4 - Integrative Practice Problems
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Enthalpy of Reaction via Formation Values
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In today's session, we're going to calculate the standard enthalpy change for the combustion of ethanol. Does anyone remember what ΔH_rxn° represents?
Isn't it the change in enthalpy for a particular reaction?
Exactly! Now, to calculate this, we use the formula ΔH_rxn° = Σ ΔH_f°(products) - Σ ΔH_f°(reactants). Can anyone tell me the values of ΔH_f° for carbon dioxide and liquid water?
ΔH_f° for CO₂ is about -393.5 kJ/mol and for H₂O it's about -285.8 kJ/mol.
Correct! Now let’s say our reaction is C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O. How do we calculate the enthalpy using those values?
We multiply those values by their stoichiometric coefficients.
Right! We calculate the enthalpy of formation for the products first. What do we get?
I think it would be -787.0 kJ for CO₂ plus -857.4 kJ for H₂O, giving us a total of -1644.4 kJ.
Perfect! Now can you do the same for the reactants and calculate the overall ΔH_rxn°?
For the reactants, it's just -277.0 kJ since O₂ has a ΔH_f° of 0. So ΔH_rxn° would be -1644.4 - (-277.0) = -1367.4 kJ/mol.
Excellent work! This shows how you can derive energy changes from formation values.
Coffee-Cup Calorimetry
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Today, we will focus on coffee-cup calorimetry. How do we measure the heat exchanged during a reaction in this setup?
We look at the temperature change of the solution, right? And then calculate the heat absorbed?
Yes, exactly! The heat is calculated using q_solution = m × c × ΔT. Can anyone recall what each variable represents?
m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature.
Great! Now, if we mixed 100 mL of H₂SO₄ with 100 mL of NaOH, both at 25°C, and the temperature rises to 32.5°C, how do we start?
First, we would calculate the total mass of the solution, which would be 200 grams.
Correct! And what's the heat absorbed?
Using the equation, it would be q_solution = 200 g × 4.18 J/g°C × (32.5 - 25)°C, which gives us around 6270 J.
Excellent! And because it’s the heat absorbed by the solution, what’s the enthalpy change for the reaction?
It would be negative because it’s exothermic, so q_reaction = -6270 J.
Yes! Now we can calculate the molar enthalpy of neutralization. If we only used 0.050 moles of H₂SO₄, how would that look?
ΔH would be -6270 J / 0.050 mol = -125400 J/mol, or about -125.4 kJ/mol.
Excellent! That shows the importance of calorimetry in determining reaction energies.
Bomb Calorimetry
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Now let's shift our focus to bomb calorimetry. Who can explain why we use a bomb calorimeter for combustion?
It’s designed to withstand high pressures and measures heat at constant volume?
Correct! By doing so, we can directly calculate ΔE. How do we calculate q_v?
We use q_v = - (C_cal × ΔT), right?
Yes! If the heat capacity of our calorimeter is 8250 J/°C and the temperature increases by 6.5°C, what does that yield?
q_v = - (8250 J/°C × 6.5°C) = -53,625 J.
Excellent job! Now, if we combusted 0.600 g of benzoic acid, how do we find ΔE for one mole?
First, we need to find the moles. That would be 0.600 g divided by its molar mass, which is about 122.12 g/mol.
Exactly! And what do we do next?
We would then divide -53,625 J by the number of moles to get ΔE per mole.
Perfect! How does ΔE compared to ΔH?
If there's a change in moles of gas, we'd need to use ΔH = ΔE + Δ(n_gas) × R × T to find the enthalpy change.
Well done! This helps underline how calorimetry can aid in calculating energy changes.
Hess’s Law
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Finally, let's discuss Hess’s Law. Who can remind us what Hess's law states?
It states that the total enthalpy change is the same regardless of the path taken?
Exactly! This principle is useful for calculating ΔH for complex reactions using known enthalpy changes of simpler steps. Let’s consider an example: How would we calculate ΔH for the combustion of methane?
We can break it down into formation reactions and use the formation enthalpies?
Correct! If we have formation reactions for CO₂ and H₂O, what do we do with those enthalpies?
We would sum the enthalpies of formation for products and subtract the sum for the reactants.
Yes! If we have 2 CH₄ + 4 O₂ → 2 CO₂ + 4 H₂O, calculate the overall ΔH.
Using Hess's law, we would find the enthalpies and apply the formula ΔH_rxn° = Σ ΔH_f°(products) - Σ ΔH_f°(reactants).
Excellent! This illustrates how Hess’s Law simplifies our ability to calculate enthalpy changes.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
Integrative Practice Problems includes exercises that require applying theoretical knowledge of thermochemistry and enthalpy changes in various contexts. It encourages two key learning methods: solving problems directly related to the theories and concepts discussed in previous sections and applying them in experimental conditions to reinforce understanding.
Detailed
Integrative Practice Problems
This section provides a comprehensive array of practice problems designed to examine and reinforce understanding of the major concepts surrounding thermochemistry, such as enthalpy changes, calorimetry, and Hess's law. Each problem encourages the application of fundamental principles described in earlier sections of this unit, emphasizing hands-on problem-solving and critical thinking. The problems range in difficulty and demand that students apply theoretical framework in practical situations and perform calculations based on standard enthalpies of formation, neutralization reactions, and bond enthalpies.
Key Topics Covered:
- Enthalpy of Reaction via Formation Values: Calculate the standard enthalpy change associated with chemical reactions, leveraging standard heats of formation.
- Coffee-Cup Calorimetry: Engage with practical aspects of measuring heat absorption during reactions, alongside theoretical calculations.
- Bomb Calorimetry: Assess combustion reactions in a calorimeter setup, comparing ΔE and ΔH values.
- Hess’s Law: Utilize enthalpy data to determine the total enthalpy change for complex reactions based on known values.
- Bond Enthalpy Estimations: Estimate overall enthalpy changes by analyzing bonds formed and broken during chemical reactions, highlighting the estimates' reliability and potential discrepancies from experimental values.
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Problem 1: Enthalpy of Reaction via Formation Values
Chapter 1 of 5
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Chapter Content
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)
Detailed Explanation
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.
Examples & Analogies
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.
Problem 2: Coffee-Cup Calorimetry
Chapter 2 of 5
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Problem 3: Bomb Calorimetry
Chapter 3 of 5
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Chapter Content
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.
Detailed Explanation
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.
Examples & Analogies
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.
Problem 4: Hess’s Law with Formation Enthalpies
Chapter 4 of 5
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Chapter Content
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)
Detailed Explanation
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.
Examples & Analogies
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.
Problem 5: Bond Enthalpies Estimate
Chapter 5 of 5
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Chapter Content
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
Detailed Explanation
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.
Examples & Analogies
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.
Key Concepts
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Enthalpy: A measure of heat content in chemical reactions.
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Standard Enthalpy of Formation: Defines the energy change when forming one mole of a substance from its elements.
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Calorimetry: A method for measuring heat transfer in chemical processes.
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Hess’s Law: The total enthalpy change for a reaction is the same regardless of the path taken.
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Bond Enthalpy: Energy required to break chemical bonds, indicative of molecular stability.
Examples & Applications
Calculating ΔH_rxn° using formation values of products and reactants in the combustion of ethanol.
Using the coffee-cup calorimeter to find the temperature change during the neutralization of H₂SO₄ and NaOH.
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
Calorimetry, oh what a breeze, measure the heat with utmost ease!
Stories
Imagine you're in a lab mixing two reactants. As they react, the calorimeter observes the heat change; this indicates whether the reaction is absorbing or releasing energy!
Memory Tools
To remember Hess’s Law: H.E.L.P. - Hess’s Enthalpy is Lawfully Path-independent.
Acronyms
C.E.R.T. - Calorimeter, Energy, Reaction, Temperature.
Flash Cards
Glossary
- Enthalpy (ΔH)
A measure of heat content in a system, especially during chemical reactions.
- Standard Enthalpy of Formation (ΔH_f°)
The change in enthalpy when one mole of a compound is formed from its elements in their standard states.
- Calorimeter
A device used for measuring the heat change in a chemical reaction.
- Hess’s Law
The principle that the total enthalpy change for a reaction is the same regardless of the number of steps taken.
- Bond Enthalpy
The energy required to break one mole of a bond in a gaseous substance.
Reference links
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