Current Attempt in Progress Your answer is partially correct. A 2.03-g sample of benzoic acid (C,H,O₂) is burned in a constant-volume calorimeter that has a heat capacity of 5.93 kJ/°C. The temperature increases from 24.5 °C to 33.5 °C. a) Determine the amount of heat released (in kJ). 53.37 b) Determine the molar energy of combustion of benzoic acid (in kJ/mol). i 8.9 kJ eTextbook and Media kJ/mol
Current Attempt in Progress Your answer is partially correct. A 2.03-g sample of benzoic acid (C,H,O₂) is burned in a constant-volume calorimeter that has a heat capacity of 5.93 kJ/°C. The temperature increases from 24.5 °C to 33.5 °C. a) Determine the amount of heat released (in kJ). 53.37 b) Determine the molar energy of combustion of benzoic acid (in kJ/mol). i 8.9 kJ eTextbook and Media kJ/mol
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter6: Thermochemistry
Section: Chapter Questions
Problem 112AE: In a bomb calorimeter, the reaction vessel is surrounded by water that must be added for each...
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![**Title: Calorimetry Calculation of Benzoic Acid Combustion**
**Experiment Overview:**
A given sample of benzoic acid (C₇H₆O₂), weighing 2.03 grams, is combusted in a constant-volume calorimeter with a heat capacity of 5.93 kJ/°C. The temperature recorded in the calorimeter increases from 24.5 °C to 33.5 °C.
**Tasks:**
a) **Determining the Amount of Heat Released (in kJ):**
To find the amount of heat (q) released, we use the formula:
\[ q = C \times \Delta T \]
Where:
- \( C \) is the heat capacity of the calorimeter.
- \( \Delta T \) is the change in temperature.
Given:
- \( C = 5.93 \) kJ/°C
- \( \Delta T = (33.5 - 24.5) °C = 9 °C \)
So,
\[ q = 5.93 \times 9 = 53.37 \text{ kJ} \]
b) **Determining the Molar Energy of Combustion of Benzoic Acid (in kJ/mol):**
We need to calculate the molar energy of combustion for benzoic acid.
- The molar mass of benzoic acid (C₇H₆O₂) is computed using the periodic table:
\[ (\text{7 Carbon} \times 12.01) + (\text{6 Hydrogen} \times 1.01) + (\text{2 Oxygen} \times 16.00) \]
\[ = 84.07 + 6.06 + 32.00 \]
\[ = 122.13 \text{ g/mol} \]
Using the heat released in the earlier calculation:
Given sample weight = 2.03 g
To find the heat released per mole:
\[ \text{Moles of benzoic acid} = \frac{2.03 \text{ g}}{122.13 \text{ g/mol}} \approx 0.0166 \text{ mol} \]
The molar heat of combustion is then:
\[ \text{Molar energy (kJ/mol)} = \frac{q}{\text{moles of benzo](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F793b6aff-f148-4523-8250-7b38a941d7e6%2Fa9f4abfa-7a9b-45f2-a5d5-0b8c98377266%2Fz8uyrq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Calorimetry Calculation of Benzoic Acid Combustion**
**Experiment Overview:**
A given sample of benzoic acid (C₇H₆O₂), weighing 2.03 grams, is combusted in a constant-volume calorimeter with a heat capacity of 5.93 kJ/°C. The temperature recorded in the calorimeter increases from 24.5 °C to 33.5 °C.
**Tasks:**
a) **Determining the Amount of Heat Released (in kJ):**
To find the amount of heat (q) released, we use the formula:
\[ q = C \times \Delta T \]
Where:
- \( C \) is the heat capacity of the calorimeter.
- \( \Delta T \) is the change in temperature.
Given:
- \( C = 5.93 \) kJ/°C
- \( \Delta T = (33.5 - 24.5) °C = 9 °C \)
So,
\[ q = 5.93 \times 9 = 53.37 \text{ kJ} \]
b) **Determining the Molar Energy of Combustion of Benzoic Acid (in kJ/mol):**
We need to calculate the molar energy of combustion for benzoic acid.
- The molar mass of benzoic acid (C₇H₆O₂) is computed using the periodic table:
\[ (\text{7 Carbon} \times 12.01) + (\text{6 Hydrogen} \times 1.01) + (\text{2 Oxygen} \times 16.00) \]
\[ = 84.07 + 6.06 + 32.00 \]
\[ = 122.13 \text{ g/mol} \]
Using the heat released in the earlier calculation:
Given sample weight = 2.03 g
To find the heat released per mole:
\[ \text{Moles of benzoic acid} = \frac{2.03 \text{ g}}{122.13 \text{ g/mol}} \approx 0.0166 \text{ mol} \]
The molar heat of combustion is then:
\[ \text{Molar energy (kJ/mol)} = \frac{q}{\text{moles of benzo
![**Current Attempt in Progress**
**❌ Incorrect.**
Determine the standard enthalpy change for the following reaction:
\[ 2 \text{NO(g)} + \text{O}_2(\text{g}) \rightarrow 2 \text{NO}_2(\text{g}) \]
\[ \Delta H = 114.14 \text{ kJ} \]
**eTextbook and Media**
[Save for Later]
### Explanation:
The image shows a question from a chemistry problem set asking students to determine the standard enthalpy change for a chemical reaction. The reaction is given as:
\[ 2 \text{NO(g)} + \text{O}_2(\text{g}) \rightarrow 2 \text{NO}_2(\text{g}) \]
The user input the value "114.14 kJ" as the enthalpy change, but the system marked this answer as incorrect.
**Note**: This transcription and analysis are provided for educational purposes to help students understand the type of feedback they might receive when solving enthalpy problems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F793b6aff-f148-4523-8250-7b38a941d7e6%2Fa9f4abfa-7a9b-45f2-a5d5-0b8c98377266%2F8jvqugb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Current Attempt in Progress**
**❌ Incorrect.**
Determine the standard enthalpy change for the following reaction:
\[ 2 \text{NO(g)} + \text{O}_2(\text{g}) \rightarrow 2 \text{NO}_2(\text{g}) \]
\[ \Delta H = 114.14 \text{ kJ} \]
**eTextbook and Media**
[Save for Later]
### Explanation:
The image shows a question from a chemistry problem set asking students to determine the standard enthalpy change for a chemical reaction. The reaction is given as:
\[ 2 \text{NO(g)} + \text{O}_2(\text{g}) \rightarrow 2 \text{NO}_2(\text{g}) \]
The user input the value "114.14 kJ" as the enthalpy change, but the system marked this answer as incorrect.
**Note**: This transcription and analysis are provided for educational purposes to help students understand the type of feedback they might receive when solving enthalpy problems.
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