7A + 5B 3C + 4D, and the standard enthalpies of forma- Using the equation shown: tion, AH'f: A: 15.7 k)/mol B: -86.4 kJ/mol C: -52.7 kJ/mol D: -71.6 kJ/mol calculate AH° pn in kJ for the hypothetical reaction above.
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 the Standard Enthalpy of Reaction (ΔH°rxn)**
Given the chemical equation:
\[ 7A + 5B \rightarrow 3C + 4D \]
we can use the standard enthalpies of formation (\(ΔH_f°\)) for the reactants and products to calculate the standard enthalpy change of the reaction. The standard enthalpies of formation are provided below:
- \( ΔH_f° \) for A: \( 15.7 \) kJ/mol
- \( ΔH_f° \) for B: \( -86.4 \) kJ/mol
- \( ΔH_f° \) for C: \( -52.7 \) kJ/mol
- \( ΔH_f° \) for D: \( -71.6 \) kJ/mol
The standard enthalpy change of the reaction (\(ΔH°_{rxn}\)) can be calculated using the formula:
\[ ΔH°_{rxn} = \sum ΔH_f° (\text{products}) - \sum ΔH_f° (\text{reactants}) \]
We need to sum the enthalpies of formation for the products and the reactants, considering their stoichiometric coefficients:
For the products:
\[ (3 \cdot ΔH_f° \text{ of C}) + (4 \cdot ΔH_f° \text{ of D}) \]
\[ (3 \cdot -52.7 \text{ kJ/mol}) + (4 \cdot -71.6 \text{ kJ/mol}) \]
\[ = -158.1 \text{ kJ} + (-286.4 \text{ kJ}) \]
\[ = -444.5 \text{ kJ} \]
For the reactants:
\[ (7 \cdot ΔH_f° \text{ of A}) + (5 \cdot ΔH_f° \text{ of B}) \]
\[ (7 \cdot 15.7 \text{ kJ/mol}) + (5 \cdot -86.4 \text{ kJ/mol}) \]
\[ = 109.9 \text{ kJ} + (-432 \text{ kJ}) \]
\[ = -322.1 \text{ kJ} \]
Finally, substitute these sums back into the ΔH°_{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4a6e4794-5644-431d-9e96-b37fbaf0b60a%2F89bcd429-55bf-481d-b1ce-790a4df1156e%2Fq7qdz4_processed.jpeg&w=3840&q=75)
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