1. a Consider an aqueous reaction of equal volumes of 2.0 M HC) 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 (gan) be for this reaction given a positive temperature change of 2.5 degrees Celsius? Assume a value of 4.18 Jg! Cfor Cutn
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.
![**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:**
\[ q_{\text{solution}} = 103 \times 4.18 \times 2.5 \]
\[ q_{\text{solution}} \approx 1077.85 \, \text{J} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b0fedb7-adbc-492c-aa14-9f5d12342848%2Fd81a28ad-d685-4fd3-b803-6cb6cdd02e73%2Fvci84jj_processed.png&w=3840&q=75)
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