Question

Enthalpy of formation of \( \text{C}_6\text{H}_6 \), \( \text{CO}_2(g) \) and \( \text{H}_2\text{O}(l) \) are -393.5, -285.8 and +48.5 KJ/mol respectively. Find the enthalpy of combustion of \( \text{C}_6\text{H}_6 \):

• A. 3267.4 KJ/mol
• B. 3218.49 KJ/mol
• C. 857.5 KJ/mol
• D. 2361 KJ/mol
• E. 2361 KJ/mol

Hint:

To calculate the enthalpy of combustion, use Hess's Law, which states that the enthalpy change of a reaction is the sum of the enthalpy changes of the products minus the reactants.

Answer 1

The Correct Option is D

The enthalpy of combustion of a substance is the enthalpy change when one mole of the substance is completely combusted in oxygen. The reaction for the combustion of benzene (\( \text{C}_6\text{H}_6 \)) is: \[ \text{C}_6\text{H}_6 + \text{O}_2 \rightarrow 6\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \] Using Hess's Law, the enthalpy of combustion can be calculated as: \[ \Delta H = \sum (\text{Enthalpies of formation of products}) - \sum (\text{Enthalpies of formation of reactants}) \] Substitute the given values for the enthalpies of formation: \[ \Delta H = [6(-393.5) + 3(-285.8)] - [-93.5 + 0] \] \[ \Delta H = [6(-393.5) + 3(-285.8)] - (-93.5) \] \[ \Delta H = 2361 \, \text{KJ/mol} \] Thus, the enthalpy of combustion of \( \text{C}_6\text{H}_6 \) is \( 2361 \, \text{KJ/mol} \).
Final Answer: (D) 2361 KJ/mol