The change in free energy will be equal to #"+173 kJ"#.
Start with the balanced chemical equation for your reaction
#N_(2(g)) + O_(2(g)) -> color(red)(2)NO_((g))#
Notice that the reaction produces #color(red)(2)# moles of nitric oxide; the standard enthlapy of formation is always given for the formation of 1 mole of a substance from its elements in their most stable form.
This means that the enthalpy change for your reaction will be
#2cancel("moles NO") * "90.25 kJ"/(1cancel("mole NO")) = "180.5 kJ"#
The main equation you'll use is
#DeltaG = DeltaH - T * DeltaS#, where
#DeltaG# - the change in Gibbs free energy;
#DeltaH# - the change in enthalpy;
#T# - the temperature in Kelvin;
#DeltaS# - the change in entropy.
To determine the change in entropy, use the standard entropies given to you
#DeltaS_"rxn" = sum(n * S_"products"^0) - sum(m * S_"rectants"^0)#
#DeltaS_"rxn" = (2cancel("moles") * 211"J"/(cancel("mole") * K)) - (1cancel("mole") * 192"J"/(cancel("mole") * K) + 1cancel("mole") * 205"J"/(cancel("mole") * K))#
#DeltaS_"rxn" = "25 J/K"#
Keep in mind, however, that the change in enthalpy is expressed in kJ, so convert the change in entropy to kJ per K
#25cancel("J")/"K" * "1 kJ"/(1000cancel("J")) = 25 * 10^(-3)"kJ/K"#
Now you have everything you need to solve for #DeltaG#. Plug your values into the main equation to get
#DeltaG = "180.5 kJ" - (273.15 + 25)cancel("K") * 25 * 10^(-3)"kJ"/cancel("K")#
#DeltaG = color(green)("+173 kJ")#
