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")#