To determine the standard enthalpy of combustion of 2 moles of benzene, we need to use the standard enthalpy of formation values provided and apply Hess's Law. Here is a step-by-step explanation:

Given Data:

Standard enthalpy of formation of benzene ($C_6H_6(l)$):

$ \Delta H_f(\text{C}_6\text{H}_6(l)) = 48.5 \, \text{kJ/mol} $

Standard enthalpy of formation of carbon dioxide ($CO_2(g)$):

$ \Delta H_f(\text{CO}_2(g)) = -393.5 \, \text{kJ/mol} $

Standard enthalpy of formation of water ($H_2O(l)$):

$ \Delta H_f(\text{H}_2O(l)) = -286 \, \text{kJ/mol} $

Combustion Reaction for Benzene:

$ \text{C}_6\text{H}_6(l) + \frac{15}{2} \text{O}_2(g) \rightarrow 6 \text{CO}_2(g) + 3 \text{H}_2O(l) $

Enthalpy Change Calculation:

Using Hess's Law, the enthalpy change for the reaction can be calculated as follows:

$ \Delta H_{\text{comb}} = \left[ 6 \Delta H_f(\text{CO}_2(g)) + 3 \Delta H_f(\text{H}_2O(l)) \right] - \Delta H_f(\text{C}_6\text{H}_6(l)) $

Substitute the given values:

$ \Delta H_{\text{comb}} = \left[ 6 \times (-393.5) + 3 \times (-286) \right] - 48.5 $

Perform the calculations:

$ \Delta H_{\text{comb}} = \left[ 6 \times (-393.5) \right] + \left[ 3 \times (-286) \right] - 48.5 $

$ \Delta H_{\text{comb}} = \left[ -2361 \right] + \left[ -858 \right] - 48.5 $

$ \Delta H_{\text{comb}} = -3267.5 \, \text{kJ/mol} $

This value is the enthalpy change for the combustion of 1 mole of benzene.

For 2 Moles of Benzene:

$ \Delta H_{\text{comb (2 moles)}} = 2 \times (-3267.5 \, \text{kJ/mol}) $

$ \Delta H_{\text{comb (2 moles)}} = -6535 \, \text{kJ} $

Conclusion:

The standard enthalpy of combustion of 2 moles of benzene is

$ x = 6535 \, \text{kJ} $

Thus, $ x = 6535 $.