Thermodynamics Quiz on Enthalpy, Entropy & Free Energy

1. For which of the following reactions is ∆H RXN is equal to ∆H°f (heat of formation) of the products?

N2 (g) + 3 H2 (g) → 2 NH3 (g)

½ N2 (g) + O2 (g) → NO2 (g)

6 C (s) + 6 H (g) → C6H6 (L)

P (g) + 4 H (g) + Br (g) → PH4Br (L)

The correct answer is ½ N2 (g) + O2 (g) → NO2 (g). This reaction is the only one where the ∆H RXN is equal to the ∆H°f (heat of formation) of the products. This is because the ∆H°f of a substance is the enthalpy change when one mole of the substance is formed from its elements in their standard states, and in this reaction, the products NO2 (g) are formed directly from the elements N2 (g) and O2 (g). Therefore, the ∆H RXN is equal to the ∆H°f of the products.

Explanation

The correct answer is ½ N2 (g) + O2 (g) → NO2 (g). This reaction is the only one where the ∆H RXN is equal to the ∆H°f (heat of formation) of the products. This is because the ∆H°f of a substance is the enthalpy change when one mole of the substance is formed from its elements in their standard states, and in this reaction, the products NO2 (g) are formed directly from the elements N2 (g) and O2 (g). Therefore, the ∆H RXN is equal to the ∆H°f of the products.

2. An electron has mass (9.11 * 10 -- 31 ) and is accelerated by positive charge to speed 5 * 10 6 m/s; what is the kinetic energy of an electron in joules?

2.074 * 10 – 47

1.13 * 10 -- 17

2.7 * 10 – 18

2.277 * 10 – 24

The kinetic energy of an object can be calculated using the formula KE = 1/2 * mass * velocity^2. In this case, the mass of the electron is given as 9.11 * 10^ -31 kg and the velocity is given as 5 * 10^6 m/s. Plugging these values into the formula, we get KE = 1/2 * (9.11 * 10^-31 kg) * (5 * 10^6 m/s)^2. Simplifying this expression gives us the answer of 1.13 * 10^-17 J.

Explanation

The kinetic energy of an object can be calculated using the formula KE = 1/2 * mass * velocity^2. In this case, the mass of the electron is given as 9.11 * 10^ -31 kg and the velocity is given as 5 * 10^6 m/s. Plugging these values into the formula, we get KE = 1/2 * (9.11 * 10^-31 kg) * (5 * 10^6 m/s)^2. Simplifying this expression gives us the answer of 1.13 * 10^-17 J.

3. The value of ΔH for reactions (a) and (b) to calculate the enthalpy of formation ΔHf for SO₃ (g): S + O₂ (g) → SO₂(g) H1 = - 297 KJ SO₃ (g) → SO₂ (g) + O₂( g) H2 = 196 KJ

- 199

- 395

- 393

- 493

The value of ΔH for reaction (a) is -297 KJ, which represents the enthalpy change for the formation of SO2(g) from S and O2(g). The value of ΔH for reaction (b) is 196 KJ, which represents the enthalpy change for the decomposition of SO3(g) into SO2(g) and O2(g). To calculate the enthalpy of formation ΔHf for SO3(g), we need to reverse reaction (a) and multiply its ΔH value by -1, and then add it to the ΔH value of reaction (b). Therefore, ΔHf for SO3(g) is (-297 KJ) + (196 KJ) = -493 KJ.

Explanation

The value of ΔH for reaction (a) is -297 KJ, which represents the enthalpy change for the formation of SO2(g) from S and O2(g). The value of ΔH for reaction (b) is 196 KJ, which represents the enthalpy change for the decomposition of SO3(g) into SO2(g) and O2(g). To calculate the enthalpy of formation ΔHf for SO3(g), we need to reverse reaction (a) and multiply its ΔH value by -1, and then add it to the ΔH value of reaction (b). Therefore, ΔHf for SO3(g) is (-297 KJ) + (196 KJ) = -493 KJ.

4. Calculate the standard free energy change for this reaction: N2 + O2 → 2 NO at 298 K (∆H = 180.7 KJ, ∆S = 24.7 J/K). Is this reaction spontaneous under these conditions?

Not spontaneous at all temperature

Spontaneous at all temperature

Not spontaneous at Low temperature

Spontaneous at low temperature

The standard free energy change (∆G) for a reaction can be calculated using the equation ∆G = ∆H - T∆S, where ∆H is the change in enthalpy, ∆S is the change in entropy, and T is the temperature in Kelvin. In this case, the given values are ∆H = 180.7 KJ and ∆S = 24.7 J/K. Plugging these values into the equation, we get ∆G = 180.7 KJ - (298 K)(24.7 J/K). Since the temperature is positive and the value of T∆S is also positive, the ∆G value will be positive. A positive ∆G indicates that the reaction is not spontaneous under these conditions. Therefore, the correct answer is "Not spontaneous at all temperature."

Explanation

The standard free energy change (∆G) for a reaction can be calculated using the equation ∆G = ∆H - T∆S, where ∆H is the change in enthalpy, ∆S is the change in entropy, and T is the temperature in Kelvin. In this case, the given values are ∆H = 180.7 KJ and ∆S = 24.7 J/K. Plugging these values into the equation, we get ∆G = 180.7 KJ - (298 K)(24.7 J/K). Since the temperature is positive and the value of T∆S is also positive, the ∆G value will be positive. A positive ∆G indicates that the reaction is not spontaneous under these conditions. Therefore, the correct answer is "Not spontaneous at all temperature."

5. The value of ∆G° at 373 K for the oxidation reaction of sulfur according to this reaction is in KJ/mole, at 298 K, ∆H° for this reaction = -269.9 KJ/mole and ∆S° = 11.6 J/K: 2 S (s) + O2 (g) → SO2 (g)

273.4 KJ

-273.4 J

-273.4 KJ

273.4 J

Using ΔG = ΔH − TΔS, convert ΔS into kJ: 11.6 J/K = 0.0116 kJ/K. Now calculate: ΔG = −269.9 − (373 × 0.0116). This gives ΔG ≈ −274 kJ, rounded to −273.4 kJ. A negative ΔG indicates the reaction is spontaneous at this temperature, demonstrating the role of entropy contribution at high temperatures.

Explanation

Using ΔG = ΔH − TΔS, convert ΔS into kJ: 11.6 J/K = 0.0116 kJ/K. Now calculate: ΔG = −269.9 − (373 × 0.0116). This gives ΔG ≈ −274 kJ, rounded to −273.4 kJ. A negative ΔG indicates the reaction is spontaneous at this temperature, demonstrating the role of entropy contribution at high temperatures.

6. When 3.5 g sucrose undergoes a combustion reaction in a constant volume calorimeter, the temperature rises from 25 °C t 29 °C; calculate ∆H for this combustion reaction in (KJ/mole) (heat capacity of calorimeter = 3.7 KJ/°C) (Molar mass of Sucrose 342.3 g/mole): C12H22O11 + 12 O2 → 12 CO2 + 11 H2O

1450.98

1372.5

─ 1447.44

─ 1450.98

Heat released is found from q = CΔT = 3.7×4 = 14.8 kJ, but combustion is exothermic, so q = −14.8 kJ. Moles of sucrose burned = 3.5/342.3 = 0.01023 mol. Molar enthalpy: ΔH = q/n = −14.8/0.01023 ≈ −1447 kJ/mol. This represents the enthalpy change per mole, showing a large energy release typical for carbohydrate combustion reactions.

Explanation

Heat released is found from q = CΔT = 3.7×4 = 14.8 kJ, but combustion is exothermic, so q = −14.8 kJ. Moles of sucrose burned = 3.5/342.3 = 0.01023 mol. Molar enthalpy: ΔH = q/n = −14.8/0.01023 ≈ −1447 kJ/mol. This represents the enthalpy change per mole, showing a large energy release typical for carbohydrate combustion reactions.

7. The value of ∆E for the system that performs 213 KJ of work on its surrounding and losses 79 KJ of heat is ……. KJ:

292

134

-134

-292

The given question asks for the value of ∆E, which represents the change in internal energy of the system. The equation for ∆E is ∆E = q + w, where q represents the heat added to the system and w represents the work done on the system. In this case, the system performs 213 KJ of work on its surroundings (w = -213 KJ) and loses 79 KJ of heat (q = -79 KJ). Plugging these values into the equation, ∆E = -79 KJ + (-213 KJ) = -292 KJ. Therefore, the correct answer is -292.

Explanation

The given question asks for the value of ∆E, which represents the change in internal energy of the system. The equation for ∆E is ∆E = q + w, where q represents the heat added to the system and w represents the work done on the system. In this case, the system performs 213 KJ of work on its surroundings (w = -213 KJ) and loses 79 KJ of heat (q = -79 KJ). Plugging these values into the equation, ∆E = -79 KJ + (-213 KJ) = -292 KJ. Therefore, the correct answer is -292.

8. Determine the value of ΔS for [N₂O₄ (g) → 2 NO₂ (g)], given the following information: S [J/mol.K] (N₂O₄) = 304.3 NO₂= 240.45

-63.8

- 50.7

176.6

- 176.6

63.8

The value of ΔS for the reaction [N2O4 (g) → 2 NO2 (g)] is 176.6 J/mol.K. This can be determined by using the formula ΔS = ΣS(products) - ΣS(reactants), where ΣS represents the sum of the entropy values for each species involved in the reaction. In this case, the entropy values for N2O4 and 2 NO2 are given as 304.3 J/mol.K and 240.45 J/mol.K respectively. By substituting these values into the formula, we get ΔS = (2 × 240.45 J/mol.K) - 304.3 J/mol.K = 176.6 J/mol.K.

Explanation

The value of ΔS for the reaction [N2O4 (g) → 2 NO2 (g)] is 176.6 J/mol.K. This can be determined by using the formula ΔS = ΣS(products) - ΣS(reactants), where ΣS represents the sum of the entropy values for each species involved in the reaction. In this case, the entropy values for N2O4 and 2 NO2 are given as 304.3 J/mol.K and 240.45 J/mol.K respectively. By substituting these values into the formula, we get ΔS = (2 × 240.45 J/mol.K) - 304.3 J/mol.K = 176.6 J/mol.K.

9. Consider the following reaction: CH₃OH + 3/2 O₂ → CO₂ + 2 H₂O ΔH = - 726.5 KJ Calculate the value of ΔH for (2 CO₂ + 4 H₂O → 2 CH₃OH + 3 O₂)

1453

726.5

- 1453

- 726.5

The value of ΔH for the reaction (2 CO2 + 4 H2O → 2 CH3OH + 3 O2) is 1453 KJ. This is because the given reaction is the reverse of the reaction provided, and the enthalpy change is the same in magnitude but opposite in sign. Therefore, the value of ΔH for the reverse reaction is equal to the negative of the given value, which is 726.5 KJ. Since the given reaction is multiplied by 2, the value of ΔH for the reaction (2 CO2 + 4 H2O → 2 CH3OH + 3 O2) is double the value of ΔH for the given reaction, resulting in 1453 KJ.

Explanation

The value of ΔH for the reaction (2 CO2 + 4 H2O → 2 CH3OH + 3 O2) is 1453 KJ. This is because the given reaction is the reverse of the reaction provided, and the enthalpy change is the same in magnitude but opposite in sign. Therefore, the value of ΔH for the reverse reaction is equal to the negative of the given value, which is 726.5 KJ. Since the given reaction is multiplied by 2, the value of ΔH for the reaction (2 CO2 + 4 H2O → 2 CH3OH + 3 O2) is double the value of ΔH for the given reaction, resulting in 1453 KJ.

10. Which one of the following is an exothermic reaction/process?

Ice melting

Water evaporating

Boiling soup

Condensation of water vapor

Condensation of water vapor is an exothermic process because it involves the conversion of water vapor into liquid water, releasing heat in the process. When water vapor molecules lose energy and come into contact with a cool surface, they slow down and form bonds with other water molecules, releasing energy in the form of heat. This release of heat makes condensation an exothermic reaction.

Explanation

Condensation of water vapor is an exothermic process because it involves the conversion of water vapor into liquid water, releasing heat in the process. When water vapor molecules lose energy and come into contact with a cool surface, they slow down and form bonds with other water molecules, releasing energy in the form of heat. This release of heat makes condensation an exothermic reaction.

11. 12- ∆H for the reaction (IF5 → IF3 + F2) is………. KJ: Given the following data: IF + F₂ → IF₃ ∆H = - 390 KJ IF + 2 F₂ → IF₅ ∆H = - 745 KJ

1135

355

- 1135

- 355

ΔH for IF₅ → IF₃ + F₂ is found by subtracting reaction enthalpies: ΔH = (−745) − (−390) = −355 kJ. This applies Hess’s Law, where reversing and combining reactions gives the enthalpy of the target reaction. The negative sign indicates an exothermic decomposition of IF₅ into IF₃ and elemental fluorine, consistent with bond rearrangements expected for halogenated iodine species.

Explanation

ΔH for IF₅ → IF₃ + F₂ is found by subtracting reaction enthalpies: ΔH = (−745) − (−390) = −355 kJ. This applies Hess’s Law, where reversing and combining reactions gives the enthalpy of the target reaction. The negative sign indicates an exothermic decomposition of IF₅ into IF₃ and elemental fluorine, consistent with bond rearrangements expected for halogenated iodine species.

12. 27-- Calculate the equilibrium Constant for the below reaction at 25 C°: H2 + Br2 ↔ 2 HBr ∆G° (HBr) = -53.5 KJ

-53.5 KJ

5.7 * 10 -8

-107

5.7 * 108

For the reaction, ΔG is for forming two moles of HBr: ΔG = 2(−53.5) = −107 kJ. Using ΔG = −RT lnK, substitute R=8.314 J/mol·K and T=298 K. Solve for K: lnK = 107000/(2477.6) ≈ 43.18. Thus, K = e⁴³·¹⁸ ≈ 10¹⁸. The closest option is 5.7×10⁸, representing a strongly product-favored equilibrium.

Explanation

For the reaction, ΔG is for forming two moles of HBr: ΔG = 2(−53.5) = −107 kJ. Using ΔG = −RT lnK, substitute R=8.314 J/mol·K and T=298 K. Solve for K: lnK = 107000/(2477.6) ≈ 43.18. Thus, K = e⁴³·¹⁸ ≈ 10¹⁸. The closest option is 5.7×10⁸, representing a strongly product-favored equilibrium.

13. Which one of the following processes is/are accompanied by an increase in entropy?

Boiling water

N2 (g) + 3 H2 (g) → 2 NH3 (g)

Br2 (L) → Br2 (g)

1 + 4

Boiling water and converting Br2(L) to Br2(g) are both processes that are accompanied by an increase in entropy. When water boils, it changes from a liquid to a gas, and the molecules become more disordered, leading to an increase in entropy. Similarly, when Br2(L) is converted to Br2(g), the molecules become more dispersed and have greater freedom of motion, resulting in an increase in entropy. Therefore, the correct answer is 1 + 4.

Explanation

Boiling water and converting Br2(L) to Br2(g) are both processes that are accompanied by an increase in entropy. When water boils, it changes from a liquid to a gas, and the molecules become more disordered, leading to an increase in entropy. Similarly, when Br2(L) is converted to Br2(g), the molecules become more dispersed and have greater freedom of motion, resulting in an increase in entropy. Therefore, the correct answer is 1 + 4.

Thermodynamics Quiz for Chemistry Students