Consider the combustion of ethane: 2C₂H6 (9) +702(g) → 4 CO2 (g) + 6H₂O(g) If the ethane is burning at the rate of 0.8 mol/L x s, at what rates are CO₂ and H₂O being produced? CO₂ = mol/L x s H₂O = mol/L x s

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**Combustion of Ethane**

Consider the combustion of ethane:

\[ 2\text{C}_2\text{H}_6(g) + 7\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 6\text{H}_2\text{O}(g) \]

If the ethane is burning at the rate of 0.8 mol/L × s, at what rates are CO\(_2\) and H\(_2\)O being produced?

- CO\(_2\) = [ ] mol/L × s
- H\(_2\)O = [ ] mol/L × s

**Explanation:**

This chemical equation represents the complete combustion of ethane (\(\text{C}_2\text{H}_6\)) in the presence of oxygen (\(\text{O}_2\)) to produce carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). 

To determine the rates of production of CO\(_2\) and H\(_2\)O, use the stoichiometry of the reaction:

- According to the reaction, 2 moles of \(\text{C}_2\text{H}_6\) yield 4 moles of \(\text{CO}_2\). Therefore, if \(\text{C}_2\text{H}_6\) is being consumed at 0.8 mol/L × s, \(\text{CO}_2\) is produced at \(0.8 \times \frac{4}{2} = 1.6\) mol/L × s.

- Similarly, 2 moles of \(\text{C}_2\text{H}_6\) yield 6 moles of \(\text{H}_2\text{O}\), leading to a rate of \(0.8 \times \frac{6}{2} = 2.4\) mol/L × s for \(\text{H}_2\text{O}\).

**Final Rates:**

- \(\text{CO}_2\) = 1.6 mol/L × s
- \(\text{H}_2\text{O}\) = 2.4 mol/L × s
Transcribed Image Text:**Combustion of Ethane** Consider the combustion of ethane: \[ 2\text{C}_2\text{H}_6(g) + 7\text{O}_2(g) \rightarrow 4\text{CO}_2(g) + 6\text{H}_2\text{O}(g) \] If the ethane is burning at the rate of 0.8 mol/L × s, at what rates are CO\(_2\) and H\(_2\)O being produced? - CO\(_2\) = [ ] mol/L × s - H\(_2\)O = [ ] mol/L × s **Explanation:** This chemical equation represents the complete combustion of ethane (\(\text{C}_2\text{H}_6\)) in the presence of oxygen (\(\text{O}_2\)) to produce carbon dioxide (\(\text{CO}_2\)) and water (\(\text{H}_2\text{O}\)). To determine the rates of production of CO\(_2\) and H\(_2\)O, use the stoichiometry of the reaction: - According to the reaction, 2 moles of \(\text{C}_2\text{H}_6\) yield 4 moles of \(\text{CO}_2\). Therefore, if \(\text{C}_2\text{H}_6\) is being consumed at 0.8 mol/L × s, \(\text{CO}_2\) is produced at \(0.8 \times \frac{4}{2} = 1.6\) mol/L × s. - Similarly, 2 moles of \(\text{C}_2\text{H}_6\) yield 6 moles of \(\text{H}_2\text{O}\), leading to a rate of \(0.8 \times \frac{6}{2} = 2.4\) mol/L × s for \(\text{H}_2\text{O}\). **Final Rates:** - \(\text{CO}_2\) = 1.6 mol/L × s - \(\text{H}_2\text{O}\) = 2.4 mol/L × s
A first-order reaction has rate constants of \(4.6 \times 10^{-2} \, \text{s}^{-1}\) and \(8.9 \times 10^{-2} \, \text{s}^{-1}\) at 0°C and 20°C, respectively. What is the value of the activation energy?

Activation energy = [ ] kJ/mol
Transcribed Image Text:A first-order reaction has rate constants of \(4.6 \times 10^{-2} \, \text{s}^{-1}\) and \(8.9 \times 10^{-2} \, \text{s}^{-1}\) at 0°C and 20°C, respectively. What is the value of the activation energy? Activation energy = [ ] kJ/mol
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