Question:medium
The standard enthalpies of formation of $C_6H_6(l)$, $CO_2(g)$ and $H_2O(l)$ are respectively $+49\text{kJ mol}^{-1}$, $-394\text{kJ mol}^{-1}$ and $-286\text{kJ mol}^{-1}$ respectively. What is the value of standard enthalpy of combustion of $C_6H_6(l)$?
The standard enthalpies of formation of $C_6H_6(l)$, $CO_2(g)$ and $H_2O(l)$ are respectively $+49\text{kJ mol}^{-1}$, $-394\text{kJ mol}^{-1}$ and $-286\text{kJ mol}^{-1}$ respectively. What is the value of standard enthalpy of combustion of $C_6H_6(l)$?
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Remember: "Combustion is always exothermic", so the final value must be negative. Always pay close attention to the stoichiometry of the combustion reaction.
Updated On: Jun 26, 2026
- -3222 $\text{kJ mol}^{-1}$
- -3173 $\text{kJ mol}^{-1}$
- -3271 $\text{kJ mol}^{-1}$
- +3173 $\text{kJ mol}^{-1}$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
The enthalpy of combustion is the heat released when one mole of a substance reacts completely with oxygen. We can use Hess's law and standard heats of formation to calculate it.
Step 2: Key Formula or Approach:
Combustion reaction: C\textsubscript{6}H\textsubscript{6}(l) + 7.5O\textsubscript{2}(g) \(\rightarrow\) 6CO\textsubscript{2}(g) + 3H\textsubscript{2}O(l).
\( \Delta H^0_{\text{comb}} = \sum \Delta H^0_f (\text{products}) - \sum \Delta H^0_f (\text{reactants}) \).
Step 3: Detailed Explanation:
\[ \Delta H^0_{\text{comb}} = [6 \times \Delta H^0_f(\text{CO}_2) + 3 \times \Delta H^0_f(\text{H}_2\text{O})] - [\Delta H^0_f(\text{C}_6\text{H}_6) + 0] \] (Enthalpy of formation of O\textsubscript{2} is zero).
\[ \Delta H^0_{\text{comb}} = [6 \times (-394) + 3 \times (-286)] - [+49] \] \[ \Delta H^0_{\text{comb}} = [-2364 - 858] - 49 \] \[ \Delta H^0_{\text{comb}} = -3222 - 49 = -3271 \text{ kJ mol}^{-1} \] Step 4: Final Answer:
The enthalpy of combustion is -3271 kJ mol{-1}.
The enthalpy of combustion is the heat released when one mole of a substance reacts completely with oxygen. We can use Hess's law and standard heats of formation to calculate it.
Step 2: Key Formula or Approach:
Combustion reaction: C\textsubscript{6}H\textsubscript{6}(l) + 7.5O\textsubscript{2}(g) \(\rightarrow\) 6CO\textsubscript{2}(g) + 3H\textsubscript{2}O(l).
\( \Delta H^0_{\text{comb}} = \sum \Delta H^0_f (\text{products}) - \sum \Delta H^0_f (\text{reactants}) \).
Step 3: Detailed Explanation:
\[ \Delta H^0_{\text{comb}} = [6 \times \Delta H^0_f(\text{CO}_2) + 3 \times \Delta H^0_f(\text{H}_2\text{O})] - [\Delta H^0_f(\text{C}_6\text{H}_6) + 0] \] (Enthalpy of formation of O\textsubscript{2} is zero).
\[ \Delta H^0_{\text{comb}} = [6 \times (-394) + 3 \times (-286)] - [+49] \] \[ \Delta H^0_{\text{comb}} = [-2364 - 858] - 49 \] \[ \Delta H^0_{\text{comb}} = -3222 - 49 = -3271 \text{ kJ mol}^{-1} \] Step 4: Final Answer:
The enthalpy of combustion is -3271 kJ mol{-1}.
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