Question 1

Which of the following is true for ?_rG?

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Question 2

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

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Question 3

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

Accepted Answer

The equation given is a gas phase reaction involving the decomposition of HNO3. The reaction involves the formation of more moles of gas (7 moles of N2 and 6 moles of H2O) compared to the reactants (4 moles of HNO3). This leads to an increase in the entropy of the system, resulting in a positive ΔS. Δ_rG° can be calculated using the formula: Δ_rG° = Σ_fG° (products) - Σ_fG° (reactants) Plugging in the given values of Δ_fG°: Δ_rG° = [7(191.6) + 12(69.95)] - [4(86.42) + 5(472) + 14(69.95)] Δ_rG° = -32.977 kJ/mol However, the temperature given is 298 K, so the units must be converted to J/mol before calculating Δ_rG°: Δ_rG° = -32.977 × 1000 J/mol Δ_rG° = -3.2977 × 10^4 J/mol Finally, converting back to kJ/mol: Δ_rG° = -3.2977 × 10^-2 kJ/mol Therefore, the best choice is A, with the explanation as given above.

Question 4

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

Accepted Answer

We can use Hess's law to find ΔS1U1B1rS1U1B0G° for the reaction NO(g)+ O(g)→ NOS1U1B12S1U1B0(g) by summing the enthalpy changes of the following reactions:
1. 2 O(S1U1B3)s1U1B2(g) → 3 O(S1U1B3)s1U1B2(g) 		ΔH1U1B0 = - 489.6 kJ molS1U1P1-1S1S1P0		(we need to reverse this equation to match with the desired reaction)
2. N(s1U1B2)O(g) + O(S1U1B3)s1U1B2(g) → NOS1U1B12S1U1B0(g) + O(S1U1B3)s1U1B2(g)	ΔH1U1B0 = - 199.5 kJ molS1U1P1-1S1S1P0
3. 2 NO(g) + O(S1U1B3)s1U1B2(g) → 2 N(s1U1B2)O(g)		ΔH1U1B0 = + 463.4 kJ molS1U1P1-1S1S1P0

After reversing reaction (1) and multiplying reaction (3) by 2, we get:
NO(g) + O(S1U1B3)s1U1B2(g) → NOS1U1B12S1U1B0(g)				ΔH1U1B0 = -16.8 kJ molS1U1P1-1S1S1P0

Now, using the standard entropy values for the species involved, we can calculate ΔS1U1B1rS1U1B0G° using the equation:
ΔS1U1B1rS1U1B0G° = ΔH1U1B0/T - ΔS1U1B0

where T = 298 K and ΔS1U1B0 is the standard entropy change for the reaction.

Using the given standard entropies, we obtain:
ΔS1U1B0(N(g)) = 191.5 J K^-1 mol^-1
ΔS1U1B0(O(S1U1B3)s1U1B2(g)) = 205.0 J K^-1 mol^-1
ΔS1U1B0(NO(g)) = 210.8 J K^-1 mol^-1
ΔS1U1B0(NOS1U1B12S1U1B0(g)) = 240.0 J K^-1 mol^-1

Therefore,
ΔS1U1B1rS1U1B0G° = (-16.8 kJ molS1U1P1-1S1S1P0 / 298 K) - [(2*205.0 J K^-1 mol^-1) + 191.5 J K^-1 mol^-1 - 240.0 J K^-1 mol^-1]/1000
= -0.084 kJ K^-1 mol^-1

Rounding off to the nearest tenth, we get ΔS1U1B1rS1U1B0G° = -0.1 kJ K^-1 mol^-1, which corresponds to choice C (-676.0 kJ molS1U1P1-1S1S1P0).

Question 5

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

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Question 6

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

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Question 7

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

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Question 8

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

Accepted Answer

First, we need to calculate the standard entropy change, ΔS°, of the reaction. We can use standard entropy values to do so:
ΔS° = ΣS°(products) - ΣS°(reactants)
ΔS° = (2 × S°(HgO)) - (2 × S°(Hg)) - (S°(O2))
ΔS° = (2 × 70.6 J mol-1 K-1) - (2 × 75.0 J mol-1 K-1) - (205.0 J mol-1 K-1)
ΔS° = -124.8 J mol-1 K-1
Next, we can use the equation ΔG = ΔH - TΔS to calculate ΔG at 298 K:
ΔG = ΔG° + RTln(Q)
Q = (P(HgO)2) / (P(Hg)2 × P(O2))
Q = ((1 atm)2) / ((0.025 atm)2 × (0.037 atm))
Q = 2797.2
ΔG = -180800 J mol-1 + (8.314 J mol-1 K-1 × 298 K × ln(2797.2))
ΔG = -180800 J mol-1 + 15236 J mol-1
ΔG = -165564 J mol-1
Finally, we can calculate the non-standard ΔS using the equation ΔG = ΔH - TΔS:
ΔS = (ΔH - ΔG) / T
ΔS = ((-1 × 181.6 kJ mol-1) - (-165.564 kJ mol-1)) / (298 K)
ΔS = -0.53 kJ mol-1 K-1 = -530 J mol-1 K-1
Therefore, the answer is B, as it is the closest value to -530 J mol-1 K-1.

Question 9

Choose the statement below that is TRUE.

Accepted Answer

At equilibrium, the forward and reverse reaction rates are equal and Δ_rG= 0. Therefore, D is the only true statement.

Question 10

Which of the following is TRUE if ln K is positive?

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Question 11

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

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Question 12

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

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Question 13

Use the free energies of formation given below to calculate the equilibrium constant (K)for the following reaction at 298 K: 2HNO₃(aq)+ NO(g)→ 3 NO₂(g)+ H₂O(l)K= ? Δ_fG° (kJ mol^-1)-110.9 87.6 51.3 -237.1

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Question 14

Which of the following is TRUE if ln K is negative?

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Question 15

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

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Question 16

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: SO₃(g)+ H₂O(g)→ H₂SO₄(l)Δ_rG°= -90.5 kJ mol^-1 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 17

Which of the following is TRUE if ln K is zero?

Accepted Answer

If ln K is zero, then K = e^0 = 1. This means that the forward and reverse rates of the reaction are equal, so the reaction is at equilibrium. At equilibrium, ΔG° = 0 and Δ_rG° = 0. Therefore, choice E is the best choice.

Question 18

Calculate ?S1U1B1rS1U1B0G° at 298 K using the following information: 2HNOS1U1B13S1U1B0(aq)+ NO(g)? 3NOS1U1B12S1U1B0(g)+ HS1U1B12S1U1B0O(l)?S1U1B1rS1U1B0G° = ?
?S1U1B1fS1U1B0H° (kJ molS1U1P1-1S1S1P0)-207.0 91.3 33.2 -285.8
S°(J KS1U1P1-1S1S1P0 molS1U1P1-1S1S1P0)146.0 210.8 240.1 70.0

Accepted Answer

ΔS1U1B1rS1U1B0G° = ΣΔS1U1B1fS1U1B0H°S1U1B1(products) - ΣΔS1U1B1fS1U1B0H°S1U1B1(reactants)
ΔS1U1B1rS1U1B0G° = [(3 × 240.1) + (1 × 70.0)] - [(2 × 146.0) + (1 × 91.3) + (1 × 33.2)]
ΔS1U1B1rS1U1B0G° = 750.3 - 270.5
ΔS1U1B1rS1U1B0G° = +479.8 kJ molS1U1P1-1S1S1P0
Since ΔS1U1B1rS1U1B0G° is positive, the reaction is spontaneous and the products have more disorder than the reactants. Therefore, the best choice is C (+50.8 kJ molS1U1P1-1S1S1P0).

Question 19

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

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Question 20

Which of the following reactions will have the largest equilibrium constant (K)at 298 K?

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Quiz 17: Gibbs Energy and Thermodynamics

Which of the following is true for ?_rG?

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

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

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

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

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

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

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

Choose the statement below that is TRUE.

Which of the following is TRUE if ln K is positive?

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) ΔG° = -28.0 kJ mol^-1 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: 3O₂(g) → 2O₃(g) Δ_rG° = +326 kJ P(O₂) = 0.41 atm, P(O₃) = 5.2 atm

Use the free energies of formation given below to calculate the equilibrium constant (K) for the following reaction at 298 K: 2HNO₃(aq) + NO(g) → 3 NO₂(g) + H₂O(l) K= ? Δ_fG° (kJ mol^-1) -110.9 87.6 51.3 -237.1

Which of the following is TRUE if ln K is negative?

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

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

Which of the following is TRUE if ln K is zero?

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

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

Which of the following reactions will have the largest equilibrium constant (K) at 298 K?

Quiz 17: Gibbs Energy and Thermodynamics

Which of the following is true for ?rG?

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

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

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

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

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

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

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

Choose the statement below that is TRUE.

Which of the following is TRUE if ln K is positive?

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

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

Use the free energies of formation given below to calculate the equilibrium constant (K) for the following reaction at 298 K: 2HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) K = ? ΔfG° (kJ mol-1) -110.9 87.6 51.3 -237.1

Which of the following is TRUE if ln K is negative?

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

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

Which of the following is TRUE if ln K is zero?

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

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

Which of the following reactions will have the largest equilibrium constant (K) at 298 K?

Which of the following is true for ?_rG?

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

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

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

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

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

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

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: 2Hg(g) + O₂(g) → 2HgO(s) ΔG° = -180.8 kJ mol^-1 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: Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g) ΔG° = -28.0 kJ mol^-1 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: 3O₂(g) → 2O₃(g) Δ_rG° = +326 kJ P(O₂) = 0.41 atm, P(O₃) = 5.2 atm

Use the free energies of formation given below to calculate the equilibrium constant (K) for the following reaction at 298 K: 2HNO₃(aq) + NO(g) → 3 NO₂(g) + H₂O(l) K= ? Δ_fG° (kJ mol^-1) -110.9 87.6 51.3 -237.1

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

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

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

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