13 temperature change for hot water: temperature change for cool water: heat leaving hot water: 2717" J heat absorbed by cool water: 3553 4. heat capacity of calorimeter: 49.17 2717 J-3553 17°C (40°C-206) = ! change of temperature: 0°C heat emitted during reaction: heat of neutralization, AH₁: HCI/NaOH
13 temperature change for hot water: temperature change for cool water: heat leaving hot water: 2717" J heat absorbed by cool water: 3553 4. heat capacity of calorimeter: 49.17 2717 J-3553 17°C (40°C-206) = ! change of temperature: 0°C heat emitted during reaction: heat of neutralization, AH₁: HCI/NaOH
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
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
Question
![**Calorimetry and Heat of Neutralization**
**Determining Heat Capacity of a Calorimeter:**
\[ \text{heat capacity of calorimeter} = \frac{\left( \text{heat leaving hot water} - \text{heat absorbed by cool water} \right)}{\text{temperature change for cool water}} \]
- Record this value on the RESULTS sheet. The units will be J/°C.
---
**Heat of Neutralization for HCl/NaOH:**
1. **Change in Temperature:**
\[ \text{change in temperature} = (\text{temperature after mixing}) - (\text{temperature before mixing}) \]
2. **Heat Emitted During Reaction (in kJ):**
\[ \text{heat emitted during reaction} = (4.184\ \text{J/(g} \cdot °\text{C)}) \times (100.\ \text{g}) \times (\text{change in temperature}) \times \left(\frac{1\ \text{kJ}}{1000\ \text{J}}\right) \]
3. **Heat Lost to Calorimeter (in kJ):**
\[ \text{heat lost to calorimeter} = (\text{heat capacity of calorimeter}) \times (\text{change in temperature}) \times \left(\frac{1\ \text{kJ}}{1000\ \text{J}}\right) \]
4. **Heat of Neutralization (\( \Delta H_n \)):**
\[ \Delta H_n = \frac{(\text{heat emitted during reaction} + \text{heat lost to calorimeter})}{(0.150\ \text{mol})} \]
*Note the negative sign. This is necessary because it is an exothermic reaction.*
---
Record the change in temperature, heat emitted during reaction, heat lost to calorimeter, and the heat of neutralization as specified for complete understanding and calculation verification.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18395f57-75e3-4bdb-980a-ed8850989594%2Fce45ab74-fa44-458d-9651-9cafc185f999%2F8u6fktg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Calorimetry and Heat of Neutralization**
**Determining Heat Capacity of a Calorimeter:**
\[ \text{heat capacity of calorimeter} = \frac{\left( \text{heat leaving hot water} - \text{heat absorbed by cool water} \right)}{\text{temperature change for cool water}} \]
- Record this value on the RESULTS sheet. The units will be J/°C.
---
**Heat of Neutralization for HCl/NaOH:**
1. **Change in Temperature:**
\[ \text{change in temperature} = (\text{temperature after mixing}) - (\text{temperature before mixing}) \]
2. **Heat Emitted During Reaction (in kJ):**
\[ \text{heat emitted during reaction} = (4.184\ \text{J/(g} \cdot °\text{C)}) \times (100.\ \text{g}) \times (\text{change in temperature}) \times \left(\frac{1\ \text{kJ}}{1000\ \text{J}}\right) \]
3. **Heat Lost to Calorimeter (in kJ):**
\[ \text{heat lost to calorimeter} = (\text{heat capacity of calorimeter}) \times (\text{change in temperature}) \times \left(\frac{1\ \text{kJ}}{1000\ \text{J}}\right) \]
4. **Heat of Neutralization (\( \Delta H_n \)):**
\[ \Delta H_n = \frac{(\text{heat emitted during reaction} + \text{heat lost to calorimeter})}{(0.150\ \text{mol})} \]
*Note the negative sign. This is necessary because it is an exothermic reaction.*
---
Record the change in temperature, heat emitted during reaction, heat lost to calorimeter, and the heat of neutralization as specified for complete understanding and calculation verification.
![**RESULTS**
*Make sure to use correct significant figures and units of measure.*
**HOT WATER/COOL WATER**
- Temperature change for hot water: \(13^\circ C\)
- Temperature change for cool water: \(17^\circ C\)
Calculations:
- Heat leaving hot water: \(2717 \, J\)
- Heat absorbed by cool water: \(3553 \, J\)
Equation:
\[
(4.18 \, \text{J/g}^\circ \text{C}) \times (50.0 \, \text{g}) \times (37^\circ C - 50^\circ C) = 3553 \, J
\]
- Heat capacity of calorimeter: \(49.17 \, J\)
Equation:
\[
(2717 \, J - 3553 \, J) = 49.17 \, J
\]
**HCl/NaOH**
- Temperature change: \(17^\circ C\)
- Change of temperature:
\[
(40^\circ C - 20^\circ C) = 20^\circ C
\]
- Heat emitted during reaction: _____
- Heat of neutralization, \(\Delta H\): _____
\[ \text{Note:} \, \text{The equations and calculations involve changes in temperature and heat transferred during mixing of hot and cool water, and an acid-base reaction.} \]
\[ \text{The use of a constant specific heat capacity of water, } (4.18 \, \text{J/g}^\circ \text{C}), \, \text{is applied in the calculations.} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18395f57-75e3-4bdb-980a-ed8850989594%2Fce45ab74-fa44-458d-9651-9cafc185f999%2Fetoarb_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**RESULTS**
*Make sure to use correct significant figures and units of measure.*
**HOT WATER/COOL WATER**
- Temperature change for hot water: \(13^\circ C\)
- Temperature change for cool water: \(17^\circ C\)
Calculations:
- Heat leaving hot water: \(2717 \, J\)
- Heat absorbed by cool water: \(3553 \, J\)
Equation:
\[
(4.18 \, \text{J/g}^\circ \text{C}) \times (50.0 \, \text{g}) \times (37^\circ C - 50^\circ C) = 3553 \, J
\]
- Heat capacity of calorimeter: \(49.17 \, J\)
Equation:
\[
(2717 \, J - 3553 \, J) = 49.17 \, J
\]
**HCl/NaOH**
- Temperature change: \(17^\circ C\)
- Change of temperature:
\[
(40^\circ C - 20^\circ C) = 20^\circ C
\]
- Heat emitted during reaction: _____
- Heat of neutralization, \(\Delta H\): _____
\[ \text{Note:} \, \text{The equations and calculations involve changes in temperature and heat transferred during mixing of hot and cool water, and an acid-base reaction.} \]
\[ \text{The use of a constant specific heat capacity of water, } (4.18 \, \text{J/g}^\circ \text{C}), \, \text{is applied in the calculations.} \]
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