Question

The enthalpy of combustion of benzene, graphite and dihydrogen at $298\,K$ are $-3260$, $-390$ and $-290\,kJ\,mol^{-1}$ respectively. Enthalpy of formation of benzene is:

• A. $-50\,kJ\,mol^{-1}$
• B. $+50\,kJ\,mol^{-1}$
• C. $+60\,kJ\,mol^{-1}$
• D. $-60\,kJ\,mol^{-1}$
• E. $+80\,kJ\,mol^{-1}$

Hint:

Use Hess law: formation = sum of reactants - product.

Answer 1

The Correct Option is B

Concept:
• Hess’s law

Step 1: Formation reaction
\[ 6C + 3H_2 \rightarrow C_6H_6 \]

Step 2: Using combustion data
\[ \Delta H_f = [6(-390) + 3(-290)] - (-3260) \] \[ = (-2340 - 870) + 3260 = 50 \] Final Conclusion:
Option (B)