Using the data provided, find the value of equilibrium constant for the following reaction at 298   K and 1 atm pressure. NO ( g ) + 1 2 O 2 (   g ) ⇌ NO 2 (   g ) Δ f H 0 NO ( g ) = 90 .4 kJ . mol − 1 Δ f H 0 NO 2 ( g ) = 32 .48 kJ ⋅ mol − 1 ΔS ∘ @ 298 K = − 70 .8 J ⋅ K − 1 ⋅ mol − 1 antilog ⁡ ( 0 .50 ) = 3162

Step-by-step solution

The given reaction is NO ( g ) + 1 2 O 2 (   g ) ⇌ NO 2 (   g ) The enthalpy of the above reaction given by the difference of enthalpies of reactant side and product side. Therefore, ∆ c H   =   ∆ f H NO 2 - ∆ f H NO + 1 2 ∆ f H O 2 ∆ c H   =   32 . 8 - 90 . 4 + 1 2 0 ∆ c H   =   - 57 . 6   KJmol - 1 Also, we know that Gibb's free energy can be determined by ∆ G   =   ∆ H - T ∆ S ∆ G   =