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The combustion of hexane takes place according to the equation C₆H₁₄(g) + 19/2 O₂(g) → 6CO₂(g) + 7H₂O(g) ΔH = -4158 kJ mol⁻¹ Consider the following reaction - VCE - SSCE Chemistry - Question 17 - 2016 - Paper 1
Question 17
The combustion of hexane takes place according to the equation C₆H₁₄(g) + 19/2 O₂(g) → 6CO₂(g) + 7H₂O(g) ΔH = -4158 kJ mol⁻¹ Consider the following reaction. 1... show full transcript
Worked Solution & Example Answer:The combustion of hexane takes place according to the equation C₆H₁₄(g) + 19/2 O₂(g) → 6CO₂(g) + 7H₂O(g) ΔH = -4158 kJ mol⁻¹ Consider the following reaction - VCE - SSCE Chemistry - Question 17 - 2016 - Paper 1
Step 1
Calculate ΔH for the given reaction
Answer
To determine the ΔH for the reaction:
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Identify the reaction and its components: The given reaction is a reverse combustion of hexane producing carbon dioxide and water.
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Use the given ΔH for the combustion of hexane: From the provided equation, the combustion of hexane has a ΔH value of -4158 kJ mol⁻¹.
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Reverse the reaction: When reversing a reaction, the sign of ΔH changes. Therefore, for the reverse reaction:
ΔH(reverse) = +4158 kJ mol⁻¹
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Scale the reaction if needed: The stoichiometry of the required reaction has a factor of 2 compared to the original. This means:
ΔH = +2 * 4158 kJ mol⁻¹ = +8316 kJ mol⁻¹
- Final answer: Thus, the value of ΔH for the reaction is:
Answer: A. +8316 kJ mol⁻¹.
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