For the reaction of oxygen and nitrogen to form nitric oxide, consider the following thermodynamic data (due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.): δh⁰rxn 180.5kj/mol δs⁰rxn 24.80j/(mol⋅k) a. calculate the temperature in kelvins above which this reaction is spontaneous b. the thermodynamic values from part a will be useful as you work through part b: δh⁰rxn 180.5kj/mol δs⁰rxn 24.80j/(mol⋅k) calculate the equilibrium constant for the following reaction at room temperature, 25 ∘c : n2(g)+o2(g)→2no(g)

The answers are: a) T= 7278.2 K ; b) K= 4.5 10⁻³¹

Explanation:

The equation which allows the calculation of the Gibbs free energy for a process is the following:

ΔGº= ΔHº - T ΔSº

ΔGº<0 for a spontaneous process, and we have both ΔHº and ΔSº higher than 0 (ΔHº>0 and ΔSº>0) so the process will be spontaneous only at high temperatures. To calculate the temperature above which the process will be spontaneous we replace ΔGº=0 in the equation:

ΔGº= ΔHº - T ΔSº

0=  ΔHº - T ΔSº

⇒ T= ΔHº/ΔSº

    T= 180.5 KJ/mol/24.80 J/K.mol

    T= 180.5 KJ/mol/0.0248 KJ/K.mol

    T= 7278.2 K

In part B, to calculate the equilibrium constant we first have to calculate the variation of Gibbs free energy at this temperature (25ºC= 298 K):

ΔGº= ΔHº - T ΔSº

ΔGº= 180.5 KJ/mol - ((298 K)x(0.0248 KJ/K.mol)

ΔGº= 173.1 KJ/mol

With this, we can calculate K (equilibrium constant) at room temperature by using the known values of ΔHº, ΔSº and ΔGº:

ΔGº= - RT ln K

⇒ln K= ΔGº/-RT

   ln K= 173100 J/mol/-(8.314 J/K.mol x 298 K)

   K=

   K= 4.5 10⁻³¹

oh it is true if thats what you are asking

fe 55.85 g/mol and h2o 18.02 g/mol.