1.
Using the data provided, find the value of the equilibrium constant for the following reaction at 298 K and 1 atm pressure.

NO(g)+12O2(g)⇌NO2(g)\text{NO}(g) + \frac{1}{2}\text{O}_2(g) \rightleftharpoons \text{NO}_2(g)

ΔfH∘(NO(g))=90.4\Delta_f H^\circ (\text{NO}(g)) = 90.4 kJ mol−1^{-1}

ΔfH∘(NO2(g))=32.48\Delta_f H^\circ (\text{NO}_2(g)) = 32.48 kJ mol−1^{-1}

ΔS∘\Delta S^\circ at 298 K = −70.8-70.8 J K−1^{-1} mol−1^{-1}

[Given: antilog(0.50)=3.162(0.50) = 3.162]