Question 1

Why can endothermic reactions be spontaneous?

Accepted Answer

The answer of Why can endothermic reactions be spontaneous?...

Question 2

For a given reaction,ΔH = -19.9 kJ/mol and ΔS = -55.5 J/Kmol.The reaction will have ΔG = 0 at ________ K.Assume that ΔH and ΔS do not vary with temperature.

Accepted Answer

The equation for Gibbs free energy (ΔG) is ΔG = ΔH - TΔS. To find the temperature at which ΔG = 0, we can set the equation equal to 0 and solve for T:

0 = ΔH - TΔS

T = ΔH/ΔS

Plugging in the values given, we get:

T = (-19.9 kJ/mol)/(-55.5 J/Kmol) = 359 K

Therefore, the answer is A) 359 K.

Question 3

The value of ΔG° at 141.0°C for the formation of phosphorous trichloride from its constituent elements, P₂(g)+ 3 Cl₂(g)→ 2 PCl₃(g) Is ________ kJ/mol.At 25.0°C for this reaction,ΔH° is -720.5 kJ/mol,ΔG° is -642.9 kJ/mol,and ΔS° is -263.7 J/K.

Accepted Answer

To find the value of ΔG° at 141.0°C for the reaction, we can use the equation ΔG° = ΔH° - TΔS°. First, convert 141.0°C to Kelvin by adding 273.15, which gives 414.15 K. Then, use the given values of ΔH° (-720.5 kJ/mol) and ΔS° (-263.7 J/K = -0.2637 kJ/K·mol) and plug them into the equation. Remember to convert ΔS° to kJ by dividing by 1000 for consistency in units with ΔH°.ΔG° = -720.5 kJ/mol - (414.15 K)(-0.2637 kJ/K·mol) = -720.5 kJ/mol + 109.2 kJ/mol = -611.3 kJ/mol.The answer is rounded to -612.3 kJ/mol, which matches choice A.

Question 4

Define entropy.

Accepted Answer

The answer of Define entropy....

Question 5

Under which of the following conditions would one mole of Ar have the highest entropy,S?

Accepted Answer

Entropy, S, increases with temperature and volume. D has the highest temperature and volume, leading to the highest entropy.

Question 6

Why is heating your home with gas more efficient than heating it with electricity?

Accepted Answer

The answer of Why is heating your home with gas...

Question 7

Calculate the ΔG°_rxn using the following information. 2 HNO₃(aq)+ NO(g)→ 3 NO₂(g)+ H₂O(l)ΔG°_rxn= ? ΔH°_f (kJ/mol)-207.0 91.3 33.2 -285.8 S°(J/mol∙K 146.0 210.8 240.1 70.0

Accepted Answer

The answer of Calculate the ΔG°_rxn using the following information....

Question 8

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

Accepted Answer

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Question 9

Calculate the ΔG°_rxn using the following information. 4 HNO₃(g)+ 5 N₂H₄(l)→ 7 N₂(g)+ 12 H₂O(l)ΔG°_rxn= ? ΔG°_f (kJ/mol)-73.5 149.3 -237.1

Accepted Answer

The ΔG°_rxn is calculated using the formula: ΔG°_rxn = ΣΔG°_f(products) - ΣΔG°_f(reactants). Plugging in the given values: ΔG°_rxn = [7(0) + 12(-237.1)] - [4(-73.5) + 5(149.3)] = -3.298 × 10³ kJ.

Question 10

In the Haber process,ammonia is synthesized from nitrogen and hydrogen: N₂(g)+ 3 H₂(g)→ 2 NH₃(g) ΔG° at 298 K for this reaction is -33.3 kJ/mol.The value of ΔG at 298 K for a reaction mixture that consists of 1.9 atm N₂,1.6 atm H₂,and 0.65 atm NH₃ is ________.

Accepted Answer

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Question 11

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

Accepted Answer

ΔG = ΔG° + RTlnQ where Q is the reaction quotient.
First, we need to determine the value of Q. Since the reaction involves gases, we should use the formula Q = (P(product)) /(P(reactant)). 
Q = [(1 bar)/(0.20 bar)] x [(0.88 bar)/(1 bar)] = 4.4
Now, we can substitute the values in the formula for ΔG:
ΔG = -90.5 kJ/mol + (8.314 J/mol.K x 298 K x ln 4.4) = -86.2 kJ/mol
Therefore, the answer is A, -86.2 kJ.

Question 12

Why can't we say that a spontaneous reaction is a fast reaction?

Accepted Answer

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Question 13

Calculate the ΔG°_rxn using the following information. 2 H₂S(g)+ 3 O₂(g)→ 2 SO₂(g)+ 2 H₂O(g)ΔG°_rxn= ? ΔG°_f (kJ/mol)-33.4 -300.1 -228.6

Accepted Answer

The ΔG°_rxn can be calculated using the formula ΔG°_rxn = ΣΔG°_f(products) - ΣΔG°_f(reactants). For the given reaction, the ΔG°_f values for the products (2 moles of SO2 and 2 moles of H2O) and the reactants (2 moles of H2S and 3 moles of O2) are plugged into this formula. Since the ΔG°_f for O2 is 0 (as it is an element in its standard state), the calculation simplifies to: ΔG°_rxn = [2(-300.1) + 2(-228.6)] - [2(-33.4) + 3(0)] = -1057.4 + 66.8 = -990.6 kJ.

Question 14

Calculate the ΔG°_rxn using the following information. 2 HNO₃(aq)+ NO(g)→ 3 NO₂(g)+ H₂O(l)ΔG°_rxn= ? ΔG°_f (kJ/mol)-110.9 87.6 51.3 -237.1

Accepted Answer

The answer of Calculate the ΔG°_rxn using the following information....

Question 15

For a given reaction,ΔH = +35.5 kJ/mol and ΔS = +83.6 J/Kmol.The reaction is spontaneous ________.Assume that ΔH and ΔS do not vary with temperature.

Accepted Answer

The spontaneity of a reaction is determined by the Gibbs free energy change (ΔG). The relationship between ΔG, ΔH, and ΔS is given by the equation:

ΔG = ΔH - TΔS

where T is the temperature in Kelvin. For a spontaneous reaction, ΔG must be negative.

Substituting the given values into the equation, we get:

ΔG = (+35.5 kJ/mol) - T(0.0836 kJ/mol·K)

At low temperatures, the positive ΔH term dominates, making ΔG positive and the reaction non-spontaneous. At high temperatures, the negative TΔS term dominates, making ΔG negative and the reaction spontaneous. Therefore, the reaction is spontaneous at T > 425 K.

Question 16

The value of ΔG° at 100.0°C for the formation of calcium chloride from its constituent elements: Ca(s)+ Cl₂(g)→ CaCl₂(s) Is ________ kJ/mol.At 25.0°C for this reaction,ΔH° is -795.8 kJ/mol,ΔG° is -748.1 kJ/mol,and ΔS° is -159.8 J/K.

Accepted Answer

To find the value of ΔG° at 100.0°C (373.15 K) for the formation of calcium chloride, we use the Gibbs free energy equation: ΔG° = ΔH° - TΔS°. Given ΔH° = -795.8 kJ/mol and ΔS° = -159.8 J/K (or -0.1598 kJ/K·mol), we convert the temperature to Kelvin and plug in the values: ΔG° = -795.8 kJ/mol - (373.15 K)(-0.1598 kJ/K·mol) = -795.8 kJ/mol + 59.7 kJ/mol = -736.1 kJ/mol.

Question 17

Phosphorous and chlorine gases combine to produce phosphorous trichloride: P₂(g)+ 3 Cl₂(g)→ 2 PCl₃(g) ΔG° at 298 K for this reaction is -642.9 kJ/mol.The value of ΔG at 298 K for a reaction mixture that consists of 1.5 atm P₂,1.6 atm Cl₂,and 0.65 atm PCl₃ is ________.

Accepted Answer

We can use the equation ΔG = ΔG° + RTlnQ to solve this problem, where Q is the reaction quotient. First, we need to calculate Q:

Q = (PCl3)^2 / (PS2Cl)^1.5 * (Cl2)^3 * (1 atm)^2

Note that we use the coefficients and concentrations of the gases in the balanced equation to determine the exponent on each term in Q. We also include atmospheric pressure in the denominator since it is not included in the concentrations given.

Q = (0.65 atm)^2 / (1.5 atm)^1.5 * (1.6 atm)^3 * (1 atm)^2

Q = 1.51

Now we can plug in the values for ΔG°, R, T, and Q:

ΔG = -642.9 kJ/mol + (8.314 J/mol-K * 298 K * ln 1.51) / 1000 J/kJ

ΔG = -649.5 kJ/mol

Therefore, the answer is E.

Question 18

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

Accepted Answer

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Question 19

Determine ΔG°_rxn using the following information. H₂(g)+ CO(g)→ CH₂O(g)ΔH°= +1.9 kJ; ΔS°= -109.6 J/K

Accepted Answer

ΔG° can be calculated using the formula ΔG° = ΔH° - TΔS°. Assuming standard temperature (298 K), ΔG° = 1.9 kJ - (298 K)(-109.6 J/K × 1 kJ/1000 J) = 1.9 kJ + 32.7 kJ = 34.6 kJ.

Question 20

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= +394.4 kJ CO(g)+ 1/2 O₂(g)→ CO₂(g)ΔG°_rxn= -257.2 kJ

Accepted Answer

The answer of Use Hess's law to calculate ΔG°_rxn using...

Quiz 18: Free Energy and Thermodynamics

Why can endothermic reactions be spontaneous?

For a given reaction,ΔH = -19.9 kJ/mol and ΔS = -55.5 J/Kmol.The reaction will have ΔG = 0 at ________ K.Assume that ΔH and ΔS do not vary with temperature.

The value of ΔG° at 141.0°C for the formation of phosphorous trichloride from its constituent elements, P₂(g) + 3 Cl₂(g) → 2 PCl₃(g) Is ________ kJ/mol.At 25.0°C for this reaction,ΔH° is -720.5 kJ/mol,ΔG° is -642.9 kJ/mol,and ΔS° is -263.7 J/K.

Define entropy.

Under which of the following conditions would one mole of Ar have the highest entropy,S?

Why is heating your home with gas more efficient than heating it with electricity?

Calculate the ΔG°_rxn using the following information. 2 HNO₃(aq) + NO(g) → 3 NO₂(g) + H₂O(l) ΔG°_rxn= ? ΔH°_f (kJ/mol) -207.0 91.3 33.2 -285.8 S°(J/mol∙K 146.0 210.8 240.1 70.0

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

Calculate the ΔG°_rxn using the following information. 4 HNO₃(g) + 5 N₂H₄(l) → 7 N₂(g) + 12 H₂O(l) ΔG°_rxn= ? ΔG°_f (kJ/mol) -73.5 149.3 -237.1

In the Haber process,ammonia is synthesized from nitrogen and hydrogen: N₂(g) + 3 H₂(g) → 2 NH₃(g) ΔG° at 298 K for this reaction is -33.3 kJ/mol.The value of ΔG at 298 K for a reaction mixture that consists of 1.9 atm N₂,1.6 atm H₂,and 0.65 atm NH₃ is ________.

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

Why can't we say that a spontaneous reaction is a fast reaction?

Calculate the ΔG°_rxn using the following information. 2 H₂S(g) + 3 O₂(g) → 2 SO₂(g) + 2 H₂O(g) ΔG°_rxn= ? ΔG°_f (kJ/mol) -33.4 -300.1 -228.6

Calculate the ΔG°_rxn using the following information. 2 HNO₃(aq) + NO(g) → 3 NO₂(g) + H₂O(l) ΔG°_rxn= ? ΔG°_f (kJ/mol) -110.9 87.6 51.3 -237.1

For a given reaction,ΔH = +35.5 kJ/mol and ΔS = +83.6 J/Kmol.The reaction is spontaneous ________.Assume that ΔH and ΔS do not vary with temperature.

The value of ΔG° at 100.0°C for the formation of calcium chloride from its constituent elements: Ca(s) + Cl₂(g) → CaCl₂(s) Is ________ kJ/mol.At 25.0°C for this reaction,ΔH° is -795.8 kJ/mol,ΔG° is -748.1 kJ/mol,and ΔS° is -159.8 J/K.

Phosphorous and chlorine gases combine to produce phosphorous trichloride: P₂(g) + 3 Cl₂(g) → 2 PCl₃(g) ΔG° at 298 K for this reaction is -642.9 kJ/mol.The value of ΔG at 298 K for a reaction mixture that consists of 1.5 atm P₂,1.6 atm Cl₂,and 0.65 atm PCl₃ is ________.

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

Determine ΔG°_rxn using the following information. H₂(g) + CO(g) → CH₂O(g) ΔH°= +1.9 kJ; ΔS°= -109.6 J/K

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= +394.4 kJ CO(g) + 1/2 O₂(g) → CO₂(g) ΔG°_rxn= -257.2 kJ

Quiz 18: Free Energy and Thermodynamics

Why can endothermic reactions be spontaneous?

For a given reaction,ΔH = -19.9 kJ/mol and ΔS = -55.5 J/Kmol.The reaction will have ΔG = 0 at ________ K.Assume that ΔH and ΔS do not vary with temperature.

The value of ΔG° at 141.0°C for the formation of phosphorous trichloride from its constituent elements, P2(g) + 3 Cl2(g) → 2 PCl3(g) Is ________ kJ/mol.At 25.0°C for this reaction,ΔH° is -720.5 kJ/mol,ΔG° is -642.9 kJ/mol,and ΔS° is -263.7 J/K.

Define entropy.

Under which of the following conditions would one mole of Ar have the highest entropy,S?

Why is heating your home with gas more efficient than heating it with electricity?

Calculate the ΔG°rxn using the following information. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -207.0 91.3 33.2 -285.8 S°(J/mol∙K 146.0 210.8 240.1 70.0

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

Calculate the ΔG°rxn using the following information. 4 HNO3(g) + 5 N2H4(l) → 7 N2(g) + 12 H2O(l) ΔG°rxn = ? ΔG°f (kJ/mol) -73.5 149.3 -237.1

In the Haber process,ammonia is synthesized from nitrogen and hydrogen: N2(g) + 3 H2(g) → 2 NH3(g) ΔG° at 298 K for this reaction is -33.3 kJ/mol.The value of ΔG at 298 K for a reaction mixture that consists of 1.9 atm N2,1.6 atm H2,and 0.65 atm NH3 is ________.

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

Why can't we say that a spontaneous reaction is a fast reaction?

Calculate the ΔG°rxn using the following information. 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g) ΔG°rxn = ? ΔG°f (kJ/mol) -33.4 -300.1 -228.6

Calculate the ΔG°rxn using the following information. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔG°f (kJ/mol) -110.9 87.6 51.3 -237.1

For a given reaction,ΔH = +35.5 kJ/mol and ΔS = +83.6 J/Kmol.The reaction is spontaneous ________.Assume that ΔH and ΔS do not vary with temperature.

The value of ΔG° at 100.0°C for the formation of calcium chloride from its constituent elements: Ca(s) + Cl2(g) → CaCl2(s) Is ________ kJ/mol.At 25.0°C for this reaction,ΔH° is -795.8 kJ/mol,ΔG° is -748.1 kJ/mol,and ΔS° is -159.8 J/K.

Phosphorous and chlorine gases combine to produce phosphorous trichloride: P2(g) + 3 Cl2(g) → 2 PCl3(g) ΔG° at 298 K for this reaction is -642.9 kJ/mol.The value of ΔG at 298 K for a reaction mixture that consists of 1.5 atm P2,1.6 atm Cl2,and 0.65 atm PCl3 is ________.

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

Determine ΔG°rxn using the following information. H2(g) + CO(g) → CH2O(g) ΔH°= +1.9 kJ; ΔS°= -109.6 J/K

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 = +394.4 kJ CO(g) + 1/2 O2(g) → CO2(g) ΔG°rxn = -257.2 kJ

The value of ΔG° at 141.0°C for the formation of phosphorous trichloride from its constituent elements, P₂(g) + 3 Cl₂(g) → 2 PCl₃(g) Is ________ kJ/mol.At 25.0°C for this reaction,ΔH° is -720.5 kJ/mol,ΔG° is -642.9 kJ/mol,and ΔS° is -263.7 J/K.

Calculate the ΔG°_rxn using the following information. 2 HNO₃(aq) + NO(g) → 3 NO₂(g) + H₂O(l) ΔG°_rxn= ? ΔH°_f (kJ/mol) -207.0 91.3 33.2 -285.8 S°(J/mol∙K 146.0 210.8 240.1 70.0

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

Calculate the ΔG°_rxn using the following information. 4 HNO₃(g) + 5 N₂H₄(l) → 7 N₂(g) + 12 H₂O(l) ΔG°_rxn= ? ΔG°_f (kJ/mol) -73.5 149.3 -237.1

In the Haber process,ammonia is synthesized from nitrogen and hydrogen: N₂(g) + 3 H₂(g) → 2 NH₃(g) ΔG° at 298 K for this reaction is -33.3 kJ/mol.The value of ΔG at 298 K for a reaction mixture that consists of 1.9 atm N₂,1.6 atm H₂,and 0.65 atm NH₃ is ________.

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

Calculate the ΔG°_rxn using the following information. 2 H₂S(g) + 3 O₂(g) → 2 SO₂(g) + 2 H₂O(g) ΔG°_rxn= ? ΔG°_f (kJ/mol) -33.4 -300.1 -228.6

Calculate the ΔG°_rxn using the following information. 2 HNO₃(aq) + NO(g) → 3 NO₂(g) + H₂O(l) ΔG°_rxn= ? ΔG°_f (kJ/mol) -110.9 87.6 51.3 -237.1

The value of ΔG° at 100.0°C for the formation of calcium chloride from its constituent elements: Ca(s) + Cl₂(g) → CaCl₂(s) Is ________ kJ/mol.At 25.0°C for this reaction,ΔH° is -795.8 kJ/mol,ΔG° is -748.1 kJ/mol,and ΔS° is -159.8 J/K.

Phosphorous and chlorine gases combine to produce phosphorous trichloride: P₂(g) + 3 Cl₂(g) → 2 PCl₃(g) ΔG° at 298 K for this reaction is -642.9 kJ/mol.The value of ΔG at 298 K for a reaction mixture that consists of 1.5 atm P₂,1.6 atm Cl₂,and 0.65 atm PCl₃ is ________.

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

Determine ΔG°_rxn using the following information. H₂(g) + CO(g) → CH₂O(g) ΔH°= +1.9 kJ; ΔS°= -109.6 J/K

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= +394.4 kJ CO(g) + 1/2 O₂(g) → CO₂(g) ΔG°_rxn= -257.2 kJ