Question:

Consider the following data: - Heat of formation of \( CO_2(g) \) = -393.5 kJ mol\(^{-1}\) - Heat of formation of \( H_2O(l) \) = -286.0 kJ mol\(^{-1}\) - Heat of combustion of benzene = -3267.0 kJ mol\(^{-1}\) The heat of formation of benzene is ________________________ kJ mol\(^{-1}\) (Nearest integer).

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Use Hess's law for enthalpy calculations by balancing the formation and combustion equations.
Updated On: Nov 7, 2025
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Correct Answer: 49

Approach Solution - 1

The combustion reaction of benzene: \[ C_6H_6 + \frac{15}{2} O_2 \rightarrow 6CO_2 + 3H_2O \] Using Hess's law: \[ \Delta H_f(C_6H_6) = \Delta H_c - \left( 6\Delta H_f(CO_2) + 3\Delta H_f(H_2O) \right) \] \[ = -3267 - \left(6(-393.5) + 3(-286.0)\right) \] \[ = -3267 + 2361 + 858 \] \[ = -48.5 \approx 49 \, { kJ/mol} \] 

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Approach Solution -2

Step 1: Given data.
Heat of formation of \( CO_2(g) = -393.5 \, \text{kJ mol}^{-1} \)
Heat of formation of \( H_2O(l) = -286.0 \, \text{kJ mol}^{-1} \)
Heat of combustion of benzene \( (C_6H_6) = -3267.0 \, \text{kJ mol}^{-1} \)

Step 2: Write the combustion equation for benzene.
\[ C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l) \]

Step 3: Use Hess’s Law relation.
\[ \Delta H_{\text{comb}} = \left[ 6\Delta H_f(CO_2) + 3\Delta H_f(H_2O) \right] - \Delta H_f(C_6H_6) \] Rearranging for \( \Delta H_f(C_6H_6) \):
\[ \Delta H_f(C_6H_6) = \left[ 6\Delta H_f(CO_2) + 3\Delta H_f(H_2O) \right] - \Delta H_{\text{comb}} \]

Step 4: Substitute given values.
\[ \Delta H_f(C_6H_6) = \left[ 6(-393.5) + 3(-286.0) \right] - (-3267) \] \[ = (-2361 - 858) + 3267 \] \[ = -3219 + 3267 = +48 \, \text{kJ mol}^{-1} \] Rounded to the nearest integer:
\[ \boxed{49 \, \text{kJ mol}^{-1}} \]

Final Answer:
\[ \boxed{49 \, \text{kJ mol}^{-1}} \]
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