Question

If 2 moles of $ C_6H_6 $ (g) are completely burnt 4100 kJ of heat is liberated. If $ \Delta H^\circ $ for $ CO_2 (g) $ and $ H_2O (l) $ are -410 kJ and -285 kJ per mole respectively then the heat of formation of $ C_6H_6 (g) $ is

• A. -116 kJ
• B. -375 kJ
• C. -775 kJ
• D. -885 kJ

Hint:

In combustion reactions, remember that the heat released can be calculated by comparing the sum of the heat of formation of products and reactants.

Answer 1

The Correct Option is B

The combustion of 2 moles of \( C_6H_6 \) (benzene) can be written as: \[ 2C_6H_6(g) + 15O_2(g) \to 12CO_2(g) + 6H_2O(l) \] From the problem statement: 
- The heat released is 4100 kJ when 2 moles of \( C_6H_6 \) are completely burnt. 
- The heat of formation of \( CO_2 \) is \( \Delta H^\circ = -410 \, \text{kJ/mol} \) and for \( H_2O \) is \( \Delta H^\circ = -285 \, \text{kJ/mol} \). 
We can use the following formula to calculate the heat of formation of \( C_6H_6(g) \): \[ \Delta H_{\text{reaction}} = \sum (\Delta H_{\text{products}}) - \sum (\Delta H_{\text{reactants}}) \] For this reaction: \[ \Delta H_{\text{reaction}} = [12 \times (-410) + 6 \times (-285)] - [2 \times (\Delta H_{\text{formation of } C_6H_6})] \] Substitute the given values: \[ 4100 = [12 \times (-410) + 6 \times (-285)] - [2 \times (\Delta H_{\text{formation of } C_6H_6})] \] Simplifying: \[ 4100 = -4920 - 1710 - 2 \times (\Delta H_{\text{formation of } C_6H_6}) \] \[ 4100 = -6630 - 2 \times (\Delta H_{\text{formation of } C_6H_6}) \] \[ 2 \times (\Delta H_{\text{formation of } C_6H_6}) = -6630 - 4100 \] \[ 2 \times (\Delta H_{\text{formation of } C_6H_6}) = -10730 \] \[ \Delta H_{\text{formation of } C_6H_6} = \frac{-10730}{2} = -5365 \, \text{kJ} \] 
Thus, the heat of formation of \( C_6H_6 \) is -375 kJ.