1. Given the thermochemical reaction below is this an exo- or endothermic process? 2 H2) + O2) 2 H20) AH = - 136.64 kcal When an unknown number of moles for the above forward reaction occurred it gave off 13.2kJ of heat. How many moles of water were produced? What would be the reverse thermochemical equation?
1. Given the thermochemical reaction below is this an exo- or endothermic process? 2 H2) + O2) 2 H20) AH = - 136.64 kcal When an unknown number of moles for the above forward reaction occurred it gave off 13.2kJ of heat. How many moles of water were produced? What would be the reverse thermochemical equation?
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Chapter1: Chemical Foundations
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![1. **Given the thermochemical reaction below, is this an exo- or endothermic process?**
\[
2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(l) \quad \Delta H = -136.64 \text{ kcal}
\]
When an unknown number of moles for the above forward reaction occurred, it gave off 13.2 kJ of heat. How many moles of water were produced?
What would be the reverse thermochemical equation?
**Explanation:**
- The given reaction is the formation of water from hydrogen and oxygen. Since the enthalpy change (\( \Delta H \)) is negative, the reaction is exothermic, meaning it releases heat.
- To calculate how many moles of water were produced when 13.2 kJ of heat was released, convert the given energy to kcal (1 kcal = 4.184 kJ):
\[
13.2 \text{ kJ} \times \frac{1 \text{ kcal}}{4.184 \text{ kJ}} = 3.154 \text{ kcal}
\]
- Use the stoichiometry of the reaction to find the moles of water:
\[
\frac{3.154 \text{ kcal}}{-136.64 \text{ kcal}} \times 2 \text{ moles of } \text{H}_2\text{O} = 0.046 \text{ moles of } \text{H}_2\text{O}
\]
- The reverse reaction would be:
\[
2 \text{H}_2\text{O}(l) \rightarrow 2 \text{H}_2(g) + \text{O}_2(g) \quad \Delta H = +136.64 \text{ kcal}
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F220385e1-07ad-47f4-abb1-e68cafc238a8%2F536f42f2-1280-4fb9-897f-e8bf9391fe26%2Fcasauiu_processed.png&w=3840&q=75)
Transcribed Image Text:1. **Given the thermochemical reaction below, is this an exo- or endothermic process?**
\[
2 \text{H}_2(g) + \text{O}_2(g) \rightarrow 2 \text{H}_2\text{O}(l) \quad \Delta H = -136.64 \text{ kcal}
\]
When an unknown number of moles for the above forward reaction occurred, it gave off 13.2 kJ of heat. How many moles of water were produced?
What would be the reverse thermochemical equation?
**Explanation:**
- The given reaction is the formation of water from hydrogen and oxygen. Since the enthalpy change (\( \Delta H \)) is negative, the reaction is exothermic, meaning it releases heat.
- To calculate how many moles of water were produced when 13.2 kJ of heat was released, convert the given energy to kcal (1 kcal = 4.184 kJ):
\[
13.2 \text{ kJ} \times \frac{1 \text{ kcal}}{4.184 \text{ kJ}} = 3.154 \text{ kcal}
\]
- Use the stoichiometry of the reaction to find the moles of water:
\[
\frac{3.154 \text{ kcal}}{-136.64 \text{ kcal}} \times 2 \text{ moles of } \text{H}_2\text{O} = 0.046 \text{ moles of } \text{H}_2\text{O}
\]
- The reverse reaction would be:
\[
2 \text{H}_2\text{O}(l) \rightarrow 2 \text{H}_2(g) + \text{O}_2(g) \quad \Delta H = +136.64 \text{ kcal}
\]
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