Calculate AHryn in kJ/mol of NaOH, when 100 mL of 1.250 M NaOH is mixed with 100 mL of 1.000 M of HCI. The initial temperature of each solution is 25.00 °C and the final temperature after mixing the two solutions is 27.69 °C. Assume the heat capacity of the solution is the same as water, 4.184 J/ °C.g. The solution density is 1.07 g/mL)
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.
![### Calculating Enthalpy Change (ΔH_rxn)
**Problem Statement:**
Calculate ΔH_rxn in kJ/mol of NaOH, when 100 mL of 1.250 M NaOH is mixed with 100 mL of 1.000 M HCl. The initial temperature of each solution is 25.00°C and the final temperature after mixing the two solutions is 27.69°C. Assume the heat capacity of the solution is the same as water, 4.184 J/(°C·g). The solution density is 1.07 g/mL.
**Key Assumptions:**
1. Heat capacity (C) of the solution is 4.184 J/(°C·g), same as water.
2. Density (ρ) of the solution is 1.07 g/mL.
3. Volume of each solution (V₁ and V₂) is 100 mL.
**Procedure:**
1. **Determine the Mass of the Solution:**
\[
\text{Total Volume} = 100 \text{ mL NaOH} + 100 \text{ mL HCl} = 200 \text{ mL}
\]
\[
\text{Density} = 1.07 \text{ g/mL}
\]
\[
\text{Mass (m)} = \text{Volume} \times \text{Density} = 200 \text{ mL} \times 1.07 \text{ g/mL} = 214 \text{ g}
\]
2. **Calculate the Temperature Change (ΔT):**
\[
\text{Initial Temperature} = 25.00°C
\]
\[
\text{Final Temperature} = 27.69°C
\]
\[
\Delta T = 27.69°C - 25.00°C = 2.69°C
\]
3. **Calculate the Heat Absorbed (q):**
\[
q = m \times C \times \Delta T
\]
\[
q = 214 \text{ g} \times 4.184 \text{ J/(°C·g)} \times 2.69°C = 2413.08 \text{ J} = 2.413](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7b0fedb7-adbc-492c-aa14-9f5d12342848%2F72d1aac8-94ca-4596-b16f-f14e0449914f%2Fj1ki85w_processed.jpeg&w=3840&q=75)
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