When the temperature-time data graph was reviewed, it was found that an error had been made in determining ΔT. Instead of 4.76 °C, ΔT was actually 4.70 °C. Based on this change only, calculate the correct ΔH neutralization for the reaction of NH3, and acetic acid.
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.
![## Educational Content: Calculation of Enthalpy Change in a Neutralization Reaction
### Problem Overview
When a neutralization reaction was carried out using 1000 mL of 0.7890 M NH₃, water, and 1000 mL of 0.7940 M acetic acid, the change in temperature (ΔT) was found to be 4.71°C. The specific heat of the reaction mixture was 4.104 J/g°C, and its density was 1.08 g/mL. The calorimeter constant was 3.336 J/K.
### Detailed Calculation
1. **Calculate ΔH (Enthalpy Change) for the Reaction of NH₃ and Acetic Acid**
Given Data:
- Volume of NH₃ solution = 1000 mL
- Molarity of NH₃ = 0.7890 M
- Volume of acetic acid solution = 1000 mL
- Molarity of acetic acid = 0.7940 M
- ΔT = 4.71°C
- Specific heat (c) = 4.104 J/g°C
- Density = 1.08 g/mL
- Calorimeter constant (C_cal) = 3.336 J/K
Steps:
- Calculate the heat absorbed by the calorimeter:
\[
\text{Heat absorbed} = C_{\text{cal}} \times \Delta T = 3.336 \, \text{J/K} \times 4.71 \, \text{K} = 15.5 \, \text{J}
\]
- Determine the limiting reagent (not explicitly stated but deduced from calculation comparisons):
\[
\text{Mols of NH}_3 = 0.7890 \, \text{mol/L} \times 1 \, \text{L} = 0.789 \, \text{mol}
\]
\[
\text{Mols of CH}_3\text{COOH} = 0.7940 \, \text{mol/L} \times 1 \, \text{L} = 0.794 \, \text{mol}
\]
- Calculate ΔH using the heat absorbed and the number of moles of the limiting reagent](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F94e81d64-5c51-48a3-9200-195d1e8f792d%2Ff0139f3d-0d18-442f-9894-88c793744051%2Fj1qur1a_processed.jpeg&w=3840&q=75)
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