The change in free energy will be equal to "+173 kJ"+173 kJ.
Start with the balanced chemical equation for your reaction
N_(2(g)) + O_(2(g)) -> color(red)(2)NO_((g))N2(g)+O2(g)→2NO(g)
Notice that the reaction produces color(red)(2)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")
