Use the data given below to find the enthalpy of reaction for the following reaction: CaCO3(s) → CaO(s) + CO2 (g) AH; (kJ/mol) СаСОз (s) -1207 CaO(s) -635.5 CO2 (g) -393.5 -2236 kJ -178 kJ O +178 kJ -1449 kJ O -2058 kJ
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 of Reaction
Use the data given below to find the enthalpy of reaction for the following reaction:
\[ \text{CaCO}_3 (\text{s}) \rightarrow \text{CaO} (\text{s}) + \text{CO}_2 (\text{g}) \]
#### Enthalpy of Formation Data (ΔH°f) in kJ/mol
| Substance | ΔH°f (kJ/mol) |
|---------------|---------------|
| CaCO₃ (s) | -1207 |
| CaO (s) | -635.5 |
| CO₂ (g) | -393.5 |
#### Possible Answers
- ○ -2236 kJ
- ○ -178 kJ
- ○ +178 kJ
- ○ -1449 kJ
- ○ -2058 kJ
### Explanation
The enthalpy change of the reaction (ΔH°rxn) can be calculated using the following formula:
\[ \Delta H°_{\text{rxn}} = \sum \Delta H°f (\text{products}) - \sum \Delta H°f (\text{reactants}) \]
For the given reaction:
\[ \text{CaCO}_3 (\text{s}) \rightarrow \text{CaO} (\text{s}) + \text{CO}_2 (\text{g}) \]
Calculate ΔH°rxn as follows:
\[
\Delta H°_{\text{rxn}} = [\Delta H°f (\text{CaO} (\text{s})) + \Delta H°f (\text{CO}_2 (\text{g}))] - \Delta H°f (\text{CaCO}_3 (\text{s}))
\]
\[
\Delta H°_{\text{rxn}} = [(-635.5) + (-393.5)] - (-1207)
\]
\[
\Delta H°_{\text{rxn}} = (-1029) - (-1207)
\]
\[
\Delta H°_{\text{rxn}} = -1029 + 1207
\]
\[
\Delta H°_{\text{rxn}} = 178 \text{ kJ}
\]
Thus, the correct answer is:
- ○ +178 kJ](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F21660ac5-28b2-4571-9555-dad5f0a9c878%2Fbc957c14-2996-47e6-934e-0b1ac75c95d0%2Fcrv29y_processed.png&w=3840&q=75)
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