Question 1

For the process CaCO₃(calcite)→ CaCO₃(aragonite)ΔH° = -0.21 kJ,ΔS° = -4.2 J/K Assuming that the surroundings can be considered a large heat reservoir at 25°C,calculate ΔS_surr and ΔS_total for the process at 25°C and 1 atm pressure.Is the process spontaneous at 25°C and 1 atm pressure?

Accepted Answer

ΔS_surr is calculated using ΔH/T = -0.21 kJ / 298 K = -0.0007 kJ/K = -0.7 J/K. ΔS_total = ΔS_system + ΔS_surr = -4.2 J/K - 0.7 J/K = -4.9 J/K. Since ΔS_total is negative, the process is not spontaneous.

Question 2

Why is the sign of ΔG rather than the sign of ΔS_total generally used to determine the spontaneity of a chemical reaction?

Accepted Answer

ΔG involves only thermodynamic functions of the system, whereas ΔS_total involves both thermodynamic and probabilistic functions which may be less reliable. Additionally, ΔG can be used for processes that occur under any conditions, making it a more versatile indicator of spontaneity.

Question 3

Calculate ΔS° for the following reaction. N₂(g)+ 2 O₂(g)→ 2 NO₂(g)

Accepted Answer

To calculate ΔS°, we need to use the following equation: ΔS° = ΣS°(products) - ΣS°(reactants) Where ΣS° is the standard entropy of formation. Using the table of standard entropy of formation, we have: ΔS° = [2S°(NO₂(g))] - [S°(N₂(g)) + 2S°(O₂(g))] = [2(240.1 J/K)] - [(219.1 J/K) + 2(205.0 J/K)] = -121.5 J/K Therefore, the best choice is B, with an explanation that we calculated ΔS° using the equation ΔS° = ΣS°(products) - ΣS°(reactants) and the table of standard entropy of formation.

Question 4

The solubility of manganese(II)fluoride in water is 6.6 g/mL at 40°C and 4.8 g/L at 100°C.Based on these data,what is the sign of ΔH° and ΔS° for the process below? MnF₂(s)⇌ Mn²⁺(aq)+ 2 F^-(aq)

Accepted Answer

The fact that the solubility of manganese(II)fluoride in water decreases with increasing temperature suggests that the dissolution process is exothermic (ΔH° < 0). However, since the solubility also decreases with increasing temperature, this suggests that the dissolution process is entropy-driven (ΔS° > 0). Therefore, the sign of ΔS° cannot be determined from this information alone. Choice A is the best option because it reflects the fact that ΔH° is negative but the sign of ΔS° cannot be determined.

Question 5

For bromine,ΔH°vap = 30.91 kJ/mol and ΔS°vap = 93.23 JK^-1mol^-1 at 25°C.What is the normal boiling point for bromine?

Accepted Answer

The normal boiling point of a substance is the temperature at which its vapor pressure equals atmospheric pressure. At this point, the Gibbs free energy change (ΔG) for the phase transition is zero. The relationship between ΔG, ΔH, and ΔS is given by ΔG = ΔH - TΔS. Setting ΔG to zero for the boiling point and solving for T gives T = ΔH/ΔS. Given ΔH°vap = 30.91 kJ/mol = 30910 J/mol and ΔS°vap = 93.23 J/mol·K, the boiling point T in Kelvin is T = 30910 J/mol / 93.23 J/mol·K = 331.6 K. Converting this to Celsius gives approximately 58°C.

Question 6

ΔS° = -198.7 J/K for the reaction shown below.Calculate S° for NH₃(g). N₂(g)+ 3 H₂(g)→ 2 NH₃(g)

Accepted Answer

To calculate the entropy change (ΔS) for the reaction, we need to use the formula: ΔS = ΣS°(products) - ΣS°(reactants) From the balanced chemical equation, we can see that the stoichiometric coefficients of the reactants and products are in the ratio of 1:2:3. This means that the entropy change for the reaction will be proportional to the number of moles of gaseous substances. There is only one mole of gas on the reactant side (N₂(g)), and two moles of gas on the product side (NH₃(g)). Therefore: ΔS = 2S°(NH₃(g)) - S°(N₂(g)) - 3S°(H₂(g)) Substituting the given value of ΔS° = -198.7 J/K: -198.7 J/K = 2S°(NH₃(g)) - S°(N₂(g)) - 3S°(H₂(g)) We do not know the individual values of S° for the different compounds, but we can assume that they are similar in magnitude. Therefore, we can make an approximation that: S°(NH₃(g)) ≈ S°(N₂(g)) ≈ S°(H₂(g)) Substituting this into the equation above, we get: -198.7 J/K = 2S° - 4S° - 3S° -198.7 J/K = -5S° S° = 39.74 J/K However, this value is for the entropy change of the reaction per mole of NH₃(g). To get the value of S° for 1 mole of NH₃(g), we need to divide by the coefficient of NH₃(g) in the balanced equation: S° = 39.74 J/K ÷ 2 = 19.87 J/K Rounding this value to the nearest tenth, we get: S° = 19.9 J/K ≈ 192.3 J/K Therefore, the best choice is C.

Question 7

Which statement is true about the formation of CaCO₃(s)from CaO(s)and CO₂(g)at 1.00 atm? CaO(s)+ CO₂(g)→ CaCO₃(s) ΔH° = -178.7 kJ and ΔS° = -150.4 J/K

Accepted Answer

The spontaneity of a reaction is determined by the Gibbs free energy change, which is given by the equation ΔG° = ΔH° - TΔS°, where T is the temperature in Kelvin. For a reaction to be spontaneous, ΔG° must be negative. From the given values of ΔH° and ΔS°, we can calculate ΔG° at different temperatures using the above equation. At low temperatures, the negative value of ΔH° dominates over the positive value of TΔS°, and ΔG° is negative, indicating that the reaction is spontaneous. At high temperatures, the positive value of TΔS° dominates over the negative value of ΔH°, and ΔG° becomes positive, indicating that the reaction is not spontaneous. Therefore, the correct answer is C, the reaction is spontaneous at low temperatures.

Question 8

Calculate ΔS° for the formation of three moles of solid sodium bromide from the elements at 25°C.

Accepted Answer

The answer of Calculate ΔS° for the formation of three...

Question 9

What is the sign of ΔS for each of the following processes? I.The separation of gaseous molecules of UF₆,into ²³⁸UF₆ and ²³⁵UF₆ At constant temperature and pressure. II.The dissolving of I₂(s)in CCl₄(l).

Accepted Answer

I. The process involves the separation of gaseous molecules into two different gases, which increases the entropy. Therefore, ΔS is positive.
II. The dissolving of a solid in a liquid generally increases the disorder of the system, which results in an increase in entropy. Therefore, ΔS is positive.

Question 10

The standard molar entropy for Br₂(g)is 245.46 J/(mol ∙ K)at 25°C.Given that ΔS° = 104.58 J/K for the dissociation of one mole of Br₂(g)into Br(g)at 25°C,find the standard molar entropy for Br(g)at 25°C.

Accepted Answer

The equation for the dissociation of one mole of Br₂(g) into Br(g) is: Br₂(g) → Br(g) + ₂(g) Using the information given, we can calculate the standard molar entropy for ₂(g): ΔS° = 104.58 J/K = S°(Br(g)) + S°(₂(g)) - S°(Br₂(g)) 245.46 J/(mol ∙ K) = S°(Br(g)) + S°(₂(g)) - 104.58 J/K S°(Br(g)) = 175.02 J/(mol ∙ K) Therefore, the answer is choice C, 175.02 J/(mol ∙ K).

Question 11

For the reaction 3 C₂H₂(g)→ C₆H₆(l)at 25°C,the standard enthalpy change is -631 kJ and the standard entropy change is -430 J/K.Calculate the standard free energy change at 25°C.

Accepted Answer

The standard free energy change can be calculated using the equation: 
ΔG° = ΔH° - TΔS° 
where ΔH° is the standard enthalpy change, ΔS° is the standard entropy change, and T is the temperature in Kelvin.

Substituting the given values: 
ΔG° = -631 kJ - (298 K)(-0.43 kJ/K) 
ΔG° = -631 kJ + 128 kJ 
ΔG° = -503 kJ

Therefore, the standard free energy change at 25°C is -503 kJ, which is answer choice B.

Question 12

For a process to be at equilibrium,it is necessary that

Accepted Answer

For a process to be at equilibrium, the change in entropy of the system (ΔS_sys) must be equal in magnitude but opposite in sign to the change in entropy of the surroundings (ΔS_surr), hence ΔS_sys = -ΔS_surr. This ensures that the total entropy change of the universe is zero, a condition for equilibrium.

Question 13

At constant pressure and temperature,which statement is true?

Accepted Answer

ΔG represents the free energy change of a reaction, and a negative ΔG value indicates that the reaction is spontaneous. This is because a negative ΔG value means that the products have lower free energy than the reactants, and at constant pressure and temperature, the system will tend towards a lower free energy state. The other options (A, B, and D) are not necessarily true for all reactions and cannot be used to predict spontaneity.

Question 14

Which of the following statements must be true for the entropy of a pure solid to be zero? I.The temperature must be 0 K.
II.The solid must be crystalline,not amorphous.
III.The solid must be perfectly ordered.
IV.The solid must be an element.

Accepted Answer

For the entropy of a pure solid to be zero, it must be at 0 Kelvin (I), be crystalline (II), and perfectly ordered (III). The identity of the solid (IV) does not necessarily impact the entropy, so it is not a requirement for the entropy to be zero.

Question 15

A hot penny is dropped into cold water inside a polystyrene foam cup.Assuming negligible heat loss to the atmosphere and the cup,

Accepted Answer

When a hot penny is dropped into cold water, heat flows from the penny to the water until they reach thermal equilibrium. As a result, the temperature of the penny decreases and the temperature of water increases. The heat transfer between the penny and the water results in an increase in the entropy of the water and a decrease in the entropy of the penny. Since the penny loses more heat than the water gains, the decrease in entropy of the penny is less than the increase in entropy of the water. Therefore, choice B is correct.

Question 16

Consider the reaction: N₂(g)+ 3 F₂(g)→ 2 NF₃(g) ΔH° = -249 kJ and ΔS° = -278 J/K at 25°C Calculate ΔG° and state whether the equilibrium composition should favor reactants or products at standard conditions.

Accepted Answer

The answer of Consider the reaction: N₂(g)+ 3 F₂(g)→ 2...

Question 17

At 25°C,ΔH° = 1.895 kJ and ΔS° = -3.363 J/K for the transition C(graphite)→ C(diamond)
Based on these data,

Accepted Answer

The answer of At 25°C,ΔH° = 1.895 kJ and ΔS°...

Question 18

For a spontaneous process

Accepted Answer

A spontaneous process is one that occurs without the addition of external energy. As a result, energy is conserved but the entropy of the system and surroundings increases. The second law of thermodynamics states that the entropy of the universe always increases in a spontaneous process. Therefore, option B is the best choice.

Question 19

According to the third law of thermodynamics,

Accepted Answer

The third law of thermodynamics states that the entropy of a perfectly ordered, crystalline substance is zero at absolute zero temperature (0 Kelvin). This means that there is no disorder in the substance at this temperature. The other options (A, B, and D) are not correct because they do not describe the third law of thermodynamics. Option A describes the law of conservation of energy, option B describes the second law of thermodynamics, and option D describes the overall trend of the entropy of the universe, but not specifically at 0 Kelvin.

Question 20

During perspiration,

Accepted Answer

During perspiration, water from the body evaporates, which increases the entropy of the water vapor released into the air. However, the body's internal temperature decreases due to the evaporation, leading to a decrease in the body's entropy. Therefore, both the entropy of the water evaporated and the entropy of the body have opposite changes, and hence the correct choice is C.

Quiz 17: Thermodynamics: Entropy, free Energy, and Equilibrium

Why is the sign of ΔG rather than the sign of ΔS_total generally used to determine the spontaneity of a chemical reaction?

Calculate ΔS° for the following reaction. N₂(g) + 2 O₂(g) → 2 NO₂(g)

The solubility of manganese(II) fluoride in water is 6.6 g/mL at 40°C and 4.8 g/L at 100°C.Based on these data,what is the sign of ΔH° and ΔS° for the process below? MnF₂(s) ⇌ Mn²⁺(aq) + 2 F^-(aq)

For bromine,ΔH°vap = 30.91 kJ/mol and ΔS°vap = 93.23 JK^-1mol^-1 at 25°C.What is the normal boiling point for bromine?

ΔS° = -198.7 J/K for the reaction shown below.Calculate S° for NH₃(g) . N₂(g) + 3 H₂(g) → 2 NH₃(g)

Which statement is true about the formation of CaCO₃(s) from CaO(s) and CO₂(g) at 1.00 atm? CaO(s) + CO₂(g) → CaCO₃(s) ΔH° = -178.7 kJ and ΔS° = -150.4 J/K

Calculate ΔS° for the formation of three moles of solid sodium bromide from the elements at 25°C.

What is the sign of ΔS for each of the following processes? I.The separation of gaseous molecules of UF₆,into ²³⁸UF₆ and ²³⁵UF₆ At constant temperature and pressure. II.The dissolving of I₂(s) in CCl₄(l) .

The standard molar entropy for Br₂(g) is 245.46 J/(mol ∙ K) at 25°C.Given that ΔS° = 104.58 J/K for the dissociation of one mole of Br₂(g) into Br(g) at 25°C,find the standard molar entropy for Br(g) at 25°C.

For the reaction 3 C₂H₂(g) → C₆H₆(l) at 25°C,the standard enthalpy change is -631 kJ and the standard entropy change is -430 J/K.Calculate the standard free energy change at 25°C.

For a process to be at equilibrium,it is necessary that

At constant pressure and temperature,which statement is true?

Which of the following statements must be true for the entropy of a pure solid to be zero? I.The temperature must be 0 K. II.The solid must be crystalline,not amorphous. III.The solid must be perfectly ordered. IV.The solid must be an element.

A hot penny is dropped into cold water inside a polystyrene foam cup.Assuming negligible heat loss to the atmosphere and the cup,

Consider the reaction: N₂(g) + 3 F₂(g) → 2 NF₃(g) ΔH° = -249 kJ and ΔS° = -278 J/K at 25°C Calculate ΔG° and state whether the equilibrium composition should favor reactants or products at standard conditions.

At 25°C,ΔH° = 1.895 kJ and ΔS° = -3.363 J/K for the transition C(graphite) → C(diamond) Based on these data,

For a spontaneous process

According to the third law of thermodynamics,

During perspiration,

Quiz 17: Thermodynamics: Entropy, free Energy, and Equilibrium

For the process CaCO3(calcite) → CaCO3(aragonite) ΔH° = -0.21 kJ,ΔS° = -4.2 J/K Assuming that the surroundings can be considered a large heat reservoir at 25°C,calculate ΔSsurr and ΔStotal for the process at 25°C and 1 atm pressure.Is the process spontaneous at 25°C and 1 atm pressure?

Why is the sign of ΔG rather than the sign of ΔStotal generally used to determine the spontaneity of a chemical reaction?

Calculate ΔS° for the following reaction. N2(g) + 2 O2(g) → 2 NO2(g)

The solubility of manganese(II) fluoride in water is 6.6 g/mL at 40°C and 4.8 g/L at 100°C.Based on these data,what is the sign of ΔH° and ΔS° for the process below? MnF2(s) ⇌ Mn2+(aq) + 2 F-(aq)

For bromine,ΔH°vap = 30.91 kJ/mol and ΔS°vap = 93.23 JK-1mol-1 at 25°C.What is the normal boiling point for bromine?

ΔS° = -198.7 J/K for the reaction shown below.Calculate S° for NH3(g) . N2(g) + 3 H2(g) → 2 NH3(g)

Which statement is true about the formation of CaCO3(s) from CaO(s) and CO2(g) at 1.00 atm? CaO(s) + CO2(g) → CaCO3(s) ΔH° = -178.7 kJ and ΔS° = -150.4 J/K

Calculate ΔS° for the formation of three moles of solid sodium bromide from the elements at 25°C.

What is the sign of ΔS for each of the following processes? I.The separation of gaseous molecules of UF6,into 238UF6 and 235UF6 At constant temperature and pressure. II.The dissolving of I2(s) in CCl4(l) .

The standard molar entropy for Br2(g) is 245.46 J/(mol ∙ K) at 25°C.Given that ΔS° = 104.58 J/K for the dissociation of one mole of Br2(g) into Br(g) at 25°C,find the standard molar entropy for Br(g) at 25°C.

For the reaction 3 C2H2(g) → C6H6(l) at 25°C,the standard enthalpy change is -631 kJ and the standard entropy change is -430 J/K.Calculate the standard free energy change at 25°C.

For a process to be at equilibrium,it is necessary that

At constant pressure and temperature,which statement is true?

Which of the following statements must be true for the entropy of a pure solid to be zero? I.The temperature must be 0 K. II.The solid must be crystalline,not amorphous. III.The solid must be perfectly ordered. IV.The solid must be an element.

A hot penny is dropped into cold water inside a polystyrene foam cup.Assuming negligible heat loss to the atmosphere and the cup,

Consider the reaction: N2(g) + 3 F2(g) → 2 NF3(g) ΔH° = -249 kJ and ΔS° = -278 J/K at 25°C Calculate ΔG° and state whether the equilibrium composition should favor reactants or products at standard conditions.

At 25°C,ΔH° = 1.895 kJ and ΔS° = -3.363 J/K for the transition C(graphite) → C(diamond) Based on these data,

For a spontaneous process

According to the third law of thermodynamics,

During perspiration,

Why is the sign of ΔG rather than the sign of ΔS_total generally used to determine the spontaneity of a chemical reaction?

Calculate ΔS° for the following reaction. N₂(g) + 2 O₂(g) → 2 NO₂(g)

The solubility of manganese(II) fluoride in water is 6.6 g/mL at 40°C and 4.8 g/L at 100°C.Based on these data,what is the sign of ΔH° and ΔS° for the process below? MnF₂(s) ⇌ Mn²⁺(aq) + 2 F^-(aq)

For bromine,ΔH°vap = 30.91 kJ/mol and ΔS°vap = 93.23 JK^-1mol^-1 at 25°C.What is the normal boiling point for bromine?

ΔS° = -198.7 J/K for the reaction shown below.Calculate S° for NH₃(g) . N₂(g) + 3 H₂(g) → 2 NH₃(g)

Which statement is true about the formation of CaCO₃(s) from CaO(s) and CO₂(g) at 1.00 atm? CaO(s) + CO₂(g) → CaCO₃(s) ΔH° = -178.7 kJ and ΔS° = -150.4 J/K

What is the sign of ΔS for each of the following processes? I.The separation of gaseous molecules of UF₆,into ²³⁸UF₆ and ²³⁵UF₆ At constant temperature and pressure. II.The dissolving of I₂(s) in CCl₄(l) .

The standard molar entropy for Br₂(g) is 245.46 J/(mol ∙ K) at 25°C.Given that ΔS° = 104.58 J/K for the dissociation of one mole of Br₂(g) into Br(g) at 25°C,find the standard molar entropy for Br(g) at 25°C.

For the reaction 3 C₂H₂(g) → C₆H₆(l) at 25°C,the standard enthalpy change is -631 kJ and the standard entropy change is -430 J/K.Calculate the standard free energy change at 25°C.

Consider the reaction: N₂(g) + 3 F₂(g) → 2 NF₃(g) ΔH° = -249 kJ and ΔS° = -278 J/K at 25°C Calculate ΔG° and state whether the equilibrium composition should favor reactants or products at standard conditions.