^-1

Accepted Answer

First, we need to use the formula: ΔG° = ΔH° - TΔS° We are given: ΔH° = -11.0 kJ/mol ΔS° = -17.4 J/mol*K = -0.0174 kJ/mol*K T = 298 K Substitute these values into the formula and solve for ΔG°: ΔG° = (-11.0 kJ/mol) - (298 K)(-0.0174 kJ/mol*K) ΔG° = -11.0 kJ/mol + 5.19 kJ/mol ΔG° = -5.81 kJ/mol Since ΔG° is negative, the reaction is spontaneous. However, the question is asking for Δ_rG°, which is the standard free energy change at 298 K. This means we need to adjust our answer to account for the standard state conditions. We know that: ΔG° = ΔG°f (products) - ΔG°f (reactants) We can look up the standard free energy of formation values for FeO(s), CO(g), Fe(s), and CO₂(g) and substitute them into the formula. Let's label them as follows: FeO(s) --> A CO(g) --> B Fe(s) --> C CO₂(g) --> D We get: ΔG° = [ΔG°f (CO₂(g))] - [ΔG°f (FeO(s))] - [ΔG°f (CO(g))] = [20.2 kJ/mol] - [ -271.9 kJ/mol] - [-137.2 kJ/mol] = 191.1 kJ/mol Note that the values we looked up are in kJ/mol, so our final answer must also be in kJ/mol. Therefore, the answer is B) -5.8 kJ/mol.

Question 2

Calculate Δ_rG° at 298 K using the following information: $\quad$$\quad$$\quad$$\quad$$\quad$2H₂S(g)+ 3O₂(g)→ 2SO₂(g)+ 2H₂O(g)Δ_rG°= ? Δ_fG°(kJ mol^-1)-33.4 $\quad$$\quad$$\quad$-300.1 $\quad$-228.6

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Question 3

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$$\quad$3O₂(g)→ 2O₃(g)$\quad$$\quad$$\quad$Δ_rG° = +326 kJ $\quad$$\quad$P(O₂)= 0.41 atm,P(O₃)= 5.2 atm

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Question 4

Use Hess's law to calculate Δ_rG°using the following information: ClO(g)+ O₃(g)→ Cl(g)+ 2O₂(g)$\quad$$\quad$$\quad$$\quad$$\quad$ Δ_rG°= ? 2O₃(g)→ 3O₂(g)$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ Δ_rG°= +489.6 kJ mol^-1 Cl(g)+ O₃(g)→ ClO(g)+ O₂(g)$\quad$$\quad$Δ_rG°= -34.5 kJ mol^-1

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Question 5

Use Hess's law to calculate Δ_rG° using the following information: CO(g)→ C(s)+ 1/2 O₂(g)$\quad$$\quad$Δ_rG°= ? CO₂(g)→ C(s)+ O₂(g)$\quad$$\quad$$\quad$Δ_rG°= +394.4 kJ mol^-1 CO(g)+ 1/2 O₂(g)→ CO₂(g)$\quad$Δ_rG°= -257.2 kJ mol^-1

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Question 6

Determine Δ_rG° at 298 K using the following information: $\quad$CaCO₃(s)→ CaO(s)+ CO₂(g)$\quad$Δ_rH°= +179.2 kJ mol^-1; $\quad$Δ_rS°= +160.2 J K^-1 mol^-1

Accepted Answer

Δ_rG°= Δ_rH° - TΔ_rS° We are given: Δ_rH°= +179.2 kJ mol^-1 Δ_rS° = +160.2 J K^-1 mol^-1 T= 298 K We need to convert Δ_rS° from J K-1 mol-1 to kJ K-1 mol-1 Δ_rS°= 160.2 J K-1 mol-1 / 1000 = 0.1602 kJ K-1 mol-1 Now we can calculate: Δ_rG°= +179.2 kJ mol^-1 - (298 K x 0.1602 kJ K-1 mol-1) Δ_rG°= 131.5 kJ mol^-1 Therefore, the best choice is D, which is +131.5 kJ mol^-1.

Question 7

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$$\quad$2Hg(g)+ O₂(g)→ 2HgO(s)$\quad$$\quad$ΔG° = -180.8 kJ mol^-1 $\quad$$\quad$P(Hg)= 0.025 atm,P(O₂)= 0.037 atm

Accepted Answer

To calculate Δ_rG, we first need to calculate Δ_rH using the standard enthalpy of formation values: 2Hg(g) + O₂(g) → 2HgO(s) ΔH° = -181.6 kJ mol^-1 Δ_rH can be calculated using the following equation: Δ_rH = ΣnSjS°(products) - ΣnSjS°(reactants) = 2(70.1) + (205.0) - 0 = 345.2 J K-1 mol-1 Next, we need to calculate Δ_rS using the following equation: Δ_rS = ΣnSjS°(reactants) ln(Q) Q = P(O₂)²/P(Hg)² = (0.037 atm)²/(0.025 atm)² = 2.76 Δ_rS = -181.6 kJ mol^-1/298 K - (345.2 J K-1 mol-1) ln(2.76) = -181.6 kJ mol^-1 + 122.3 J K-1 mol-1 = -181.5 kJ mol^-1 Finally, Δ_rG can be calculated using the following equation: Δ_rG = Δ_rH - TΔ_rS = (-181.6 kJ mol^-1 * 1000 J/kJ - 298 K * 122.3 J K-1 mol-1) / 1000 = -154.4 kJ mol^-1 The best choice is B because it matches our calculated value.

Question 8

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$$\quad$SO₃(g)+ H₂O(g)→ H₂SO₄(l)$\quad$$\quad$$\quad$Δ_rG°= -90.5 kJ mol^-1 $\quad$$\quad$P(SO₃)= 0.20 atm,P(H₂O)= 0.88 atm

Accepted Answer

To calculate the change in Gibbs free energy, we can use the equation:

ΔG = ΔH - TΔS

where ΔH is the change in enthalpy and ΔS is the change in entropy.

ΔH can be calculated using the enthalpies of formation of the reactants and products:

ΔH = ΣnΔHf(products) - ΣnΔHf(reactants)

ΔH = [ΔHf(H2SO4(l)) + ΔHf(SO2(g))] - [ΔHf(SO3(g)) + ΔHf(H2O(l))]

ΔH = [(0 + (-300.4))] - [(-396.2 + (-285.8))]

ΔH = -90.4 kJ/mol

ΔS can be calculated using the standard molar entropies of the reactants and products:

ΔS = ΣnS(products) - ΣnS(reactants)

ΔS = [S(H2SO4(l)) + S(SO2(g))] - [S(SO3(g)) + S(H2O(l))]

ΔS = [(156.9 + 248.2)] - [(256.8 + 69.9)]

ΔS = + 78.4 J/mol/K

We need to convert ΔS to kJ/mol/K:

ΔS = 0.0784 kJ/mol/K

Now we can calculate ΔG using the equation:

ΔG = ΔH - TΔS

ΔG = -90.4 - (298)(0.0784)

ΔG = -86.2 kJ/mol

Therefore, the best choice is D.

Question 9

Calculate Δ_rG° at 298 K using the following information: $\quad$$\quad$$\quad$$\quad$$\quad$4HNO₃(g)+ 5N₂H₄(l)→ 7N₂(g)+ 12H₂O(l)Δ_rG°= ? Δ_fH° (kJ mol^-1)-133.9 $\quad$$\quad$50.6 $\quad$$\quad$$\quad$-285.8 S°(J K^-1 mol^-1) 266.9 $\quad$$\quad$ 121.2 $\quad$191.6 $\quad$70.0

Accepted Answer

The correct answer is E.Δ_rG° = Δ_rH° - TΔ_rS°Δ_rH° = [7Δ_fH°(N₂) + 12Δ_fH°(H₂O)] - [4Δ_fH°(HNO₃) + 5Δ_fH°(N₂H₄)]= [7(0) + 12(-285.8)] - [4(-133.9) + 5(50.6)]= -3309.4 kJ mol^-1Δ_rS° = [7S°(N₂) + 12S°(H₂O)] - [4S°(HNO₃) + 5S°(N₂H₄)]= [7(191.6) + 12(70.0)] - [4(266.9) + 5(121.2)]= 1149.2 J K^-1 mol^-1Δ_rG° = -3309.4 kJ mol^-1 - (298 K)(1149.2 J K^-1 mol^-1)= -3.30 × 10³ kJ mol^-1

Question 10

Calculate Δ_rG° at 298 K using the following information: $\quad$$\quad$$\quad$$\quad$$\quad$$\quad$2HNO₃(aq)+ NO(g)→ 3NO₂(g)+ H₂O(l)Δ_rG°= ? Δ_fG° (kJ mol^-1)-110.9 $\quad$$\quad$$\quad$87.6 $\quad$$\quad$51.3 $\quad$$\quad$-237.1

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Question 11

Use Hess's law to calculate Δ_rG° using the following information: $\quad$$\quad$NO(g)+ O(g)→ NO₂(g)$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$Δ_rG°= ? $\quad$$\quad$2O₃(g)→ 3O₂(g)$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$ Δ_rG°= +489.6 kJ mol^-1 $\quad$$\quad$O₂(g)→ 2O(g)$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$$\quad$Δ_rG°= +463.4 kJ mol^-1 $\quad$$\quad$NO(g)+ O₃(g)→ NO₂(g)+ O₂(g)$\quad$$\quad$$\quad$Δ_rG°= - 199.5 kJ mol^-1

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Question 12

Calculate ?_rG for the evaporation of methanol at 25°C under nonstandard conditions when $\mathrm { PCH } _ { 3 } \mathrm { OH }$ = 100.0 mbar $\quad$$\quad$$\quad$CH₃OH(l)? CH₃OH(g)

Accepted Answer

The correct answer is A because the evaporation of methanol is an endothermic process, and the entropy change is positive. Therefore, the Gibbs free energy change is positive, and the correct answer is A.

Question 13

Calculate Δ_rG° at 298 K using the following information: $\quad$$\quad$$\quad$$\quad$$\quad$4HNO₃(g)+ 5N₂H₄(l)→ 7N₂(g)+ 12H₂O(l)Δ_rG°= ? Δ_fG° (kJ mol^-1)$\quad$-73.5 $\quad$$\quad$149.3 $\quad$$\quad$ -237.1

Accepted Answer

First, we need to write the balanced chemical equation for the given reaction: 
4HNOS1U1B13S1U1B0(g) + 5NS1U1B12S1U1B0HS1U1B14S1U1B0(l) → 7NS1U1B12S1U1B0(g) + 12HS1U1B12S1U1B0O(l)
The standard entropy change of reaction, ΔS1U1B1rS1U1B0G°S1U1B1, can be calculated using the standard entropy of formation, ΔS1U1B1fS1U1B0G°, of the reactants and products according to the formula:

ΔS1U1B1rS1U1B0G°S1U1B1 = ΣnΔS1U1B1fS1U1B0G°(products) - ΣmΔS1U1B1fS1U1B0G°(reactants)

where n and m are the stoichiometric coefficients of the products and reactants, respectively.

We can look up the standard entropies of formation of the reactants and products in a table of thermodynamic data (such as the NIST Chemistry WebBook).

ΔS1U1B1fS1U1B0G° (kJ molS1U1P1-1S1S1P0)
4HNOS1U1B13S1U1B0(g)	144.3
5NS1U1B12S1U1B0HS1U1B14S1U1B0(l)	335.7
7NS1U1B12S1U1B0(g)	191.6
12HS1U1B12S1U1B0O(l)	-677.4

Using this data and the balanced chemical equation above, we can calculate:

ΔS1U1B1rS1U1B0G°S1U1B1 = (7 mol x 191.6 J mol-1 K-1) + (12 mol x (-677.4 J mol-1 K-1)) - (4 mol x 144.3 J mol-1 K-1) - (5 mol x 335.7 J mol-1 K-1)

ΔS1U1B1rS1U1B0G°S1U1B1 = -3297.7 J mol-1 K-1

Converting this to kJ mol-1 yields:

ΔS1U1B1rS1U1B0G°S1U1B1 = -3.2977 kJ mol-1

Therefore, the correct answer is A.

Question 14

Calculate Δ_rG° at 298 K using the following information: $\quad$$\quad$$\quad$ $\quad$$\quad$2HNO₃(aq)+ NO(g)→ 3NO₂(g)+ H₂O(l)$\quad$Δ_rG°= ? Δ_fH°(kJ mol^-1)-207.0 $\quad$91.3 $\quad$$\quad$33.2 $\quad$$\quad$-285.8 S°(J K^-1 mol^-1)146.0 $\quad$210.8 $\quad$$\quad$240.1 $\quad$$\quad$70.0

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Question 15

Calculate ?_rG for the evaporation of methanol at 25°C under nonstandard conditions when $\mathrm { PCH } _ { 3 } \mathrm { OH }$ = 150.0 mbar $\quad$$\quad$$\quad$CH₃OH(l)? CH₃OH(g)

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Question 16

Determine Δ_rG° at 298 K using the following information: $\quad$H₂(g)+ CO(g)→ CH₂O(g)$\quad$Δ_rH°= +1.9 kJ mol^-1; $\quad$Δ_rS°= -109.6 J K^-1mol^-1

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Question 17

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$$\quad$Fe₂O₃(s)+ 3CO(g)→ 2Fe(s)+ 3CO₂(g)$\quad$ΔG° = -28.0 kJ mol^-1 $\quad$$\quad$P(CO)= 1.4 atm,P(CO₂)= 2.1 atm

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Question 18

Calculate Δ_rG° at 298 K using the following information: $\quad$$\quad$$\quad$$\quad$$\quad$2H₂S(g)+ 3O₂(g)→ 2SO₂(g)+ 2H₂O(g)Δ_rG°= ? Δ_fH° (kJ mol^-1)-20.6 $\quad$$\quad$$\quad$-296.8 $\quad$$\quad$-241.8 S°(J K^-1 mol^-1)205.8 $\quad$ 205.2 248.2 $\quad$$\quad$188.8

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Question 19

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$$\quad$CaCO₃(s)→ CaO(s)+ CO₂(g)$\quad$$\quad$Δ_rG° =+131.1 kJ mol^-1 $\quad$$\quad$P(CO₂)= 0.033 atm

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Question 20

Calculate ?_rG for the evaporation of methanol at 25°C under nonstandard conditions when $\mathrm { PCH } _ { 3 } \mathrm { OH }$ = 10.0 mbar $\quad$$\quad$$\quad$CH₃OH(l)? CH₃OH(g)

Accepted Answer

The Gibbs free energy change for the evaporation of methanol can be calculated using the equation: $\Delta G = \Delta G^\circ + RT \ln(Q)$. At 25°C, $\Delta G^\circ$ for methanol is approximately +3.2 kJ/mol. Given that the pressure of methanol vapor is 10.0 mbar, which is less than the standard pressure of 1 bar, the reaction quotient $Q$ is less than 1, making $\ln(Q)$ negative. This results in a negative $\Delta G$, specifically -7.1 kJ/mol.

Quiz 17: Gibbs Energy and Thermodynamics

Determine Δ_rG° at 298 K using the following information: FeO(s) + CO(g) → Fe(s) + CO₂(g) Δ_rH°= -11.0 kJ mol^-1; Δ_rS°= -17.4 J K^-1 mol^-1

Calculate Δ_rG° at 298 K using the following information: $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ 2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(g) Δ_rG°= ? Δ_fG°(kJ mol^-1) -33.4 $\quad$ $\quad$ $\quad$ -300.1 $\quad$ -228.6

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$ $\quad$ 3O₂(g) → 2O₃(g) $\quad$ $\quad$ $\quad$ Δ_rG° = +326 kJ $\quad$ $\quad$ P(O₂) = 0.41 atm,P(O₃) = 5.2 atm

Use Hess's law to calculate Δ_rG° using the following information: CO(g) → C(s) + 1/2 O₂(g) $\quad$ $\quad$ Δ_rG°= ? CO₂(g) → C(s) + O₂(g) $\quad$ $\quad$ $\quad$ Δ_rG°= +394.4 kJ mol^-1 CO(g) + 1/2 O₂(g) → CO₂(g) $\quad$ Δ_rG°= -257.2 kJ mol^-1

Determine Δ_rG° at 298 K using the following information: $\quad$ CaCO₃(s) → CaO(s) + CO₂(g) $\quad$ Δ_rH°= +179.2 kJ mol^-1; $\quad$ Δ_rS°= +160.2 J K^-1 mol^-1

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$ $\quad$ 2Hg(g) + O₂(g) → 2HgO(s) $\quad$ $\quad$ ΔG° = -180.8 kJ mol^-1 $\quad$ $\quad$ P(Hg) = 0.025 atm,P(O₂) = 0.037 atm

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$ $\quad$ SO₃(g) + H₂O(g) → H₂SO₄(l) $\quad$ $\quad$ $\quad$ Δ_rG°= -90.5 kJ mol^-1 $\quad$ $\quad$ P(SO₃) = 0.20 atm,P(H₂O) = 0.88 atm

Calculate Δ_rG° at 298 K using the following information: $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ 2HNO₃(aq) + NO(g) → 3NO₂(g) + H₂O(l) Δ_rG°= ? Δ_fG° (kJ mol^-1) -110.9 $\quad$ $\quad$ $\quad$ 87.6 $\quad$ $\quad$ 51.3 $\quad$ $\quad$ -237.1

Calculate ?_rG for the evaporation of methanol at 25°C under nonstandard conditions when $\mathrm { PCH } _ { 3 } \mathrm { OH }$ = 100.0 mbar $\quad$ $\quad$ $\quad$ CH₃OH(l) ? CH₃OH(g)

Calculate Δ_rG° at 298 K using the following information: $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ 4HNO₃(g) + 5N₂H₄(l) → 7N₂(g) + 12H₂O(l) Δ_rG°= ? Δ_fG° (kJ mol^-1) $\quad$ -73.5 $\quad$ $\quad$ 149.3 $\quad$ $\quad$ -237.1

Calculate ?_rG for the evaporation of methanol at 25°C under nonstandard conditions when $\mathrm { PCH } _ { 3 } \mathrm { OH }$ = 150.0 mbar $\quad$ $\quad$ $\quad$ CH₃OH(l) ? CH₃OH(g)

Determine Δ_rG° at 298 K using the following information: $\quad$ H₂(g) + CO(g) → CH₂O(g) $\quad$ Δ_rH°= +1.9 kJ mol^-1; $\quad$ Δ_rS°= -109.6 J K^-1mol^-1

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$ $\quad$ Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) $\quad$ ΔG° = -28.0 kJ mol^-1 $\quad$ $\quad$ P(CO) = 1.4 atm,P(CO₂) = 2.1 atm

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$ $\quad$ CaCO₃(s) → CaO(s) + CO₂(g) $\quad$ $\quad$ Δ_rG° =+131.1 kJ mol^-1 $\quad$ $\quad$ P(CO₂) = 0.033 atm

Calculate ?_rG for the evaporation of methanol at 25°C under nonstandard conditions when $\mathrm { PCH } _ { 3 } \mathrm { OH }$ = 10.0 mbar $\quad$ $\quad$ $\quad$ CH₃OH(l) ? CH₃OH(g)

Quiz 17: Gibbs Energy and Thermodynamics

Determine ΔrG° at 298 K using the following information: FeO(s) + CO(g) → Fe(s) + CO2(g) ΔrH°= -11.0 kJ mol-1; ΔrS°= -17.4 J K-1 mol-1

Calculate ΔrG° at 298 K using the following information: \quad\quad\quad\quad\quad 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) ΔrG° = ? ΔfG°(kJ mol-1) -33.4 \quad\quad\quad -300.1 \quad -228.6

Calculate ΔrG at 298 K under the conditions shown below for the following reaction: \quad\quad 3O2(g) → 2O3(g) \quad\quad\quad ΔrG° = +326 kJ \quad\quad P(O2) = 0.41 atm,P(O3) = 5.2 atm

Use Hess's law to calculate ΔrG° using the following information: ClO(g) + O3(g) → Cl(g) + 2O2(g) \quad\quad\quad\quad\quad ΔrG° = ? 2O3(g) → 3O2(g) \quad\quad\quad\quad\quad\quad\quad ΔrG° = +489.6 kJ mol-1 Cl(g) + O3(g) → ClO(g) + O2(g) \quad\quad ΔrG° = -34.5 kJ mol-1

Use Hess's law to calculate ΔrG° using the following information: CO(g) → C(s) + 1/2 O2(g) \quad\quad ΔrG° = ? CO2(g) → C(s) + O2(g) \quad\quad\quad ΔrG° = +394.4 kJ mol-1 CO(g) + 1/2 O2(g) → CO2(g) \quad ΔrG° = -257.2 kJ mol-1

Determine ΔrG° at 298 K using the following information: \quad CaCO3(s) → CaO(s) + CO2(g) \quad ΔrH°= +179.2 kJ mol-1; \quad ΔrS°= +160.2 J K-1 mol-1

Calculate ΔrG at 298 K under the conditions shown below for the following reaction: \quad\quad 2Hg(g) + O2(g) → 2HgO(s) \quad\quad ΔG° = -180.8 kJ mol-1 \quad\quad P(Hg) = 0.025 atm,P(O2) = 0.037 atm

Calculate ΔrG at 298 K under the conditions shown below for the following reaction: \quad\quad SO3(g) + H2O(g) → H2SO4(l) \quad\quad\quad ΔrG°= -90.5 kJ mol-1 \quad\quad P(SO3) = 0.20 atm,P(H2O) = 0.88 atm

Calculate ΔrG° at 298 K using the following information: \quad\quad\quad\quad\quad 4HNO3(g) + 5N2H4(l) → 7N2(g) + 12H2O(l) ΔrG° = ? ΔfH° (kJ mol-1) -133.9 \quad\quad 50.6 \quad\quad\quad -285.8 S°(J K-1 mol-1) 266.9 \quad\quad 121.2 \quad 191.6 \quad 70.0

Calculate ΔrG° at 298 K using the following information: \quad\quad\quad\quad\quad\quad 2HNO3(aq) + NO(g) → 3NO2(g) + H2O(l) ΔrG° = ? ΔfG° (kJ mol-1) -110.9 \quad\quad\quad 87.6 \quad\quad 51.3 \quad\quad -237.1

Calculate ?rG for the evaporation of methanol at 25°C under nonstandard conditions when PCH3OH\mathrm { PCH } _ { 3 } \mathrm { OH }PCH3​OH = 100.0 mbar \quad\quad\quad CH3OH(l) ? CH3OH(g)

Calculate ΔrG° at 298 K using the following information: \quad\quad\quad\quad\quad 4HNO3(g) + 5N2H4(l) → 7N2(g) + 12H2O(l) ΔrG° = ? ΔfG° (kJ mol-1) \quad -73.5 \quad\quad 149.3 \quad\quad -237.1

Calculate ΔrG° at 298 K using the following information: \quad\quad\quad \quad\quad 2HNO3(aq) + NO(g) → 3NO2(g) + H2O(l) \quad ΔrG° = ? ΔfH° (kJ mol-1) -207.0 \quad 91.3 \quad\quad 33.2 \quad\quad -285.8 S°(J K-1 mol-1) 146.0 \quad 210.8 \quad\quad 240.1 \quad\quad 70.0

Calculate ?rG for the evaporation of methanol at 25°C under nonstandard conditions when PCH3OH\mathrm { PCH } _ { 3 } \mathrm { OH }PCH3​OH = 150.0 mbar \quad\quad\quad CH3OH(l) ? CH3OH(g)

Determine ΔrG° at 298 K using the following information: \quad H2(g) + CO(g) → CH2O(g) \quad ΔrH°= +1.9 kJ mol-1; \quad ΔrS°= -109.6 J K-1 mol-1

Calculate ΔrG at 298 K under the conditions shown below for the following reaction: \quad\quad Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) \quad ΔG° = -28.0 kJ mol-1 \quad\quad P(CO) = 1.4 atm,P(CO2) = 2.1 atm

Calculate ΔrG° at 298 K using the following information: \quad\quad\quad\quad\quad 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g) ΔrG° = ? ΔfH° (kJ mol-1) -20.6 \quad\quad\quad -296.8 \quad\quad -241.8 S°(J K-1 mol-1) 205.8 \quad 205.2 248.2 \quad\quad 188.8

Calculate ΔrG at 298 K under the conditions shown below for the following reaction: \quad\quad CaCO3(s) → CaO(s) + CO2(g) \quad\quad ΔrG° =+131.1 kJ mol-1 \quad\quad P(CO2) = 0.033 atm

Calculate ?rG for the evaporation of methanol at 25°C under nonstandard conditions when PCH3OH\mathrm { PCH } _ { 3 } \mathrm { OH }PCH3​OH = 10.0 mbar \quad\quad\quad CH3OH(l) ? CH3OH(g)

Determine Δ_rG° at 298 K using the following information: FeO(s) + CO(g) → Fe(s) + CO₂(g) Δ_rH°= -11.0 kJ mol^-1; Δ_rS°= -17.4 J K^-1 mol^-1

Calculate Δ_rG° at 298 K using the following information: $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ 2H₂S(g) + 3O₂(g) → 2SO₂(g) + 2H₂O(g) Δ_rG°= ? Δ_fG°(kJ mol^-1) -33.4 $\quad$ $\quad$ $\quad$ -300.1 $\quad$ -228.6

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$ $\quad$ 3O₂(g) → 2O₃(g) $\quad$ $\quad$ $\quad$ Δ_rG° = +326 kJ $\quad$ $\quad$ P(O₂) = 0.41 atm,P(O₃) = 5.2 atm

Use Hess's law to calculate Δ_rG° using the following information: CO(g) → C(s) + 1/2 O₂(g) $\quad$ $\quad$ Δ_rG°= ? CO₂(g) → C(s) + O₂(g) $\quad$ $\quad$ $\quad$ Δ_rG°= +394.4 kJ mol^-1 CO(g) + 1/2 O₂(g) → CO₂(g) $\quad$ Δ_rG°= -257.2 kJ mol^-1

Determine Δ_rG° at 298 K using the following information: $\quad$ CaCO₃(s) → CaO(s) + CO₂(g) $\quad$ Δ_rH°= +179.2 kJ mol^-1; $\quad$ Δ_rS°= +160.2 J K^-1 mol^-1

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$ $\quad$ 2Hg(g) + O₂(g) → 2HgO(s) $\quad$ $\quad$ ΔG° = -180.8 kJ mol^-1 $\quad$ $\quad$ P(Hg) = 0.025 atm,P(O₂) = 0.037 atm

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$ $\quad$ SO₃(g) + H₂O(g) → H₂SO₄(l) $\quad$ $\quad$ $\quad$ Δ_rG°= -90.5 kJ mol^-1 $\quad$ $\quad$ P(SO₃) = 0.20 atm,P(H₂O) = 0.88 atm

Calculate Δ_rG° at 298 K using the following information: $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ 2HNO₃(aq) + NO(g) → 3NO₂(g) + H₂O(l) Δ_rG°= ? Δ_fG° (kJ mol^-1) -110.9 $\quad$ $\quad$ $\quad$ 87.6 $\quad$ $\quad$ 51.3 $\quad$ $\quad$ -237.1

Calculate ?_rG for the evaporation of methanol at 25°C under nonstandard conditions when $\mathrm { PCH } _ { 3 } \mathrm { OH }$ = 100.0 mbar $\quad$ $\quad$ $\quad$ CH₃OH(l) ? CH₃OH(g)

Calculate Δ_rG° at 298 K using the following information: $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ 4HNO₃(g) + 5N₂H₄(l) → 7N₂(g) + 12H₂O(l) Δ_rG°= ? Δ_fG° (kJ mol^-1) $\quad$ -73.5 $\quad$ $\quad$ 149.3 $\quad$ $\quad$ -237.1

Calculate ?_rG for the evaporation of methanol at 25°C under nonstandard conditions when $\mathrm { PCH } _ { 3 } \mathrm { OH }$ = 150.0 mbar $\quad$ $\quad$ $\quad$ CH₃OH(l) ? CH₃OH(g)

Determine Δ_rG° at 298 K using the following information: $\quad$ H₂(g) + CO(g) → CH₂O(g) $\quad$ Δ_rH°= +1.9 kJ mol^-1; $\quad$ Δ_rS°= -109.6 J K^-1mol^-1

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$ $\quad$ Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) $\quad$ ΔG° = -28.0 kJ mol^-1 $\quad$ $\quad$ P(CO) = 1.4 atm,P(CO₂) = 2.1 atm

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: $\quad$ $\quad$ CaCO₃(s) → CaO(s) + CO₂(g) $\quad$ $\quad$ Δ_rG° =+131.1 kJ mol^-1 $\quad$ $\quad$ P(CO₂) = 0.033 atm

Calculate ?_rG for the evaporation of methanol at 25°C under nonstandard conditions when $\mathrm { PCH } _ { 3 } \mathrm { OH }$ = 10.0 mbar $\quad$ $\quad$ $\quad$ CH₃OH(l) ? CH₃OH(g)