At `25^(@)C` the standard heat of formation of liquid benzene is +49.0 `kJ*mol^(-1)`-what does it mean?

Explanation

At `25^(@)C`, the standard enthalpy of formation of liquid benzene is +49.0 kJ`*mol^(-1)`. This means that at `25^(@)C` and 1 atm when 1 mol liquid benzene forms from its constituent elements, the enthalpy change that occurs is +49.0 kJ.
In other words, at `25^(@)C` and 1 atm pressure, the change in enthalpy in the following reaction is +49.0 kJ.
6C(s, graphite)`+3H_(2)(g)toC_(6)H_(6)(l)`