(SHOW CALCULATIONS) How much energy is released when 0.750 mol of diethyl ether is cooled from 75.0°C to 25.0°C? boiling point 34.5°C Standard enthalpy of vaporization 27.53 kJ/mol Heat capacity, (CH3)2O(l) 172.0 J/(mol-K) specific heat capacity, (CH3)2O(g) 119.46 J/(mol-K)
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.
![**Energy Calculation for Cooling Diethyl Ether**
In this educational module, we will calculate the energy released when 0.750 mol of diethyl ether is cooled from 75.0°C to 25.0°C. The properties of diethyl ether required for this calculation are outlined below:
- **Boiling Point:** 34.5°C
- **Standard Enthalpy of Vaporization:** 27.53 kJ/mol
- **Heat Capacity, \((CH_3)_2O(l)\):** 172.0 J/(mol·K)
- **Specific Heat Capacity, \((CH_3)_2O(g)\):** 119.46 J/(mol·K)
We assume no phase changes other than boiling. Let's break down the calculation into phases:
1. **Cooling from 75.0°C to 34.5°C (Gas Phase):**
\[
q_1 = n \cdot C_p(g) \cdot \Delta T_1
\]
where:
- \( n \) = 0.750 mol (amount of diethyl ether)
- \( C_p(g) \) = 119.46 J/mol·K (specific heat capacity of diethyl ether in gas phase)
- \( \Delta T_1 \) = (75.0 - 34.5)°C = 40.5 K
Therefore:
\[
q_1 = 0.750 \cdot 119.46 \cdot 40.5 = 3,626.565 J
\]
2. **Condensation at 34.5°C (Phase Change from Gas to Liquid):**
\[
q_2 = n \cdot \Delta H_{vap}
\]
where:
- \( \Delta H_{vap} \) = 27.53 kJ/mol = 27,530 J/mol (enthalpy of vaporization)
Therefore:
\[
q_2 = 0.750 \cdot 27,530 = 20,647.5 J
\]
3. **Cooling from 34.5°C to 25.0°C (Liquid Phase):**
\[
q_3 = n \cdot C_p(l) \cdot \Delta T_2
\]
where:
- \( C](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd5b1881-d329-47e7-a65d-eb7760c3b964%2F24518c4d-1a8c-4a42-8dfb-4fde61dffb8a%2F3bvolaf_processed.jpeg&w=3840&q=75)
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