Interactive Audio Lesson
Listen to a student-teacher conversation explaining the topic in a relatable way.
Enthalpy of Reaction via Formation Values
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
In today's session, we're going to calculate the standard enthalpy change for the combustion of ethanol. Does anyone remember what ฮH_rxnยฐ represents?
Isn't it the change in enthalpy for a particular reaction?
Exactly! Now, to calculate this, we use the formula ฮH_rxnยฐ = ฮฃ ฮH_fยฐ(products) - ฮฃ ฮH_fยฐ(reactants). Can anyone tell me the values of ฮH_fยฐ for carbon dioxide and liquid water?
ฮH_fยฐ for COโ is about -393.5 kJ/mol and for HโO it's about -285.8 kJ/mol.
Correct! Now letโs say our reaction is CโHโ OH + 3 Oโ โ 2 COโ + 3 HโO. 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 COโ plus -857.4 kJ for HโO, giving us a total of -1644.4 kJ.
Perfect! Now can you do the same for the reactants and calculate the overall ฮH_rxnยฐ?
For the reactants, it's just -277.0 kJ since Oโ has a ฮH_fยฐ of 0. So ฮH_rxnยฐ 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
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
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 ร c ร ฮT. Can anyone recall what each variable represents?
m is the mass of the solution, c is the specific heat capacity, and ฮT is the change in temperature.
Great! Now, if we mixed 100 mL of HโSOโ with 100 mL of NaOH, both at 25ยฐC, and the temperature rises to 32.5ยฐC, 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 ร 4.18 J/gยฐC ร (32.5 - 25)ยฐC, which gives us around 6270 J.
Excellent! And because itโs the heat absorbed by the solution, whatโs the enthalpy change for the reaction?
It would be negative because itโs exothermic, so q_reaction = -6270 J.
Yes! Now we can calculate the molar enthalpy of neutralization. If we only used 0.050 moles of HโSOโ, how would that look?
ฮH 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
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
Now let's shift our focus to bomb calorimetry. Who can explain why we use a bomb calorimeter for combustion?
Itโs designed to withstand high pressures and measures heat at constant volume?
Correct! By doing so, we can directly calculate ฮE. How do we calculate q_v?
We use q_v = - (C_cal ร ฮT), right?
Yes! If the heat capacity of our calorimeter is 8250 J/ยฐC and the temperature increases by 6.5ยฐC, what does that yield?
q_v = - (8250 J/ยฐC ร 6.5ยฐC) = -53,625 J.
Excellent job! Now, if we combusted 0.600 g of benzoic acid, how do we find ฮE 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 ฮE per mole.
Perfect! How does ฮE compared to ฮH?
If there's a change in moles of gas, we'd need to use ฮH = ฮE + ฮ(n_gas) ร R ร T to find the enthalpy change.
Well done! This helps underline how calorimetry can aid in calculating energy changes.
Hessโs Law
Unlock Audio Lesson
Signup and Enroll to the course for listening the Audio Lesson
Finally, let's discuss Hessโs 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 ฮH for complex reactions using known enthalpy changes of simpler steps. Letโs consider an example: How would we calculate ฮH 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 COโ and HโO, 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 CHโ + 4 Oโ โ 2 COโ + 4 HโO, calculate the overall ฮH.
Using Hess's law, we would find the enthalpies and apply the formula ฮH_rxnยฐ = ฮฃ ฮH_fยฐ(products) - ฮฃ ฮH_fยฐ(reactants).
Excellent! This illustrates how Hessโs Law simplifies our ability to calculate enthalpy changes.
Introduction & Overview
Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.
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 ฮE and ฮH values.
- Hessโs 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.
Audio Book
Dive deep into the subject with an immersive audiobook experience.
Problem 1: Enthalpy of Reaction via Formation Values
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
Given Standard Heats of Formation (all at 298.15 K):
- ฮH_fยฐ[CโHโ(g)] = โ84.0 kJ/mol
- ฮH_fยฐ[NOโ(g)] = +33.2 kJ/mol
- ฮH_fยฐ[HโO(l)] = โ285.8 kJ/mol
- ฮH_fยฐ[COโ(g)] = โ393.5 kJ/mol
- ฮH_fยฐ[NHโ(g)] = โ45.9 kJ/mol
- ฮH_fยฐ[CโHโ
OH(l)] = โ277.0 kJ/mol
Calculate ฮH_rxnยฐ for the combustion of ethanol:
CโHโ
OH(l) + 3 Oโ(g) โ 2 COโ(g) + 3 HโO(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 (ฮH_rxnยฐ) 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 COโ and 3 HโO), we calculate:
- 2 COโ: 2 ร (โ393.5 kJ) = โ787.0 kJ
- 3 HโO: 3 ร (โ285.8 kJ) = โ857.4 kJ
Total for products = โ787.0 + (โ857.4) = โ1,644.4 kJ.
3. For the reactants, which include ethanol and oxygen, we need:
- CโHโ
OH: โ277.0 kJ (as Oโ has an enthalpy of formation of 0)
Total for reactants = โ277.0 kJ.
4. Finally, we substitute these values into the equation:
ฮH_rxnยฐ = (โ1,644.4 kJ) - (โ277.0 kJ) = โ1,367.4 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
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
In a coffee-cup calorimeter, 100.0 mL of 1.00 M HโSOโ(aq) at 25.0 ยฐC is mixed with 100.0 mL of 1.00 M NaOH(aq) at 25.0 ยฐC. The final temperature of the mixture after complete neutralization is 32.5 ยฐC. Assume the density of the final solution is 1.00 g/mL and its specific heat capacity is 4.18 J/(gยทยฐC).
(a) Write the balanced neutralization reaction.
(b) Calculate the heat absorbed by the solution.
(c) Calculate the molar enthalpy of neutralization per mole of HโSOโ consumed.
Detailed Explanation
This exercise involves a coffee-cup calorimetry experiment, where we calculate the heat exchanged during a neutralization reaction:
1. Balanced Reaction: HโSOโ and NaOH react to form NaโSOโ and water. The balanced equation will show that sulfuric acid (HโSOโ) can neutralize two moles of NaOH.
2. Heat Calculation: The heat absorbed by the solution can be calculated using the formula: q_solution = m ร c ร ฮT, where
m is the total mass of the solution, c is the specific heat capacity, and ฮT is the temperature change.
- You have the initial and final temperatures (25ยฐC to 32.5ยฐC) to find ฮT = 7.5ยฐC. The mass of 200 mL of solution is roughly 200 g. Thus, q_solution = 200 g ร 4.18 J/(gยทยฐC) ร 7.5 ยฐC = 6,270 J.
3. Molar Enthalpy: The neutralization reaction produced water (2 HโO), which must be accounted for. With the amount of HโSOโ consumed known, we can find the enthalpy per mole. Since 0.050 moles of HโSOโ react (as NaOH is limiting), we divide the total heat by the moles, giving ฮH = โ6,270 J รท 0.050 mol = โ125,400 J/mol โ โ125.4 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
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
A 0.600 g sample of benzoic acid (CโHโ COOH) is combusted in a bomb calorimeter. The calorimeterโs heat capacity (determined earlier) is 8250 J/ยฐC. The temperature of the calorimeter increases from 22.00 ยฐC to 28.50 ยฐC. Assume the combustion at constant volume yields no change in moles of gas (ฮn_gas = 0).
(a) Calculate q_v (heat released by reaction).
(b) Calculate ฮE_combustion per mole of benzoic acid.
(c) Given ฮH_cยฐ for benzoic acid is โ3,261 kJ/mol, compare to your ฮE 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 (ฮT = 28.50 ยฐC - 22.00 ยฐC = 6.50 ยฐC). The heat released by the reaction (q_v) can be calculated using the formula q_v = โ (C_calorimeter ร ฮT), where C_calorimeter is the heat capacity. Plugging in the numbers gives q_v = โ(8,250 J/ยฐC ร 6.50 ยฐC) = โ53,625 J.
2. ฮE_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, ฮE_combustion = q_v = โ53,625 J per 0.00491 mol gives about โ10,919,000 J/mol or โ10,919 kJ/mol.
3. Comparison: This value is significantly larger than the known ฮH_cยฐ = โ3,261 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: Hessโs Law with Formation Enthalpies
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
Use the following ฮH_fยฐ values (all at 298.15 K):
- ฮH_fยฐ[NHโ(g)] = โ45.9 kJ/mol
- ฮH_fยฐ[NO(g)] = +90.3 kJ/mol
- ฮH_fยฐ[HโO(l)] = โ285.8 kJ/mol
- ฮH_fยฐ[Nโ(g)] = 0
- ฮH_fยฐ[Oโ(g)] = 0
Calculate ฮH_rxnยฐ for:
NHโ(g) + Oโ(g) โ NO(g) + HโO(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 NHโ + 5 Oโ โ 2 NO + 3 HโO.
2. We can then use the formation enthalpies to calculate the reaction's enthalpy:
- For products, we use: (2 NO and 3 HโO).
- ฮH_fยฐ = (2 ร 90.3 kJ) + (3 ร โ285.8 kJ) = 180.6 โ 857.4 = -676.8 kJ (total for products).
- For reactants, ฮH_fยฐ is (2 ร -45.9 kJ) + (5 ร 0 kJ) = -91.8 kJ.
3. Finally, we calculate the entire reaction's enthalpy change:
ฮH_rxnยฐ = (-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
Unlock Audio Book
Signup and Enroll to the course for listening the Audio Book
Estimate ฮH for the following gas-phase reaction using average bond enthalpies:
CโHโ(g) + 2 Hโ(g) โ CโHโ(g)
Use these average bond enthalpies (kJ/mol):
- CโกC (triple bond) = 839
- CโH (single) = 413
- HโH = 436
- CโC (single) = 347
Detailed Explanation
In this estimation:
1. Bonds to Break: For the reactants CโHโ and Hโ, we need to break:
- 1 CโกC bond = 839 kJ and
- 2 HโH bonds = 2 ร 436 kJ = 872 kJ.
Total energy to break: 839 kJ + 872 kJ = 1711 kJ.
2. Bonds to Form: In CโHโ, we have
- 1 CโC bond= 347 kJ and
- 6 CโH bonds, but 4 were already in the reactant so we add only 2 more, yielding 2 ร 413 kJ = 826 kJ.
Total energy released: 347 kJ + 826 kJ = 1173 kJ.
3. Estimate ฮH: Using the bond energy estimation equation:
ฮH_estimate = ฮฃ D(bonds broken) - ฮฃ D(bonds formed), we have
ฮH = 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.
Definitions & Key Concepts
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
-
Enthalpy: A measure of heat content in chemical reactions.
-
Standard Enthalpy of Formation: Defines the energy change when forming one mole of a substance from its elements.
-
Calorimetry: A method for measuring heat transfer in chemical processes.
-
Hessโs Law: The total enthalpy change for a reaction is the same regardless of the path taken.
-
Bond Enthalpy: Energy required to break chemical bonds, indicative of molecular stability.
Examples & Real-Life Applications
See how the concepts apply in real-world scenarios to understand their practical implications.
Examples
-
Calculating ฮH_rxnยฐ 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 HโSOโ and NaOH.
Memory Aids
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
๐ต Rhymes Time
-
Calorimetry, oh what a breeze, measure the heat with utmost ease!
๐ Fascinating 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!
๐ง Other Memory Gems
-
To remember Hessโs Law: H.E.L.P. - Hessโs Enthalpy is Lawfully Path-independent.
๐ฏ Super Acronyms
C.E.R.T. - Calorimeter, Energy, Reaction, Temperature.
Flash Cards
Review key concepts with flashcards.
Glossary of Terms
Review the Definitions for terms.
-
Term: Enthalpy (ฮH)
Definition:
A measure of heat content in a system, especially during chemical reactions.
-
Term: Standard Enthalpy of Formation (ฮH_fยฐ)
Definition:
The change in enthalpy when one mole of a compound is formed from its elements in their standard states.
-
Term: Calorimeter
Definition:
A device used for measuring the heat change in a chemical reaction.
-
Term: Hessโs Law
Definition:
The principle that the total enthalpy change for a reaction is the same regardless of the number of steps taken.
-
Term: Bond Enthalpy
Definition:
The energy required to break one mole of a bond in a gaseous substance.