Calculate the AH for the reaction: CH4 (g) +4 Cl2 (g) → CCl4 (g) + 4HC1 (g) using the standard enthalpies of formation: CH4 (g) = -75 kJ/mol CC14 (g) = -96 kJ/mol HCl (g) = -92 kJ/mol O 113 kJ O 389 kJ O -389 kJ O 254 kJ O-113 kJ < Previous No now du
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.
![### Question 4
#### Calculate the ΔH for the reaction:
\[ \text{CH}_4 (\text{g}) + 4 \text{Cl}_2 (\text{g}) \rightarrow \text{CCl}_4 (\text{g}) + 4 \text{HCl} (\text{g}) \]
Using the standard enthalpies of formation:
- \(\text{CH}_4 (\text{g}) = -75 \, \text{kJ/mol}\)
- \(\text{CCl}_4 (\text{g}) = -96 \, \text{kJ/mol}\)
- \(\text{HCl} (\text{g}) = -92 \, \text{kJ/mol}\)
### Answer choices:
- 113 kJ
- 389 kJ
- -389 kJ
- 254 kJ
- -113 kJ
### Explanation
To calculate the enthalpy change (\(\Delta H\)) for the reaction, use the standard enthalpies of formation (\(\Delta H_f^\circ\)) of the products and reactants:
\[ \Delta H = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \]
Identify the enthalpies for the products and reactants:
- Products: \(\text{CCl}_4 (\text{g})\) and \(4 \text{HCl} (\text{g})\)
- Reactants: \(\text{CH}_4 (\text{g})\) and \(4 \text{Cl}_2 (\text{g})\)
Calculate the total enthalpy for the products and the reactants using the given standard enthalpies of formation.
Next, substitute into the ΔH formula to find the enthalpy change for the reaction.
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