SO₂Cl2 (g) + 2H₂O(l)→ 2HCI(g) + H₂SO4(1) AHf (kJ/mol): SO₂Cl₂(g) -364 H₂O(l) -286 HCI(g) 92 H₂SO4(1) -814 +161 kJ O-62 kJ O-422 kJ -256 kJ O +800. kJ AHrxn° = ?
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 Enthalpy Change for a Reaction
#### Chemical Reaction:
\[ \text{SO}_2\text{Cl}_2 (g) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{HCl}(g) + \text{H}_2\text{SO}_4(l) \]
\[ \Delta H_{rxn}^\circ = ? \]
#### Standard Enthalpy of Formation (\(\Delta H_f^\circ\)) [kJ/mol]:
- SO\(_2\)Cl\(_2\)(g): -364
- H\(_2\)O(l): -286
- HCl(g): 92
- H\(_2\)SO\(_4\)(l): -814
To calculate the enthalpy change for the reaction (\(\Delta H_{rxn}^\circ\)), use the following formula:
\[ \Delta H_{rxn}^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \]
Plug in the values for the reactants and products:
**Reactants:**
- \(\Delta H_f^\circ\) of SO\(_2\)Cl\(_2\)(g) = -364 kJ/mol
- \(\Delta H_f^\circ\) of 2 H\(_2\)O(l) = 2 \times (-286) = -572 kJ/mol
**Products:**
- \(\Delta H_f^\circ\) of 2 HCl(g) = 2 \times 92 = 184 kJ/mol
- \(\Delta H_f^\circ\) of H\(_2\)SO\(_4\)(l) = -814 kJ/mol
Now, sum the values:
**Products Total:**
\[ 184 + (-814) = -630 \, \text{kJ/mol} \]
**Reactants Total:**
\[ -364 + (-572) = -936\, \text{kJ/mol} \]
Finally, calculate \(\Delta H_{rxn}^\circ\):
\[ \Delta H_{rxn}^\circ = -630 - (-936) = -630 + 936 = 306 \, \text{kJ/mol} \]
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