Question:
One mole of ethanol is produced reacting graphite, $H_2$ and $O_2$ together. The standard enthalpy of formation is $-277.7\,\text{kJ mol}^{-1}$. Calculate the standard enthalpy of the reaction when 4 moles of graphite is involved
One mole of ethanol is produced reacting graphite, $H_2$ and $O_2$ together. The standard enthalpy of formation is $-277.7\,\text{kJ mol}^{-1}$. Calculate the standard enthalpy of the reaction when 4 moles of graphite is involved
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Always check the stoichiometric coefficient of the reactant mentioned in the question. Here, the "formation" definition uses 2 carbons for every 1 ethanol.
Updated On: Apr 29, 2026
- \(-277.7\)
- \(-555.4\)
- \(-138.85\)
- \(-69.42\)
- \(-1110.8\)
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The Correct Option is B
Solution and Explanation
Concept:
The enthalpy of formation (\(\Delta H_f^\circ\)) is the heat change when 1 mole of a substance is formed from its elements in their standard states. We must use the balanced chemical equation to relate the amount of graphite to the total heat.
Step 1: Write the balanced formation equation for ethanol (\(C_2H_5OH\)).
\[ 2C (\text{graphite}) + 3H_2 (g) + \frac{1}{2}O_2 (g) \rightarrow C_2H_5OH (l) \] From this equation, we see that 2 moles of graphite produce 1 mole of ethanol.
Step 2: Relate enthalpy to graphite amount.
The given \(\Delta H = -277.7 \, \text{kJ}\) corresponds to the reaction involving 2 moles of graphite.
Step 3: Calculate for 4 moles of graphite.
If 2 moles release \(277.7 \, \text{kJ}\), then 4 moles (which is double the amount) will release double the energy: \[ \Delta H_{\text{total}} = 2 \times (-277.7 \, \text{kJ}) = -555.4 \, \text{kJ} \]
Step 1: Write the balanced formation equation for ethanol (\(C_2H_5OH\)).
\[ 2C (\text{graphite}) + 3H_2 (g) + \frac{1}{2}O_2 (g) \rightarrow C_2H_5OH (l) \] From this equation, we see that 2 moles of graphite produce 1 mole of ethanol.
Step 2: Relate enthalpy to graphite amount.
The given \(\Delta H = -277.7 \, \text{kJ}\) corresponds to the reaction involving 2 moles of graphite.
Step 3: Calculate for 4 moles of graphite.
If 2 moles release \(277.7 \, \text{kJ}\), then 4 moles (which is double the amount) will release double the energy: \[ \Delta H_{\text{total}} = 2 \times (-277.7 \, \text{kJ}) = -555.4 \, \text{kJ} \]
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