Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![**Problem 1: Thermochemistry Reaction**
**Question:**
**a.** Consider an aqueous reaction of equal volumes of 2.0 M HCl and 2.0 M NaOH, with a final volume of 100.0 mL and a density of 1.03 g/mL. What will the change in heat (q_solution) be for this reaction given a positive temperature change of 2.5 degrees Celsius? Assume a value of 4.18 J·g⁻¹·ºC⁻¹ for c_solution.
**Components:**
1. **Molarity (M):**
- HCl: 2.0 M
- NaOH: 2.0 M
2. **Volumes:**
- Final volume: 100.0 mL
3. **Density:**
- 1.03 g/mL
4. **Temperature Change (ΔT):**
- 2.5 °C (increase)
5. **Specific Heat Capacity (c_solution):**
- 4.18 J·g⁻¹·°C⁻¹
**Steps for Calculation:**
1. **Calculate the mass of the solution using volume and density:**
\[ \text{Mass (m)} = \text{Volume} \times \text{Density} \]
\[ m = 100.0 \, \text{mL} \times 1.03 \, \text{g/mL} \]
\[ m = 103 \, \text{g} \]
2. **Use the specific heat capacity and temperature change to find the heat change:**
\[ q_{\text{solution}} = m \times c_{\text{solution}} \times \Delta T \]
\[ q_{\text{solution}} = 103 \, \text{g} \times 4.18 \, \text{J·g}^{-1}·\text{°C}^{-1} \times 2.5 \, \text{°C} \]
3. **Perform the multiplication:**
\[ q_{\text{solution}} = 103 \times 4.18 \times 2.5 \]
\[ q_{\text{solution}} \approx 1077.85 \, \text{J} \]](https://content.bartleby.com/qna-images/question/7b0fedb7-adbc-492c-aa14-9f5d12342848/d81a28ad-d685-4fd3-b803-6cb6cdd02e73/vci84jj_processed.png)
Transcribed Image Text:**Problem 1: Thermochemistry Reaction**
**Question:**
**a.** Consider an aqueous reaction of equal volumes of 2.0 M HCl and 2.0 M NaOH, with a final volume of 100.0 mL and a density of 1.03 g/mL. What will the change in heat (q_solution) be for this reaction given a positive temperature change of 2.5 degrees Celsius? Assume a value of 4.18 J·g⁻¹·ºC⁻¹ for c_solution.
**Components:**
1. **Molarity (M):**
- HCl: 2.0 M
- NaOH: 2.0 M
2. **Volumes:**
- Final volume: 100.0 mL
3. **Density:**
- 1.03 g/mL
4. **Temperature Change (ΔT):**
- 2.5 °C (increase)
5. **Specific Heat Capacity (c_solution):**
- 4.18 J·g⁻¹·°C⁻¹
**Steps for Calculation:**
1. **Calculate the mass of the solution using volume and density:**
\[ \text{Mass (m)} = \text{Volume} \times \text{Density} \]
\[ m = 100.0 \, \text{mL} \times 1.03 \, \text{g/mL} \]
\[ m = 103 \, \text{g} \]
2. **Use the specific heat capacity and temperature change to find the heat change:**
\[ q_{\text{solution}} = m \times c_{\text{solution}} \times \Delta T \]
\[ q_{\text{solution}} = 103 \, \text{g} \times 4.18 \, \text{J·g}^{-1}·\text{°C}^{-1} \times 2.5 \, \text{°C} \]
3. **Perform the multiplication:**
\[ q_{\text{solution}} = 103 \times 4.18 \times 2.5 \]
\[ q_{\text{solution}} \approx 1077.85 \, \text{J} \]
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