The standard enthalpy change for the reaction that represents breaking all bonds in gaseous benzene is 1129 kJ/mol. This value can be calculated using the bond enthalpy values of the C-C and C-H bonds in benzene. The bond enthalpy of the C-C bond is 348 kJ/mol, and the bond enthalpy of the C-H bond is 413 kJ/mol. There are six C-C bonds and six C-H bonds in benzene, so the total bond enthalpy of benzene is:

$$6(348\text{ kJ/mol})+6(413 \text{ kJ/mol)}=4554\text{ kJ/mol}$$

The standard enthalpy change for the reaction that represents breaking all bonds in gaseous benzene is the negative of the total bond enthalpy, so:

$$\Delta H°= -4554 \text{ kJ/mol} = \boxed{-1129 \text{ kJ/mol}}$$