4. Integrative Practice Problems

Given Standard Heats of Formation (all at 298.15 K):

• ΔH_f°[C₂H₆(g)] = –84.0 kJ/mol
• ΔH_f°[NO₂(g)] = +33.2 kJ/mol
• ΔH_f°[H₂O(l)] = –285.8 kJ/mol
• ΔH_f°[CO₂(g)] = –393.5 kJ/mol
• ΔH_f°[NH₃(g)] = –45.9 kJ/mol
• ΔH_f°[C₂H₅OH(l)] = –277.0 kJ/mol

Calculate ΔH_rxn° for the combustion of ethanol: C₂H₅OH(l) + 3 O₂(g) → 2 CO₂(g) + 3 H₂O(l)

This problem requires using the standard enthalpy of formation values to calculate the enthalpy change for the combustion of ethanol.

1. First, we need to recognize that the enthalpy change for a reaction (ΔH_rxn°) can be calculated as the difference between the total enthalpy of the products and the total enthalpy of the reactants.
2. For the products (2 CO₂ and 3 H₂O), we calculate:
   • 2 CO₂: 2 × (–393.5 kJ) = –787.0 kJ
   • 3 H₂O: 3 × (–285.8 kJ) = –857.4 kJ Total for products = –787.0 + (–857.4) = –1,644.4 kJ.
3. For the reactants, which include ethanol and oxygen, we need:
   • C₂H₅OH: –277.0 kJ (as O₂ has an enthalpy of formation of 0) Total for reactants = –277.0 kJ.
4. Finally, we substitute these values into the equation: ΔH_rxn° = (–1,644.4 kJ) - (–277.0 kJ) = –1,367.4 kJ per mole of ethanol burned.

Think of this process like measuring the energy produced when you burn wood in a fireplace. Just as you would account for how much wood (reactants) you have and how much heat (enthalpy change) is produced by the fire (products), we are calculating how much energy is absorbed or released during a chemical reaction using the values for each material involved.

In a coffee-cup calorimeter, 100.0 mL of 1.00 M H₂SO₄(aq) at 25.0 °C is mixed with 100.0 mL of 1.00 M NaOH(aq) at 25.0 °C. The final temperature of the mixture after complete neutralization is 32.5 °C. Assume the density of the final solution is 1.00 g/mL and its specific heat capacity is 4.18 J/(g·°C).

(a) Write the balanced neutralization reaction. (b) Calculate the heat absorbed by the solution. (c) Calculate the molar enthalpy of neutralization per mole of H₂SO₄ consumed.

This exercise involves a coffee-cup calorimetry experiment, where we calculate the heat exchanged during a neutralization reaction:

1. Balanced Reaction: H₂SO₄ and NaOH react to form Na₂SO₄ and water. The balanced equation will show that sulfuric acid (H₂SO₄) can neutralize two moles of NaOH.
2. Heat Calculation: The heat absorbed by the solution can be calculated using the formula: q_solution = m × c × ΔT, where m is the total mass of the solution, c is the specific heat capacity, and ΔT is the temperature change.
   • You have the initial and final temperatures (25°C to 32.5°C) to find ΔT = 7.5°C. The mass of 200 mL of solution is roughly 200 g. Thus, q_solution = 200 g × 4.18 J/(g·°C) × 7.5 °C = 6,270 J.
3. Molar Enthalpy: The neutralization reaction produced water (2 H₂O), which must be accounted for. With the amount of H₂SO₄ consumed known, we can find the enthalpy per mole. Since 0.050 moles of H₂SO₄ react (as NaOH is limiting), we divide the total heat by the moles, giving ΔH = –6,270 J ÷ 0.050 mol = –125,400 J/mol ≈ –125.4 kJ/mol.

Imagine making a big pot of soup: you mix hot stock (like sulfuric acid) with pasta (sodium hydroxide). As they cook together, the temperature of the soup increases. If we could measure that increase, we could calculate how much heat was generated, similar to how we measure heat in the calorimeter.

A 0.600 g sample of benzoic acid (C₆H₅COOH) is combusted in a bomb calorimeter. The calorimeter’s heat capacity (determined earlier) is 8250 J/°C. The temperature of the calorimeter increases from 22.00 °C to 28.50 °C. Assume the combustion at constant volume yields no change in moles of gas (Δn_gas = 0).

(a) Calculate q_v (heat released by reaction). (b) Calculate ΔE_combustion per mole of benzoic acid. (c) Given ΔH_c° for benzoic acid is –3,261 kJ/mol, compare to your ΔE value and discuss differences.

In this case, we measure the heat released during the combustion of a compound using a bomb calorimeter:

1. Heat Released Calculation (q_v): We find the temperature change (ΔT = 28.50 °C - 22.00 °C = 6.50 °C). The heat released by the reaction (q_v) can be calculated using the formula q_v = – (C_calorimeter × ΔT), where C_calorimeter is the heat capacity. Plugging in the numbers gives q_v = –(8,250 J/°C × 6.50 °C) = –53,625 J.
2. ΔE_combustion: To find the energy change per mole, we first determine how many moles of benzoic acid were combusted. The molar mass (122.12 g/mol) gives us approximately 0.00491 moles. So, ΔE_combustion = q_v = –53,625 J per 0.00491 mol gives about –10,919,000 J/mol or –10,919 kJ/mol.
3. Comparison: This value is significantly larger than the known ΔH_c° = –3,261 kJ/mol, indicating that there may be an inconsistency in the problem's provided values or setup.

Think about using a pressure cooker: when you seal and heat food inside, the temperature rises rapidly because energy gets trapped. Similarly, a bomb calorimeter traps gases during combustion, allowing us to measure the heat released effectively.

Use the following ΔH_f° values (all at 298.15 K):

• ΔH_f°[NH₃(g)] = –45.9 kJ/mol
• ΔH_f°[NO(g)] = +90.3 kJ/mol
• ΔH_f°[H₂O(l)] = –285.8 kJ/mol
• ΔH_f°[N₂(g)] = 0
• ΔH_f°[O₂(g)] = 0

Calculate ΔH_rxn° for: NH₃(g) + O₂(g) → NO(g) + H₂O(l)

In this exercise:

1. We first balance the chemical equation for the reaction of ammonia and oxygen to produce nitric oxide and water. The balanced version is 2 NH₃ + 5 O₂ → 2 NO + 3 H₂O.
2. We can then use the formation enthalpies to calculate the reaction's enthalpy:
   • For products, we use: (2 NO and 3 H₂O).
   • ΔH_f° = (2 × 90.3 kJ) + (3 × –285.8 kJ) = 180.6 – 857.4 = -676.8 kJ (total for products).
   • For reactants, ΔH_f° is (2 × -45.9 kJ) + (5 × 0 kJ) = -91.8 kJ.
3. Finally, we calculate the entire reaction's enthalpy change: ΔH_rxn° = (-676.8 kJ) - (-91.8 kJ) = -676.8 + 91.8 = -585 kJ. This value represents the enthalpy change for the reaction based on the enthalpy of formation values.

Imagine baking an intricate cake where you mix different ingredients, but you also keep account of how much each ingredient will contribute to the final result. Similarly, Hess's law allows you to sum up the energy contributions from various reactants and products to find the overall energy change of the reaction.

Estimate ΔH for the following gas-phase reaction using average bond enthalpies: C₂H₂(g) + 2 H₂(g) → C₂H₆(g)

Use these average bond enthalpies (kJ/mol):

• C≡C (triple bond) = 839
• C–H (single) = 413
• H–H = 436
• C–C (single) = 347

In this estimation:

1. Bonds to Break: For the reactants C₂H₂ and H₂, we need to break:
   • 1 C≡C bond = 839 kJ and
   • 2 H–H bonds = 2 × 436 kJ = 872 kJ. Total energy to break: 839 kJ + 872 kJ = 1711 kJ.
2. Bonds to Form: In C₂H₆, we have
   • 1 C–C bond= 347 kJ and
   • 6 C–H bonds, but 4 were already in the reactant so we add only 2 more, yielding 2 × 413 kJ = 826 kJ. Total energy released: 347 kJ + 826 kJ = 1173 kJ.
3. Estimate ΔH: Using the bond energy estimation equation: ΔH_estimate = Σ D(bonds broken) - Σ D(bonds formed), we have ΔH = 1711 kJ - 1173 kJ = -288 kJ. This indicates the reaction is exothermic.

Think of Hydrogen as a tight knit team needing energy to break through barriers (bonds) created by their tight bond to move forward. The energy input needed to separate them is similar to calculating bond enthalpies. In contrast, when new bonds in the products are formed, energy is released, comparable to the team celebrating their success after reaching their goal.

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