2 C5H10 (1)+15 O₂ (g) → 10 CO₂ (g) + 10 H₂O(l) a. Calculate 4H of the reaction. b. Calculate 4S of the reaction.

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### Thermochemistry and Spontaneity of Reactions

Consider the following chemical reaction:
\[ 2 \text{C}_5\text{H}_{10} (\text{l}) + 15 \text{O}_2 (\text{g}) \rightarrow 10 \text{CO}_2 (\text{g}) + 10 \text{H}_2\text{O} (\text{l}) \]

**a. Calculate \( \Delta H^\circ \) of the reaction.**

**b. Calculate \( \Delta S^\circ \) of the reaction.**

**c. Show by calculation whether or not the reaction is spontaneous at 298.15 K.**

**d. At what temperature is this reaction in equilibrium? In one brief sentence, explain the temperature dependence of the spontaneity of this reaction.**

### Explanation of Components:

1. **Enthalpy Change (\( \Delta H^\circ \)) Calculation**:
   - To find \( \Delta H^\circ \), use standard enthalpies of formation (\( \Delta H_f^\circ \)) for each compound involved.
   - The equation is: 
     \[
     \Delta H^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})
     \]

2. **Entropy Change (\( \Delta S^\circ \)) Calculation**:
   - Similar to enthalpy, calculate \( \Delta S^\circ \) using standard entropy values (\( S^\circ \)).
   - The equation is:
     \[
     \Delta S^\circ = \sum S^\circ (\text{products}) - \sum S^\circ (\text{reactants})
     \]

3. **Spontaneity at a Given Temperature**:
   - Determine spontaneity using Gibbs free energy change (\( \Delta G^\circ \)).
   - The equation is:
     \[
     \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ
     \]
   - If \( \Delta G^\circ < 0 \), the reaction is spontaneous.

4. **Equilibrium Temperature**:
   - At equilibrium, \( \Delta G^\circ = 0 \).
   - The rearranged equation is:
     \[
     T = \frac{\Delta H^\circ}{\Delta S^\circ
Transcribed Image Text:### Thermochemistry and Spontaneity of Reactions Consider the following chemical reaction: \[ 2 \text{C}_5\text{H}_{10} (\text{l}) + 15 \text{O}_2 (\text{g}) \rightarrow 10 \text{CO}_2 (\text{g}) + 10 \text{H}_2\text{O} (\text{l}) \] **a. Calculate \( \Delta H^\circ \) of the reaction.** **b. Calculate \( \Delta S^\circ \) of the reaction.** **c. Show by calculation whether or not the reaction is spontaneous at 298.15 K.** **d. At what temperature is this reaction in equilibrium? In one brief sentence, explain the temperature dependence of the spontaneity of this reaction.** ### Explanation of Components: 1. **Enthalpy Change (\( \Delta H^\circ \)) Calculation**: - To find \( \Delta H^\circ \), use standard enthalpies of formation (\( \Delta H_f^\circ \)) for each compound involved. - The equation is: \[ \Delta H^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants}) \] 2. **Entropy Change (\( \Delta S^\circ \)) Calculation**: - Similar to enthalpy, calculate \( \Delta S^\circ \) using standard entropy values (\( S^\circ \)). - The equation is: \[ \Delta S^\circ = \sum S^\circ (\text{products}) - \sum S^\circ (\text{reactants}) \] 3. **Spontaneity at a Given Temperature**: - Determine spontaneity using Gibbs free energy change (\( \Delta G^\circ \)). - The equation is: \[ \Delta G^\circ = \Delta H^\circ - T \Delta S^\circ \] - If \( \Delta G^\circ < 0 \), the reaction is spontaneous. 4. **Equilibrium Temperature**: - At equilibrium, \( \Delta G^\circ = 0 \). - The rearranged equation is: \[ T = \frac{\Delta H^\circ}{\Delta S^\circ
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