At $$25°$$C and $$1$$ atm pressure, the enthalpy of combustion of benzene $$1$$ and acetylene $$g$$ are $$-3268$$ kJ mol$$^{-1}$$ and $$-1300$$ kJ mol$$^{-1}$$, respectively. The change in enthalpy for the reaction
$$3C_2H_2(g) \to C_6H_6(l)$$, is

Enthalpy of combustion of benzene: $$\Delta H_1 = -3268$$ kJ/mol

Enthalpy of combustion of acetylene: $$\Delta H_2 = -1300$$ kJ/mol

We need to find $$\Delta H$$ for: $$3C_2H_2(g) \to C_6H_6(l)$$

Combustion of benzene:

$$C_6H_6(l) + \frac{15}{2}O_2(g) \to 6CO_2(g) + 3H_2O(l), \quad \Delta H_1 = -3268 \text{ kJ/mol}$$

Combustion of acetylene:

$$C_2H_2(g) + \frac{5}{2}O_2(g) \to 2CO_2(g) + H_2O(l), \quad \Delta H_2 = -1300 \text{ kJ/mol}$$

The target reaction can be obtained by:

3 × (combustion of acetylene) − 1 × (combustion of benzene)

This works because:

$$3C_2H_2 + \frac{15}{2}O_2 \to 6CO_2 + 3H_2O$$ ... (multiply reaction 2 by 3)

$$6CO_2 + 3H_2O \to C_6H_6 + \frac{15}{2}O_2$$ ... (reverse reaction 1)

Adding: $$3C_2H_2 \to C_6H_6$$

$$\Delta H = 3 \times \Delta H_2 - \Delta H_1$$

$$= 3 \times (-1300) - (-3268)$$

$$= -3900 + 3268$$

$$= -632 \text{ kJ/mol}$$

Hence, the correct answer is Option C.